Maths Study Material - Three Dimentional Geometry

THREE DIMENSIONAL GEOMETRYCoordinate of a point in spaceCoordinate of a point in spaceCoordinate of a point in spaceCoordinate of a point in space

There are infinite number of points in space. We want to identify each and every point of space with the help ofthree mutually perpendicular coordinate axes OX, OY and OZ.Three mutually perpendicular lines OX, OY, OZ are considered as the three axes.

The plane formed with the help of x and y axes is called x-y plane, similarly y & z axes form y-z plane and z andx axes form z - x plane.Consider any point P in the space, Drop a perpendicular from that point to x -y plane, then the algebraic lengthof this perpendicular is considered as z-coordinate and from foot of the perpendicular drop perpendiculars to xand y axes. These algebraic lengths of perpendiculars are considered as y and x coordinates respectively.

1.1 .1 .1 . Vector representation of a point in spaceVector representation of a point in spaceVector representation of a point in spaceVector representation of a point in spaceIf coordinate of a point P in space is (x, y, z) then the position vector of the point P with respect to the

same origin is x i + y j + z k .

2.2 .2 .2 . Distance formulaDistance formulaDistance formulaDistance formula Distance between any two points (x1, y

1, z

1) and (x

2, y

2, z

2) is given as

221

221

221 )zz()yy()xx( −+−+−

Vector method We know that if position vector of two points A and B are given as OA and OB then

AB = | OB – OA |

⇒ AB = |(x2i + y

2 j + z

2k) – (x

1i + y

1j + z

1k)| ⇒AB = 2

122

122

12 )zz()yy()xx( −+−+−

3.3 .3 .3 . Distance of a point P from coordinate axesDistance of a point P from coordinate axesDistance of a point P from coordinate axesDistance of a point P from coordinate axesLet PA, PB and PC are distances of the point P(x, y, z) from the coordinate axes OX, OY and OZrespectively then

PA =22 zy + , PB = 22 xz + , PC =

22 yx +

Example : Show that the points (0, 7, 10), (– 1, 6, 6) and (– 4, 9, 6) form a right angled isosceles triangle.Solution: Let A ≡ (0, 7, 10), B ≡ (–1, 6, 6), C ≡ (– 4, 9, 6)

AB2 = (0 + 1)2 + (7 – 6)2 + (10 – 6)2 = 18 ∴ AB = 3 2

Similarly ∴ BC = 3 2 ,& AC = 6

Clearly AB2 +BC2 = AC2 ∴ ∠ ABC = 90°Also AB = BCHence ∆ABC is right angled isosceles.

Example : Show by using distance formula that the points (4, 5, –5), (0, –11, 3) and (2, –3, –1) are collinear.Solution Let A ≡ (4, 5, –5), B ≡ (0, –11, 3), C ≡ (2, –3, –1).

AB = 842844336)35()115()04( 222 =×==−−+++−

BC = 84)13()311()20( 222 =+++−+− AC = 84)15()35()24( 222 =+−+++−

BC + AC = ABHence points A, B, C are collinear and C lies between A and B.

Example : Find the locus of a point which moves such that the sum of its distances from points A(0, 0, – α)and B(0, 0, α) is constant.

Solution. Let the variable point whose locus is required be P(x, y, z)Given PA + PB = constant = 2a (say)

∴ 222 )z()0y()0x( α++−+− + 222 )z()0y()0x( α−+−+− = 2a

⇒2

1 = 2a – 222 )z(yx α−++

⇒ x2 + y2 + z2 + α2 + 2zα = 4a2 + x2 + y2 + z2 + α2 – 2zα – 4a 222 )z(yx α−++

⇒ 4zα – 4a2 = – 4a 222 )z(yx α−++

⇒ 2

22

a

z α + a2 – 2zα = x2 + y2 + z2 + α2 – 2zα

or, x2 + y2 + z2

α−

2

2

a1 = a2 – α2 or, 2

2

22

22

a

z

a

yx+

α−

+ = 1 This is the required locus.

Self practice problems :1. One of the vertices of a cuboid is (1, 2, 3) and the edges from this vertex are along the +ve x-axis,

+ve y-axis and +z axis respectively and are of length 2, 3, 2 respectively find out the vertices.Ans. (1, 2, 5), (3, 2, 5), (3, 2, 3), (1, – 1, 3), (1, – 1, 5), (3, – 1, 5), (3, – 1, 3).

2. Show that the points (0, 4, 1), (2, 3, –1), (4, 5, 0) and (2, 6, 2) are the vertices of a square.3. Find the locus of point P if AP2 – BP2 = 18, where A ≡ (1, 2, – 3) and B ≡ (3, – 2, 1)

Ans. 2x – 4y + 4z – 9 = 0

4.4 .4 .4 . Section FormulaSection FormulaSection FormulaSection FormulaIf point P divides the distance between the points A (x

1, y

1, z

1) and B (x

2, y

2, z

2) in the ratio of m : n, then

coordinates of P are given as

+

+

+

+

+

+

nm

nzmz,

nm

nymy,

nm

nxmx 121212

Note :- Mid point

+++

2

zz,

2

yy,

2

xx 212121

5.5 .5 .5 . Centroid of a triangleCentroid of a triangleCentroid of a triangleCentroid of a triangle

G ≡

++++++

3

zzz,

3

yyy,

3

xxx 321321321

6.6 .6 .6 . Incentre of triangle ABC:Incentre of triangle ABC:Incentre of triangle ABC:Incentre of triangle ABC:

++

++

++

++

++

++

cba

czbzaz,

cba

cybyay,

cba

cxbxax 321321321

Where AB = c, BC = a, CA = b

7.7.7.7. Centroid of a tetrahedronCentroid of a tetrahedronCentroid of a tetrahedronCentroid of a tetrahedron A (x1, y

1, z

1) B (x

2, y

2, z

2) C (x

3, y

3, z

3) and D (x

4, y

4, z

4) are the

vertices of a tetrahedron then coordinate of its centroid (G) is given as

∑∑∑4

z,

4

y,

4

x iii

Example : Show that the points A(2, 3, 4), B(–1, 2, –3) and C(–4, 1, –10) are collinear. Also find the ratio inwhich C divides AB.

Solution: Given A ≡ (2, 3, 4), B ≡ (–1, 2, –3), C ≡ (– 4, 1, –10).

A (2, 3, 4) B (–1, 2, –3)Let C divide AB internally in the ratio k : 1, then

C ≡

+

+−

+

+

+

+−

1k

4k3,

1k

3k2,

1k

2k

∴1k

2k

+

+− = – 4 ⇒ 3k = – 6 ⇒ k = –2

For this value of k, 1k

3k2

+

+ = 1, and

1k

4k3

+

+− = –10

Since k < 0, therefore k divides AB externally in the ratio 2 : 1 and points A, B, C are collinear.

Example : The vertices of a triangle are A(5, 4, 6), B(1, –1, 3) and C(4, 3, 2). The internal bisector of ∠ BACmeets BC in D. Find AD.

Solution AB = 25354 222 =++

AC = 23411 222 =++

A(5, 4, 6)

B D C(1, –1, 3) (4, 3, 2)

Since AD is the internal bisector of ∠ BAC

∴3

5

AC

AB

DC

BD==

∴ D divides BC internally in the ratio 5 : 3

∴ D ≡

+

×+×

+

−+×

+

×+×

35

3325,

35

)1(335,

35

1345or, D =

8

19,

8

12,

8

23

∴ AD =

222

8

196

8

124

8

235

−+

−+

−

= 8

1530 unit

Example :

If the points P, Q, R, S are (4, 7, 8), (– 1, – 2, 1), (2, 3, 4) and (1, 2, 5) respectively, show that PQ andRS intersect. Also find the point of intersection.

SolutionLet the lines PQ and RS intersect at point A.

P(4, 7, 8) S(1, 2, 5)

R(2, 3, 4) Q(–1, –2, 1)

AK

1

λ l

Let A divide PQ in the ratio λ : 1, then

A ≡

+λ

+λ

+λ

+λ−

+λ

+λ−

1

8,

1

72,

1

4. .... (1)

Let A divide RS in the ratio k : 1, then

A ≡

+

+

+

+

+

+

1k

4k5,

1k

3k2,

1k

2k..... (2)

From (1) and (2), we have,

1k

2k

1

4

+

+=

+λ

+λ−..... (3)

1k

3k2

1

72

+

+=

+λ

+λ−..... (4)

1k

4k5

1

8

+

+=

+λ

+λ..... (5)

From (3), – λk – λ + 4k + 4 = λk + 2λ + k + 2or 2λk + 3λ – 3k – 2 = 0 ..... (6)From (4), –2λk – 2λ + 7k + 7 = 2λk + 3λ + 2k + 3or 4λk + 5λ – 5k – 4 = 0 ..... (7)Multiplying equation (6) by 2, and subtracting from equation (7), we get

– λ + k = 0 or , λ = kPutting λ = k in equation (6), we get

2λ2 + 3λ – 3λ – 2 = 0or, λ = ± 1.But λ ≠ –1, as the co-ordinates of P would then be underfined and in this case

PQ || RS, which is not true.∴ λ = 1 = k.Clearly λ k = 1 satisfies eqn. (5).Hence our assumption is correct

∴ A ≡

++−+−

2

81,

2

72,

2

41or, A ≡

2

9,

2

5,

2

3.

Self practice problems:1. Find the ratio in which xy plane divides the line joining the points A (1, 2, 3) and B (2, 3, 6).

Ans. – 1 : 22. Find the co-ordinates of the foot of perpendicular drawn from the point A(1, 2, 1) to the line joining the

point B(1, 4, 6) and C(5, 4, 4).Ans. (3, 4, 5)

3. Two vertices of a triangle are (4, –6, 3) and (2, –2, 1) and its centroid is

− 2,1,

3

8. Find the third vertex.

Ans. (2, 5, 2)

4. If centroid of the tetrahedron OABC, where co-ordinates of A, B, C are (a, 2, 3), (1, b, 2) and (2, 1, c)respectively be (1, 2, 3), then find the distance of point (a, b, c) from the origin.

Ans. 107

5. Show that

− 0,2,

2

1 is the circumcentre of the triangle whose vertices are A (1, 1, 0), B (1, 2, 1) and

C (– 2, 2, –1) and hence find its orthocentre. Ans. (1, 11, 0)

8.8 .8 .8 . Direction Cosines And Direction RatiosDirection Cosines And Direction RatiosDirection Cosines And Direction RatiosDirection Cosines And Direction Ratios(i) Direction cosines: Let α, β, γ be the angles which a directed

line makes with the positive directions of the axes of x, y andz respectively, then cos α, cosβ, cos γ are called the directioncosines of the line. The direction cosines are usually denotedby (�, m, n).Thus � = cos α, m = cos β, n = cos γ.

(ii) If �, m, n be the direction cosines of a line, then �2 + m2 + n2 = 1

(iii) Direction ratios: Let a, b, c be proportional to the direction cosines �, m, n then a, b, c arecalled the direction ratios.

If a, b, c, are the direction ratios of any line L then kcjbia ++ will be a vector parallel to the line L.

If �, m, n are direction cosines of line L then � i + m j + n k is a unit vector parallel to the line L.

(iv) If �, m, n be the direction cosines and a, b, c be the direction ratios of a vector, then

++=

++=

++=

222222222 cba

cn,

cba

bm,

cba

a�

or � = 222 cba

a

++

−, m = 222 cba

b

++

−, n = 222 cba

c

++

−

(v) If OP = r, when O is the origin and the direction cosines of OP are �, m, n then the coordinatesof P are (�r, mr, nr).I f d i r e c t i o n c o s i n e s o f t h e l i n e A B a r e I f d i r e c t i o n c o s i n e s o f t h e l i n e A B a r e I f d i r e c t i o n c o s i n e s o f t h e l i n e A B a r e I f d i r e c t i o n c o s i n e s o f t h e l i n e A B a r e �, m,n, |AB| = r, and the coordinates of A is (x

1, y

1, z

1)

then the coordinates of B is given as (x1 + r�, y

1 + rm, z

1 + rn)

(vi) If the coordinates P and Q are (x1, y

1, z

1) and (x

2, y

2, z

2) then the direction ratios of line PQ are,

a = x2 − x

1, b = y

2 − y

1 & c = z

2 − z

1 and the direction cosines of line PQ are � =

|PQ|

xx 12 −,

m =|PQ|

yy 12 − and n =

|PQ|

zz 12 −

(vii) Direction cosines of axes: Since the positive x−axis makes angles 0º, 90º, 90º with axes of x,y and z respectively. ThereforeDirection cosines of x−axis are (1, 0, 0)Direction cosines of y−axis are (0, 1, 0)Direction cosines of z−axis are (0, 0, 1)

Example : If a line makes angles α, β, γ with the co-ordinate axes, prove that sin2α + sin2β + sinγ2 = 2.Solution Since a line makes angles α, β, γ with the co-ordinate axes,

hence cosα, cosβ, cosγ are its direction cosines∴ cos2α + cos2β + cos2γ = 1⇒ (1 – sin2α) + (1 – sin2β) + (1 – sin2γ) = 1⇒ sin2α + sin2β + sin2γ = 2.

Example : Find the direction cosines �, m, n of a line which are connected by the relations � + m + n = 0,2mn + 2m� – n� = 0

Solution Given, � + m + n = 0 ..... (1)2mn + 2m� – n� = 0 ..... (2)

From (1), n = – (� + m).Putting n = – (� + m) in equation (2), we get,

– 2m(� + m) + 2m� + (� + m) � = 0or, – 2m� – 2m2 + 2m� + �2 + m� = 0or, �2 + m� – 2m2 = 0

or,

+

mm

2��

– 2 = 0 [dividing by m2]

or2

31

2

811

m

±−=

+±−=�

= 1, –2

Case I. when m

� = 1 : In this case m = �

From (1), 2� + n = 0 ⇒ n = – 2�∴ � : m : n = 1 : 1 : – 2∴ Direction ratios of the line are 1, 1, – 2∴ Direction cosines are

± 222 )2(11

1

−++, ±

222 )2(11

1

−++, ±

222 )2(11

2

−++

−

or,6

1,

6

1,

6

2−or –

6

1, –

6

1,

6

2

Case II. When m

� = – 2 : In this case � = – 2m

From (1), – 2m + m + n = 0 ⇒ n = m∴ � : m : n = – 2m : m : m

= – 2 : 1 : 1∴ Direction ratios of the line are – 2, 1, 1.∴ Direction cosines are

222 11)2(

2

++−

−,

222 11)2(

1

++−

−,

222 11)2(

1

++−

− or,6

2−,

6

1,

6

1.

Self practice problems:1. Find the direction cosine of a line lying in the xy plane and making angle 30° with x-axis.

Ans. m = ± 2

1, � =

2

3, n = 0

2. A line makes an angle of 60° with each of x and y axes, find the angle which this line makes withz-axis. Ans. 45°

3. A plane intersects the co-ordinates axes at point A(a, 0, 0), B(0, b, 0), C(0, 0, c) O is origin. Find thedirection ratio of the line joining the vertex B to the centroid of face AOC.

Ans.3

a, – b,

3

c

4. A line makes angles α, β, γ, δ with the four diagonals of a cube, prove that

cos2α + cos2β + cos2γ + cos2δ = 3

4.

9.9 .9 .9 . Angle Between Two Line Segments:Angle Between Two Line Segments:Angle Between Two Line Segments:Angle Between Two Line Segments:If two lines have direction ratios a

1, b

1, c

1 and a

2, b

2, c

2 respectively then we can consider two vectors

parallel to the lines as a1i + b

1j + c

1k and a

2i + b

2j + c

2k and angle between them can be given as.

cos θ =22

22

22

21

21

21

212121

cbacba

ccbbaa

++++

++.

(i)The line will be perpendicular if a1a

2 + b

1b

2 + c

1c

2 = 0 (ii)The lines will be parallel if

2

1

a

a =

2

1

b

b =

2

1

c

c

(iii) Two parallel lines have same direction cosines i.e. �1 = �

2, m

1 = m

2, n

1 = n

2

Example : What is the angle between the lines whose direction cosines are 2

3,

4

1,

4

3;

2

3,

4

1,

4

3−−− ?

Solution Let θ be the required angle, thencosθ = �

1�

2 + m

1m

2 + n

1n

2

=

−+

+

−

−

2

3.

2

3

4

1

4

1

4

3

4

3

= 2

1

4

3

16

1

16

3−=−+ ⇒ θ = 120°,

Example : Find the angle between any two diagonals of a cube.Solution The cube has four diagonals

Y

G(0, a, 0) F(a, a, 0)

A X(a, 0, 0)

B(a, 0, a)C(0, 0, a)

(0, a, a)D

E

O(0, 0, 0)

Z’

(a, a, a)

OE, AD, CF and GBThe direction ratios of OE are

a, a, a or, 1, 1, 1

∴ its direction cosines are 3

1,

3

1,

3

1.

Direction ratios of AD are – a, a, a. or, – 1, 1, 1.

∴ its direction cosines are 3

1,

3

1,

3

1−.

Similarly, direction cosines of CF and GB respectively are

3

1,

3

1,

3

1 − and

3

1,

3

1,

3

1 −.

We take any two diagonals, say OE and ADLet θ be the acute angle between them, then

cosθ = 3

1

3

1.

3

1

3

1.

3

1

3

1

3

1=

+

+

−

or, θ = cos–1

3

1.

Example : If two pairs of opposite edges of a tetrahedron are mutually perpendicular, show that the third pairwill also be mutually perpendicular.

Solution: Let OABC be the tetrahedron where O is the origin and co-ordinates of A, B, C be (x1, y

1, z

1),

(x2, y

2, z

2), (x

3, y

3, x

3) respectively.

O (0, 0, 0)

CB

A (x , y1 1 1, z )

(x , y2 2 2, z ) (x , y3 3 3, z )

Let BCOA ⊥ and CAOB ⊥ .We have to prove that

BAOC ⊥ .Now, direction ratios of OA are x

1 – 0, y

1 – 0, z

1 – 0 or, x

1, y

1, z

1direction ratios of BC are (x

3 – x

2), (y

3 – y

2), (z

3 – z

2).

∵ BCOA ⊥ .∴ x

1(x

3 – x

2) + y

1(y

3 – y

2) + z

1(z

3 – z

2) = 0 ..... (1)

Similarly,

∵ CAOB ⊥∴ x

2(x

1 – x

3) + y

2(y

1 – y

3) + z

2(z

1 – z

3) = 0 ..... (2)

Adding equations (1) and (2), we getx

3(x

1 – x

2) + y

3(y

1 – y

2) + z

3(z

1 – z

2) = 0

∴ BAOC ⊥ [∵ direction ratios of OC are x3, y

3, z

3 and that of BA are (x

1 – x

2), (y

1 – y

2), (z

1 – z

2)]

Self practice problems:1. Find the angle between the lines whose direction cosines are given by � + m + n = 0 and

�2 + m2 – n2 = 0 Ans. 60°2. P (6, 3, 2)

Q (5, 1, 4)R (3, 3, 5)are vertices of a ∆ find ∠Q. Ans. 90°

3. Show that the direction cosines of a line which is perpendicular to the lines having directions cosines�

1 m

1 n

1 and �

2 m

2 n

2 respectively are proportional to

m1n

2 – m

2n

1 , n

1�

2 – n

2�

1, �

1m

2 – �

2m

1

10 .10 .10 .10 . Projection of a line segment on a lineProjection of a line segment on a lineProjection of a line segment on a lineProjection of a line segment on a line(i) If the coordinates P and Q are (x

1, y

1, z

1) and (x

2, y

2, z

2) then the projection of the line segments

PQ on a line having direction cosines �, m, n is )zz(n)yy(m)xx( 121212 −+−+−�

(ii) Vector form: projection of a vector a�

on another vector b�

is a�

. b =|b|

b.a�

��

In the above case we can consider PQ→

as (x2 – x

1) i + (y

2 – y

1) j + (z

2 – z

1) k in place of a

� and

� i + m j + n k in place of b�

. (iii) � | r�

|, m | r� | & n | r

� | are the projection of r� in OX, OY &

OZ axes. (iv) r�

= | r�

| (� i + m j + n k )Solved Example : Find the projection of the line joining (1, 2, 3) and (–1, 4, 2) on the line having direction

ratios 2, 3, – 6.Solution Let A ≡ (1, 2, 3), B ≡ (–1, 4, 2)

P L M Q

90° 90°

AB

Direction ratios of the given line PQ are 2, 3, – 6

222 )6(32 −++ = 7 ∴ direction cosines of PQ are

7

2,

7

3, –

7

6

Projection of AB on PQ= � (x

2 – x

1) + m(y

2 – y

1) + n(z

2 – z

1)

= 7

2 (–1 – 1) +

7

3 (4 – 2) –

7

6 (2 – 3) =

7

8

7

664=

++−

Self practice problems:1. A (6, 3, 2), B (5, 1, 1,), C(3, –1, 3) D (0, 2, 5)

Find the projection of line segment AB on CD line. Ans. 5/7

2. The projections of a directed line segment on co-ordinate axes are – 2, 3, – 6. Find its length and

direction cosines. Ans. 13 ; 13

12,

13

4,

13

3

3. Find the projection of the line segment joining (2, – 1, 3) and (4, 2, 5) on a line which makes equal

acute angles with co-ordinate axes. Ans.3

7

A PLANEA PLANEA PLANEA PLANEIf line joining any two points on a surface lies completely on it then the surface is a plane.

ORIf line joining any two points on a surface is perpendicular to some fixed straight line. Then this surfaceis called a plane. This fixed line is called the normal to the plane.

11 .11 .11 .11 . Equation Of A PlaneEquation Of A PlaneEquation Of A PlaneEquation Of A Plane(i) Normal form of the equation of a plane is �x + my + nz = p, where, �,m n are the direction

cosines of the normal to the plane and p is the distance of the plane from the origin.(ii) General form: ax + by + cz + d = 0 is the equation of a plane, where a, b, c are the

direction ratios of the normal to the plane.(iii) The equation of a plane passing through the point (x

1, y

1, z

1) is given by

a (x − x1) + b( y − y

1) + c (z − z

1) = 0 where a, b, c are the direction ratios of the normal

to the plane.(iv) Plane through three points: The equation of the plane through three non−collinear points

(x1, y

1, z

1), (x

2, y

2, z

2), (x

3, y

3, z

3) is

1zyx

1zyx

1zyx

1zyx

333

222

111 = 0

(v) Intercept Form: The equation of a plane cutting intercept a, b, c on the axes is 1c

z

b

y

a

x=++

(vi) Vector form: The equation of a plane passing through a point having position vector a�

&

normal to vector n�

is ( r�

− a�

). n�

= 0 or r�

. n�

= a�

. n�

Note: (a) Vector equation of a plane normal to unitvector n and at a distance d from the origin is

r� . n� = d

(b) Coordinate planes (i) Equation of yz−plane is x = 0 (ii) Equation of xz−plane is y = 0(iii) Equation of xy−plane is z = 0

(c) Planes parallel to the axes:If a = 0, the plane is parallel to x−axis i.e. equation of the plane parallel to the x−axis isby + cz + d = 0.Similarly, equation of planes parallel to y−axis and parallel to z−axis are ax + cz +d = 0and ax + by + d = 0 respectively.

(d) Plane through origin: Equation of plane passing through origin is ax + by + cz = 0.

(e) Transformation of the equation of a plane to the normal form: To reduce any equationax + by + cz − d = 0 to the normal form, first write the constant term on the right hand

side and make it positive, then divide each term by 222 cba ++ , where a, b, c are

coefficients of x, y and z respectively e.g.

222 cba

ax

++± +

222 cba

by

++± +

222 cba

cz

++± =

222 cba

d

++±Where (+) sign is to be taken if d > 0 and (−) sign is to be taken if d < 0.

(f) Any plane parallel to the given plane ax + by + cz + d = 0 is ax + by + cz + λ = 0.Distance between two parallel planes ax + by + cz + d

1 = 0 and ax + by + cz + d

2 = 0 is

given as 222

21

cba

|dd|

++

−

(g) Equation of a plane passing through a given point & parallel to the given vectors:

The equation of a plane passing through a point having position vector a�

and parallel to

b�

& c�

is r�

+ λ b�

+ µ c�

(parametric form) where λ & µ are scalars.

or r�

. )cb(��

× = a�

. )cb(��

ratio

+++

+++−

dczbyax

dczbyax

222

111

× (non parametric form)

(h) A plane ax + by + cz + d = 0 divides the line segment joining (x1, y

1, z

1) and (x

2, y

2, z

2). in the

= a�

(i) The xy−plane divides the line segment joining the points (x1, y

1, z

1) and (x

2, y

2, z

2) in the ratio

−2

1

z

z. Similarly yz−plane in −

2

1

x

x and zx−plane in −

2

1

y

y

(j) Coplanarity of four pointsThe points A(x

1 y

1 z

1), B(x

2 y

2 z

2) C(x

3 y

3 z

3) and D(x

4 y

4 z

4) are coplaner then

141414

131313

121212

zzyyxx

zzyyxx

zzyyxx

−−−

−−−

−−−

= 0

very similar in vector method the points A ( 1r�

), B( 2r�

), C( 3r�

) and D( 4r�

) are coplanar if

[ 4r�

– 1r�

, 4r�

– 2r�

, 4r�

– 3r�

] = 0Example : Find the equation of the plane upon which the length of normal from origin is 10 and direction ratios

of this normal are 3, 2, 6.Solution If p be the length of perpendicular from origin to the plane and �, m, n be the direction cosines

of this normal, then its equation is�x + my + nz = p ..... (1)

Here p = 10; Direction ratios of normal to the plane are 3, 2, 6

222 623 ++ = 7 ∴ Direction cosines of normal to the required plane are

� = 7

3, m =

7

2, n =

7

6

Putting the values of �, m, n, p in (1), equation of required plane is

z7

6y

7

2x

7

3++ = 10 or, 3x + 2y + 6z = 70

Example : Show that the points (0, – 1, 0), (2, 1, – 1), (1, 1, 1), (3, 3, 0) are coplanar.Solution Let A ≡ (0, – 1, 0), B ≡ (2, 1, – 1), C ≡ (1, 1, 1) and D ≡ (3, 3, 0)

Equation of a plane through A (0, – 1, 0) isa (x – 0) + b (y + 1) + c (z – 0) = 0

or, ax + by + cz + b = 0 ..... (1)If plane (1) passes through B (2, 1, – 1) and C (1, 1, 1)Then 2a + 2b – c = 0 ..... (2)and a + 2b + c = 0 ..... (3)From (2) and (3), we have

24

c

21

b

22

a

−=

−−=

+

or,2

c

3

b

4

a=

−= = k (say)

Putting the value of a, b, c, in (1), equation of required plane is4kx – 3k(y + 1) + 2kz = 0

or, 4x – 3y + 2z – 3 = 0 ..... (2)Clearly point D (3, 3, 0) lies on plane (2)Thus points D lies on the plane passing through A, B, C and hence points A, B, C and D are coplanar.

Example : If P be any point on the plane �x + my + nz = p and Q be a point on the line OP such thatOP . OQ = p2, show that the locus of the point Q is p(�x + my + nz) = x2 + y2 + z2.

Solution Let P ≡ (α, β, γ), Q ≡ (x1, y

1, z

1)

Direction ratios of OP are α, β, γ and direction ratios of OQ are x1, y

1, z

1.

Since O, Q, P are collinear, we have

111 zyx

γ=

β=

α = k (say) ..... (1)

As P (α, β, γ) lies on the plane �x + my + nz = p,�α + mβ + nγ = p or k(x

1 + my

1 + nz

1) = p ..... (2)

Given, OP . OQ = p2

∴ 21

21

21

222 zyx ++γ+β+α = p2

or, 21

21

21

21

21

21

2 zyx)zyx(k ++++ = p2

or, k )zyx( 21

21

21 ++ = p2 ..... (3)

On dividing (2) by (3), we get,

p

1

zyx

nzmyx21

21

21

111 =++

++�

or, p (�x1 + my

1 + nz

1) =

21

21

21 zyx ++

Hence the locus of point Q isp (�x + my + nz) = x2 + y2 + z2.

Example : A point P moves on a plane c

z

b

y

a

x++ = 1. A plane through P and perpendicular to OP meets the

co-ordinate axes in A, B and C. If the planes through A, B and C parallel to the planes x = 0, y = 0,z = 0 intersect in Q, find the locus of Q.

Solution Given plane is

1c

z

b

y

a

x=++ ..... (1)

Let P ≡ (h, k, �)

Then cb

k

a

h �++ = 1 ..... (2)

OP = 222 kh �++

Direction cosines of OP are 222222222 kh,

kh

k,

kh

h

�

�

�� ++++++∴ Equation of the plane through P and normal to OP is

222

222222222kh

khy

kh

kx

kh

h�

�

�

��

++=++

+++

+++

or, hx + ky + �z = (h2 + k2 + �2)

∴ A ≡

++0,0,

h

kh 222�

, B ≡

++0,

k

kh,0

222�

,

C ≡

++

�

�222 kh

,0,0

Let Q ≡ (α, β, γ), then

α = h

kh 222�++

, β = k

kh 222�++

, γ = �

�222 kh ++

..... (3)

Now)kh(

1

)kh(

kh1112222222

222

222��

�

++=

++

++=

γ+

β+

α..... (4)

From (3), h = α

++ 222 kh �

∴α

++=

a

kh

a

h 222�

Similarly β

++=

b

kh

b

k 222�

and γ

++=

c

kh

c

222��

∴cb

k

a

h

c

kh

b

kh

a

kh 222222222��

++=γ

+++

β

+++

α

++ = 1 [from (2)]

or, 222222

111

kh

1

c

1

b

1

a

1

γ+

β+

α=

++=

γ+

β+

α �[from (4)]

∴ Required locus of Q (α, β, γ) is

222 z

1

y

1

x

1

cz

1

by

1

ax

1++=++ .

Self practice problems :1. Check wether this point are coplanar if yes find the equation of plane containing them

A ≡ (1, 1, 1)B ≡ (0, – 1, 0)C ≡ (2, 1, –1)D ≡ (3, 3, 0) Ans. yes, 4x – 3y + 2z = 3

2. Find the plane passing through point (– 3, – 3, 1) and perpendicular to the line joining the points(2, 6, 1) and (1, 3, 0). Ans. x + 3y + z + 11 = 0

3. Find the equation of plane parallel to x + 5y – 4z + 5 = 0 and cutting intercepts on the axes whose rum

is 150. Ans. x + 5y – 4z = 19

3000

4. Find the equation of plane passing through (2, 2, 1) and (9, 3, 6) and perpendicular to the planex + 3y + 3z = 8. Ans. 3x + 4y – 5z = 9

5. Find the equation of the plane | | to kji ++ and ji − and passing through (1, 1, 2).

Ans. x + y – 2z + 2 = 06. Find the equation of the plane passing through the point (1, 1, – 1) and perpendicular to the planes

x + 2y + 3z – 7 = 0 and 2x – 3y + 4z = 0. Ans. 17x + 2y – 7z = 26

12 .12 .12 .12 . Sides of a plane:Sides of a plane:Sides of a plane:Sides of a plane:A plane divides the three dimensional space in two equal parts. Two points A (x

1 y

1 z

1)

and B (x2 y

2 z

2) are on the same side of the plane ax + by + cz + d = 0 if ax

1 + by

1 + cz

1 + d and

TEKO CLASSES, H.O.D. MATHS : SUHAG R. KARIYA (S. R. K. Sir) PH: (0755)- 32 00 000, 0 98930 58881 , BHOPAL, (M.P.)

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ax2 + by

2 + cz

2 + d are both positive or both negative and are opposite side of plane if both of these

values are in opposite sign.Example : Show that the points (1, 2, 3) and (2, – 1, 4) lie on opposite sides of the plane x + 4y + z – 3 = 0.Solution Since the numbers 1+ 4 × 2 + 3 – 3 = 9 and 2 – 4 + 4 – 3 = – 1 are of opposite sign., the points are

on opposite sides of the plane.

13 .13 .13 .13 . A Plane & A PointA Plane & A PointA Plane & A PointA Plane & A Point

(i) Distance of the point (x′, y′, z′) from the plane ax + by + cz+ d = 0 is given by 222

cba

d'cz'by'ax

++

+++.

(ii) The length of the perpendicular from a point having position vector a�

to plane r�

. n�

= d is

given by p =|n|

|dn.a|�

��−

.

(iii) The coordinates of the foot of perpendicular from the point (x, y, z) to the plane

ax + by + cz + d = 0 aregain by c

z'z

b

y'y

a

x'x 111 −=

−=

− = –

222

111

cba

)dczbyax(

++

+++

(iv) To find image of a point w.r.t. a plane.Let P (x

1, y

1, z

1) is a given point and ax + by + cz + d = 0 is given plane Let (x′, y′, z′) is the

image point. then(a) x′ – x

1 = λa, y′ – y

1 = λb, z′ – z

1 = λc

⇒ x′ = λa + x1, y′ = λb + y

1, z′ = λc + z

1

(b) a

+′

2

xx 1 + b

+′

2

yy 1 + c

+′

2

zz 1 = 0

from (i) put the values of x′, y′, z′ in (ii) and get the values of λ and resubtitute in (i) to get(x′ y′ z′).The coordinate of the image of point (x

1 , y

1 , z

1) w.r.t the plane ax + by + cz + d = 0 are given

by c

z'z

b

y'y

a

x'x 111 −=

−=

− = – 2

222

111

cba

)dczbyax(

++

+++

(v) The distance between two parallel planes ax + by + cx + d = 0 and ax + by + cx + d’ = 0 is

given by 222 cba

|'dd|

++

−

Example : Find the image of the point P (3, 5, 7) in the plane 2x + y + z = 16.Solution Given plane is 2x + y + z = 16 ..... (1)

P ≡ (3, 5, 7)Direction ratios of normal to plane (1) are 2, 1, 1Let Q be the image of point P in plane (1). Let PQ meet plane (1) in Rthen PQ ⊥ plane (1)Let R ≡ (2r + 3, r + 5, r + 7)Since R lies on plane (1)∴ 2(2r + 3) + r + 5 + r + 7 = 0 or, 6r + 18 = 0 ∴ r = – 3∴ R ≡ (– 3, 2, 4)Let Q ≡ (α, β, γ)Since R is the middle point of PQ

∴ – 3 = 2

3+α ⇒ α = – 9

2 = 2

5+β ⇒ β = – 1

4 = 2

7+γ ⇒ γ = 1 ∴ Q = (– 9, – 1, 1).

Example : Find the distance between the planes 2x – y + 2z = 4 and 6x – 3y + 6z = 2.Solution Given planes are

2x – y + 2z – 4 = 0 ..... (1)and 6x – 3y + 6z – 2 = 0 ..... (2)

We find that 2

1

2

1

2

1

c

c

b

b

a

a== Hence planes (1) and (2) are parallel.

Plane (2) may be written as 2x – y + 2z – 3

2 = 0 ..... (3)

∴ Required distance between the planes

= 9

10

3.3

10

2)1(2

3

24

222==

+−+

−

Example : A plane passes through a fixed point (a, b, c). Show that the locus of the foot of perpendicularto it from the origin is the sphere x2 + y2 + z2 – ax – by – cz = 0

Vec&3D/Page : 12 of 77


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Solution Let the equation of the variable plane beO(0, 0, 0)

P( , , )α β γ

�x + my + nz + d = 0 ..... (1)Plane passes through the fixed point (a, b, c)∴ �a + mb + nc + d = 0 ..... (2)Let P (α, β, γ) be the foot of perpendicular from origin to plane (1).Direction ratios of OP are

α – 0, β – 0, γ – 0 i.e. α, β, γFrom equation (1), it is clear that the direction ratios of normal to the plane i.e. OP are �, m, n ;α, β, γ and �, m, n are the direction ratios of the same line OP

∴�

α =

m

β =

n

γ =

k

1 (say)

∴ � = kα, m = kβ, n = kγ ..... (3)Putting the values of

�, m, n in equation (2), we getkaα + kbβ + kcγ + d = 0 ..... (4)

Since α, β, γ lies in plane (1)∴ �α + mβ + nγ + d = 0 ..... (5)Putting the values of �, m, n from (3) in (5), we get

kα2 + kβ2 + kγ2 + d = 0 ..... (6)or kα2 + kβ2 + kγ2 – kaα – kbβ – kcγ = 0

[putting the value of d from (4) in (6)]or α2 + β2 + γ2 – aα – bβ – cγ = 0Therefore, locus of foot of perpendicular P (α, β, γ) is

x2 + y2 + z2 – ax – by – cz = 0 ..... (7)Self practice problems:1. Find the intercepts of the plane 3x + 4y – 7z = 84 on the axes. Also find the length of perpendicular

from origin to this line and direction cosines of this normal.

Ans. a = 28, b = 21, c = – 12, p = 74

7,

74

4,

74

3;

74

1 −

2. Find : (i) perpendicular distance(ii) foot of perpendicular(iii) image of (1, 0, 2) in the plane 2x + y + z = 5

Ans. (i) 6

1(ii)

6

13,

6

1,

3

4(iii)

3

7,

3

1,

3

5

14 .14 .14 .14 . Angle Between Two Planes:Angle Between Two Planes:Angle Between Two Planes:Angle Between Two Planes:(i) Consider two planes ax + by + cz + d = 0 and a′x + b′y + c′z + d′ = 0. Angle between these

planes is the angle between their normals. Since direction ratios of their normals are (a, b, c)and (a′, b′, c′) respectively, hence θ, the angle between them, is given by

cos θ = 2'2'2'222 cbacba

'cc'bb'aa

++++

++

Planes are perpendicular if aa′ + bb′ + cc′ = 0 and planes are parallel if 'a

a=

'b

b =

'c

c

(ii) The angle θ between the planes r�

. n�

= d1 and r

�. 2n�

= d2 is given by, cos θ =

|n|.|n|

n.n

21

21��

��

Planes are perpendicular if1n�

.2n�

= 0 & planes are parallel if 1n�

= λ 2n�

.

15 .15 .15 .15 . Angle BisectorsAngle BisectorsAngle BisectorsAngle Bisectors(i) The equations of the planes bisecting the angle between two given planes

a1x + b

1y + c

1z + d

1 = 0 and a

2x + b

2y + c

2z + d

2 = 0 are

21

21

21

1111

cba

dzcybxa

++

+++ = ±

22

22

22

2222

cba

dzcybxa

++

+++

(ii) Equation of bisector of the angle containing origin: First make both the constant terms positive.

Then the positive sign in 21

21

21

1111

cba

dzcybxa

++

+++= ±

22

22

22

2222

cba

dzcybxa

++

+++ gives the bisector of

the angle which contains the origin.(iii) Bisector of acute/obtuse angle: First make both the constant terms positive. Then

a1a

2 + b

1b

2 + c

1c

2 > 0 ⇒ origin lies on obtuse angle



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a1a

2 + b

1b

2 + c

1c

2 < 0 ⇒ origin lies in acute angle

16 .16 .16 .16 . Family of PlanesFamily of PlanesFamily of PlanesFamily of Planes(i) Any plane passing through the line of intersection of non−parallel planes or equation of

the plane through the given line in serval form.a

1x + b

1y + c

1z + d

1 = 0 & a

2x + b

2y + c

2z + d

2 = 0 is

a1x + b

1y + c

1z + d

1 + λ (a

2x + b

2y + c

2z + d

2) = 0

(ii) The equation of plane passing through the intersection of the planes r�

. 1n�

= d1 &

r�

. 2n�

= d2 is r�

. (n1 + λ 2n�

) = d1 + λd

2 where λ is arbitrary scalar

Example : The plane x – y – z = 4 is rotated through 90° about its line of intersection with the planex + y + 2z = 4. Find its equation in the new position.

Solution Given planes arex – y – z = 4 ..... (1)

and x + y + 2z = 4 ..... (2)Since the required plane passes through the line of intersection of planes (1) and (2)∴ its equation may be taken as

x + y + 2z – 4 + k (x – y – z – 4) = 0or (1 + k)x + (1 – k)y + (2 – k)z – 4 – 4k = 0 ..... (3)Since planes (1) and (3) are mutually perpendicular,∴ (1 + k) – (1 – k) – (2 – k) = 0

or, 1 + k – 1 + k – 2 + k = 0 or, k = 3

2

Putting k = 3

2 in equation (3), we get,

5x + y + 4z = 20This is the equation of the required plane.

Example : Find the equation of the plane through the point (1, 1, 1) which passes through the line ofintersection of the planes x + y + z = 6 and 2x + 3y + 4z + 5 = 0.

Solution Given planes arex + y + z – 6 = 0 ..... (1)

and 2x + 3y + 4z + 5 = 0 ..... (2)Given point is P (1, 1, 1).Equation of any plane through the line of intersection of planes (1) and (2) is

x + y + z – 6 + k (2x + 3y + 4z + 5) = 0 ..... (3)If plane (3) passes through point P, then

1 + 1 + 1 – 6 + k (2 + 3 + 4 + 5) = 0 or, k = 14

3

From (1) required plane is20x + 23y + 26z – 69 = 0

Example : Find the planes bisecting the angles between planes2x + y + 2z = 9 and 3x – 4y + 12z + 13 = 0.

Which of these bisector planes bisects the acute angle between the given planes. Does origin lie in theacute angle or obtuse angle between the given planes ?

Solution Given planes are– 2x – y – 2z + 9 = 0 ..... (1)

and 3x – 4y + 12z + 13 = 0 ..... (2)Equations of bisecting planes are

222222 )12()4(3

13z12y4x3

)2()1()2(

9z2yx2

+−+

++−±=

−+−+−

+−−−

or, 13 [– 2x – y – 2z + 9] = ± 3 (3x – 4y + 12z + 13)or, 35x + y + 62z = 78, ..... (3) [Taking +ve sign]and 17x + 25y – 10z = 156 ..... (4) [Taking – ve sign]Now a

1a

2 + b

1b

2 + c

1c

2 = (– 2) (3) + (– 1) (– 4) + (– 2) (12)

= – 6 + 4 – 24 = – 26 < 0∴ Bisector of acute angle is given by 35x + y + 62z = 78∵ a

1a

2 + b

1b

2 + c

1c

2 < 0, origin lies in the acute angle between the planes.

Example : If the planes x – cy – bz = 0, cx – y + az = 0 and bx + ay – z = 0 pass through a straight line,then find the value of a2 + b2 + c2 + 2abc.

Solution Given planes arex – cy – bz = 0 ..... (1)cx – y + az = 0 ..... (2)bx + ay – z = 0 ..... (3)

Equation of any plane passing through the line of intersection of planes (1) and (2) may be taken asx – cy – bz + λ (cx – y + az) = 0

or, x (1 + λc) – y (c + λ) + z (– b + aλ) = 0 ..... (4)If planes (3) and (4) are the same, then equations (3) and (4) will be identical.

∴1

ab

a

)c(

b

c1

−

λ+−=

λ+−=

λ+

(i) (ii) (iii)From (i) and (ii), a + acλ = – bc – bλ



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or, λ = – )bac(

)bca(

+

+..... (5)

From (ii) and (iii),

c + λ = – ab + a2λ or λ = 2a1

)cab(

−

+..... (6)

From (5) and (6), we have,

)a1(

)cab(

bac

)bca(2−

+−=

+

+−.

or, a – a3 + bc – a2bc = a2bc + ac2 + ab2 + bcor, a2bc + ac2 + ab2 + a3 + a2bc – a = 0or, a2 + b2 + c2 + 2abc = 1.

Self practice problems:1. A tetrahedron has vertices at O(0, 0, 0), A(1, 2, 1), B(2, 1, 3) and C(–1, 1, 2). Prove that the angle

between the faces OAB and ABC will be cos–1

35

19.

2. Find the equation of plane passing through the line of intersection of the planes 4x – 5y – 4z = 1 and

2x + y + 2z = 8 and the point (2, 1, 3). Ans. 32x – 5y + 8z – 83 = 0, λ = 3

10

3. Find the equations of the planes bisecting the angles between the planesx + 2y + 2z – 3 = 0, 3x + 4y + 12z + 1 = 0 and sepecify the plane which bisects the acute anglebetween them. Ans. 2x + 7y – 5z = 21, 11x + 19y + 31z = 18; 2x + 7y – 5z = 21

4. Show that the origin lies in the acute angle between the planesx + 2y + 2z – 9 = 0 and 4x – 3y + 12z + 13 = 0

5. Prove that the planes 12x – 15y + 16z – 28 = 0, 6x + 6y – 7z – 8 = 0 and 2x + 35y – 39z + 12 = 0 havea common line of intersection.

17.17.17.17. Area of a triangle:Area of a triangle:Area of a triangle:Area of a triangle:

Let A (x1, y

1, z

1), B (x

2, y

2, z

2), C (x

3, y

3, z

3) be the vertices of a triangle, then ∆ = )( 2

z2y

2x ∆+∆+∆

where ∆x =

2

1

1zy

1zy

1zy

33

22

11

, ∆y =

2

1

1xz

1xz

1xz

33

22

11

and ∆z =

1yx

1yx

1yx

33

22

11

Vector Method − From two vector→

AB and→

AC . Then area is given by

2

1 |

→

AB x→

AC | =2

1

131313

121212

zzyyxx

zzyyxx

kji

−−−

−−−

Example : Through a point P (h, k, �) a plane is drawn at right angles to OP to meet the co-ordinate axes in A,

B and C. If OP = p, show that the area of ∆ABC is �hk2

p5

.

Solution OP = 222 kh �++ = p

Direction cosines of OP are

222222222 kh,

kh

k,

kh

h

�

�

�� ++++++Since OP is normal to the plane, therefore, equation of the plane will be,

222

222222222khz

khy

kh

kx

kh

h�

�

�

��

++=++

+++

+++

or, hx + ky + �z = h2 + k2 + �2 = p2 ..... (1)

∴ A ≡

0,0,

h

p2

, B ≡

0,

k

p,0

2

, C ≡

�

2p,0,0

Now area of ∆ABC, ∆ = A2xy

+ A2yz

+ A2zx

Now Axy

= area of projection of ∆ABC on xy-plane = area of ∆AOB

= Mod of 2

1

100

1k

p0

10h

p

2

2

= |hk|

p

2

1 4



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Similarly, Ayz

= |k|

p

2

1 4

� and AA

zx =

|h|

p

2

1 4

�

∴ ∆2 = 22

8

22

8

22

8

h

p

4

1

k

p

4

1

kh

p

4

1

��++

= 222

8

kh4

p

� (�2 + k2 + h2) = 222

10

kh4

p

�or, ∆ =

�hk2

p5

.

18 .18 .18 .18 . Volume Of A TetrahedronVolume Of A TetrahedronVolume Of A TetrahedronVolume Of A Tetrahedron:Volume of a tetrahedron with vertices A (x

1, y

1, z

1), B( x

2, y

2, z

2), C (x

3, y

3, z

3) and

D (x4, y

4, z

4) is given by V =

6

1

1zyx

1zyx

1zyx

1zyx

444

333

222

111

A LINEA LINEA LINEA LINE19 .19 .19 .19 . Equation Of A LineEquation Of A LineEquation Of A LineEquation Of A Line(i) A straight line in space is characterised by the intersection of two planes which are not parallel and

therefore, the equation of a straight line is a solution of the system constituted by the equations of thetwo planes, a

1x + b

1y + c

1z + d

1 = 0 and a

2x + b

2y + +c

2z + d

2 =0. This form is also known as non−

symmetrical form.

(ii) The equation of a line passing through the point (x1, y

1, z

1) and having direction ratios a, b, c is

a

xx 1−

=b

yy 1− =

c

zz 1− = r. This form is called symmetric form. A general point on the line is given by (x

1 +

ar, y1 + br, z

1 + cr).

(iii) Vector equation: Vector equation of a straight line passing through a fixed point with position

vector a�

and parallel to a given vector b�

is r�

= a�

+ λ b�

where λ is a scalar..(iv) The equation of the line passing through the points (x

1, y

1, z

1) and (x

2, y

2, z

2) is

12

1

xx

xx

−

− =

12

1

yy

yy

−

− =

12

1

zz

zz

−

−

(v) Vector equation of a straight line passing through two points with position vectors a�

& b�

is r�

= a�

+ λ ( b�

− a�

).(vi) Reduction of cartesion form of equation of a line to vector form & vice versa

a

xx 1− =

b

yy 1− =

c

zz 1−⇔ r

� = (x1 i + y

1 j + z1 k ) + λ (a i + b j + c k ).

Note: Straight lines parallel to co-ordinate axes:Straight lines Equation Straight lines Equation(i) Through origin y = mx, z = nx (v) Parallel to x−axis y = p, z = q(ii) x−axis y = 0, z = 0 (vi) Parallel to y−axis x = h, z = q(iii) y−axis x = 0, z = 0 (vii) Parallel to z−axis x = h, y = p(iv) z−axis x = 0, y = 0

Example : Find the equation of the line through the points (3, 4, –7) and (1, – 1, 6) in vector form as well as incartesian form.

Solution Let A ≡ (3, 4, – 7), B ≡ (1, – 1, 6)

Now→a =

→

OA = →

i3 + →

j4 – →

k7 ,

= →

b = →

OB = →

i – →

j + 6→

k

Equation of the line through A(→

a ) and B(→

b ) is→

r = →

a + t (→

b – →a )

or→

r = 3→

i + 4→

j – 7→

k + t (–2→

i – 5→

j + 13→

k ) ..... (1)Equation in cartesian form :

Equation of AB is 67

7z

14

4y

13

3x

−−

+=

+

−=

−

−or,

13

7z

5

4y

2

3x

−

+=

−=

−

Example : Find the co-ordinates of those points on the line 6

3z

3

2y

2

1x −=

+=

− which is at a distance of

3 units from point (1, –2, 3).



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Solution Given line is 6

3z

3

2y

2

1x −=

+=

−..... (1)

Let P ≡ (1, –2, 3) Direction ratios of line (1) are 2, 3, 6

∴ Direction cosines of line (1) are 7

6,

7

3,

7

2

Equation of line (1) may be written as

7

6

3z

7

3

2y

7

2

1x −=

+=

−..... (2)

Co-ordinates of any point on line (2) may be taken as

+−+ 3r

7

6,2r

7

3,1r

7

2

Let Q ≡

+−− 3r

7

6,2r

7

3,1r

7

2

Distance of Q from P = | r |According to question | r | = 3 ∴ r = ± 3 Putting the value of r, we have

Q ≡

−−

7

39,

7

5,

7

1or Q ≡

−−

7

3,

7

23,

7

13

Example : Find the equation of the line drawn through point (1, 0, 2) to meet at right angles the line

1

1z

2

2y

3

1x

−

+=

−

−=

+

Solution Given line is

1

1z

2

2y

3

1x

−

+=

−

−=

+..... (1)

Let P ≡ (1, 0, 2)Co-ordinates of any point on line (1) may be taken as

Q ≡ (3r – 1, – 2r + 2, – r – 1)Direction ratios of PQ are 3r – 2, – 2r + 2, – r – 3Direction ratios of line AB are 3, – 2, – 1

Since PQ ⊥ AB∴ 3 (3r – 2) – 2 (– 2r + 2) – 1 (– r – 3) = 0

⇒ 9r – 6 + 4r – 4 + r + 3 = 0 ⇒ 14r = 7 ⇒ r = 2

1

Therefore, direction ratios of PQ are

– 2

1, 1, –

2

7 or, – 1, 2, – 7

Equation of line PQ is

7

2z

2

0y

1

1x

−

−=

==

−

− or,

7

2z

2

y

1

1x −=

−=

−

Example : Show that the two lines 4

3z

3

2y

2

1x −=

−=

− and

2

1y

5

4x −=

− = z intersect. Find also the

point of intersection of these lines.

Solution Given lines are4

3z

3

2y

2

1x −=

−=

−..... (1)

and1

0z

2

1y

5

4x −=

−=

−..... (2)

Any point on line (1) is P (2r + 1, 3r + 2, 4r +3)and any point on line (2) is Q (5λ + 4, 2λ + 1, λ)Lines (1) and (2) will intersect if P and Q coincide for some value of λ and r.∴ 2r + 1 = 5λ + 4 ⇒ 2r – 5λ = 3 ..... (1)

3r + 2 + 2λ + 1 ⇒ 3r – 2λ = – 1 ..... (2)4r + 3 = λ ⇒ 4r – λ = – 3 ..... (3)

Solving (1) and (2), we get r = – 1, λ = – 1Clearly these values of r and λ satisfy eqn. (3)Now P ≡ (– 1, – 1, – 1) Hence lines (1) and (2) intersect at (– 1, – 1, – 1).

Self practice problems:1. Find the equation of the line passing through point (1, 0, 2) having direction ratio 3, – 1, 5. Prove that

this line passes through (4, – 1, 7). Ans.5

2z

1

y

3

1x −=

−−

−

2. Find the equation of the line parallel to line 9

7z

1

1y

3

2x −=

+=

− and passing through the point (3, 0, 5).



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Ans.3

5z

1

y

3

3x −==

−

3. Find the coordinates of the point when the line through (3, 4, 1) and (5, 1, 6) crosses the xy plane.

Ans.

0,

5

23,

5

13

20 .20 .20 .20 . Reduction Of Non-Symmetrical Form To Symmetrical Form:Reduction Of Non-Symmetrical Form To Symmetrical Form:Reduction Of Non-Symmetrical Form To Symmetrical Form:Reduction Of Non-Symmetrical Form To Symmetrical Form:Let equation of the line in non−symmetrical form be a

1x + b

1y + c

1z + d

1 = 0, a

2x + b

2y + c

2z + d

2 = 0.

To find the equation of the line in symmetrical form, we must know (i) its direction ratios (ii) coordinateof any point on it.(i) Direction ratios: Let �, m, n be the direction ratios of the line. Since the line lies in both the

planes, it must be perpendicular to normals of both planes. So a1� + b

1m + c

1n = 0,

a2� + b

2m + c

2n = 0. From these equations, proportional values of �, m, n can be found by

cross−multiplication as 1221 cbcb −

� =

1221 acac

m

− =

1221 baba

n

−

Alternative method The vector

222

111

cba

cba

kji

= i (b1c

2 – b

2c

1) + j (c

1a

2 – c

2a

1) + k (a

1b

2 – a

2b

1) will be parallel

to the line of intersection of the two given planes. hence � : m: n = (b1c

2 – b

2c

1): (c

1a

2 – c

2a

1):

(a1b

2 – a

2b

1)

(ii) Point on the line − Note that as �, m, n cannot be zero simultaneously, so at least one mustbe non−zero. Let a

1b

2 − a

2b

1 ≠ 0, then the line cannot be parallel to xy plane, so it intersect it.

Let it intersect xy−plane in (x1, y

1, 0). Then a

1x

1 + b

1y

1 + d

1 = 0 and a

2x

1 + b

2y

1 + d

2 = 0.Solving

these,we get a point on the line. Then its equation becomes.

1221

1

cbcb

xx

−

−=

1221

1

acac

yy

−

−=

1221 baba

0z

−

−or

1221

1221

1221

cbcb

baba

dbdbx

−

−

−−

= 1221

1221

1221

acac

baba

adady

−

−

−−

= 1221 baba

0z

−

−

Note: If � ≠ 0, take a point on yz−plane as (0, y1, z

1) and if m ≠ 0, take a point on xz−plane as (x

1, 0, z

1).

Alternative method

If 2

1

a

a≠

2

1

b

b Put z = 0 in both the equations and solve the equations a

1x + b

1y + d

1 = 0, a

2x + b

2y + d

2 =0

otherwise Put y = 0 and solve the equations a1x

+ c

1z

+ d

1 = 0 and a

2x + c

2z + d

2 = 0

Example : Find the equation of the line of intersection of planes 4x + 4y – 5z = 12, 8x + 12y – 13z = 32 inthe symmetric form.

Solution Given planes are 4x + 4y – 5z – 12 = 0 ..... (1)and 8x + 12y – 13z – 32 = 0 ..... (2)Let �, m, n be the direction ratios of the line of intersection :then 4� + 4m – 5n = 0 ..... (3)and 8� + 12m – 13n = 0

∴3248

n

5240

m

6052 −=

+−=

+−

�or,

16

n

12

m

8==

�or,

4

n

3

m

2==

�

Hence direction ratios of line of intersection are 2, 3, 4.Here 4 ≠ 0, therefore line of intersection is not parallel to xy-plane.Let the line of intersection meet the xy-plane at P (α, β, 0).Then P lies on planes (1) and (2)∴ 4α + 4β + 12 = 0or, α + β – 3 = 0 ..... (5)and 8α + 12β – 32 = 0or, 2α + 3β – 8 = 0 ..... (6)Solving (5) and (6), we get

23

1

8698 −=

+−

β=

+−

αor,

1

1

21=

β=

α

∴ α = 1, β = 2

Hence equation of line of intersection in symmetrical form is 4

0z

3

2y

2

1x −=

−=

−.

Example : Find the angle between the lines x – 3y – 4 = 0, 4y – z + 5 = 0 and x + 3y – 11 =0, 2y – z + 6 = 0.Solution Given lines are

=+−

=−−

05zy4

04y3x..... (1)

and

=+−

=−+

06zy2

011y3x

..... (2)

Let �1, m

1, n

1 and �

1, m

2, n

2 be the direction cosines of lines (1) and (2) respectively



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∵ line (1) is perpendicular to the normals of each of the planesx – 3y – 4 = 0 and 4y – z + 5 = 0

∴ �1 – 3m

1 + 0.n

1 = 0 ..... (3)

and 0�1 + 4m

1 – n

1 = 0 ..... (4)

Solving equations (3) and (4), we get,

031

−

� =

04

n

)1(0

m 11

−=

−−or,

4

n

1

m

3111 ==

� = k (let).

Since line (2) is perpendicular to the normals of each of the planesx + 3y – 11 = 0 and 2y – z + 6 = 0,

∴ �2 + 3m

2 = 0 ..... (5)

and 2m2 – n

2 = 0 ..... (6)

∴ �2 = – 3m

2or,

32

−

� = m

2

and n2 = 2m

2or,

2

n2 = m

2.

∴2

n

1

m

3232 ==

−

� = t (let).

If θ be the angle between lines (1) and (2), thencosθ = �

1�

2 + m

1m

2 + n

1n

2= (3k) (– 3t) + (k) (t) + (4k) (2t)= – 9kt + kt + 8kt = 0 ∴ θ = 90°.

Self practice problems:1. Find the equation of the line of intersection of the plane

4x + 4y – 5z = 12

8x + 12y – 13z = 32 Ans.2

1x − =

3

2y − =

4

0z −

2. Show that the angle between the two lines defined by the equations x = y and xy + yz + zx = 0 is

cos–1

3

1

3. Prove that the three planes 2x + y – 4z – 17 = 0, 3x + 2y – 2z – 25 = 0, 2x – 4y + 3z + 25 = 0 intersectat a point and find its co-ordinates. Ans. (3, 7, – 1)

21 .21 .21 .21 . Foot, Length And Equation Of Perpendicular From A Point To A Line:Foot, Length And Equation Of Perpendicular From A Point To A Line:Foot, Length And Equation Of Perpendicular From A Point To A Line:Foot, Length And Equation Of Perpendicular From A Point To A Line:

(i)Cartesian form: Let equation of the line be �

ax − =

m

by − =

n

cz − = r (say) ..........(i)

and A (α, β, γ) be the point.Any point on line (i) is P (�r + a, mr + b, nr + c) ......... (ii)If it is the foot of the perpendicular from A on the line, then AP is perpendicular to the line. So � (�r + a − α) +m (mr + b − β) + n (nr + c − γ) = 0 i.e. r = (α − a) � + (β − b) m + (γ − c)n since �2 + m2 + n2 = 1. Putting thisvalue of r in (ii), we get the foot of perpendicular from point A on the given line. Since foot of perpendicular P is

known, then the length of perpendicular is given by AP =222 )cnr()bmr()ar( γ−++β−++α−+� the

equation of perpendicular is given byα−+

α−

ar

x

� =

β−+

β−

bmr

y =

γ−+

γ−

cnr

z (ii) Vector Form: Equation of a line

passing through a point having position vector α�

and perpendicular to the lines r�

=1a�

+ λ 1b�

and r�

=2a�

+ λ2b�

is parallel to1b�

x2b�

. So the vector equation of such a line is r�

= α�

+ λ ( 1b�

x 2b�

). Position vector β�

of the

image of a point α�

in a straight line r�

= a�

+ λ b�

is given by β�

= 2 a�

−

α−2|b|

b.)a(2�

��

b�

− α�

. Position vector of

the foot of the perpendicular on line is f�

= a�

−

α−2|b|

b.)a(�

��

b�

. The equation of the perpendicular is r�

= α�

+ µ

α−−α− b

|b|

b.)a()a(

2

��

��

.

22. To find image of a point w. r. t a line22. To find image of a point w. r. t a line22. To find image of a point w. r. t a line22. To find image of a point w. r. t a line

Let L ≡a

xx 2− =

b

yy 2− =

c

zz 2− is a given line

Let (x′, y′, z′) is the image of the point P (x1, y

1, z

1) with respect to the line L. Then

(i) a (x1 – x′) + b (y

1 – y′) + c (z

1 – z′) = 0

(ii)a

x2

xx2

1 −′+

=b

y2

yy2

1 −′+

=c

z2

zz2

1 −′+

= λ

from (ii) get the value of x′, y′, z′ in terms of λ asx′ = 2aλ + 2x

2 – x

1, y′ = 2bα – 2y

2 – y

1,

z′ = 2cλ + 2z2 – z

1now put the values of x′, y′, z′ in (i) get λ and resubtitute the value of λ to to get (x′ y′ z′).

Example : Find the length of the perpendicular from P (2, – 3, 1) to the line 1

2z

3

3y

2

1x

−

+=

−=

+.

Solution Given line is1

2z

3

3y

2

1x

−

+=

−=

+..... (1)

P ≡ (2, – 3, 1)Co-ordinates of any point on line (1) may be taken as

Q ≡ (2r – 1, 3r + 3, – r – 2)Direction ratios of PQ are 2r – 3, 3r + 6, – r – 3Direction ratios of AB are 2, 3, – 1Since PQ ⊥ AB∴ 2 (2r – 3) + 3 (3r + 6) – 1 (– r – 3) = 0

or, 14r + 15 = 0 ∴ r = 14

15−

∴ Q ≡

−−−

14

13,

14

3,

7

22∴ PQ =

14

531 units.

Second method : Given line is

1

2z

3

3y

2

1x

−

+=

−=

+

P ≡ (2, – 3, 1)

Direction ratios of line (1) are 14

2,

14

3, –

14

1

P (2, –3, 1)

A R Q B(–1, 3, –2)

RQ = length of projection of RP on AB

= 14

15)21(

4

1)33(

14

3)12(

14

2=+−−−++

PR2 = 32 + 62 + 32 = 54

∴ PQ = 22 RQPR − = 14

531

14

22554 =−

Self practice problems:

1. Find the length and foot of perpendicular drawn from point (2, –1, 5) to the line 11

8z

4

2y

10

11x

−

+=

−

+=

−.

Also find the image of the point in the line. Ans. 14 , N ≡ (1, 2, 3), Ι ≡ (0, 5, 1)

2. Find the image of the point (1, 6, 3) in the line 3

2z

2

1y

1

x −=

−= . Ans. (1, 0 , 7)

3. Find the foot and hence the length of perpendicular from (5, 7, 3) to the line 5

5z

8

29y

3

15x

−

−=

−=

−.

Find also the equation of the perpendicular. Ans. (9, 13, 15) ; 14 ; 6

3z

3

7y

2

5x −=

−=

−

23 .23 .23 .23 . Angle Between A Plane And A Line:Angle Between A Plane And A Line:Angle Between A Plane And A Line:Angle Between A Plane And A Line:

(i) If θ is the angle between line�

1xx − =

m

yy 1− =

n

zz 1− and the plane ax + by + cz + d = 0, then

sin θ =

++++

++

222222nmcba

ncmba

)( �

�.

(ii) Vector form: If θ is the angle between a line r� = ( a

� + λ b�

) and r� . n� = d then sin θ =

|n||b|

n.b��

��

.

(iii) Condition for perpendicularitya

� =

b

m =

c

n b�

x n�

= 0

(iv) Condition for parallel a� + bm + cn = 0 b�

. n�

= 0

24. 24. 24. 24. Condition For A Line To Lie In A PlaneCondition For A Line To Lie In A PlaneCondition For A Line To Lie In A PlaneCondition For A Line To Lie In A Plane

(i) Cartesian form: Line�

1xx − =

m

yy 1− =

n

zz 1− would lie in a plane

ax + by + cz + d = 0, if ax1 + by

1 + cz

1 + d = 0 & a� + bm + cn = 0.

(ii) Vector form: Line r�

= a�

+ λ b�

would lie in the plane r�

. n�

= d if b�

. n�

= 0 & a�

. n�

= d

25 .25 .25 .25 . Coplanar Lines:Coplanar Lines:Coplanar Lines:Coplanar Lines:

(i) If the given lines are�

α−x =

m

y β− =

n

z γ− and '

'x

�

α− =

'

'

m

y β−=

'

'

n

z γ−, then condition

for intersection/coplanarity is

'n'm'

nm

'''

�

�

γ−γβ−βα−α

= 0 & plane containing the above

two lines is

'n'm'

nm

zyx

�

�

γ−β−α−

= 0

(ii) Condition of coplanarity if both the lines are in general form Let the lines beax + by + cz + d = 0 = a′x + b′y + c′z + d′ &αx + βy + γz + δ = 0 = α′x + β′y + γ′z + δ′

They are coplanar if

''''

'd'c'b'a

dcba

δγβα

δγβα = 0

Alternative method: get vector along the line of shortest distance as ′′′ nm

nm

kji

�

�

Now get unit vector along this vector

u = �i + mj + nk

Let v�

= (α – α′ ) i + (B – B′) j + (y – y′)

S. D. = u. vExample : Find the distance of the point (1, 0, – 3) from the plane x – y – z = 9 measured parallel to the

line 6

6z

3

2y

2

2x

−

−=

+=

−.

Solution Given plane is x – y – z = 9 ..... (1)

Given line AB is 6

6z

3

2y

2

2x

−

−=

+=

−..... (2)

Equation of a line passing through the point Q(1, 0, – 3) and parallel to line (2) is

6

3z

3

y

2

1x

−

+==

− = r.. ..... (3)

Co-ordinates of point on line (3) may be taken asP (2r + 1, 3r, – 6r – 3)

If P is the point of intersection of line (3) and plane (1), then P lies on plane (1),∴ (2r + 1) – (3r) – (– 6r – 3) = 9

r = 1or, P ≡ (3, 3, – 9)Distance between points Q (1, 0, – 3) and P (3, 3, – 9)

PQ = 222 ))3(9()03()13( −−+−+− = 3694 ++ = 7.

Q (1, 0, – 3)

B

A

P

Example: Find the equation of the plane passing through (1, 2, 0) which contains the line 2

2z

4

1y

3

3x

−

−=

−=

+.

Solution Equation of any plane passing through (1, 2, 0) may be taken asa (x – 1) + b (y – 2) + c (z – 0) = 0 ..... (1)

where a, b, c are the direction ratios of the normal to the plane. Given line is

2

2z

4

1y

3

3x

−

−=

−=

+..... (2)

If plane (1) contains the given line, then3a + 4b – 2c = 0 ..... (3)

Also point (– 3, 1, 2) on line (2) lies in plane (1)∴ a (– 3 – 1) + b (1 – 2) + c (2 – 0) = 0or, – 4a – b + 2c = 0 ..... (4)Solving equations (3) and (4), we get,

163

c

68

b

28

a

+−=

−=

−

or,13

c

2

b

6

a== = k (say). ..... (5)

Substituting the values of a, b and c in equation (1), we get,6 (x – 1) + 2 (y – 2) + 13 (z – 0) = 0.

or, 6x + 2y + 13z – 10 = 0. This is the required equation.

Example : Find the equation of the projection of the line 4

3z

1

1y

2

1x −=

−

+=

− on the plane x + 2y + z = 9.

SolutionLet the given line AB be

4

3z

1

1y

2

1x −=

−

+=

−..... (1)

Given plane isx + 2y + z = 9 ..... (2) D C

AB

Let DC be the projection of AB on plane (2)Clearly plane ABCD is perpendicular to plane (2).Equation of any plane through AB may be taken as (this plane passes through the point (1, – 1, 3) online AB)

a (x – 1) + b (y + 1) + c (z – 3) = 0 ..... (3)where 2a – b + 4c = 0 ..... (4)

[∵ normal to plane (3) is perpendicular to line (1)]Since plane (3) is perpendicular to plane (2),∴ a + 2b + c = 0 ..... (5)Solving equations (4) & (5), we get,

5

c

2

b

9

a==

−.

Substituting these values of a, b and c in equation (3), we get9 (x – 1) – 2 (y + 1) – 5 (z – 3) = 0

or, 9x – 2y – 5z + 4 = 0 ...... (6)Since projection DC of AB on plane (2) is the line of intersection of plane ABCD and plane (2), thereforeequation of DC will be

and

=−++

=+−−

)ii.....(09zy2x

)i.....(04z5y2x9..... (7)

Let �, m, n be the direction ratios of the line of intersection of planes (i) and (ii)∴ 9� – 2m – 5n = 0 ..... (8)and � + 2m + n = 0 ..... (9)

∴218

n

95

m

102 +=

−−=

+−

�

Example : Show that the lines 1

2z

3

1y

2

3x +=

−

+=

− and

2

2z

1

y

3

7x +==

−

− are coplanar. Also find the

equation of the plane containing them.Solution Given lines are

1

2z

3

1y

2

3x +=

−

+=

− = r (say) ..... (1)

and2

7z

1

y

3

7x +==

−

− = R (say) ..... (2)

If possible, let lines (1) and (2) intersect at P.Any point on line (1) may be taken as

(2r + 3, – 3r – 1, r – 2) = P (let).Any point on line (2) may be taken as

(– 3R + 7, R, 2R – 7) = P (let).∴ 2r + 3 = – 3R + 7or, 2r + 3R = 4 ..... (3)Also – 3r – 1 = Ror, – 3r – R = 1 ..... (4)

and r – 2 = 2R – 7or, r – 2R = – 5. ..... (5)Solving equations (3) and (4), we get,

r = – 1, R = 2Clearly r = – 1, R = 2 satisfies equation (5).Hence lines (1) and (2) intersect. ∴ lines (1) and (2) are coplanar.Equation of the plane containing lines (1) and (2) is

213

132

2z1y3x

−

−

++−

= 0

or, (x – 3) (– 6 – 1) – (y + 1) (4 + 3) + (z + 2) (2 – 9) = 0or, – 7 (x – 3) – 7 (y + 1) – 7 (z + 2) = 0or, x – 3 + y + 1 + z + 2 = 0 or, x + y + z = 0.


1. Find the values of a and b for which the line 2

6z

4

3y

a

2x

−

−=

+=

− is perpendicular to the plane

3x – 2y + bz + 10 = 0. Ans. a = 3, b = – 2

2. Prove that the lines 3

3z

3

2y

2

1x −=

−=

− and

5

4z

4

3y

3

2x −=

−=

− are coplanar. Also find the equation

of the plane in which they lie. Ans. x – 2y + z = 0

3. Find the plane containing the line 2

2x − =

3

3y − =

5

4z − and parallel to the line

1

1x + =

2

1y

−

− =

1

1z +−

Ans. 13x + 3y – 72 – 7 = 0

4. Show that the line 2

1x − =

3

2y − =

4

3z − &

5

4x − =

2

1y − = z are intersecting each other. Find thire

intersection and the plane containing the line. Ans. (– 1, – 1, – 1) & 5x – 8y + 11z – 2 = 0

5. Show that the lines r�

= (– i – 3 j – 5 k ) + λ (–3 i – 5 j – 7 k ) & r�

(2 i + 4 j + 6 k ) + µ ( i +4 j + 7 k ) are

coplanar and find the plane containing the line. Ans. r . ( )kj2i +− = 0

26 .26 .26 .26 . Skew LinesSkew LinesSkew LinesSkew Lines:(i) The straight lines which are not parallel and non−coplanar i.e. non−intersecting are called

skew lines. If ∆ =

'n'm'

nm

'''

�

�


≠ 0, then lines are skew..

(ii) Shortest distance: Suppose the equation of the lines are

�

α−x =

m

y β− =

n

z γ− and

'n

'z

'm

'y

'

'x γ−=

β−=

α−

�

S.D. =

∑ −

−γ−γ+−β−β+−α−α2)n'm'mn(

)m''m()'()'nn()'()n'm'mn()'( ��

=

'n'm'

nm

'''

�

�


∑ ′−′÷ 2)nmnm(

(iii) Vector Form: For lines 1a�

+ λ1b�

& 2a�

+ λ2b�

to be skew

(1b�

x 2b�

). (2a�

− 1a�

) ≠ 0 or [1b�

2b�

(2a�

− 1a�

)] ≠ 0.

(iv) Shortest distance between the two parallel lines r� = 1a�

+ λ b�

&

r�

= 2a�

+ µ b�

is d =|b|

bx)aa( 12 �

��−

.

Example : Find the shortest distance and the vector equation of the line of shortest distance between thelines given by

+−λ+++=

→→→→→→→kjr3k3j8r3r and

++−µ++−−=

→→→→→→→k4j2i3k6j7r3r

Solution Given lines are

+−λ+++=

→→→→→→→kji3k3j8r3r ..... (1)

and

++−µ++−−=

→→→→→→→k4j2i3k6j7r3r ..... (2)

Equation of lines (1) and (2) in cartesian form is

A

90°

90°

B

C DM

L

AB : 1

3z

1

8y

3

3x −=

−

−=

− = λ

and CD :

3

1= µ

Let L ≡ (3λ + 3, – λ + 8, λ + 3)and M ≡ (– 3µ – 3, 2µ – 7, 4µ + 6)Direction ratios of LM are

3λ + 3µ + 6, – λ – 2µ + 15, λ – 4µ – 3.

Since LM ⊥ AB∴ 3 (3λ + 3µ + 6) – 1 (– λ – 2µ + 15) + 1 (λ – 4µ – 3) = 0or, 11λ + 7µ = 0 ..... (5)

Again LM ⊥ CD∴ – 3 (3λ + 3µ + 6) + 2 (– λ – 2µ + 15) + 4 (λ – 4µ – 3) = 0or, – 7λ – 29µ – 0 ..... (6)Solving (5) and (6), we get λ = 0, µ = 0∴ L ≡ (3, 8, 3), M ≡ (– 3, – 7, 6)

Hence shortest distance LM = 222 )63()78()33( −++++

= 270 = 3 30 units

Vector equation of LM is

−++++=

→→→→→→→

k3j15i6tk3j8i3r

Note : Cartesian equation of LM is 3

3z

15

8y

6

3x

−

−=

−=

−.

Example : Prove that the shortest distance between any two opposite edges of a tetrahedron formed by

the planes y + z = 0, x + z = 0, x + y = 0, x + y + z = 3 a is 2 a.Solution Given planes are y + z = 0 ..... (i) x + z = 0 ..... (ii)

x + y = 0 ..... (iii) x + y + z = 3 a ..... (iv)Clearly planes (i), (ii) and (iii) meet at O(0, 0, 0)Let the tetrahedron be OABC

O (0, 0, 0)

(0, 0, a)3

CQ D

APLet the equation to one of the pair of opposite edges OA and BC be

y + z = 0, x + z = 0 ..... (1)

x + y = 0, x + y + z = 3 a ..... (2)equation (1) and (2) can be expressed in symmetrical form as

1

0z

1

0y

1

0x

−

−=

−=

−..... (3)

and,0

a3z

1

0y

1

0x −=

−

−=

−..... (4)

d. r. of OA and BC are (1, – 1) and (1, – 1, 0).Let PQ be the shortest distance between OA and BC having direction cosine (�, m, n)∴ PQ is perpendicular to both OA and BC.∴ � + m – n = 0and � – m = 0Solving (5) and (6), we get,

2

n

1

m

1==

� = k (say)

also, �2 + m2 + n2 = 1

∴ k2 + k2 + 4k2 = 1 ⇒ k = 6

1A

O

CB∴ � = 6

1, m =

6

1, n =

6

2

Shortest distance between OA and BC

C Q

O AP

90°

90°

B

i.e. PQ = The length of projection of OC on PQ= | (x

2 – x

1) � + (y

2 – y

1) m + (z

2 – z

1) n |

= 6

2.a3

6

1.0

6

1.0 ++ = 2 a.


1. Find the shortest distance between the lines 4

3z

3

2y

2

1x −=

−=

− and

5

5z

4

4y

3

2x −=

−=

−. Find also

its equation. Ans.6

1, 6x – y = 10 – 3y = 6z – 25

2. Prove that the shortest distance between the diagonals of a rectangular parallelopiped whose sides are

a, b, c and the edges not meeting it are 222222 ba

ab,

ac

ca,

cb

bc

+++

27.27.27.27. SphereSphereSphereSphere: General equation of a sphere is given by x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 (−u,

−v, −w) is the centre and dwvu 222 −++ is the radius of the sphere.

Example : Find the equation of the sphere having centre at (1, 2, 3) and touching the plane x + 2y + 3z = 0.Solution: Given plane is x + 2y + 3z = 0 ..... (1)

Let H be the centre of the required sphere.H

P

Given H ≡ (1, 2, 3)Radius of the sphere,

HP = length of perpendicular from H to plane (1)

= 14

|33221| ×+×+ = 14

Equation of the required sphere is (x – 1)2 + (y – 2)2 + (z – 3)2 = 14or x2 + y2 + z2 – 2x – 4y – 6z = 0

Example : Find the equation of the sphere if it touches the plane )kj2i2.(r→→→→

−− = 0 and the position

vector of its centre is →→→

−+ k4j6i3

Solution Given plane is )kj2i2.(r→→→→

−− = 0 ..... (1)Let H be the centre of the sphere, then

→→→→→

=−+= ck4j6i3OH (say)Radius of the sphere = length of perpendicular from H ot plane (1)

=

|kj2i2|

|)kj2i2.(c|→→→

→→→→

−−

−−

= |kj2i2|

|)kj2i2).(k4j6i3(|→→→

→→→→→→

−−

−−−+

= 3

2

3

|4126|=

+− = a (say)

Equation of the required sphere is

|cr|→→

− = a

or3

2|)k4j6i3(kzjyix| =−+−++

→→→→→→

or | (x – 3) →

i + (y – 6)→

j + (z + 4) →

k |2 = 9

4

or (x – 3)2 + (y – 6)2 + (z + 4)2 = 9

4or 9 (x2 + y2 + z2 – 6x – 12y + 8z + 61) = 4

or 9x2 + 9y2 + 9z2 – 54x – 108y + 72z + 545 = 0Example : Find the equation of the sphere passing through the points (3, 0, 0), (0, – 1, 0), (0, 0, – 2) and

whose centre lies on the plane 3x + 2y + 4z = 1Solution Let the equation of the sphere be

x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 ..... (1)Let A ≡ (3, 0, 0), B ≡ (0, – 1, 0), C ≡ (0, 0, – 2)Since sphere (1) passes through A, B and C,∴ 9 + 6u + d = 0 ..... (2)

1 – 2v + d = 0 ..... (3)4 – 4w + d = 0 ..... (4)

Since centre (– u, – v, – w) of the sphere lies on plane3x + 2y + 4z = 1

∴ – 3u – 2v – 4w = 1 ..... (5)(2) – (3) ⇒ 6u + 2v = – 8 ..... (6)(3) – (4) ⇒ – 2v + 4w = 3 ..... (7)

From (6), u = 6

8v2 −−..... (8)

From (7), 4w = 3 + 2v ..... (9)



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om

Putting the values of u, v and w in (5), we get

1v2–3–v22

8v2=−

+

⇒ 2v + 8 – 4v – 6 – 4v = 2⇒ v = 0

From (8), u = 3

4

6

80−=

−

From (9), 4w = 3 ∴ w = 4

3

From (3), d = 2v – 1 = 0 – 1 = – 1From (1), equation of required sphere is

x2 + y2 + z2 – 3

8

6

80−

− x +

2

3z – 1 = 0

or 6x2 + 6y2 + 6z2 – 16x + 9z – 6 = 0

Example :

Find the equation of the sphere with the points (1, 2, 2) and (2, 3, 4) as the extremities of a diameter.Find the co-ordinates of its centre.

SolutionLet A ≡ (1, 2, 2), B ≡ (2, 3, 4)Equation of the sphere having (x

1, y

1, z

1) and (x

2, y

2, z

2) as the extremities of a diameter is

(x – x1) (x – x

2) + (y – y

1) (y – y

2) + (z – z

1) (z – z

2) = 0

Here x1 = 1, x

2 = 2, y

1 = 2, y

2 = 3, z

1 = 2, z

2 = 4

∴ required equation of the sphere is(x – 1) (x – 2) + (y – 2) (y – 3) + (z – 2) (z – 4) = 0

or x2 + y2 + z2 – 3x – 5y – 6z + 16 = 0Centre of the sphere is middle point of AB

∴ Centre is

3,

2

5,

2

3


1. Find the value of k for which the plane x + y + z = k3 touches the sphere

x2 + y2 + z2 – 2x – 2y – 2z – 6 = 0.

Ans. 3 ± 3

2. Find the equation to the sphere passing through (1, – 3, 4), (1, – 5, 2) and (1, – 3, 0) which has itscentre in the plane x + y + z = 0Ans. x2 + y2 + z2 – 2x + 6y – 4z + 10 = 0

3. Find the equation of the sphere having centre on the line 2x – 3y = 0, 5y + 2z = 0 and passing throughthe points (0, – 2, – 4) and (2, – 1, – 1).Ans. x2 + y2 + z2 – 6x – 4y + 10 z + 12 = 0

4. Find the centre and radius of the circle in which the plane 3x + 2y – z – 7 14 = 0 intersects the spherex2 + y2 + z2 = 81.

Ans. 4 2 units

5. A plane passes through a fixed point (a, b, c) and cuts the axes in A, B, C. Show that the locus of thecentre of the sphere OABC is

z

c

y

b

x

a++ = 2.

1.1 .1 .1 . Vectors & Their Representat ion:Vectors & Their Representat ion:Vectors & Their Representat ion:Vectors & Their Representat ion:Vector quantities are specified by definite magnitude and definite directions. A vector is generally

represented by a directed line segment, say AB . A is called the initial point & B is called the

terminal point. The magnitude of vector AB is expressed by AB .Zero VectorZero VectorZero VectorZero Vector: A vector of zero magnitude is a zero vector. i.e. which has the same initial & terminalpoint, is called a Zero Vector. It is denoted by O. The direction of zero vector is indeterminate.Unit VectorUnit VectorUnit VectorUnit Vector: A vector of unit magnitude in the direction of a vector a

� is called unit vector along a

�

and is denoted by a symbolically, |a|

aa �

�

= .

Equal VectorsEqual VectorsEqual VectorsEqual Vectors: Two vectors are said to be equal if they have the same magnitude, direction &represent the same physical quantity.Collinear Vectors: Two vectors are said to be collinear if their directed line segments are parallelirrespective of their directions. Collinear vectors are also called parallel vectors. If they have thesame direction they are named as like vectors otherwise unlike vectors.

Symbolically, two non zero vectors�a and

�b are collinear if and only if, bKa

��= , where K ∈ R

Vectors a�

= a1 i + ja2

+ ka3 and b

� = b

1 i + jb2 + kb3

are collinear if 1

1

b

a =

2

2

b

a =

3

3

b

a

Coplanar VectorsCoplanar VectorsCoplanar VectorsCoplanar Vectors: A given number of vectors are called coplanar if their line segments are allparallel to the same plane. Note that “TWO VECTORS ARE ALWAYS COPLANAR”.

Solved Example Find unit vector of k3j2i +−

Solution a�

= k3j2i +− if a�

= iax + jay + kaz

then |a|�

= 2

z2

y2

x aaa ++ ∴ |a|�

= 14

a = |a|

a�

�

= 14

1 i –

14

2 j +

14

3 k

Solved Example Find values of x & y for which the vectors a�

= (x + 2) i – (x – y) j + k

b�

= (x – 1) i + (2x + y) j + 2 k are parallel.

Solution a�

and b�

are parallel if 1x

2x

−+

= yx2

xy

+−

= 2

1

x = – 5, y = – 20

2.2 .2 .2 . Angle Between two VectorsAngle Between two VectorsAngle Between two VectorsAngle Between two VectorsIt is the smaller angle formed when the initial points or the terminal points of the two vectors arebrought together. It should be noted that 0º ≤ θ ≤ 180º .

3.3 .3 .3 . Addition Of Vectors:Addition Of Vectors:Addition Of Vectors:Addition Of Vectors:

� If two vectors � �a b& are represented by OA OB

→ →& , then their sum

� �a b+ is a vector represented

by OC→

, where OC is the diagonal of the parallelogram OACB.

�� a b b a+ = + (commutative) � ( ) ( )

� � � � � �a b c a b c+ + = + + (associativity)

�� a a a+ = = +0 0 �

� � � � �a a a a+ − = = − +( ) ( )0 � |b||a||ba|

��+≤+

� ||b||a|||ba|��

−≥− � ba��

± = θ±+ cos|b||a|2|b||a| 22��

where θ is the angle between the vectors

� A vector in the direction of the bisector of the angle between the two vectors � �a b& is

�

�

�

�a

a

b

b+ . Hence

bisector of the angle between the two vectors bndaa��

is ( )λ � �a b+ , where λ ∈ R+. Bisector of the exterior

angle between � �a b& is ( )λ � �a b− , λ ∈ R+.

4.4 .4 .4 . Multiplication Of A Vector By A Scalar:Multiplication Of A Vector By A Scalar:Multiplication Of A Vector By A Scalar:Multiplication Of A Vector By A Scalar:If a�

is a vector & m is a scalar, then m a�

is a vector parallel to a�

whose modulus is m times that of

a�

. This multiplication is called SCALAR MULTIPLICATION. If a�

and b�

are vectors & m, n are scalars, then:

Vector

amm)a()a(m��

== a)mn()am(n)an(m��

==

anama)nm(��

+=+ bmam)ba(m��

+=+

Solved Example: If k3j2ia ++=�

and k5j4i2b −+=�

represent two adjacent sides of a parallelogram, find

unit vectors parallel to the diagonals of the parallelogram.

Solution. Let ABCD be a parallelogram such that AB = a�

and BC = b�

.

Then, AB + BC = AC ⇒ AC = ba��

+ = k2j6i3 −+

and AB + BD = AD ⇒ AD + AD = AB

⇒ BD = ABAD − = ab��

−Now, AC = k2j6i3 −+ ⇒ |AC| = 4369 ++ = 7

and, BD = k8j2i −+ ⇒ |BD| = 6441 ++ = 69

∴ Unit vector along AC = |AC|

AC =

7

1 ( )k2j6i3 −+

and, Unit vector along BD = |BD|

BD =

69

1 ( )k8j2i −+

Solved Example ABCDE is a pentagon. Prove that the resultant of the forces AB , AE , BC , DC , ED

and AC is 3 AC .Solution. Let R be the resultant force

∴ R = AB + AE + BC + DC + ED + AC

∴ R = ( AB + BC ) + ( AE + ED + DC ) + AC

= AC + AC + AC

= 3 AC . Hence proved.Self Practice Problems :

1. Express : (i) The vectors BC CA and AB in terms of the vectors OA , OB and OC

(ii) The vectors OA , OB and in terms of the vectors OC , OB and OC .

Ans. (i) OBOCBC −= , OCOACA −= , OAOBAB −=2. Given a regular hexagon ABCDEF with centre O, show that

(i) OB – OA = OC – OD (ii) OD + OA = 2 OB + OF (iii) AD + EB + PC = 4 AB

3. The vector kji −+− bisects the angle between the vectors c�

and j4i3 + . Determine the unit vector

along c�

. Ans. k15

14j

15

2i

3

1−+−

4. The sum of the two unit vectors is a unit vector. Show that the magnitude of the their difference is 3 .

5.5 .5 .5 . Position Vector Of A PointPosition Vector Of A PointPosition Vector Of A PointPosition Vector Of A Point:

let O be a fixed origin, then the position vector of a point P is the vector OP . If a�

and b�

are positionvectors of two points A and B, then,

AB = ab��

− = pv of B − pv of A.A.

DISTDISTDISTDISTANCE FANCE FANCE FANCE FORMULAORMULAORMULAORMULA

Distance between the two points A )a(�

and B )b(�

is AB = ba��

−SECTION FORMULASECTION FORMULASECTION FORMULASECTION FORMULA

If a�

and b�

are the position vectors of two points A & B then the p.v. of

a point which divides AB in the ratio m: n is given by: nm

bmanr

++

=

��

.

Note p.v. of mid point of AB =2

ba��

+.

Solved Example: ABCD is a parallelogram. If L, M be the middle point of BC and CD, express AL and

AM in terms of AB and AD , also show that AL + AM = 2

3 AC .

Solution.Let the position vectors of points B and D be respectively b�

and a�

referred to A as origin of reference.

Then AC = AD + DC = AD + AB [∵ DC = AB ]

= d�

+ b�

∵ AB = b�

, AD = d�

i.e. position vector of C referred to A is d�

+ b�

∴ AL = p.v. of L, the mid point of BC .

= 2

1 [p.v. of D + p.v. of C] =

2

1 ( )bdb

��++ = AB +

2

1 AD

AM = 2

1 [ ]bdd

��++ = AD +

2

1 AB

∴ AL + AM = b�

+ 2

1 d�

+ d�

+ 2

1b�

= 2

3b�

+ 2

3d�

= 2

3( b�

+ d�

) = 2

3AC .

Solved Example If ABCD is a parallelogram and E is the mid point of AB, show by vector method thatDE trisects and is trisected by AC.

Solution. Let AB = a�

and AD = b�

Then BC = AD = b�

and AC = AB + AD = a�

+ b�

Also let K be a point on AC, such that AK : AC = 1 : 3

or, AK = 3

1 AC ⇒ AK =

3

1 ( a�

+ b�

) .........(i)

Again E being the mid point of AB, we have

AE = 2

1a�

Let M be the point on DE such that DM : ME = 2 : 1

∴ AM = 21

AE2AD

++

= 3

ab��

+..........(ii)

From (i) and (ii) we find that : AK = 3

1 ( a�

+ b�

) = AM , and so we conclude that K and M coincide. i.e. DE

trisect AC and is trisected by AC. Hence proved.Self Practice Problems

1. If b,a��

are position vectors of the points (1, –1), (–2, m), find the value of m for which a�

and b�

are

collinear. Ans. m = 2

2. The position vectors of the points A, B, C, D are kji ++ , j5i2 + , k3j2i3 −+ , kj6i −− respectively..

Show that the lines AB and CD are parallel and find the ratio of their lengths. Ans. 1 : 2

3. The vertices P, Q and S of a triangle PQS have position vectors q,p��

and s�

respectively..

(i) Find m�

, the position vector of M, the mid-point of PQ, in terms of p�

and q�

.

(ii) Find t�

, the position vector of T on SM such that ST : TM = 2 : 1, in terms of q,p��

and s�

.

(iii) If the parallelogram PQRS is now completed. Express r�

, the position vector of the point R in

terms of q,p��

and s�

Prove that P, T and R are collinear.

Ans. m�

= 2

1 )qp(

��+ , t

� =

2

1 )sqp(

��++ , r

� =

2

1 spq

��−+

4. D, E, F are the mid-points of the sides BC, CA, AB respectively of a triangle. Show FE = 1/2 BC and

that the sum of the vectors AD , BE , CF is zero.5. The median AD of a triangle ABC is bisected at E and BE is produced to meet the side AC in F; show

that AF = 1/3 AC and EF = 1/4 BF.6. Point L, M, N divide the sides BC, CA, AB of ∆ABC in the ratios 1 : 4, 3 : 2, 3 : 7 respectively. Prove

that AL + BM + CN is a vector parallel to CK , when K divides AB in the ratio 1 : 3.

6.6 .6 .6 . Scalar Product Of Two Vectors:Scalar Product Of Two Vectors:Scalar Product Of Two Vectors:Scalar Product Of Two Vectors:Geometrical interpretation of Scalar Product

Let a�

and b�

be vectors represented by OA and OB respectively. Let θ be the angle between OA and

OB . Draw BL ⊥ OA and AM ⊥ OB.From ∆s OBL and OAM, we have OL = OB cos θ and OM = OA cos θ. Here OL and OM are known as

projections of b�

on a�

and a�

on b�

respectively..

Now, b.a��

= | a�

| | b�

| cos θ= | a�

| (OB cos θ )

= | a�

| (OL)

= (Magnitude of a�

) (Projection of b�

on a�

) ........(i)

Again b.a��

= | a�

| | b�

| cos θ = | b�

| (| a�

| cos θ )

= | b�

| (OA cos θ)

= | b�

| (OM)

= (magnitude of b�

) (Projection of a�

on b�

) ........(ii)Thus geometrically interpreted, the scalar product of two vectors is the product of modulus of eithervector and the projection of the other in its direction.

1. i.i = j.j = k.k = 1; i.j = j.k = k.i = 0 � projection of |b|

b.abona �

��

=

2. if�a = a

1i + a

2j + a

3k &�b = b

1i + b

2j + b

3k then

� �a b. = a

1b

1 + a

2b

2 + a

3b

3

�a a a a= + +1

22

23

2 ,

�b b b b= + +1

22

23

2

3. the angle φ between� �a b& is given by

|b||a|

b.acos ��

��

=φ 0 ≤ φ ≤ π

4. )0(cosbab.a π≤θ≤θ=��

, note that if θ is acute then � �a b. > 0 & if θ is obtuse then

� �a b. < 0

5. a.bb.a,aaa.a 22 ��=== (commutative) � c.ab.a)cb(.a

��+=+ (distributive)

6. ba0b.a��

⊥⇔= )0b0a( ≠≠��

7. ( )m a�

.�b =�a .( )m b

� = m( ).

� �a b (associative) where m is scalar..

Note : (i) Maximum value of�a .�b is

�a �b (ii) Minimum value of

�a .�b is –

�a �b

(iii) Any vector�a can be written as,

�a = ( ) ( ) ( )� � �

a i i a j j a k k. � � . � � . � �+ + .

Solved Example Find the value of p for which the vectors k9j2i3a ++=�

and k3jpib ++=�

are

(i) perpendicular (ii) parallel

Solution. (i) ba��

⊥ ⇒ b.a��

= 0 ⇒ ( )k9j2i3 ++ . ( )k3jpi ++ = 0

⇒ 3 + 2p + 27 = 0 ⇒ p = – 15

(ii) We know that the vectors a�

= kajaia 321 ++ and b�

= kbjbib 321 ++ are parallel iff

ba��

λ= ⇔ ( )kajaia 321 ++ = λ ( )kbjbib 321 ++ ⇔ a1 = λb

1, a

2 = λb

2, a

3 = λb

3

⇔ 1

1

b

a =

2

2

b

a =

3

3

b

a ( =λ)

So, vectors a�

= k9j2i3 ++ and b�

= k3jpi ++ are parallel iff

1

3 =

p

2 =

3

9⇒ 3 =

p

2 ⇒ p = 2 / 3

Solved Example: If a�

+ b�

+ c�

= 0�

, |a|�

= 3, |b|�

= 5 and |c|�

= 7, find the angle between a�

and b�

.

Solution. We have, 0cba��

=++⇒ ba

��+ = – c

�⇒ ( )ba

��+ . ( )ba

��+ = ( )c

�− . ( )c

�−

⇒2

ba��

+ = 2|c|�

⇒2

a�

+ 2

b�

+ b.a2��

= 2

c�

⇒2

a�

+ 2

b�

+ 2 a�

b�

cos θ = 2

c�

⇒ 9 + 25 + 2 (3) (5) cos θ = 49 ⇒ cos θ = 2

1⇒ θ =

3

π.

Solved Example Find the values of x for which the angle between the vectors a�

= 2x2 i + 4x j + k and

b�

= 7 i – 2 j + x k is obtuse.

Solution. The angle q between vectors a�

and b�

is given by cos θ = |b||a|

b.a��

��

Now, θ is obtuse ⇒ cos θ < 0 ⇒|b||a|

b.a��

��

< 0

⇒ b.a��

< 0 [∵, 0|b||,a| >��

]

⇒ 14x2 – 8x + x < 0

⇒ 17x (2x – 1) < 0 ⇒ x(2x – 1) < 0 ⇒ 0 < x < 2

1

Hence, the angle between the given vectors is obtuse if x ∈ (0, 1/2)Solved Example:D is the mid point of the side BC of a triangle ABC, show that AB2 + AC2 = 2 (AD2 + BD2)Solution. We have

AB = AD + DB

⇒ AB2 = 2)DBAD( +

= AD2 + DB + AD2 . DB .............(i)Also we have

AC = AD + DC ⇒ AC2 = 2)DCAD( +

= AD2 + DC2 + AD2 . DC .............(ii)Adding (i) and (ii), we get

AB2 + AC2 = 2AD2 + 2BD2 + 2 AD . )DCDB( +

= 2(DA2 + DB2), for DB + DC = 0Solved Example

If a�

= i + j + k and b�

= 2 i – j + 3 k , then find

(i) Component of b�

along a�

. (ii) Component of b�

perpendicular to along a�

.

Solution. (i) Component of b�

along a�

is

2|a|

b.a�

��

a�

Here a�

. b�

= 2 – 1 + 3 = 42|a|�

= 3

Hence

2|a|

b.a�

��

a�

= 3

4a�

= 3

4( i + j + k )

(ii) Component of b�

perpendicular to along a�

is b�

–

2|a|

b.a�

��

a�

. = 3

1 ( )k5j7i2 +−

Self Practice Problems :1. If a�

and b�

are unit vectors and θ is angle between them, prove that tan 2

θ =

|ba|

|ba|��

��

+

−. 2.Find the values of x and y is the vectors a

� = kjxi3 −+ and b

� = kyji2 ++ are mutually

perpendicular vectors of equal magnitude. Ans. x = – 12

31, y =

12

41

3. Let a�

= k2j2ix2 −+ , b�

= kji +− and c = k4j5ix2 −+ be three vectors. Find the values of x for which the

angle between a�

and b�

is acute and the angle between b�

and c�

is obtuse. Ans.(–3, –2) ∪ (2, 3)

4. The points O, A, B, C, D, are such that aOA�

= , bOB�

= , b3a2OC��

+= , b2aOD��

+= . Give that the

length of OA is three times the length of OB show that BD and AC are perpendicular..5. ABCD is a tetrahedron and G is the centroid of the base BCD. Prove that

AB2 + AC2 + AD2 = GB2 + GC2 + GD2 + 3GA2

7.7.7.7. Vector Product Of Two Vectors:Vector Product Of Two Vectors:Vector Product Of Two Vectors:Vector Product Of Two Vectors:

1. If� �a b& are two vectors & θ is the angle between them then

� � � � �a x b a b n= sin θ , where

�n is the unit

vector perpendicular to both� �a b& such that

� � �a b n, & forms a right handed screw system.

2. Geometrically� �a x b = area of the parallelogram whose two adjacent sides are represented by

� �a b& .

3. 0kkjjii�

=×=×=× ; jik,ikj,kji =×=×=×

4. If�a = a

1 i +a2

j + a3 k &

�b = b

1 i + b2

j + b3 k then

321

321

bbb

aaa

kji

ba =×��

5.� � � �a x b b x a≠ (not commutative)

6. ( )m a�

×�b =�a ×( )m b

� = m( )� �

a b× (associative) where m is a scalar..

7.� � � � � � �a x b c a x b a x c( ) ( ) ( )+ = + (distributive)

8.� � � �a x b a b= ⇔0 & are parallel (collinear) ( , )

� �a b≠ ≠0 0 i.e.

� �a K b= , where K is a scalar..

9. Unit vector perpendicular to the plane of � �

� �

� �a b is na x b

a x b& � = ±

� A vector of magnitude ‘r’ & perpendicular to the palne of ( )� ��

� �a b isr a x b

a x b& ±

� If θ is the angle between � �

� �

� �a b thena x b

a b& sinθ =

� If� � �a b c, & are the pv’s of 3 points A, B & C then the vector area of triangle ABC =

[ ]1

2

� � � � � �a x b bx c cxa+ + . The points A, B & C are collinear if 0axccxbbxa

��=++

� Area of any quadrilateral whose diagonal vectors are� �d d1 2& is given by

1

21 2

� �d xd

� Lagrange's Identity: for any two vectors� � � � � � � �

� � � �

� � � �a b a x b a b a ba a a b

a b b b& ;( ) ( . )

. .

. .

2 2 22= − =

Solved Example

Find a vector of magnitude 9, which is perpendicular to both the vectors k3ji4 ++ and k2ji2 −+− .

Solution. Let a�

= k3ji4 +− and b�

= k2ji2 −+− . Then,

ba��

× = 212

314

kji

−−−

= (2 – 3) i – (–8 + 6) j + (4 – 2) k = k2j2i ++−

⇒ |ba|��

× = 222 22)1( ++− = 3

∴ Required vector = 9

×

×

|ba|

ba��

��

= 3

9 )k2j2i( ++− = k6j6i3 ++−

Solved Example

For any three vectors c,b,a��

. Show that 0)ba(c)ac(b)cb(a��

=+×++×++× .

Solution. We have, a�

× )cb(��

+ + b�

× )ac(��

+ + c�

× )ba(��

+= bcacabcbcaba

��×+×+×+×+×+× [Using distributive law]

= cbcabacbcaba��

×−×−×−×+×+× [∵ abba��

×−=× etc]

Solved Example: For any vector a�

, prove that 2|ia| ×�

+ 2|ja| ×�

+ 2|ka| ×�

= 2 2|a|�

Solution. Let a�

= kajaia 321 ++ . Then

ia ×�

= )kajaia( 321 ++ × i = a1 )ii( × + a

2 )ij( × + a3 )ik( × = –a

2 iak 3+

⇒ 2|ia| ×�

= a2

2 + a3

2

ja×�

= )kajaia( 321 ++ × j = iaka 31 −

⇒ 2|ja| ×�

= a21 + a

32 ⇒ 2|ka| ×

� = a

12 + a

22

∴ 2|ia| ×�

+ 2|ja| ×�

+ 2|ka| ×�

= a2

2 + a3

3 + a1

2 + a3

2 + a1

2 + a2

2

2 (d1

2 + a2

2 + a3

2) = 2 2|a|�

Solved Example: Let OA = a�

, OB = 10 a�

+ b2�

and OC = b�

where O is origin. Let p denote the area of thequadrilateral OABC and q denote the area of the parallelogram with OA and OC as adjacent sides.Prove that p = 6q.

Solution. We have, p = Area of the quadrilateral OABC

= 2

1 |ACOB| ×

= 2

1 |)OAOC(OB| −×

= 2

1 |)ab()b2a10(|

��−×+

= 2

1 |)ab(2)bb(2)aa(10ba(10|

��×−×+×−×

= 2

1 |)ba(200)ba(10|

��×++−×

and, q = Area of the parallelogram with OA and OC as adjacent sides

= |OCOA| × = |ba|��

× ........(ii)

From (i) and (ii), we get p = 6qSelf Practice Problems :

1. If p�

and q�

are unit vectors forming an angle of 30º; find the area of the parallelogram having q2pa��

+=

and qp2b��

+= as its diagonals. Ans. 3/4 sq. units

2. Show that {( a�

+ b�

+ c�

) × ( c�

– b�

)} . a�

= 2 [ a�

b�

c�

].

3. Prove that the normal to the plane containing the three points whose position vectors are c,b,a��

lies in

the direction baaccb��

×+×+×4. ABC is a triangle and EF is any straight line parallel to BC meeting AC, AB in E, F respectively. If BR

and CQ be drawn parallel to AC, AB respectively to meet EF in R and Q respectively, prove that∆ ARB = ∆ACQ.

8.8 .8 .8 . Scalar Triple Product:Scalar Triple Product:Scalar Triple Product:Scalar Triple Product:

� The scalar triple product of three vectors� � �a b c, & is defined as:

� � � � � �a x b c a b c. = sin cosθ φ where

θ is the angle between� �a b& & φ is the angle between

� � �a x b c& . It is also written as[ ]

� � �a b c and

spelled as box product.

� Scalar triple product geometrically represents the volumeof the parallelopiped whose three coterminous edges are

represented by � � � � � �a b c i e V a b c, & . . [ ]=

� In a scalar triple product the position of dot & cross can be interchanged i.e.� � � � � � � � � � � � � � �a b x c a x b c OR a b c b c a c a b. ( ) ( ). [ ] [ ] [ ]= = =

�� a bx c a cx b i e a b c a c b. ( ) .( ) . . [ ] [ ]= − = −

� If �a = a

1i+a

2j+a

3k;�b = b

1i+b

2j+b

3k &�c = c

1i+c

2j+c

3k then [ ]

� � �a b c

a a a

b b b

c c c

=1 2 3

1 2 3

1 2 3

.

In general, if� � � �a a l a m a n= + +1 2 3 ;

� � � �b b l b m b n= + +1 2 3

& � � � �c c l c m c n= + +1 2 3

then [ ] [ ]� � � �a a a

b b b

c c c

l mn=1 2 3

1 2 3

1 2 3

; where��

, &m n are non coplanar vectors.

� If� � �a b c, , are coplanar ⇔ =[ ]

� � �a b c 0 .

� Scalar product of three vectors, two of which are equal or parallel is 0 i.e. [ ]� � �a b c = 0 ,

� If� � �a b c, , are non − coplanar then [ ]

� � �a b c > 0 for right handed system & [ ]

� � �a b c < 0 for left handed

system.

� [i j k] = 1 � [ ] [ ]Ka b c K a b c� � � � � �

= � [( ) ] [ ] [ ]� � � � � � � � � �a b c d a c d b c d+ = +

� The volume of the tetrahedron OABC with O as origin & the pv’s of A, B and C being� � �a b c, & respectively

is given by V a b c=1

6[ ]� � �

� �a b c

� The positon vector of the centroid of a tetrahedron if the pv’s of its vertices are� � � �a b c d, , & are given by

1

4[ ]� � � �a b c d+ + + .

Note that this is also the point of concurrency of the lines joining the vertices to the centroids of theopposite faces and is also called the centre of the tetrahedron. In case the tetrahedron is regular it isequidistant from the vertices and the four faces of the tetrahedron.

Remember that: [ ]� � � � � �a b b c c a− − − = 0 & [ ]� � � � � �

a b b c c a+ + + = 2 [ ]� � �a b c .

Solved Example

Find the volume of a parallelopiped whose sides are given by k5j7i3 ++− , k3j7i5 −+− and k3j5i7 −−

Solution. Let k5j7i3a ++−=�

, k3j7i5b ++−=�

and k3j5i7c −−=�

.

We know that the volume of a parallelopiped whose three adjacent edges are c,b,a��

is ]c,b,a[��

.

Now, ]cba[��

= 357

375

573

−−−−

−

= –3 (–21 – 15) – 7 (15 + 21) + 5 (25 – 49)

= 108 – 252 – 120 = –264

So, required volume of the parallelopiped = ]c,b,a[��

= | – 264 | = 264 cubic units

Solved Example: Simplify ]accbba[��

−−−Solution. We have :

]accbba[��

−−− = )}cb()ba{(��

−×− . )ac(��

− [by def.]

= )cbbbcaba(��

×+×−×−× . )ac(��

− [by dist. law]

= )cbacba(��

×+×+× . )ac(��

− [ ∵ 0bb =×��

]

= )ba(��

× . c�

– )ba(��

× . a�

+ )ac(��

× . c�

– )ac(��

× . a�

+ )cb(��

× . c�

– )cb(��

× . a�

[by dist. law]

= ]cba[��

– ]aba[��

+ ]cac[��

– ]aac[��

+ ]ccb[��

– ]acb[��

= ]cba[��

– ]acb[��

[∵ scalar triple product when any two vectors are equal is zero ]

= ]cba[��

– ]cba[��

= 0 [∵ ]acb[��

= ]cba[��

]

Solved Example: Find the volume of the tetrahedron whose four vertices have position vectors a�

b�

c�

and d�

.

Solution. Let four vertices be A, B, C, D with p. v. a�

b�

c�

and d�

. respectively..

∴ DA = ( a�

– d�

)

DB = ( b�

– d�

)

DC = ( c�

– d�

)

Hence volume = 6

1 [ a�

– d�

b�

– d�

c�

– d�

]

= 6

1( a�

– d�

) . [( b�

– d�

) × ( c�

– d�

)]

= 6

1( a�

– d�

) . [ b�

× c�

– b�

× d�

+ c�

× d�

]

= 6

1 {[ a�

b�

c�

] – [ a�

b�

d�

] + [ a�

c�

d�

] – [ d�

b�

c�

]}

= 6

1 {[ a�

b�

c�

] – [ a�

b�

d�

] + [ a�

c�

d�

] – [ b�

c�

d�

]}

Solved Example: Show that the vectors k2j2i4b,k2j4i2a��

−−=−+−= and k4j2i2c��

+−−= are coplanar..

Solution: The vectors are coplanar since ]cba[��

=

422

224

242

−−−−−−

= 0

Self Practice Problems : 1. Show that 0)cba()cb(.a =++×+��

2. One vertex of a parallelopiped is at the point A (1, –1, –2) in the rectangular cartesian co-ordinate. If threeadjacent vertices are at B(–1, 0, 2), C(2, –2, 3) and D(4, 2, 1), then find the volume of the parallelopiped.Ans. 72

3. Find the value of m such that the vectors kji2 +− , k3j2i −+ and k5jmi3 ++ are coplanar..

Ans. – 4

4. Show that the vector c,b,a��

, are coplanar if and only if cb��

+ , ac��

+ , ba��

+ are coplanar..

9.9 .9 .9 . Vector Triple Product:Vector Triple Product:Vector Triple Product:Vector Triple Product:

Let� � �a b c, , be any three vectors, then the expression

� � �a x b x c( ) is a vector & is called a vector

triple product.

Geometrical Interpretation of � � �a x b x c( )

Consider the expression� � �a x b x c( ) which itself is a vector, since it is a cross product of two

vectors� � �a bx c& ( ) . Now

� � �a x b x c( ) is a vector perpendicular to the plane containing

� � �a bx c& ( )

but� �b x c is a vector perpendicular to the plane containing

� �b c& , therefore

� � �a x b x c( ) is a vector

which lies in the plane of� �b c& and perpendicular to

�a . Hence we can express

� � �a x b x c( ) in terms

of� �b c& i.e.

� � �a x b x c( ) = xb yc

� �+ where x & y are scalars.

�� a x b x c( ) = ( . ) ( . )

� � � � � �a c b a b c− � ( )

� � �a x b x c = ( . ) ( . )

� � � � � �a c b b c a−

� ( ) ( )� � � � � �a x b x c a x b x c≠ , in general

Solved Example For any vector a�

, prove that )ka(k)ja(j)ia(i��

××+××+×× = a2�

Solution. Let kajaiaa 321 ++=�

. Then, )ka(k)ja(j)ia(i ××+××+××��

= }i)a.i(a)i.i{(��

− + }j)a.j(a)j.j{(��

− + }k)a.k(a)k.k{(��

−

= }j)a.j(a{}i)a.i(a{(��

−+− + }k)a.k(a{��

−

= k)a.k(j)a.j(i)a.i{(a3��

++−

= )kajaia(a3 321 ++−�

= a2aa3��

=−

Solved Example Prove that )}ac(b{a��

××× = )ca)(d.b(��

× – )c.b(��

)da(��

×Solution. We have,

)}ac(b{a��

××× = }d)c.b(c)d.b{(a��

−×

= }d)c.b{(a}c)d.b{(a��

×−× [by dist. law]

= )da()c.b()c.b()ca()d.b(��

×−−×

Solved Example: Let a�

= iα + j2 – k3 , b�

= i + j2α – k2 and c�

= i2 – jα + k . Find the value(s) of α, if

any, such that ( ) ( ){ }cbba��

××× × ( )ac��

× = 0. Find the vector product when α = 0.

Solution.: ( ) ( ){ }cbba��

××× × ( )ac��

×

= [ ]cba��

b�

× ( )ac��

× = [ ]cba��

( ) ( ){ }ac.bcb.a��

−

which vanishes if (i) ( )b.a��

c�

= ( )c.b��

a�

(ii) [ ]cba��

= 0

(i) ( )b.a��

c�

= ( )c.b��

a�

leads to the equation 2 α3 + 10 α + 12 = 0, α2 + 6α = 0 and 6α – 6 = 0, which do

not have a common solution. (ii) [ ]cba��

= 0

⇒12

221

32

α−−α−α

= 0 ⇒ 3α = 2 ⇒ α = 3

2

when α = 0, [ ]cba��

= – 10, b.a��

= 6, c.b��

= 0 and the vector product is – 60 ( )ki2 + .

Sol ExaIf BA��

+ = a�

, A�

. a�

= 1 and BA��

× = b�

, then prove that A�

= 2|a|

aba�

��+×

and B�

= ( )

2

2

|a|

1|a|aab�

��

−+×.

Solution.: Given aBA��

=+ .....(i)

⇒ ( )BA.a��

+ = a.a��

⇒ aA.a��

+ . B�

= a.a��

⇒ 1 + B.a��

= 2|a|�

⇒ B.a��

= 2|a|�

– 1 Given bBA��

=× ⇒ ( )BAa��

×× = a�

× b�

⇒ ( )B.a��

A�

– ( )A.a��

baB��

×=

⇒ ( )1|a| 2 −�

BA��

− = ba��

× [using equation (2)]

solving equation (1) and (5), simultaneously, we get

A�

= 2|a|

aba�

��+×

and B�

= ( )

2

2

|a|

1|a|aab�

��−+×

Sol. Ex. Solve for r�

, the simultaneous equations bcbr��

×=× , 0a.r =��

provided a�

is not perpnedicular to b�

.

Solution )cr(��

− × b�

= 0 ⇒ cr��

− and b�

are collinear

∴ bkcr��

=− ⇒ r = bkc��

+ ........(i)

a.r��

= 0 ⇒ )bkc(��

+ . a�

= 0

⇒ k = – b.a

c.a��

��

putting in (i) we getb.a

c.acr ��

��

−= b�

Solved Example :If bxkax��

=+× , where k is a scalar and b,a��

are any two vectors, then determine x�

in

terms of b,a��

and k.

Solution: bxkax��

=+× ..........(i)

Premultiple the given equation vectorially by a�

)ax(a��

×× + k )xa(��

× = ba��

×⇒ ba)xa(ka)x.a(x)a.a(

��×=×+− ..........(ii)

Premultiply (i) scalarly by a�

]axa[��

+ )x.a(k��

= b.a��

b.a)x.a(k��

= .......(iii)

Substituting ax��

× from (i) and x.a��

from (iii) in (ii) we get

x�

= 22 ka

1

+

+×+ a

k

)b.a()ba(bk

��

��

Self Practice Problems : 1. Prove that 0)ba(c)ac(b)cb(a =××+××+××��

.

2. Find the unit vector coplanar with i + j + 2 k and i + 2 j + k and perpendicular to i + j + k .

Ans.2

1( – j + k ) or,

2

1( j – k )

3. Prove that )ab()a.a()}ba(a{a��

×=××× .

4. Given that 2p

1x ��

+ )x.p(��

qp��

= , show that q.p2

1x.p

��= and find x

� in terms of p

� and q

�.

5. If a.x��

= 0, b.x��

= 0 and c.x��

= 0 for some non-zero vector x� , then show that ]cba[

�� = 0

6. Prove that r�

= ]abc[

)cb()a.r(��

× +

]abc[

)ac()b.r(��

× +

]abc[

)ba()c.r(��

× where c,b,a

�� are three non-coplanar vectors

10 .10 .10 .10 . Reciprocal System Of Vectors:Reciprocal System Of Vectors:Reciprocal System Of Vectors:Reciprocal System Of Vectors: If � � �a b c, , &

� � �a' , b' , c' are two sets of non coplanar

vectors such that � � � � � �a.a' = b.b' = c.c' = 1 then the two systems are called Reciprocal System of vectors.

Note: [ ] [ ] [ ]cba

bxac

cba

axcb

cba

cxb=a ��

��

��

��

��

��

==

Solved Example If cba��

and c,b,a ′′′��

be the reciprocal system of vectors, prove that

(i) 3c.cb.ba.a =′+′+′��

(ii) 0ccbbaa��

=′×+′×+′×

Solution. (i) We have : a.a ′��

= b.b ′��

= c.c ′��

= 1

a.a ′��

+ b.b ′��

+ c.c ′��

= 1 + 1 + 1 = 3

(ii) We have : a′�

= λ )cb(��

× , b′�

= λ )ac(��

× and c′�

= λ ( )ba(��

× , where λ = ]cba[

1��

}c)b.a(b)c.a{()}cb(a{)cb(aaa��

−λ=××λ=×λ×=′×

}a)c.b(c)a.b{()}ac(b{)ac(bbb��

−λ=××λ=×λ×=′×

and }b)a.c(a)b.c{()}ba(c{)ba(ccc��

−λ=××λ=×λ×=′×∴ ccbbaa ′×+′×+′×

��

= λ }b)a.c(a)b.c{(}a)c.b(c)a.b{(}c)b.a(b)c.a{(��

−λ+−λ+−

= λ ]b)a.c(a)b.c(a)c.b(c)a.b(c)b.a(b)c.a[(��

−+−+−

= λ ]b)c.a(a)c.b(a)c.b(c)b.a(c)b.a(b)c.a[(��

−+−+−

= 00��

=λ

11 .11 .11 .11 . Linear Combinations:Linear Combinations:Linear Combinations:Linear Combinations:

Given a finite set of vectors� � �a b c, , ,...... then the vector

� � � �r xa yb zc= + + + ........ is called a linear

combination of� � �a b c, , ,...... for any x, y, z..... ∈ R. We have the following results:

(a) If� �a b, are non zero, non−collinear vectors then xa yb x a y b x x y y

� � � �+ = + ⇒ = =' ' ' ; '

(b) FundamentalTheorem: Let� �a b, be non zero, non collinear vectors. Then any vector

�r coplanar

with� �a b, can be expressed uniquely as a linear combination of

� �a b,

i.e. There exist some uniquly x, y ∈ R such that xa yb r� � �

+ = .

(c) If� � �a b c, , are non−zero, non−coplanar vectors then:

xa yb zc x a y b z c x x y y z z� � � � � �

+ + = + + ⇒ = = =' ' ' ' , ' , '

(d) Fundamental Theorem In Space: Let� � �a b c, , be non−zero, non−coplanar vectors in space. Then any

vector�r , can be uniquly expressed as a linear combination of

� � �a b c, , i.e. There exist some unique x,y

∈ R such that xa yb zc r� � � �

+ + = .

(e) I f� � �x x xn1 2, ,...... are n non zero v ectors, & k

1, k

2, . . . . .k

n are n scalars & i f the l inear

combinat ion k x k x k x k k kn n n1 1 2 2 1 20 0 0 0� � �

+ + = ⇒ = = =........ , ..... then we say that

vectors� � �x x xn1 2, ,...... are LINEARLY INDEPENDENT VECTORS.

(f) If� � �x x xn1 2, ,...... are not LINEARLY INDEPENDENT then they are said to be LINEARLY DEPENDENT vectors. i.e.

if k x k x k xn n1 1 2 2 0� � �

+ + + =........ & if there exists at least one kr ≠ 0 then

� � �x x xn1 2, ,...... are said to

beLLLL INEARLINEARLINEARLINEARLYYYY D D D DEPENDENTEPENDENTEPENDENTEPENDENT.

Note 1: If kr ≠ 0; k x k x k x k x k xr r n n1 1 2 2 3 3 0

� � � � �+ + + + + + =....... ......

− = + + + + + +− − + +k x k x k x k x k x k xr r r r r r n n

� � � � � �1 1 2 2 1 1 1 1....... . . ......

− = + + + + +− −kk

x kk

x kk

x kk

x kk

xr

r

r

r r

r

r

r n

r

n

1 1 1 1 11 1 2 2 1 1

� � � � �..... . .....

� � � � � �x c x c x c x c x c xr r r r r n n= + + + + + +− − −1 1 2 2 1 1 1...... ......

i.e.�x r is expressed as a linear combination of vectors.

� � � � �x x x x xr r n1 2 1 1, ,.......... , ,...........− +

Hence� � � � � �x with x x x x xr r r n1 2 1 1, ,.... , ....− + forms a linearly dependent set of vectors.

Note 2: � If�a = 3 i + 2 j + 5 k then

�a is expressed as a LINEAR COMBINATION of vectors i , j , k Also,Also,

�a , i , j ,

k form a linearly dependent set of vectors. In general, every set of four vectors is a linearly dependentsystem.

� i , j , k are Linearly Independent set of vectors. For K1 i + K

2j + K

3k = 0 ⇒ K

1= K

2= K

3 = 0

� Two vectors� �a b& are linearly dependent ⇒

�a is parallel to

�b i.e.

� �a x b = 0 ⇒ linear dependence

of� �a b& . Conversely if

� �a x b ≠ 0 then

� �a b& are linearly independent.

� If three vectors� � �a b c, , are linearly dependent, then they are coplanar i.e. [ , , ]

� � �a b c = 0 , conversely,,

if [ , , ]� � �a b c ≠ 0 , then the vectors are linearly independent.

Solved Example: Given A that the points a�

– 2 b�

+ 3 c�

, 2 a�

+ 3 b�

– 4 c�

, – 7 b�

+ 10 c�

, A, B, C have

position vector prove that vectors AB and AC are linearly dependent.Solution. Let A, B, C be the given points and O be the point of reference then

OA = a�

– 2 b�

+ 3 c�

, OB = 2 a�

+ 3 b�

– 4 c�

and OC = – 7 b�

+ 10 c�

Now AB = p.v. of B – p.v. of AA

= OB – OA = ( a�

+ 5 b�

– 7 c�

) = – AB

∴ AC = λ AB where λ = – 1. Hence AB and AC are linearly dependent

Solved Example: Prove that the vectors 5 a�

+ 6 b�

+ 7 c�

, 7 a�

– 8 b�

+ 9 c�

and 3 a�

+ 20 b�

+ 5 c�

are linearly

dependent a�

, b�

, c�

being linearly independent vectors.Solution. We know that if these vectors are linearly dependent , then we can express one of

them as a linear combination of the other two.

Now let us assume that the given vector are coplanar, then we can write

5 a�

+ 6 b�

+ 7 c�

= �( 7 a�

– 8 b�

+ 9 c�

) + m (3 a�

+ 20 b�

+ 5 c�

)where �, m are scalars

Comparing the coefficients of a�

, b�

and c�

on both sides of the equation5 = 7� + 3 ..........(i) 6 = – 8� + 20 m ..........(ii)7 = 9� + 5m ..........(iii)From (i) and (iii) we get

4 = 8� ⇒ � = 2

1 = m which evidently satisfies (ii) equation too.

Hence the given vectors are linearly dependent .Self Practice Problems :

1. Does there exist scalars u, v, w such that ieweveu 321

��=++ where ke1

��= , kje2

��+= , k2je3

��+−= ?

Ans. No

2. Consider a base c,b,a��

and a vector cb3a2��

−+− . Compute the co-ordinates of this vector relatively to

the base p, q, r where b3a2p��

−= , cb2aq��

+−= , c2ba3r��

++−= . Ans. (0, –7/5, 1/5)

3. If a�

and b�

are non-collinear vectors and A�

=(x + 4y) a�

+ (2x + y + 1) b�

and B�

= (y – 2x + 2) a�

+

(2x – 3y – 1) b�

, find x and y such that B2A3��

= . Ans. x = 2, y = –1

4. If vectors c,b,a��

be linearly independent, then show that :(i) c3b2a��

+− , c4b3a2��

−+− , c2b��

+− are

linearly dependent (ii) c2b3a��

+− , cb4a2��

−−− , cb2a3��

−+ are linearly independent.

5. Given that ji − , j2i − are two vectors. Find a unit vector coplanar with these vectors and perpendicular

to the first vector ji − . Find also the unit vector which is perpendicular to the plane of the two given

vectors. Do you thus obtain an orthonormal triad? Ans.2

1 )ji( + ; k; Yeses

6. If with reference to a right handed system of mutually perpendicular unit vectors k,j,i ji3��

−=α ,

k3ji2��

−+=β express β�

in the form 21 β+β=β��

where 1β�

is parallel to α�

& 2β�

is perpendicular to α�

.

Ans. j2

1i

2

31

��−=β , k3j

2

3i

2

12

��−+=β

7. Prove that a vector r�

in space can be expressed linearly in terms of three non-coplanar, non-null

vectors c,b,a��

in the form ]cba[

c]bar[b]acr[a]cbr[r ��

�� ++

=

Note: Test Of Collinearity: Three points A,B,C with position vectors� � �a b c, , respectively are collinear, if &

only if there exist scalars x , y, z not all zero simultaneously such that; xa yb zc� � �

+ + = 0 , where x + y+ z = 0.

Note: Test Of Coplanarity: Four points A, B, C, D with position vectors � � � �a b c d, , , respectively are coplanar

if and only if there exist scalars x, y, z, w not all zero simultaneously such that xa + yb + zc + wd = 0� � � �

where, x + y + z + w = 0.

Solved Example Show that the vectors c3ba2��

+− , c2ba��

−+ and c3ba��

−+ are non-coplanar vectors.Solution. Let, the given vectors be coplanar.

Then one of the given vectors is expressible in terms of the other two.

Let c3ba2��

+− = x ( )c2ba��

−+ + y ( )c3ba��

−+ , for some scalars x and y..

⇒ c3ba2��

+− = (x + y) a�

(x + y) b�

+ (–2x – 3y) c�

⇒ 2 = x + y, –1 = x + y and 3 = 2x – 3y.Solving, first and third of these equations, we get x = 9 and y = –7.Clearly, these values do not satisfy the thrid equation.Hence, the given vectors are not coplanar.

Solved Example: Prove that four points cb3a2��

−+ , c3b2a��

+− , c2b4a3��

−+ and c6b6a��

+− are coplanar..

Solution. Let the given four points be P, Q, R and S respectively. These points are coplanar if the vectors PQ ,

PR and PS are coplanar. These vectors are coplanar iff one of them can be expressed as a linearcombination of other two. So, let

PQ = x PR + y PS

⇒ c4b5a��

+−− = x ( )cba��

−+ + y ( )c7b9a��

−−− ⇒ c4b5a��

+−− = (x – y) a�

+ (x – 9y) b�

+ (–x + 7y) c�

⇒x – y = –1, x – 9y = –5, –x + 7y = 4 [Equating coeff. of c,b,a��

on both sides]

Solving the first of these three equations, we get x = – 2

1, y =

2

1.

These values also satisfy the third equation. Hence, the given four points are coplanar.Self Practice Problems :

1. If, d,c,b,a��

are any four vectors in 3-dimensional space with the same initial point and such that

0d2cb2a3��

=−+− , show that the terminal A, B, C, D of these vectors are coplanar. Find the point atwhich AC and BD meet. Find the ratio in which P divides AC and BD.

2. Show that the vector cba��

+− , acb��

−− and c4b3a2��

−− are non-coplanar, where c,b,a��

, are any non-

coplanar vectors.

3. Find the value of λ for which the four points with position vectors kj −− , kj5i4 λ++ . k4j9i3 ++ and

k4j4i4 ++− are coplanar.. Ans. λ = 1

12 .12 .12 .12 . Application Of Vectors:Application Of Vectors:Application Of Vectors:Application Of Vectors:(a) Work done against a constant force�F over a displacement

�s

is defined as � � �

W F s= . (b) The tangential velocity�V of a body moving in a circle is given

by� � �V w x r= where

�r is the pv of the point P..

(c) The moment of�F about ’O’ is defined as

� � � �M r x F where r= is the pv of P wrt ’O’. The direction of

�M

is along the normal to the plane OPN such that� � �r F M, & form a right handed system.

(d) Moment of the couple = ( )� � �r r x F1 2− where r r

� �1 2& are pv’s of the point of the application of the forces

� �F F& .−

Solved Example: Forces of magnitudes 5 and 3 units acting in the directions k3j2i6 ++ and k6j2i3 ++respectively act on a particle which is displaced from the point (2, 2, –1) to (4, 3, 1). Find the workdone by the forces.

Solution. Let F�

be the resultant force and d�

be the displacement vector. Then,

F�

= 9436

)k3j2i6(5

++

++ + 3

3649

)k6j2i3(

++

++ =

7

1 )k33j4i39( ++

and, d�

= )kj3i4( ++ – )kj2i2( −+ = k2ji2 ++

∴ Total work done = F�

. d�

= 7

1 )k33j4i39( ++ . )k2ji2( ++

= 7

1 (78 + 4 + 66) =

7

148 units.

Self Practice Problems :1. A point describes a circle uniformly in the i , j plane taking 12 seconds to

complete one revolution. If its initial position vector relative to the centre is i , and the rotation is from

i to j , find the position vector at the end of 7 seconds. Also find the velocity vector.. Ans. 1 / 2

( )i3j − , p/12 ( )j3i −

2. The force represented by k2i3 + is acting through the point k3j4i5 −+ . Find its moment about the

point kj3i ++ . Ans. k3j20i2 −−

3. Find the moment of the comple formed by the forces ki5 + and ki5 −− acting at the points

(9, –1, 2) and (3, –2, 1) respectively Ans. k5ji −−

Miscellaneous Solved Examples

Solved Example: Show that the points A, B, C with position vectors kji2 +− , k5j3i −− and k4j4i3 −−respectively, are the vertices of a right angled triangle. Also find the remaining angles of the triangle.

Solution. We have, AB = Position vector of B – Position vector of A

= )k5j3i( −− – )kji2( +− = k6j2i −−−

BC = Position vector of C – Position vector of B

= )k4j4i3( −− – )k5j3i( −− = kji2 +−

and, CA = Position vector of A – Position vector of C

= )kji2( +− – )k4j4i3( −− = k5j3i ++−

Since AB + BC + CA = )k6j2i( −−− + )kji2( +− + )k5j3i( ++− = 0�

So, A, B and C are the vertices of a triangle.

Now, BC . CA = )kji2( +− . )k5j3i( ++− = –2 – 3 + 5 = 0

⇒ BC ⊥ CA ⇒ ∠BCA = 2

πHence, ABC is a right angled triangle.

Since a is the angle between the vectors AB and AC . Therefore

cos A = |AC||AB|

AC.AB =

222222 )5()3(1)6()2()1(

)k5j3i(.)k6j2i(

−+−+−+−+−

−−−−−

= 25913641

3061

++++

++− =

3541

35 =

41

35A = cos–1

41

35

cos B = |BC||BA|

BA = 222222 )1()1(2621

)kji2()k6j2i(

+−+++

+−++ ⇒ cos B =

641

622 +− =

41

6 ⇒ B = cos–1

41

6

Solved Example: If c,b,a��

are three mutually perpendicular vectors of equal magnitude, prove that cba��

++

is equally inclined with vectors b,a��

and c�

.

Solution.: Let |a|�

= |b|�

= |c|�

= λ (say). Since c,b,a��

are mutually

perpendicular vectors, therefore b.a��

= c.b��

= a.c��

= 0 ..............(i)

Now,2

cba��

++

= a.a��

+ b.b��

+ c.c��

+ b.a2��

+ c.b2��

+ a.c2��

= 2|a|�

| + 2|b|�

+ 2|c|�

[Using (i) ]

= 3λ2 [∵ |a|�

= |b|�

= |c|�

= λ]

∴ |cba|��

++ = λ3 ..............(ii)

Suppose cba��

++ makes angles θ1, θ

2, θ

3 with b,a

�� and c

� respectively. Then,

cosθ1 =

|cba||a|

)cba(.a��

��

++

++ =

|cba||a|

c.ab.aa.a��

��

++

++

= |cba||a|

|a| 2

��

�

++ =

|cba|

|a|��

�

++ =

λ

λ

3 =

3

1[Using (ii)]

∴ θ1 cos–1

3

1

Similarly, θ2 = cos–1

3

1 and θ

3 = cos–1

3

1∴ θ

1 = θ

2 = θ

3.

Hence, cba��

++ is equally inclineded with b,a��

and c�

Solved Example: Prove using vectors : If two medians of a triangle are equal, then it is isosceles.Solution. : Let ABC be a triangle and let BE and CF be two equal medians. Taking A as the origin,

let the position vectors of B and C be b�

and c�

respectively. Then,

P.V. of E = 2

1 c�

and, P.V. of F = 2

1 b�

∴ BE = 2

1 )b2c(

��−

CF = 2

1 )c2b(

��−

Now, BE = CF ⇒ |BE| = |CF|

⇒ 2|BE| = 2|CF| ⇒

2

)b2c(2

1 ��− =

2

)c2b(2

1 ��

−

⇒4

1 2|b2c|

��− =

4

1 2|c2b|

��− ⇒ 2|b2c|

��− = 2|c2b|

��−

⇒ )b2c(��

− . )b2c(��

− = )c2b(��

− . )c2b(��

−

⇒ c.c��

– c.b4��

+ b.b4��

= b.b��

– c.b4��

+ c.c4��

⇒ 2|c|�

– c.b4��

+ 4 2|b|�

= 2|b|�

– c.b4��

+ 2|c|4�

⇒ 3 2|b| = 3 2|c|�

⇒ 2|b| = 2|c|�

⇒ AB = AC Hence, triangle ABC is an isosceles triangle.Solved Example: Using vectors : Prove that cos (A + B) = cos A cos B – sin A sin B

Solution. Let OX and OY be the coordinate axes and let i and j be unit vectors along OX and OY

respectively. Let ∠XOP = A and ∠XOQ = B. Drawn PL ⊥ OX and QM ⊥ OX.

Clearly angle between OP and OQ is A + B

In ∆OLP, OL = OP cos A and LP = OP sin A. Therefore OL = (OP cos A) i and LP = (OP sin A) A) ( )j−

Now. OL + LP = OP

⇒ OP = OP [(cos A ) i – (sin A)A) j ]

In ∆OMQ, OM = OQ cos B and MQ = OQ sin B.Therefore,

OM = (OQ cos B) i , MQ = (OQ sin B) j

Now, OM + MQ= OQFrom (i) and (ii), we get

OP . OQ = OP [(cos A) i – (sin A)A) j ] . OQ [(cos B) i + (sin B) j ]

= OP . OQ [cos A cos B – sin A sin B]

But, OP . OQ = |OP| |OQ| cos (A + B) = OP . OQ cos (A + B)

∴ OP . OQ cos (A + B) = OP . OQ [cos A cos B – sin A sin B]⇒ cos (A + B) = cos A cos B – sin A sin B

Solved Example: Prove that in any triangle ABC(i) c2 = a2 + b2 – 2ab cos C (ii) c = bcosA + acosB.

Solution. (i) In ∆ABC, AB + BC + CA = 0

or, BC + CA = – AB ......(i)Squaring both sides

( BC )2 + ( CA )2 + ( BC ). CA + ( AB )2

⇒ a2 + b2 + 2 ( BC . CA ) = c2 ⇒ c2 = a2 + b2 = 2 ab cos (π – C)⇒ c2 = a2 + b2 – 2ab cosC

(ii) ( BC + CA ). AB = – AB . AB

BC . AB + CA . AB = – c2

– ac cosB – bc cos A = – c2

acosB + bcosA = c.Solved Ex.: If D, E, F are the mid-points of the sides of a triangle ABC, prove by vector method that area of

∆DEF = 4

1 (area of ∆ABC)

Solution. Taking A as the origin, let the position vectors of B and C be b�

and c�

respectively. Then, the

position vectors of D, E and F are 2

1 )cb(

��+ ,

2

1c�

and 2

1b�

respectively..

Now, DE = 2

1 c�

– 2

1 )cb(

��+ =

2

b�

−

and DF = 2

1 b�

– 2

1 ( )cb(

��+ =

2

c�

−

∴ Vector area of ∆DEF 2

1 )DFDE( × =

2

1

−×

−

2

c

2

b��

= 8

1 )cb(

��

× = 4

1

× )ACAB(2

1

1(vector area of ∆ABC) Hence, area of ∆DEF =

4

1 area of ∆ABC.

Solved Example: P, Q are the mid-points of the non-parallel sides BC and AD of a trapezium ABCD. Showthat ∆APD = ∆CQB.

Solution. Let AB = b�

and AD = d�

Now DC is parallel to AB ⇒ there exists a acalar t sush that DC = t DB = t b�

∴ AC = AD + DC = btd��

+

= 4



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The position vectors of P and Q are 2

1 )btdb(

��

++ and 2

1 d�

respectively..

Now 2∆ APD = AP × AD

= 2

1 )btdb(

��

++ × d�

= 2

1 (1 + t) )db(

��

×

Also 2∆ CQB = CQ × CB =

+− btd(d

2

1 ��

× )]btd(b[��

+−

= [ ]b)t1(dbtd2

1−+−×

−−�

= dbtbd)t1(2

1 ��

×+×−−

= db)t2t1(2

1 ��

×+− = db)t1(2

1 ��

×+ = APD2∆ Hence the result.

Solved Example: Let u�

and v�

are unit vectors and w�

is a vector such that uvu��

+× = w�

and vuw��

=× then

find the value of [ ]wvu��

.

Solution. Given uvu��

+× = w�

and vuw��

=×

⇒ ( )uvu��

+× × uwu��

×= ⇒ ( )vu��

× × uuu��

×+ = v�

(as, vuw��

=× )

⇒ ( )u.u��

( )u.vv��

− vuuu��

=×+ (using u�

. u�

= 1 and uu��

× = 0, since unit vector)

⇒ vu)u.v(v��

=− ⇒ ( ) 0uv.u��

=

⇒ v.u��

= 0 (as; u�

≠ 0) .............(i)

⇒ )wv(.u��

×

= ))uvu(v(.u��

+×× (given vuw��

×= + u)

= )uv)vu(v(.u��

×+×× = )uvv)u.v(u)v.v((.u��

×+−

= )uv0u|v(|.u 2 ��×+− (as; 0v.u =

�� from (i))

= 2|v|�

)u.u(��

– u�

. )uv(��

× = 2|v|�

2|u|�

– 0 (as, [ ]uvu��

= 0)

= 1 (as; |u|�

= |v|�

= 1) ∴ [ ]wvu��

= 1Sol. Ex.: In any triangle, show that the perpendicular bisectors of the sides are concurrent.Solution. Let ABC be the triangle and D, E and F are respectively middle points of sides BC, CA and AB.

Let the perpendicular of D and E meet at O join OF. We are required to prove that OF is ⊥ to AB. Let the

position vectors of A, B, C with O as origin of reference be a�

, b�

and c�

respectively..

∴ OD = 2

1 ( b�

+ c�

), OE = 2

1 ( c�

+ a�

) and OF = 2

1 ( a�

+ b�

)

Also BC = c�

– b�

, CA = a�

– c�

and AB = b�

– a�

Since OD ⊥ BC, 2

1 ( b�

+ c�

) . ( c�

– b�

) = 0

⇒ b2 = c2 ............(i)

Similarly 2

1 ( c�

+ a�

) . ( a�

+ c�

) = 0

⇒ a2 = c2 ............(ii)from (i) and (ii) we have a2 – b2 = 0

⇒ ( a�

+ b�

) . ( b�

+ a�

) = 0 ⇒ 2

1( b�

+ a�

) . ( b�

– a�

) = 0

Solved Example: A, B, C, D are four points in space. using vector methods, prove thatAC2 + BD2 + AD2 + BC2 ≥ AB2 + CD2 what is the implication of the sign of equaility.

Solution.: Let the position vector of A, B, C, D be c,b,a��

and d�

respectively then

AC2 + BD2 + AD2 + BC2 = ( )ac��

− . ( )ac��

− + ( )bd��

− . ( )bd��

− + ( )ad��

− . ( )ad��

− + ( )bc��

− . ( )bc��

−

= 2|c|�

+ 2|a|�

– 2 c.a��

+ 2|d|�

+ 2|b|�

– 2 b.d��

+ 2|d|�

+ 2|a|�

– 2 d.a��

+ 2|c|�

+ 2|b|�

– c.b2��

= 2|a|�

+ 2|b|�

– 2 b.a��

+ 2|c|�

+ 2|d|�

– 2 d.c��

+ 2|a| +

2|b|�

+ 2|c|�

+ 2|d|�

+ 2 b.a��

+ d.c2��

– c.a2��

– d.b2��

– d.a2��

– c.b2��

= ( )ba��

− . ( )ba��

− + ( )dc��

− . ( )dc��

− + ( )dcba��

−−− ≥ ABAB2 + CD2

= AB2 + CD2 + ( )dcba��

−−+ . ( )dcba��

−−+ ≤ ABAB2 + CD2

∴ AC2 + BD2 + AD2 + BC2 ≥ AB2 + CD2

for the sign of equality to hold, dcba��

−−+ = 0

bdca��

−=−

⇒ AC and BD are collinear the four points A, B, C, D are collinear

SHORT REVISION1. DEFINITIONS: A VECTOR may be described as a quantity having both magnitude & direction. A

vector is generally represented by a directed line segment, say →

AB . A is called the initial point & B is

called the terminal point. The magnitude of vector →

AB is expressed by →

AB .ZERO VECTOR a vector of zero magnitude i.e.which has the same initial & terminal point, is called aZERO VECTOR. It is denoted by O.

UNIT VECTOR a vector of unit magnitude in direction of a vector �a is called unit vector along a

� and is

denoted by a symbolically a

aa �

�

= .

EQUAL VECTORS two vectors are said to be equal if they have the same magnitude, direction & representthe same physical quantity.

COLLINEAR VECTORS two vectors are said to be collinear if their directed line segments are paralleldisregards to their direction. Collinear vectors are also called PARALLEL VECTORS. If they have the samedirection they are named as like vectors otherwise unlike vectors.

Simbolically, two non zero vectors �a and

�b are collinear if and only if, bKa

��= ,

where K ∈ RCOPLANAR VECTORS a given number of vectors are called coplanar if their line segments are all parallelto the same plane. Note that “TWO VECTORS ARE ALWAYS COPLANAR”.

POSITION VECTOR let O be a fixed origin, then the position vector of a point P is the vector →

OP . If

b&a��

& position vectors of two point A and B, then ,→

AB = ab��

− = pv of B − pv of A . .

2. VECTOR ADDITION :� If two vectors b&a��

are represented by →→

OB&OA , then their

sum ba��

+ is a vector represented by →

OC , where OC is the diagonal of the parallelogram OACB.

� abba��

+=+ (commutative) � ( ) ( )� � � � � �a b c a b c+ + = + + (associativity)

� a0a0a��

+==+ � a)a(0)a(a��

+−==−+3. MULTIPLICATION OF VECTOR BY SCALARS :

If �a is a vector & m is a scalar, then m

�a is a vector parallel to

�a whose modulus is m times that of

�a . This multiplication is called SCALAR MULTIPLICATION. If b&a

�� are vectors & m, n are scalars, then:

amm)a()a(m��

== a)mn()am(n)an(m��

==anama)nm(��

+=+ bmam)ba(m��

+=+4. SECTION FORMULA :

If � �a b& are the position vectors of two points A & B then the p.v. of a point which divides AB in the

ratio m : n is given by : nm

bmanr

++

=��

�. Note p.v. of mid point of AB =

2

ba��

+ .

5. DIRECTION COSINES :

Let kajaiaa 321 ++=�

the angles which this vector makes with the +ve directions OX,OY & OZ are

called DIRECTION ANGLES & their cosines are called the DIRECTION COSINES .

a

acos 1

�=α , a

acos 2

�=β , a

acos 3

�=Γ . Note that, cos² α α α α + cos² ββββ + cos² ΓΓΓΓ = 1

6. VECTOR EQUATION OF A LINE :

Parametric vector equation of a line passing through two point )b(B&)a(A��

is given by, )ab(tar��

−+=where t is a parameter. If the line passes through the point )a(A

� & is parallel to the vector b

� then its

equation is, btar��

+=

�r = �a + λ

�b &

�r = �a + µ �c is :

�r =

�a + t ( )� �b c+ &

�r =

�a + p ( )� �c b− .

7. TEST OF COLLINEARITY : Three points A,B,C with position vectors c,b,a��

respectively are

collinear, if & only if there exist scalars x , y , z not all zero simultaneously such that ; 0czbyax =++��

,where x + y + z = 0.

8. SCALAR PRODUCT OF TWO VECTORS :

� )0(cosbab.a π≤θ≤θ=��

,

note that if θ is acute then b.a��

> 0 & if θ is obtuse then b.a��

< 0

� a.bb.a,aaa.a 22 ��=== (commutative) �

� � � � � � �a b c a b a c. ( ) . .+ = + (distributive)

� ba0b.a��

⊥⇔= )0b0a( ≠≠��

� 1k.kj.ji.i === ; 0i.kk.jj.i === � projection of b

b.abona �

��

= .

Note: That vector component of a�

along b�

= � �

��a b

bb

⋅

2

and perpendicular to b�

= a�

–

� �

��a b

bb

⋅

2.

� the angle φ between a�

& b�

is given by cos.

φ =� �

� �a b

a b 0 ≤ φ ≤ π

� if kajaiaa 321 ++=�

& kbjbibb 321 ++=�

then b.a��

= a1b

1 + a

2b

2 + a

3b

3

2

3

2

2

2

1aaaa ++=

�,

2

3

2

2

2

1bbbb ++=

�

Note : (i) Maximum value of a�

. b�

= a�

b�

(ii) Minimum values of a

� . b�

= a�

. b�

= − a�

b�

(iii) Any vector a

� can be written as , a

� = ( ) ( ) ( )� � �

a i i a j j a k k. � � . � � . � �+ + .

(iv) A vector in the direction of the bisector of the angle between the two vectors � �a b& is

�

�

�

�a

a

b

b+ . Hence

bisector of the angle between the two vectors � �a b&

is ( )λ � �a b+ , where λ ∈ R+. Bisector of the

exterior angle between � �a b& is ( )λ � �a b− , λ ∈ R+ .

9. VECTOR PRODUCT OF TWO VECTORS :

(i) � If � �a b& are two vectors & θ is the angle between them then nsinbaba

��θ=× ,

where n�

is the unit vector perpendicular to both � �a b& such that n&b,a

�� forms a

right handed screw system .

(ii) � Lagranges Identity : for any two vectors � � � � � � � �

� � � �

� � � �a b a x b a b a ba a a b

a b b b& ;( ) ( . )

. .

. .

2 2 22= − =

(iii) � Formulation of vector product in terms of scalar product:

The vector product bxa��

is the vector �c , such that

(i) |�c | =

� � � �a b a b2 2 2− ⋅( ) (ii)

� �c a⋅ = 0;

� �c b⋅ =0 and

(iii) � � �a b c, , form a right handed system

(iv) � b&a0ba��

⇔=× are parallel (collinear) )0b,0a( ≠≠��

i.e. bKa��

= , where K is a scalar..

� abba��

×≠× (not commutative)

� )ba(m)bm(ab)am(��

×=×=× where m is a scalar .

� )ca()ba()cb(a��

×+×=+× (distributive)

� 0kkjjii =×=×=× � jik,ikj,kji =×=×=×

(v) � If kajaiaa 321 ++=�

& kbjbibb 321 ++=�

then

321

321

bbb

aaa

kji

ba =×��

(vi) � Geometrically ba��

× = area of the parallelogram whose two adjacent sides are

represented by b&a��

.

(vii) � Unit vector perpendicular to the plane of ba

banisb&a ��

��

×

×±=

� A vector of magnitude ‘r’ & perpendicular to the palne of ( )

ba

barisb&a ��

��

×

×±

� If θ is the angle between ba

basinthenb&a ��

�� ×

=θ

(viii) Vector area

� If c&b,a��

are the pv’s of 3 points A, B & C then the vector area of triangle ABC =

[ ]1

2

� � � � � �a x b bx c cxa+ + . The points A, B & C are collinear if

� � � � � �a x b bx c cxa+ + = 0

� Area of any quadrilateral whose diagonal vectors are 21

d&d��

is given by 1

21 2

� �d xd

10. SHORTEST DISTANCE BETWEEN TWO LINES :If two lines in space intersect at a point, then obviously the shortest distance between them is zero. Lineswhich do not intersect & are also not parallel are called SKEW LINES. For Skew lines the direction ofthe shortest distance would be perpendicular to both the lines. The magnitude of the shortest distance

vector would be equal to that of the projection of →

AB along the direction of the line of shortest distance,→

LM is parallel to qxp��

i.e. LM ojection of AB on LM→ → →

= Pr = Projection of AB on pxq→ � �

= AB pxq

p x q

b a pxq

p x q

→

=−. ( ) ( ) . ( )

� �

� �

� � � �

� �

1. The two lines directed along � �p q& will intersect only if shortest distance = 0 i.e.

( ).( )� � � �b a pxq− = 0 i.e. ( )� �

b a− lies in the plane containing � �p q& . ⇒ ( )[ ]� � � �

b a p q− = 0 .

2. If two lines are given by � � � � � �r a Kb r a Kb1 1 2 2= + = +&

i.e. they are parallel then , dbx a a

b=

−� � �

�( )2 1

11. SCALAR TRIPLE PRODUCT / BOX PRODUCT / MIXED PRODUCT :

� The scalar triple product of three vectors c&b,a��

is defined as :

cbac.bxa��

= sin cosθ φ where θ is the angle between � �a b& & φ is the angle between c&ba

��× .

It is also defined as ]cba[��

, spelled as box product .� Scalar triple product geometrically represents the volume of the parallelopiped whose three couterminous

edges are represented by ]cba[V.e.ic&b,a��

=� In a scalar triple product the position of dot & cross can be interchanged i.e.

]bac[]acb[]cba[ORc.)bxa()cxb(.a��

===�

� � � � � � � � � � � �a bx c a cx b i e a b c a c b. ( ) .( ) . . [ ] [ ]= − = −

� If kajaiaa 321 ++=�

; kbjbibb 321 ++=�

& kcjcicc 321 ++=�

then [ ]� � �a b c

a a a

b b b

c c c

=1 2 3

1 2 3

1 2 3

.

In general , if � � � �a a l a m a n= + +1 2 3 ;

� � � �b b l b m b n= + +1 2 3

& � � � �c c l c m c n= + +1 2 3

then [ ] [ ]� � � � � �a b c

a a a

b b b

c c c

l mn=1 2 3

1 2 3

1 2 3

; where ��

, &m n are non coplanar vectors .

� If � � �a b c, , are coplanar ⇔ =[ ]

� � �a b c 0 .

� Scalar product of three vectors, two of which are equal or parallel is 0 i.e. [ ]� � �a b c = 0 ,

Note : If � � �a b c, , are non − coplanar then [ ]

� � �a b c > 0 for right handed system &

[ ]� � �a b c < 0 for left handed system .

� [i j k] = 1 � [ ] [ ]Ka b c K a b c� � � � � �

= � [( ) ] [ ] [ ]� � � � � � � � � �a b c d a c d b c d+ = +

� The volume of the tetrahedron OABC with O as origin & the pv’s of A, B and C being � � �a b c, &

respectively is given by V a b c=1

6[ ]� � �

� The positon vector of the centroid of a tetrahedron if the pv’s of its angular vertices are � � � �a b c d, , & are

given by 1

4[ ]� � � �a b c d+ + + .

Note that this is also the point of concurrency of the lines joining the vertices to the centroids of theopposite faces and is also called the centre of the tetrahedron. In case the tetrahedron is regular it isequidistant from the vertices and the four faces of the tetrahedron .

Remember that : [ ]� � � � � �a b b c c a− − − = 0 & [ ]� � � � � �

a b b c c a+ + + = 2 [ ]� � �a b c .

*12. VECTOR TRIPLE PRODUCT : Let c,b,a��

be any three vectors, then the expression

)cb(a��

×× is a vector & is called a vector triple product .

GEOMETRICAL INTERPRETATION OF )cb(a��

××Consider the expression )cb(a

��×× which itself is a vector, since it is a cross product of two vectors

� � �a bx c& ( ) . Now

� � �a x b x c( ) is a vector perpendicular to the plane containing

� � �a bx c& ( ) but

� �b x c

is a vector perpendicular to the plane c&b��

, therefore � � �a x b x c( ) is a vector lies in the plane of

c&b��

and perpendicular to a�

. Hence we can express � � �a x b x c( ) in terms of

� �b c&

i.e. � � �a x b x c( ) = xb yc

� �+ where x & y are scalars .

�� a x b x c( ) = ( . ) ( . )

� � � � � �a c b a b c− � ( )

� � �a x b x c = ( . ) ( . )

� � � � � �a c b b c a−

� ( ) ( )� � � � � �a x b x c a x b x c≠

13. LINEAR COMBINATIONS / Linearly Independence and Dependence of Vectors :

Given a finite set of vectors � � �a b c, , ,...... then the vector ........czbyaxr +++=

��is called a linear

combination of � � �a b c, , ,...... for any x, y, z ...... ∈ R. We have the following results :

(a) FUNDAMENTALTHEOREM IN PLANE : Let b,a��

be non zero , non collinear vectors . Then any vector r�

coplanar with b,a��

can be expressed uniquely as a linear combination of b,a��

i.e. There exist some unique x,y ∈ R such that rbyax��

=+ .

(b) FUNDAMENTAL THEOREM IN SPACE : Let c,b,a��

be non−zero, non−coplanar vectors in space. Then

any vector r�

, can be uniquily expressed as a linear combination of c,b,a��

i.e. There exist some unique

x,y ∈ R such that rczbyax��

=++ .

(c) If n21x......,x,x��

are n non zero vectors, & k1, k

2, .....k

n are n scalars & if the linear combination

0k.....0k,0k0xk........xkxkn21nn2211

===⇒=++��

then we say that vectors n21x......,x,x��

are LINEARLY INDEPENDENT VECTORS .

(d) If n21x......,x,x��

are not LINEARLY INDEPENDENT then they are said to be LINEARLY DEPENDENT

vectors . i.e. if 0xk........xkxknn2211

=+++��

& if there exists at least one kr ≠ 0 then

n21x......,x,x��

are said to be LINEARLY DEPENDENT .

Note :� If a�

= 3i + 2j + 5k then a�

is expressed as a LINEAR COMBINATION of vectors k,j,i . Also , a�

,

k,j,i form a linearly dependent set of vectors. In general , every set of four vectors is a linearly

dependent system.

� k,j,i are LINEARLY INDEPENDENT set of vectors. For 0kKjKiK 321 =++ ⇒ K1 = 0 = K

2 = K

3.

� Two vectors b&a��

are linearly dependent ⇒ a�

is parallel to �b i.e. 0bxa =

�� ⇒ linear dependence of

b&a��

. Conversely if 0bxa ≠��

then b&a��

are linearly independent .

� If three vectors c,b,a��

are linearly dependent, then they are coplanar i.e. [ , , ]� � �a b c = 0 , conversely, if

[ , , ]� � �a b c ≠ 0 , then the vectors are linearly independent.

14. COPLANARITY OF VECTORS :

Four points A, B, C, D with position vectors d,c,b,a��

respectively are coplanar if and only if there exist

scalars x, y, z, w not all zero simultaneously such that 0=dw+cz+by+ax��

where, x + y + z + w = 0.15. RECIPROCAL SYSTEM OF VECTORS :

If c,b,a��

& 'c,'b,'a��

are two sets of non coplanar vectors such that 1='c.c='b.b='a.a��

then the

two systems are called Reciprocal System of vectors.

Note : [ ] [ ] [ ]a'=bx c

a b c a b c a b c

� �

� � �

� �

� � �

� �

� � �; ' ; 'bcxa

ca x b

= =

16. EQUATION OF A PLANE :

(a) The equation 0n.)rr(0

=−��

represents a plane containing the point with p.v. nwherer0

�� is a

vector normal to the plane . dn.r =��

is the general equation of a plane.(b) Angle between the 2 planes is the angle between 2 normals drawn to the planes and the angle between

a line and a plane is the compliment of the angle between the line and the normal to the plane.17. APPLICATION OF VECTORS :

(a) Work done against a constant force �F over a

displacement �s is defined as s.FW

��=

(b) The tangential velocity V�

of a body moving in a

circle is given by rwV��

×= where �r is the pv of the

point P.

(c) The moment of F�

about ’O’ is defined as rwhereFrM��

×=is the pv of P wrt ’O’. The direction of M

� is along the

normal to the plane OPN such that M&F,r��

form a

right handed system.

(d) Moment of the couple = F)rr( 21

��×− 21

r&rwhere��

are pv’s of the

point of the application of the forces .F&F��

−3 -D COORDINATE GEOMETRY USEFUL RESULTSA General :(1) Distance (d) between two points (x

1 , y

1 , z

1) and (x

2 , y

2 , z

2)

d = 212

212

212 )zz()yy()xx( −+−+−

(2) Section Fomula

x = 21

2112

mm

xmxm

++

; y = 21

2112

mm

ymym

++

; z = 21

2112

mm

zmzm

++

( For external division take –ve sign )Direction Cosine and direction ratio's of a line

(3) Direction cosine of a line has the same meaning as d.c's of a vector.(a) Any three numbers a, b, c proportional to the direction cosines are called the direction ratios i.e.

222cba

1

c

n

b

m

a ++±===

l

same sign either +ve or –ve should be taken through out.note that d.r's of a line joining x

1 , y

1 , z

1 and x

2 , y

2 , z

2 are proportional to x

2 – x

1 , y

2 – y

1 and z

2 – z

1

(b) If θ is the angle between the two lines whose d.c's are l1 , m

1 , n

1 and l

2 , m

2 , n

2cosθ = l

1 l

2 + m

1 m

2 + n

1 n

2hence if lines are perpendicular then l

1 l

2 + m

1 m

2 + n

1 n

2 = 0

if lines are parallel then 2

1

2

1

2

1

n

n

m

m==

l

l

note that if three lines are coplanar then

333

222

111

nm

nm

nm

l

l

l

= 0

(4) Projection of the join of two points on a line with d.c's l, m, n arel (x

2 – x

1) + m(y

2 – y

1) + n(z

2 – z

1)

B PLANE: (i) General equation of degree one in x, y, z i.e. ax + by + cz + d = 0 represents a plane.(ii) Equation of a plane passing through (x

1 , y

1 , z

1) is

a (x – x1) + b (y – y

1) + c (z – z

1) = 0

where a, b, c are the direction ratios of the normal to the plane.(iii) Equation of a plane if its intercepts on the co-ordinate axes are x

1 , y

1 , z

1 is

1z

z

y

y

x

x

111

=++ .

(iv) Equation of a plane if the length of the perpendicular from the origin on the plane is p and d.c's of theperpendicular as l , m, , n is l x + m y + n z = p

(v) Parallel and perpendicular planes – Two planesa

1 x + b

1 y + c

1z + d

1 = 0 and a

2x + b

2y + c

2z + d

2 = 0 are

perpendicular if a1 a

2 + b

1 b

2 + c

1 c

2 = 0

parallel if2

1

2

1

2

1

c

c

b

b

a

a== and

coincident if2

1

2

1

2

1

2

1

d

d

c

c

b

b

a

a===

(vi) Angle between a plane and a line is the compliment of the angle between the normal to the plane and the

line . If |n|.|b|

n.bsin)90cos(then

dn.r:Planebar:Line

��

��

��

��

=θ=θ−

=

λ+=.

where θ is the angle between the line and normal to the plane.

(vii) Length of the perpendicular from a point (x1 , y

1 , z

1) to a plane ax + by + cz + d = 0 is

p = 222

111

cba

dczbyax

++

+++

(viii) Distance between two parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d

2 = 0 is

222

21

cba

dd

++

−

(ix) Planes bisecting the angle between two planesa

1x + b

1y + c

1z + d

1 = 0 and a

2 + b

2y + c

2z + d

2 = 0 is given by

21

21

21

1111

cba

dzcybxa

++

+++ =

22

22

22

2222

cba

dzcybxa

++

+++±

Of these two bisecting planes , one bisects the acute and the other obtuse angle between the givenplanes.

(x) Equation of a plane through the intersection of two planes P1 and P

2 is given by P

1 + λP

2 = 0

C STRAIGHT LINE IN SPACE

(i) Equation of a line through A (x1 , y

1 , z

1) and having direction cosines l ,m , n are

n

zz

m

yyxx 111 −=

−=

−l

and the lines through (x1 , y

1 ,z

1) and (x

2 , y

2 ,z

2)

12

1

12

1

12

1

zz

zz

yy

yy

xx

xx

−−

=−

−=

−−

(ii) Intersection of two planes a1x + b

1y + c

1z + d

1 = 0 and a

2x + b

2y + c

2z + d

2 = 0

together represent the unsymmetrical form of the straight line.

(iii) General equation of the plane containing the line n

zz

m

yyxx 111 −=

−=

−l

is

A (x – x1) + B(y – y

1) + c (z – z

1) = 0 where Al + bm + cn = 0 .

LINE OF GREATEST SLOPEAB is the line of intersection of G-plane and H is the horizontalplane. Line of greatest slope on a given plane, drawn through agiven point on the plane, is the line through the point 'P'perpendicular to the line of intersetion of the given plane withany horizontal plane.

EXERCISE–1Q.1 If

� �a b& are non collinear vectors such that , b)1yx2(a)y4x(p

��++++= &

b)1y3x2(a)2x2y(q��

−−++−= , find x & y such that q2p3��

= .

Q.2 (a) Show that the points c10b7&c4b3a2;c3b2a��

+−−++− are collinear .(b) Prove that the points A = (1,2,3), B (3,4,7), C (−3,−2,−5) are collinear & find the ratio in which

B divides AC.

Q.3 Points X & Y are taken on the sides QR & RS , respectively of a parallelogram PQRS, so that QX XR→ →

= 4

& RY YS→ →

= 4 . The line XY cuts the line PR at Z . Prove that →→

= PR25

21PZ .

Q.4 Find out whether the following pairs of lines are parallel, non-parallel & intersecting, or non-parallel &non-intersecting.

(i)( )

( )�

�

r i j k i j k

r i j k i j k

1

2

2 3 2 4

2 3 6 4 8

= + + + − +

= + + + − + −

� � � � � �

� � � � � �

λ

µ(ii)

( )( )

�

�

r i j k i j k

r i j k i j k

1

2

3

2 4 6 2 3

= − + + − +

= + + + + +

� � � � � �

� � � � � �

λ

µ

(iii)( )

( )�

�

r i k i j k

r i j i j k

1

2

3 4

2 3 4

= + + + +

= + + − +

� � � � �

� � � � �

λ

µQ.5 Let OACB be paralelogram with O at the origin & OC a diagonal. Let D be the mid point of OA.

Using vector method prove that BD & CO intersect in the same ratio. Determine this ratio.

Q.6 A line EF drawn parallel to the base BC of a ∆ ABC meets AB & AC in F & E respectively. BE & CFmeet in L. Use vectors to show that AL bisects BC.

Q.7 ‘O’is the origin of vectors and A is a fixed point on the circle of radius‘a’with centre O. The vector →

OA

is denoted by a�

. A variable point ‘P’ lies on the tangent at A & →

OP = r�

. Show that 2ar.a =

��. Hence

if P ≡ (x,y) & A ≡ (x1,y

1) deduce the equation of tangent at A to this circle.

Q.8 (a) By vector method prove that the quadrilateral whose diagonals bisect each other at right anglesis a rhombous.

(b) By vector method prove that the right bisectors of the sides of a triangle are concurrent.

Q.9 The resultant of two vectors b&a��

is perpendicular to �a . If a2b

��= show that the resultant of

2� �a b& is perpendicular to

�b .

Q.10 c,b,a��

and d�

are the position vectors of the points A A ≡ (x, y, z) ; B ≡ (y, – 2z, 3x) ; C ≡ (2z, 3x, – y)

and D≡(1,–1, 2) respectively. If |a|�

= 32 ;( )ba�� = ( )ca

��; ( )da�� =

2

π and ( )ja

^� is obtuse, then find x, y, z., z.

Q.11 If r�

and s�

are non zero constant vectors and the scalar b is chosen such that sbr��

+ is minimum, then

show that the value of 22|sbr|sb��

++ is equal to 2|r|�

.Q.12 Use vectors to prove that the diagonals of a trapezium having equal non parallel sides are equal &

conversely.Q.13(a) Find a unit vector â which makes an angle (π/4) with axis of z & is such that â + i + j is a unit vector.

(b) Prove that

22

22 |b||a|

ba

b

b

a

a

−=

− ��

��

�

�

�

�

Q.14 Given four non zero vectors � � � �a b c and d, , . The vectors c&b,a

�� are coplanar but not collinear pair by

pair and vector �d is not coplanar with vectors c&b,a

��and β=

∧α=

∧π=

∧=

∧)bd(,)ad(,

3)cb()ba(

�� then

prove that ( ) cos (cos cos )� �d c∧

= −−1 β α .

Q.15 (a) Use vectors to find the acute angle between the diagonals of a cube .(b) Prove cosine & projection rule in a triangle by using dot product .

Q.16 In the plane of a triangle ABC, squares ACXY, BCWZ are described , in the order given, externally to

the triangle on AC & BC respectively. Given that CX b CA a→ →

= =� �

, , CW x CB y→ →

= =� �

, . Prove that

� � � �a y x b. .+ = 0 . Deduce that AW BX

→ →=. 0 .

Q.17 A ∆ OAB is right angled at O ; squares OALM & OBPQ are constructed on the sides OA and OBexternally. Show that the lines AP & BL intersect on the altitude through 'O' .

Q.18 Given that k3j2iu +−=�

; k4ji2v ++=�

; k3j3iw ++=�

and

k)20R·w(j)20R·v(i)10R·u( −+−+−��

= 0. Find the unknown vector R�

.

Q.19 If O is origin of reference, point )cba(G);ac(F;)cb(E;)ba(D;)c(C;)b(B;)a(A��

+++++ where

kajaiaa321

++=�

; kbjbibb321

++=�

and kcjcicc321

++=�

then prove that these points are

vertices of a cube having length of its edge equal to unity provided the matrix.

321

321

321

ccc

bbb

aaa

is orghogonal. Also find the length XY such that X is the point of intersection of CM and

GP ; Y is the point of intersection of OQ and DN where P, Q, M, N are respectively the midpoint ofsides CF, BD, GF and OB.

Q.20 (a) If � � �a b c+ + = 0 , show that axccxbbxa

��== . Deduce the Sine rule for a ∆ ABC.

(b) Find the minimum area of the triangle whose vertices are A(–1, 1, 2); B(1, 2, 3) and C(t, 1, 1)where t is a real number.

Q.21 (a) Determine vector of magnitude 9 which is perpendicular to both the vectors :

k3ji4 +− & k2ji2 −+−(b) A triangle has vertices (1, 1, 1) ; (2, 2, 2), (1, 1, y) and has the area equal to csc

π4

sq. units.Find the value of y.

Q.22 The internal bisectors of the angles of a triangle ABC meet the opposite sides in D, E, F ; use vectors toprove that the area of the triangle DEF is given by

)ac()cb()ba(

)abc2(

+++∆

where ∆ is the area of the triangle.

Q.23 If d,c,b,a��

are position vectors of the vertices of a cyclic quadrilateral ABCD prove that :� � � � � �

� � � �

� � � � � �

� � � �

a x b bxd dxa

b a d a

bx c cxd dx b

b c d c

+ +

− −+

+ +

− −=

( ) . ( ) ( ) . ( )0

Q.24 T h e l e n g t h o f t h e e d g e o f t h e r e g u l a r t e t r a h e d r o n DT h e l e n g t h o f t h e e d g e o f t h e r e g u l a r t e t r a h e d r o n DT h e l e n g t h o f t h e e d g e o f t h e r e g u l a r t e t r a h e d r o n DT h e l e n g t h o f t h e e d g e o f t h e r e g u l a r t e t r a h e d r o n D − ABC is 'a' . Point E and F are taken on the edges

AD and BD respectively such that E divides DA→

and F divides BD→

in the ratio 2:1 each . Then find thearea of triangle CEF.

Q.25 Let ji3a −=�

and j2

3i

2

1b +=�

and b)3q(ax 2��

−+= , bqapy��

+−= . If yx��

⊥ , then express p

as a function of q, say p = f (q), (p ≠ 0 & q ≠ 0) and find the intervals of monotonicity of f (q).

EXERCISE–2Q.1 )a(A

� ; )b(B�

; )c(C�

are the vertices of the triangle ABC such that )k7ri2(2

1a −−=

�� ; k4jr3b −+=

��;

r9j11i22c��

−−= . A vector kj2p −=�

is such that )pr(��

+ is parallel to i and )i2r( −�

is parallel to

p�

. Show that there exists a point )d(D�

on the line AB with k)4t(j)t21(it2d −+−+=�

. Also find the

shortest distance C from AB.

Q.2 The position vectors of the points A, B, C are respectively (1, 1, 1) ; (1, −1, 2) ; (0, 2, −1). Find a unitvector parallel to the plane determined by ABC & perpendicular to the vector (1, 0, 1) .

Q.3 Let 2

12

12

1

21

21

21

21

21

21

)cc()bc()ac(

)cb()bb()ab(

)ca()ba()aa(

−−−−−−−−−

= 0 and if the vectors kajai 2++=α�

; kbjbi 2++=β�

;

kcjci 2++=γ�

are non coplanar, show that the vectors kbjbi;kajai 2111

2111

++=β++=α��

and

kcjci 2111

++=γ�

are coplaner..Q.4 Given non zero number x

1, x

2, x

3 ; y

1, y

2, y

3 and z

1, z

2 and z

3 such that x

i > 0 and y

i < 0 for all i = 1, 2, 3.

(i) Can the given numbers satisfy

321

321

321

zzz

yyy

xxx

= 0 and

=++=++

=++

0zzyyxx0zzyyxx

0zzyyxx

131313

323232

212121

(ii) If P = (x1, x

2, x

3) ; Q (y

1, y

2, y

3) and O (0, 0, 0) can the triangle POQ be a right angled triangle?

Q.5 The pv's of the four angular points of a tetrahedron are : ( )A j k� �+ 2 ; ( )B i k3� �+ ; ( )C i j k4 3 6� � �+ +

& ( )D i j k2 3 2� � �+ + . Find :

(i) the perpendicular distance from A to the line BC. (ii) the volume of the tetrahedron ABCD.(iii) the perpendicular distance from D to the plane ABC.(iv) the shortest distance between the lines AB & CD.

Q.6 The length of an edge of a cube ABCDA1B

1C

1D

1 is equal to unity. A point E taken on the edge AA

1

→ is

such that →AE =

1

3 . A point F is taken on the edge BC

→ such that

→BF =

1

4. If O

1 is the centre of

the cube, find the shortest distance of the vertex B1 from the plane of the ∆ O

1EF.

Q.7 The vector →

OP = k2j2i ++ turns through a right angle, passing through the positive x-axis on the way..Find the vector in its new position.

Q.8 Find the point R in which the line AB cuts the plane CDE where

a�

= kj2i ++ , �b = k2ji2 ++ , c

� = k4j4 +− , d

� = k2j2i2 +− & e

� = k2ji4 ++ .

Q.9 If kajaiaa321

++=�

; kbjbibb321

++=�

and kcjcicc321

++=�

then show that the value of the

scalar triple product ]acncbnban[��

+++ is (n3 + 1)

k·cj·ci·c

k·bj·bi·b

k·aj·ai·a

��

��

��

Q.10 F i n d t h e s c a l a r s F i n d t h e s c a l a r s F i n d t h e s c a l a r s F i n d t h e s c a l a r s α & β if � � � � � � � � � � � �a x bx c a b b b c c c a c( ) ( . ) ( sin ) ( ) & ( . )+ = − − + − =4 2 12β α β while

� �b c&

are non zero non collinear vectors.

Q.11 If the vectors d,c,b��

are not coplanar, then prove that the vector

)cb()da()bd()ca()dc()ba(��

×××+×××+××× is parallel to a�

.

Q.12 � , � , �a b c are non−coplanar unit vectors . The angle between � & �b c is α, between � & �c a is β and between

� & �a b is γ . If A ( )� cosa α , B ( )� cosb β , C ( )� cosc γ , then show that in ∆ ABC,ABC,

( )� � �

sin

a x b x c

A =

( )� � �

sin

b x c x a

B =

( )� � �

sin

c x a x b

C=

( )� � �

sin cos cos �

a x b x c

n

∏∑ α β γ 1

where

�n1 = � �

� �

b x c

b x c , �n2 =

� �

� �

c x a

c x a & �n3 =

� �

� �

a x b

a x b.

Q.13 Given that q,p,b,a��

are four vectors such that 1)b(&0q.b,pba 2 ==µ=+��

, where µ is a scalar then

prove that q.pa)q.p(p)q.a(��

=− .

Q.14 Show that � � � �a px qx r= ( ) ;

� � � �b q x r x p= ( ) &

� � � �c r x p x q= ( ) represents the sides of a triangle. Further

prove that a unit vector perpendicular to the plane of this triangle is

± ( ) ( ) ( )( ) ( ) ( )

� tan ^ � tan ^ � tan ^

� tan ^ � tan ^ � tan ^

n p q n q r n r p

n p q n q r n r p

1 2 3

1 2 3

� � � � � �

� � � � � �+ +

+ + where

� � � � �a b c p q, , , , are non zero vectors and

no two of � � �p q r, , are mutually perpendicular & �n

pxq

pxq1 =� �

� � ; �nqx r

qx r2 =� �

� � & �nr x p

r x p3 =� �

� �

Q.15 Given four points P1, P

2, P

3 and P

4 on the coordinate plane with origin O which satisfy the condition

1nPO − + 1nPO + = nPO

2

3, n = 2, 3

(i) If P1, P

2 lie on the curve xy = 1, then prove that P

3 does not lie on the curve.

(ii) If P1, P

2, P

3 lie on the circle x2 + y2 = 1, then prove that P

4 lies on this circle.

Q.16 Let �a i j k= + −α � � �2 3 ,

�b i j k= + −� � �2 2α and

�c i j k= − +2 � � �α . Find the value(s) of α, if any, such that

( ) ( ){ } ( )� � � � � �a b b c c a× × × × × = 0. Find the vector product when α = 0.

Q.17 Prove the result (Lagrange’s identity) s.qr.q

s.pr.p)sr(·)qp( ��

��

=×× & use it to prove the following. Let

(ab)denote the plane formed by the lines a,b. If (ab) is perpendicular to (cd) and (ac) is perpendicular to(bd) prove that (ad) is perpendicular to (bc).

Q.18 (a) If )0p(;b)ax(xp ≠=×+��

prove that �

� � � � � �

�xp b b a a p bxa

p p a=

+ −+

2

2 2

( . ) ( )

( ).

(b) Solve the following equation for the vector �p ; ( )� � � � � � �

pxa p b c bx c+ =. where � � �a b c, , are non

zero non coplanar vectors and �a is neither perpendicular to

�b nor to

�c , hence show that

[ ]

+× c

c·a

cbaap

��

��

is perpendicular to � �b c− .

Q.19 Find a vector �v which is coplanar with the vectors � � �i j k+ − 2 & � � �i j k− +2 and is orthogonal to the

vector − + +2 � � �i j k . It is given that the projection of�v along the vector � � �i j k− + is equal to 6 3 .

Q.20 Consider the non zero vectors � � � �a b c d, , & such that no three of which are coplanar then prove that

[ ] [ ] [ ] [ ]� � � � � � � � � � � � � � � �a b cd c a b d b a cd d a b c+ = + . Hence prove that

� � � �a b c d, , & represent the position vectors of

the vertices of a plane quadrilateral if [ ] [ ][ ] [ ]

� � � � � �

� � � � � �

b cd a bd

a cd a b c

+

+= 1 .

Q.21 The base vectors 321a,a,a��

are given in terms of base vectors 321

b,b,b��

as, 3211

bb3b2a��

−+= ;

3212b2b2ba��

+−= & 3213

b2bb2a��

−+−= . If 321

b2bb3F��

+−= , then express�F in terms of

321a&a,a��

.

Q.22 If ( ) ( ) ( )A a B b C c� � �

; & are three non collinear points , then for any point ( )P p�

in the plane of the

∆ ABC , prove that ; (i) [ ] ( )� � � � � � � � � �a b c p a x b bx c cxa= + +.

(ii) The vector �v perpendicular to the plane of the triangle ABC drawn from the origin 'O' is given by

�v = ±

[ ] ( )2

4

accbbacba

∆×+×+×��

where �∆ is the vector area of the triangle ABC.

Q.23 Given the points P (1, 1, –1), Q (1, 2, 0) and R (–2, 2, 2). Find

(a) RPQP ×(b) Equation of the plane in

(i) scalar dot product form (ii) parametric form (iii) cartesian form(iv) if the plane through PQR cuts the coordinate axes at A, B, C then the area of the ∆ABC

Q.24 Let � � �a b c, & be non coplanar unit vectors, equally inclined to one another at an angle θ. If

� � � � � � �a x b bx c pa q b r c+ = + + . Find scalars p , q & r in terms of θ.

Q.25 Solve the simultaneous vector equations for the vectors � �x and y .

� � � � � � � �x c y a and y c x b+ × = + × = where

�c is a non zero vector..

EXERCISE–3Q.1 Find the angle between the two straight lines whose direction cosines l, m, n are given by

2l + 2m – n = 0 and mn + nl + lm = 0.Q.2 If two straight line having direction cosines l, m, n satisfy al + bm + cn = 0 and f m n + g n l + h l m = 0

are perpendicular, then show that c

h

b

g

a

f++ = 0.

Q.3 P is any point on the plane lx + my + nz = p. A point Q taken on the line OP (where O is the origin) suchthat OP. OQ = p2. Show that the locus of Q is p( lx + my + nz ) = x2 + y2 + z2.

Q.4 Find the equation of the plane through the points (2, 2, 1), (1, –2, 3) and parallel to the x-axis.

Q.5 Through a point P (f, g, h), a plane is drawn at right angles to OP where 'O' is the origin, to meet the

coordinate axes in A, B, C. Prove that the area of the triangle ABC is hgf2

r5

where OP = r..

Q.6 The plane lx + my = 0 is rotated about its line of intersection with the plane z = 0 through an angle θ.

Prove that the equation to the plane in new position is lx + my + θ+ tanmlz 22 = 0Q.7 Find the equations of the straight line passing through the point (1, 2, 3) to intersect the straight line

x + 1 = 2 (y – 2) = z + 4 and parallel to the plane x + 5y + 4z = 0.

Q.8 Find the equations of the two lines through the origin which intersect the line 1

z

1

3y

2

3x=

−=

− at an

angle of 3

π.

Q.9 A variable plane is at a constant distance p from the origin and meets the coordinate axes in points A, Band C respectively. Through these points, planes are drawn parallel to the coordinates planes. Find thelocus of their point of intersection.

Q.10 Find the distance of the point P (– 2, 3, – 4) from the line 5

4z3

4

3y2

3

2x +=

+=

+ measured parallel to

the plane 4x + 12y – 3z + 1 = 0.Q.11 Find the equation to the line passing through the point (1, –2, –3) and parallel to the line

2x + 3y – 3z + 2 = 0 = 3x – 4y + 2z – 4.Q.12 Find the equation of the line passing through the point (4, –14, 4) and intersecting the line of intersection

of the planes : 3x + 2y – z = 5 and x – 2y – 2z = –1 at right angles.Q.13 Let P = (1, 0, – 1) ; Q = (1, 1, 1) and R = (2, 1, 3) are three points.(a) Find the area of the triangle having P, Q and R as its vertices.(b) Give the equation of the plane through P, Q and R in the form ax + by + cz = 1.(c) Where does the plane in part (b) intersect the y-axis.(d) Give parametric equations for the line through R that is perpendicular to the plane in part (b).

Q.14 Find the point where the line of intersection of the planes x – 2y + z = l and x + 2y – 2z = 5, intersectsthe plane 2x + 2y + z + 6 = 0.

Q.15 Feet of the perpendicular drawn from the point P (2, 3, –5) on the axes of coordinates are A, B and C.Find the equation of the plane passing through their feet and the area of ∆ABC.

Q . 1 6Q . 1 6Q . 1 6Q . 1 6 F i n d t h e e q u a t i o n s t o t h e l i n e w h i c h c a n b e d r a w n f r o m t h e p o i n t ( 2 , – 1 , 3 ) p e r p e n d i c u l a r t o t h e l i n e sF i n d t h e e q u a t i o n s t o t h e l i n e w h i c h c a n b e d r a w n f r o m t h e p o i n t ( 2 , – 1 , 3 ) p e r p e n d i c u l a r t o t h e l i n e sF i n d t h e e q u a t i o n s t o t h e l i n e w h i c h c a n b e d r a w n f r o m t h e p o i n t ( 2 , – 1 , 3 ) p e r p e n d i c u l a r t o t h e l i n e sF i n d t h e e q u a t i o n s t o t h e l i n e w h i c h c a n b e d r a w n f r o m t h e p o i n t ( 2 , – 1 , 3 ) p e r p e n d i c u l a r t o t h e l i n e s

4

3z

3

2y

2

1x −=

−=

− and

3

3z

5

y

4

4x +==

− at right angles.

Q.17 Find the equation of the plane containing the straight line 5

z

3

2y

2

1x=

−+

=−

and perpendicular to the

plane x – y + z + 2 = 0.

Q.18 Find the value of p so that the lines 1

2z

2

py

3

1x +=

−=

−+

and 2

7z

3

7y

1

x +=

−−

= are in the same

plane. For this value of p, find the coordinates of their point of intersection and the equation of the planecontaining them.

Q.19 Find the equations to the line of greatest slope through the point (7, 2 , –1) in the planex – 2y + 3z = 0 assuming that the axes are so placed that the plane 2x + 3y – 4z = 0 is horizontal.

Q.20 Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let thearea of triangles ABC, ACD and ADB be denoted by x, y and z sq. units respectively. Find the area ofthe triangle BCD.

Q.21 The position vectors of the four angular points of a tetrahedron OABC are (0, 0, 0); (0, 0, 2); (0, 4, 0)and (6, 0, 0) respectively. A point P inside the tetrahedron is at the same distance 'r' from the four planefaces of the tetrahedron. Find the value of 'r'.

Q.22 The line 8

14z

3

10y

5

6x +=

+=

+ is the hypotenuse of an isosceles right angled triangle whose opposite

vertex is (7, 2, 4). Find the equation of the remaining sides.Q.23 Find the foot and hence the length of the perpendicular from the point (5, 7, 3) to the line

5

z5

8

29y

3

15x −=

−=

−. Also find the equation of the plane in which the perpendicular and the given

straight line lie.

Q.24 Find the equation of the line which is reflection of the line 3

3z

1

2y

9

1x

−+

=−−

=−

in the plane3x – 3y + 10z = 26.

Q.25 Find the equation of the plane containing the line 2

z

3

y

2

1x==

− and parallel to the line

4

2z

5

y

2

3x −==

−.

Find also the S.D. between the two lines.

EXERCISE–4

Q.1(a) Let OA a OB a b and OC b→ → → → → → →

= = + =, 10 2 where O, A & C are non−collinear points . Let p denote thearea of the quadrilateral OABC, and let q denote the area of the parallelogram with OA and OC asadjacent sides. If p = kq, then k = _______ .

(b) If � � �A B C, & are vectors such that | | | |

� �B C= , Prove that ;

( ) ( )[ ] ( ) ( )� � � � � � � �A B x A C x BxC B C+ + + =. 0 [ JEE ' 97, 2 + 5 ]

Q.2(a) Vectors � � �x y z, & each of magnitude 2 , make angles of 600 with each other. If

cyxandb)xz(y,a)zy(x��

=×=××=×× then find � � �x y and z, in terms of

� � �a b and c, .

(b) The position vectors of the points P & Q are 5 7 2 3 3 6� � � � � �i j k and i j k+ − − + + respectively. The

vector �A i j k= − +3 � � � passes through the point P & the vector

�B i j k= − + +3 2 4� � � passes through the

point Q. A third vector 2 7 5� � �i j k+ − intersects vectors � �A B& . Find the position vectors of the points

of intersection. [ REE ' 97, 6 + 6 ]Q.3(a) Select the correct alternative(s)

(i) If � � �a i j k b i j k and c i j k= + + = + + = + +� � � , � � � � � �4 3 4 α β are linearly dependent vectors &

�c = 3 , then:

(ii) For three vectors �

v , �w which of the following expressions is not equal to any of the remaining three?

(A) �u . (�v x �

v x �w ) .�u (C)

�v . (�u x �w ) (D) (

�u x �v ) .�w

(iii) Which of the following expressions are meaningful ?w ) (B) (

�u , �

(A) α = 1, β = −1 (B) α = 1, β = ±1 (C) α = −1, β = ±1 (D) α = ±1, β = 1

(A) �u . (�v x �w ) (B) (

�u . �v ) .

�w (C) (

�u . �v ) �w (D)

�u x (

�v .�w )

(b) Prove, by vector methods or otherwise, that the point of intersection of the diagonals of a trapezeum lieson the line passing through the mid−points of the parallel sides.(You may assume that the trapezeum is not a parallelogram.)

(c) For any two vectors � �u v& , prove that [ JEE ' 98 , 2 + 2 + 2 + 8 + 8 ]

(i) ( . ) | | | | | |� � � � � �u v u v u v2 2 2 2+ × = & (ii) ( | | )( | | ) ( . ) | ( )|1 1 12 2 2 2+ + = − + + + ×

� � � � � � � �u v u v u v u v

Q.4(a) If � � � � � � � � � � � �x y a y z b x b x y and y z× = × = = = =, , . , . .γ 1 1 then find

� � �x y z, & in terms of

� �a b and, γ .

(b) Vectors AB→

= 3� � �i j k− + & CD→

= − + +3 2 4� � �i j k are not coplanar. The position vectors of points

A and C are 6 7 4 9 2� � � � �i j k and j k+ + − + respectively . Find the position vectors of a point P on the

line AB & a point Q on the line CD such that PQ→

is perpendicular to AB→

and CD→

both.

Q.5(a) Let k2ji2a −+=�

& jib +=�

. If �c is a vector such that |c|c·a

��= , 22ac =−

�� and the angle

between ( )ba��

× and c�

is 30º, then ( ) cba��

×× =

(A) 2/3 (B) 3/2 (C) 2 (D) 3

(b) Let kji2a ++=�

, �b i j k= + −� � �2 and a unit vector

�c be coplanar. If

�c is perpendicular to

�a , then

�c =

(A) ( )1

2− +� �j k (B) ( )1

3− − −� � �i j k (C) ( )1

52� �i j− (D) ( )1

3

� � �i j k− −

(c) Let � �a b& be two non-collinear unit vectors . If ( )� � � � �

u a a b b= − . & � � �v a x b= , then

�v is :

(A) �u (B)

� � �u u a+ . (C)

� � �u u b+ . (D) ( )� � � �

u u a b+ +.

(d) Let � �u v& be unit vectors . If

�w is a vector such that ( )� � � �

w w x u v+ = , then prove that

( )� � �u x v w. ≤

1

2 and the equality holds if and only if

�u is perpendicular to

�v .

Q.6(a) An arc AC of a circle subtends a right angle at the centre O. The point B divides the arc in the ratio 1 : 2.

If aOA�

=→

& bOB�

=→

, then calculate →

OC in terms of b&a��

.

(b) If � � �a b c, , are non-coplanar vectors and

�d is a unit vector, then find the value of,

( ) ( ) ( ) ( ) ( ) ( )� � � � � � � � � � � �a d b x c b d c x a c d a x b. . .+ + independent of

�d . [ REE '99, 6 + 6 ]

Q.7(a) Select the correct alternative :

(i) If the vectors � � �a b c, & form the sides BC, CA & AB respectively of a triangle ABC, then

(A) � � � � � �a b b c c a. . .+ + = 0 (B)

� � � � � �a b b c c a× = × = ×

(C) � � � � � �a b b c c a. . .= = (D)

� � � � � �a b b c c a× + × + × = 0

(ii) Let the vectors � � � �a b c d, , & be such that ( ) ( )dcba

��××× =

�0 . Let P

1 & P

2 be planes determined by

the pairs of vectors � � � �a b c d, & , respectively . Then the angle between P

1 and P

2 is :

(A) 0 (B) π/4 (C) π/3 (4) π/2

(iii) If � � �a b c, & are unit coplanar vectors, then the scalar triple product

[ ]2 2 2� � � � � �a b b c c a− − − = [ JEE ,2000 (Screening) 1 + 1 + 1 out of 35 ]

(A) 0 (B) 1 (C) − 3 (D) 3(b) Let ABC and PQR be any two triangles in the same plane . Assume that the perpendiculars from the

points A, B, C to the sides QR, RP, PQ respectively are concurrent . Using vector methods or otherwise,prove that the perpendiculars from P, Q, R to BC, CA, AB respectively are also concurrent .

[ JEE '2000 (Mains) 10 out of 100 ]

Q.8. (i) If �a = � � �i j k+ −

, �b = − + +� � �i j k2 2 &

�c = − + −� � �i j k2

, find a unit vector normal to the vectors

�a +

�b and

�b −

�c .

(ii) Given that vectors �a &

�b are perpendicular to each other, find vector

�υ in terms of

�a &

�b satisfying

υ . �a = 0 ,

�υ . �b = 1 and [ ]� � �

υ , ,a b = 1the equations, �

(iii)�a , �b &

�c are three unit vectors such that

�a × ( )� �

b c× = 1

2 ( )� �

b c+ . Find angle between vectors

�a &

�b given that vectors

�b &

�c are non-parallel.

(iv) A particle is placed at a corner P of a cube of side 1 meter . Forces of magnitudes 2, 3 and 5 kg weightact on the particle along the diagonals of the faces passing through the point P . Find the moment of theseforces about the corner opposite to P . [ REE '2000 (Mains) 3 + 3 + 3 + 3 out of 100 ]

Q.9(a) The diagonals of a parallelogram are given by vectors 2 3 6� � �i j k+ − and 3 4� � �i j k− − . Determine its sides

and also the area. (b) Find the value of λ such that a, b, c are all non-zero and

( )− + + − + + + +4 5 3 3 3� � ( � � � ) (� � � )i j a i j k b i j k c = λ ( � � � )ai bj ck+ + [ REE '2001 (Mains) 3 + 3]

Q.10(a) Find the vector �r which is perpendicular to

�a = � � �i j k− +2 5 and

�b i j k= + −2 3� � � and

( )�r i j k⋅ + +2� � � + 8 = 0.

(b) Two vertices of a triangle are at − +� �i j3 and 2 5� �i j+ and its orthocentre is at � �i j+ 2 . Find the positionvector of third vertex. [ REE '2001 (Mains) 3 + 3]

Q.11 (a) If � � �a b and c, are unit vectors, then

� � � � � �a b b c c a− + − + −

2 2 2 does NOT exceed

(A) 4 (B) 9 (C) 8 (D) 6

(b) Let k)yx1(jxiycandk)x1(jixb,kia −+++=−++=−=��

. Then [ , , ]� � �a b c depends on

(A) only x (B) only y (C) NEITHER x NOR y (D) both x and y[ JEE '2001 (Screening) 1 + 1 out of 35]

Q.12(a) Show by vector methods, that the angular bisectors of a triangle are concurrent and find an expressionfor the position vector of the point of concurrency in terms of the position vectors of the vertices.

(b) Find 3–dimensional vectors � � �v v v1 2 3, , satisfying

� �v v1 1⋅ = 4,

� �v v1 2⋅ = –2,

� �v v1 3⋅ = 6,

� �v v2 2⋅ = 2,

� �v v2 3⋅ = –5,

� �v v3 3⋅ = 29.

(c) Let �A t( ) = f t i f t j1 2( )� ( )�+ and

�B t g t i g t j( ) ( )� ( )�= +1 2

, t ∈ [0, 1], where f1, f

2, g

1, g

2 are continuous

functions. If �A t( ) and

�B t( ) are nonzero vectors for all t and

�A( )0 = 2 3� �i j+ ,

�A( )1 = 6 2� �i j+ ,

�B( )0 = 3 2� �i j+ and

�B( )1 = 2 6� �i j+ , then show that

�A t( ) and

�B t( ) are parallel for

some t. [ JEE '2001 (Mains) 5 + 5 + 5 out of 100 ]

Q.13(a) If �a and

�b are two unit vectors such that

�a + 2

�b and 5

�a – 4

�b are perpendicular to each other then

the angle between �a and

�b is

(A) 450 (B) 600 (C) cos–11

3

(D) cos–1

2

7

(b) Let kji2V −+=�

and k3iW +=�

. If �U is a unit vector, then the maximum value of the scalar triple

product [ ]� � �U V W is [JEE 2002(Screening), 3 + 3]

(A) –1 (B) 10 6+ (C) 59 (D) 60

Q.14 Let V be the volume of the parallelopiped formed by the vectors kajaiaa 321 ++=�

,

kbjbibb 321 ++=�

, kcjcicc 321 ++=�

. If ar , b

r , c

r , where r = 1, 2, 3, are non-negative real

numbers and ( )∑=

++3

1r

rrr cba = 3L, show that V < L3. [JEE 2002(Mains), 5]

Q.15 If a�

= kjai ++ , b�

= kaj + , c�

= kia + , then find the value of ‘a’ for which volume of

parallelopiped formed by three vectors as coterminous edges, is minimum, is

(A) 3

1(B) –

3

1(C) ±

3

1(D) none [JEE 2003(Scr.), 3]

Q.16(i) Find the equation of the plane passing through the points (2, 1, 0) , (5, 0, 1) and (4, 1, 1).

(ii) If P is the point (2, 1, 6) then find the point Q such that PQ is perpendicular to the plane in (i) and the mid

point of PQ lies on it. [ JEE 2003, 4 out of 60]

Q.17 If w,v,u��

are three non-coplanar unit vectors and α, β, γ are the angles between vandu��

,

wandv��

, uandw��

respectively and z,y,x��

are unit vectors along the bisectors of the angles

α, β, γ respectively. Prove that [ ] [ ]2

sec2

sec2

secwvu16

1xzzyyx 222

2 γβα=×××

��.

[ JEE 2003, 4 out of 60 ]

Q.18(a) If the lines 4

1z

3

1y

2

1x −=

+=

− and

1

z

2

ky

1

3x=

−=

− intersect, then k =

(A) 9

2(B)

2

9(C) 0 (D) – 1

(b) A unit vector in the plane of the vectors kji2 ++ , kji +− and orthogonal to k6j2i5 ++

(A) 61

k5i6 −(B)

10

kj3 −(C)

29

k5i2 −(D)

3

k2ji2 −+

(c) If kjia ++=��

, 1b·a =��

and kjba −=×��

, then b�

= [ JEE 2004 (screening)]

(A) i (B) kji +− (C) kj2 − (D) i2

Q.19(a) Let d,c,b,a��

are four distinct vectors satisfying ba��

× = dc��

× and dbca��

×=× . Show that

d·bc·ad·cb·a��

+≠+ .

(b) T is a parallelopiped in which A, B, C and D are vertices of one face. And the face just above it hascorresponding vertices A', B', C', D'. T is now compressed to S with face ABCD remaining same andA', B', C', D' shifted to A., B., C., D. in S. The volume of parallelopiped S is reduced to 90% of T. Provethat locus of A. is a plane.

(c) Let P be the plane passing through (1, 1, 1) and parallel to the lines L1 and L

2 having direction ratios

1, 0, –1 and –1, 1, 0 respectively. If A, B and C are the points at which P intersects the coordinate axes,find the volume of the tetrahedron whose vertices are A, B, C and the origin.

Q.20(a) If c,b,a��

are three non-zero, non-coplanar vectors and a|a|

a·bbb

21

��

��

−= , a|a|

a·bbb

22

��

��

+= ,

1221b

|c|

c·ba

|a|

a·ccc

�

�

��

�

��

+−= , 12

1

1

22b

|b|

c·ba

|a|

a·ccc

��

��

��

��

−−= , 1223b

|c|

c·ba

|c|

a·ccc

�

�

��

�

��

+−= , 1224b

|b|

c·ba

|c|

a·ccc

��

��

�

��

−−=

then the set of orthogonal vectors is

(A) ( )31

c,b,a��

(B) ( )21

c,b,a��

(C) ( )11

c,b,a��

(D) ( )22

c,b,a��

(b) A variable plane at a distance of 1 unit from the origin cuts the co-ordinate axes at A, B and C. If the

centroid D (x, y, z) of triangle ABC satisfies the relation 222 z

1

y

1

x

1++ = k, then the value of k is

(A) 3 (B) 1 (C) 1/3 (D) 9 [JEE 2005 (Screening), 3] (c) Find the equation of the plane containing the line 2x – y + z – 3 = 0, 3x + y + z = 5 and at a distance of

6

1 from the point (2, 1, – 1).

(d) Incident ray is along the unit vector v and the reflected ray is along the

unit vector w . The normal is along unit vector a outwards. Express

w in terms of a and v . [ JEE 2005 (Mains), 2 + 4 out of 60 ]

Q.21(a) A plane passes through (1, –2, 1) and is perpendicular to two planes 2x – 2y + z = 0 andx – y + 2z = 4. The distance of the plane from the point (1, 2, 2) is

(A) 0 (B) 1 (C) 2 (D) 22

(b) Let kj2ia ++=�

, kjib +−=�

and kjic −+=�

. A vector in the plane of banda��

whose projection

on c�

is 3

1, is [JEE 2006,3 marks each]

(A) k4ji4 +− (B) k3ji3 −+ (C) k2ji2 −+ (D) k4ji4 −+ (c) Let A

�be vector parallel to line of intersection of planes P

1 and P

2 through origin. P

1 is parallel to the

vectors 2 j + 3 k and 4 j – 3 k and P2 is parallel to j – k and 3 i + 3 j , then the angle between

vector A�

and 2 i + j – 2 k is

(A) 2

π(B)

4

π(C)

6

π(D)

4

3π[JEE 2006, 5]

(d) Match the following(i) Two rays in the first quadrant x + y = | a | and ax – y = 1 intersects each other in the interval a ∈ (a

0, ∞), the value of a

0 is (A) 2

(ii) Point (α, β, γ) lies on the plane x + y + z = 2.

Let kjia γ+β+α=�

, )ak(k�

×× = 0, then γ = (B) 4/3

(iii) ∫ −1

0

2dy)y1( + ∫ −

0

1

2dy)1y( (C) ∫ −

1

0

dxx1 + ∫−

+0

1

dxx1

(iv) If sinA sinB sinC + cos A cosB = 1, then the value of sin C = (D) 1 [JEE 2006, 6]

(e) Match the following

(i) ∑∞

=

− =

1i2

1 ti2

1tan , then tan t = (A) 0

(ii) Sides a, b, c of a triangle ABC are in A.P.

and ca

bcos,

cb

acos

21 +=θ

+=θ ,

ba

ccos

3 +=θ ,

then 2

tan2

tan 3212 θ+

θ = (B) 1

(iii) A line is perpendicular to x + 2y + 2z = 0 and passes through (0, 1, 0). The perpendicular

distance of this line from the origin is (C) 3

5

(D) 2/3 [JEE 2006, 6]

ANSWER KEY

EXERCISE–1Q.1 x = 2 , y = −1 Q.2 (b) externally in the ratio 1 : 3

Q.4 (i) parallel (ii) the lines intersect at the point p.v. − +2 2� �i j (iii) lines are skew

Q.5 2 : 1 Q.7 xx1 + yy

1 = a2 Q.10 x = 2, y = – 2, z = – 2

Q.13 (a) −

− +1

2

1

2

1

2i j k Q.15 (a) arc cos

1

3Q.18 k5j2i ++−

Q.193

11Q.20 (b)

2

3Q.21 (a) ± 3( k2j2i −− ), (b) y = 3 or y = – 1 Q.24

5

12 3

2asq. units

Q.25 p = 4

)3q(q 3 −; decreasing in q ∈ (–1, 1), q ≠ 0

EXERCISE–2

Q.1 172 Q.2 )kj5i(33

1−+± Q.4 NO, NO

Q.5 (i) 6

714 (ii) 6 (iii)

3

510 (iv) 6 Q.6

11

170Q.7 k

2

1j

2

1i

2

4−−

Q.8 p.v. of �R = r = 3i + 3k Q.10 1&In,

2

)1(n

n

=β∈π−

+π=α

Q.16 α = 2/3 ; if α = 0 then vector product is − 60 ( )2 � �i k+

Q.18 (b)[ ]

( )( ) ( ) ( )( )

( )( )

−+×+=b.a

cb.b

b.a

bc.bbca

b.ac.a

cbap ��

��

��

��

��

��

Q.19 9 ( )− +� �j k

Q.21 F = 321a3a5a2��

++

Q.23 (a) k3j3i2 +− , (b) (i) – 4, (ii) kjir −+= + )k3ji3()kj( ++−µ++λ , (iii) – 4, (iv) 9

224

Q.24 p = −+

1

1 2cosθ ; q = 2

1 2

cos

cos

θθ+

; r = −+

1

1 2cosθ

or p = 1

1 2+ cos θ ; q = −

+2

1 2

cos

cos

θθ

; r = 1

1 2+ cos θ

Q.25 ��

�

� � � � � �

�xa c a c b c

cy

b c b c a c

c=

+ + ×+

=+ + ×

+( . )

,( . )

1 12 2

EXERCISE–3

Q.1 θ = 900 Q.4 y + 2z = 4 Q.73

3z

2

2y

2

1x

−−

=−

=−

Q.81

z

2

y

1

x

−== or

2

z

1

y

1

x

−==

−Q.9 2222 p

1

z

1

y

1

x

1=++ Q.10

2

17

Q.1117

3z

13

2y

6

1x +=

+=

−Q.12

4

4z

10

14y

3

4x −=

+=

−

Q.13 (a) 2

3 ; (b)

3

z

3

y2

3

x2−+ = 1; (c)

0,

2

3,0 ; (d) x = 2t + 2 ; y = 2t + 1 and z = – t + 3

Q.14 (1, –2, – 4) Q.15 15

z

3

y

2

x=

−++ , Area =

2

19 sq. units Q.16

2

3z

10

1y

11

2x −=

−+

=−

Q.17 2x + 3y + z + 4 = 0 Q.18 p = 3, (2, 1, –3) ; x + y + z = 0

Q.19 4

1z

5

2y

22

7x

−+

=−

=−

Q.20 )zyx( 222 ++ Q.213

2

Q.222

4z

6

2y

3

7x −=

−=

− ;

6

4z

3

2y

2

7x −=

−−

=−

Q.23 (9, 13, 15) ; 14 ; 9x – 4y – z = 14 Q.243

7z

1

1y

9

4x

−−

=−+

=−

Q.25 x – 2y + 2z – 1 = 0; 2 units

EXERCISE–4

Q.1 (a) 6

Q.2 (a) � � �x a c= × ;

� � �y b c= × ;

� � � �z b a c= + × or

� � �b c a× − (b) (2, 8, − 3) ; (0, 1, 2)

Q.3 (a) (i) D (ii) C (iii) A, C

Q.4 (a) x = 2

ba

babaa

×−

γ×

γ×

γ×

��

��

; ya b

=×� �

γ ; z =

� � � �

� �

�a b a b

a b

b× ×

×

+ ×

γ γ

γ

2 (b) P ≡ (3, 8, 3) & Q ≡ (−3, −7, 6)

Q.5 (a) B (b) A (c) A, C Q.6 (a) b2a3c��

+−= (b) [ ]� � �a b c

Q.7 (a) (i) B (ii) A (iii) A

Q.8 (i) + �i ; (ii) 22 )ba(

bxa

b

b��

��

�

�

×+ ; (iii)

2

3

π; (iv) | |

�M = 7

Q.9 (a) ( )1

25 7

1

27 5

1

21274� � � , ( � � �);i j k i j k− − − + − sq. units (b) λ = 0, λ = –2 + 29

Q.10 (a) �r i j k= − + +13 11 7� � � ; (b) j

7

17i

7

5+

Q.11 (a) B (b) C Q.12 (b) � � �v i v i j v i j k1 2 32 3 2 4= = − ± = ± ±�, � �, � � �

Q.13 (a) B ; (b) C Q.15 D Q.16 (i) x + y – 2z = 3 ; (ii) (6, 5, –2)

Q.18 (a) B, (b) B, (c) A Q.19 (c) 9/2 cubic units

Q.20 (a) B, (b) D ; (c) 2x – y + z – 3 = 0 and 62x + 29y + 19z – 105 = 0, (d) w = v – 2( a · v ) a

Q.21 (a) D; (b) A; (c) B, D; (d) (i) D, (ii) A, (iii) B, C, (iv) D; (e) (i) B, (ii) D, (iii) C

EXERCISE–5Part : (A) Only one correct option1. The locus of a point P which moves such that PA2 – PB2 = 2k2 where A and B are (3, 4, 5) and

(– 1, 3 – 7) respectively is(A) 8x + 2y + 24z – 9 + 2k2 = 0 (B) 8x + 2y + 24z – 2k2 = 0(C) 8x + 2y + 24z + 9 + 2k2 = 0 (D) none of these

2. The position vectors of three points A, B, C are � � �i j k+ +2 3 , 2 3� � �i j k+ + & 3 2� � �i j k+ + . A unit vectorperpendicular to the plane of the triangle ABC is:

(A) ( )−

+ +

1

3

� � �i j k (B) ( )1

3

− +� � �i j k (C) ( )1

3

+ −� � �i j k (D) none

3. The square of the perpendicular distance of a point P (p, q, r) from a line through A(a, b, c) and whosedirection cosine are �, m, n is(A) Σ {(q – b) n – (r – c) m}2 (B) Σ {(q + b) n – (r + c) m}2

(C) Σ {(q – b) n + (r – c) m}2 (D) none of these4. A variable plane passes through a fixed point (1, 2, 3). The locus of the foot of the perpendicular drawn

from origin to this plane is:(A) x2 + y2 + z2 − x − 2y − 3z = 0 (B) x2 + 2y2 + 3z2 − x − 2y − 3z = 0(C) x2 + 4y2 + 9z2 + x + 2y + 3 = 0 (D) x2 + y2 + z2 + x + 2y + 3z = 0

5. The equation of the plane which bisects the angle between the planes 3x − 6y + 2z + 5 = 0 and4x − 12y + 3z − 3 = 0 which contains the origin is(A) 33x − 13y + 32z + 45 = 0 (B) x − 3y + z − 5 = 0 (C) 33x + 13y + 32z + 45 = 0 (D) None

6. The distance of the point of intersection of the line x – 3 = (1/2) (y–4) = (1/2) (z–5) and the planex + y + z = 17 from the point (3, 4, 5)(A) 2 (B) 3 (C) 1/3 (D) 1/2

7. The lines x = ay + b, z = cy + d and x = a’y + b’, z = c’y + d’ will be mutually perpendicular provided(A) (a + a’)(b + b’) (c + c’) (B) aa’ + cc’ + 1 = 0(C) aa’ + bb’ + cc’ + 1 = 0 (D) (a + a’) (b + b’) (c + c’) + 1 = 0

8. A straight line r�

= a�

+ λ b�

meets the plane r�

. n = p in the point P whose position vector is

(A) a�

+

n.b

n.a�

�

b�

(B) a�

+

−n.b

n.ap�

�

b�

(C) a�

−

n.b

n.a�

�

b�

(D) a�

−

−n.b

n.ap�

�

b�

9. Equation of the angle bisector of the angle between the lines 1

1x − =

1

2y − =

1

3z − &

1

1x − =

1

2y − =

1

3z

−−

is

(A) 2

1x − =

2

2y −; z – 3 = 0 (B)

1

1x − =

2

2y − =

3

3z −

(C) x – 1 = 0 ; 1

2y − =

1

3z −(D) None of these

10. The distance of the point, (− 1, − 5, − 10) from the point of intersection of the line, x y z−

=+

=−2

3

1

4

2

12and the plane, x − y + z = 5, is:(A) 10 (B) 11 (C) 12 (D) 13

11. If a plane cuts off intercepts OA = a, OB = b, OC = c from the coordinate axes, then the area of thetriangle ABC =

(A)2

1 222222 baaccb ++ (B)

2

1 (bc + ca + ab) (C)

2

1 abc (D)

2

1222 )ba()ac()cb( −+−+−

12. The angle between the lines whose direction cosines satisfy the equations � + m + n = 0 and�2 = m2 + n2 is

(A) 6

π(B)

2

π(C)

3

π(D)

4

π

13. If a1, b

1, c

1 and a

2 , b

2, c

2 are the direction ratios of two lines and θ is the angle between the lines then

tan θ is equal to

(A) 212211

21221

++

−Σ(B)

212121

21221

ccbbaa

)cbcb(

++

−Σ(C)

212121

21221

ccbbaa

)cbcb(

++

+Σ(D) none of these

14. A point moves so that the sum of the squares of its distances from the six faces of a cube given byx = ± 1, y = ± 1, z = ± 1 is 10 units. The locus of the point is(A) x2 + y2 + z2 = 1 (B) x2 + y2 + z2 = 2 (C) x + y + z = 1 (D) x + y + z = 2

15. In the adjacent figure ‘P’ is any arbitrary interior point of the triangle ABC such that the lines AA1, BB

1 and

CC1 are concurrent at P. Value of

1

1

AA

PA +

1

1

BB

PB +

1

1

CC

PC is always equal to .

(A) 1 (B) 2 (C) 3 (D) None of these

16. The plane ax + by + cz = d, meets the coordinate axes at the points A, B and C respectively. Area of triangleABC is equal to

(A) |abc|

cbad 2222 ++(B)

|abc|2

cbad 2222 ++(C)

|abc|4

cbad 2222 ++(D) None of these

a b a b c c

(b c b c )

17. The length of projection, of the line segment joining the points (1, –1, 0) and (–1, 0, 1), to the plane2x + y + 6z = 1, is equal to

(A) 61

255(B)

61

237(C)

61

137(D)

61

155

18. Two systems of rectangular axes have the same origin. If a plane cuts them at distances a, b, c anda

1, b

1, c

1 from the origin, then

(A) 222 c

1

b

1

a

1++ = 2

121

21 c

1

b

1

a

1++ (B) 222 c

1

b

1

a

1+− = 2

121

21 c

1

b

1

a

1+−

(C) a2 + b2 + c2 = 21

21

21 cba ++ (D) a2 – b2 + c2 = 2

121

21 cba +−

19. The angle between the plane 2x – y + z = 6 and a plane perpendicular to the planes x + y + 2z = 7 andx – y = 3 is :

(A) 4

π(B)

3

π(C)

6

π(D)

2

π

20. The non zero value of ‘a’ for which the lines 2x – y + 3z + 4 = 0 = ax + y – z + 2 andx – 3y + z = 0 = x + 2y + z + 1 are co-planar is :(A) – 2 (B) 4 (C) 6 (D) 0

21. The equation of the plane through the point (–1, 2 , 0) and parallel to the lines

3

x =

0

1y + =

1

2z

−−

and 1

1x − =

2

1y2 + =

1

1z

−+

is -

(A) x + 2y + 3z - 1 = 0 (B) x – 2y + 3z + 5 = 0(C) x + y – 3z + 1 = 0 (D) x + y + 3z – 1 = 0

22. The equation of the plane bisecting the acute angle between the planes 2x + y + 2z = 9 and3x – 4y + 12z + 13 = 0 is :(A) 11x + 33y – 34z – 172 = 0 (B) 11x + 33y – 34z – 182 = 0(C) 41x – 7y + 86z – 52 = 0 (D) 41x – 7y + 86z – 62 = 0

23. The base of the pyramid AOBC is an equilateral triangle OBA with each side equal to 4 2 , ' O

' is the

origin of reference, AO is perpendicular to the plane of ∆ OBC and | |AO→

= 2 . Then the cosine of theangle between the skew straight lines one passing through A and the mid point of OB and the otherpassing through O and the mid point of BC is :

(A) − 1

2 (B) 0 (C)

1

6 (D)

1

224. The coplanar points A , B , C , D are (2 − x , 2 , 2) , (2 , 2 − y , 2) , (2 , 2 , 2 − z) and (1 , 1 , 1)

respectively . Then :

(A) z

1

y

1

x

1++ = 1 (B) x + y + z = 1 (C)

x1

1

− +

y1

1

− +

z1

1

− = 1(D) none of these

25. Let the centre of the parallelopiped formed by PA i j k→

= + +� � �2 2 ; PB i j k→

= − +4 3� � � ;

PC i j k→

= + −3 5� � � is given by the position vector (7, 6, 2). Then the position vector of the point P is:(A) (3, 4, 1) (B) (6, 8, 2) (C) (1, 3, 4) (D) (2, 6, 8)

26. Taken on side AC→

of a triangle ABC, a point M such that AM→

=1

3AC

→. A point N is taken on the

side CB→

such that BN→

= CB→

then, for the point of intersection X of AB→

& MN→

which of the followingholds good?

(A) XB→

=1

3 AB

→(B) A X

→ =

1

3AB

→(C) XN

→ =

3

4MN

→(D) XM

→ = 3 XN

→

27. If the acute angle that the vector, α β γ� � �i j k+ + makes with the plane of the two vectors

2 3� � � i j k− + 2 is cot −1 2 then:(A) α (β + γ) = β γ (B) β (γ + α) = γ α (C) γ (α + β) = α β (D) α β + β γ + γ α = 0

28. Locus of the point P, for which OP

→ represents a vector with direction cosine cos α =

1

2( '

O

' is the origin) is:

(A) A circle parallel to y z plane with centre on the x − axis

(B) a cone concentric with positive x − axis having vertex at the origin and the slantheight equal to the magnitude of the vector

(C) a ray emanating from the origin and making an angle of 60º with x − axis

(D) a disc parallel to y z plane with centre on x − axis & radius equal to| |O P

→ sin 60º

29. Equation of the plane passing through A(x1, y

1, z

1) and containing the line

1

2

d

xx − =

2

2

d

yy − =

3

2

d

zz − is

i + j − k & � � �

(A)

321

121212

111

ddd

zzyyxx

zzyyxx

−−−−−−

= 0 (B)

321

212121

222

ddd

zzyyxx

zzyyxx

−−−−−−

= 0

(C)

222

111

321

zyx

zyx

dzdydx −−−

= 0 (D)

321

212121

ddd

zzyyxx

zyx

−−− = 0

30. The equations of the line of shortest distance between the lines

1

z

3

y

2

x=

−= and

3

2x − =

5

1y

−−

= 2

2z − are

(A) 3(x – 21) = 3y + 92 = 3z – 32 (B) 3/1

)3/62(x −=

3/1

31y + =

3/1

)3/31(z −

(C) 3/1

21x − =

3/1

)3/92(y + =

3/1

)3/32(z −(D)

3/1

2x − =

3/1

3y + =

3/1

1z −

31. A line passes through a point A with p.v. 3� � �i j k+ − & is parallel to the vector 2 2� � �i j k− + . If P is a point onthis line such that AP = 15 units, then the p.v. of the point P is:

(A) 13 4 9� � �i j k+ − (B) 13 4 9� � �i j k− + (C) 7 6 11� � �i j k− + (D) − + −7 6 11� � �i j k32. The equations of the planes through the origin which are parallel to the line

2

1x − =

1

3y

−+

= 2

1z

−+

and distant 3

5 from it are

(A) 2x + 2y + z = 0 (B) x + 2y + 2z = 0 (C) 2x – 2y + z = 0 (D) x – 2y + 2z = 033. The value(s) of k for which the equation x2 + 2y2 – 5z2 + 2kyz + 2zx + 4xy = 0 represents a pair of

planes passing through origin is/are(A) 2 (B) – 2 (C) 6 (D) – 6

34. The equation of lines AB is 2

x =

3

y

− =

6

2. Through a point P(1, 2, 5), line PN is drawn perpendicular

to AB and line PQ is drawn parallel to the plane 3x + 4y + 5z = 0 to meet AB is Q. Then

(A) coordinate of N is

−49

156,

49

78,

49

52 (B) the coordinates of Q is

− 9,2

9,3

(C) the equation of PN is 3

1x − =

176

2y

−−

= 89

5z

−−

(D) the equation of PQ is 4

1x − =

13

2y

−−

= 8

5z −

35. Let a perpendicular PQ be drawn from P (5, 7, 3) to the line 3

15x − =

8

29y − =

5

5z

−−

when Q is the

foot. Then(A) Q is (9, 13, – 15) (B) PQ = 14(C) the equation of plane containing PQ and the given line is 9x – 4y – z – 14 = 0 (D) none

EXERCISE–61. Find the equation of the plane which contains the origin and the line of intersection of the

planes r�

. a�

= p and r�

. b�

= q

2. If the lines a

ax

′−

= b

by

′−

= c

cz

′−

and a

ax ′−=

b

by ′− =

c

cz ′− intersect at a point then the coordinate of the

point of intersection.3. The locus of a point which is a equidistant from the two given points with position vectors

a�

and b�

is the plane

+− )ba(2

1r

��

. ( a�

– b�

) = 0 bisecting the line joining the points normally..

5. Match the following :Column A Column B

(a) Sum of the square of the direction (P) 0cosines of line is

(b) All the points on the z-axis have (Q) 1their x and y coordinate equal to

(c) Distance between the points (1, 3, 2) (R) 9and (2, 3, 1) is

(d) Shortest distance between the lines (S) 2

1

6x − =

2

2y

−−

= 2

2z − and

3

4x + =

2

y

− =

2

1z

−+

is

6. Show that the angle between the straight lines whose direction cosines are given by the equations

� + m + n = 0 and amn + bn� + c�m = 0 is3

π if

a

1 +

b

1 +

c

1 = 0.

4. The foot of the perpendicular from (a, b, c) on the line x = y = z is the point (r, r, r) where3r = a + b + c.

7. Prove that the two lines whose direction cosines are given by the relations.p� + qm + rn = 0 &a�2 + bm2 + cn2 = 0 are perpendicular if, p2(b + c) + q2 (c + a) + r2 (a + b) = 0 and parallel if

0c

r

b

q

a

p 222

=++ .

8. Find the plane π passing through the points of intersection of the planes 2x + 3y− z +1= 0 andx + y − 2z + 3 = 0 and is perpendicular to the plane 3x − y − 2z = 4. Find the image of point (1, 1, 1) inplane π.

9. Given parallel planes r�

. (2 i − λ j + k ) = 3 and r�

. (4 i + j − µ k ) = 5 for what values of α, planes

r�

. (µ i − α j + 3 k ) = 0 & r�

. (α i − 3 j + 2λ k ) = 0 would be perpendicular..10. The edges of a rectangular parallelepiped are a, b, c; show that the angles between the four diagonals

are given by cos−1222

222

cba

cba

++±±

.

11. Prove that the line of intersection of the planes r�

. ( i + 2 j + 3 k ) = 0 and r�

. (3 i + 2 j + k ) = 0 is

r�

= t( i − 2 j + k ). Show that the line is equally inclined to i and k and makes an angle

(1/2) sec−1 3 with. j .

12. Find the shortest distance between the lines2

1x − =

3

1y + = z &

3

1x + = (y − 2); z = 2

13. Show that the line L whose equation is, r�

= (2 i − 2 j + 3 k ) + λ ( i − j + 4 k ) is parallel to the plane π

whose vector r�

. ( i + 5 j + k ) = 5. Find the distance between them.

14. A sphere has an equation2

ar��

− +2

br��

− = 72 where a�

= i + 3 j − 6 k and b�

= 2 i + 4 j + 2 k . Find:

(i) the centre of the sphere (ii) the radius of the sphere

(iii) perpendicular distance from the centre of the sphere to the plane r�

. ( )kj2i2 −+ = − 3.

15. Find the equation of the sphere which is tangential to the plane x − 2y − 2z = 7 at (3, −1, −1) andpasses through the point (1, 1, −3).

16. P1 and P2 are planes passing through origin. L1 and L2 also passes through origin. L1 lies on P1 not on P2 andL2 lies on P2 but not on P1. Show that there exists points A, B, C and whose permutation A′ .B′.C′ can bechosen such that [IIT - 2004]

(i)A is on L1, B on P1 but not on L1 and C not on P1.(ii)A′ in on L2, B′ on P2 but not on L2 and C′ not on P2.

17. A parallelopiped ‘S’ has base points A, B, C and D and upper face points A′, B′, C′ and D′. This parallelopipedis compressed by upper face A′B′C′D′ to form a new parallelopiped ‘T’ having upper face points A′′, B′′, C′′and D′′. Volume of parallelpiped T is 90 percent of the volume of parallelopiped S. Prove that the locus of ‘A′′’is a plane. [IIT - 2004]

EXERCISE–51. C 2. A 3. A

4. A 5. D 6. B

7. B 8. B 9. A

10. D 11. A 12. C

13. B 14. B 15. A

16. B 17. B 18. A

19. D 20. A 21. A

22. C 23. D 24. A

25. A 26. C 27. A

28. B 29. AB 30. ABC

31. AB 32. AD 33. BC

34. ABCD 35. BC

EXERCISE–61. r

�. ( a�

q − p b�

) = 0

2. ( cc,bb,aa ′+′+′+ )

3. True 4. True

5. (a) → (Q), (b) → (P), (c) → (S), (d) → (R)

8. 7x + 13y + 4z – 9 = 0 ;

−117

57,

117

78,

117

12

9. α = + 3 12.3

5913.

33

10

14. (i) (0, 5, 5) (ii) 9 (iii) 3

8

15. (x – 2)2 + (y – 1)2 + (z – 1)2 = 5

EXERCISE–7Part : (A) Only one correct option

1. The lengths of the diagonals of a parallelogram constructed on the vectors� � �p a b= +2 &

� � �q a b= − 2 ,

where�a &�b are unit vectors forming an angle of 60º are:

(A) 3 & 4 (B) 7 13& (C) 5 11& (D) none

2.

�

�

�

�a

a

b

b2 2

2

−

=

(A)� �a b

2 2

− (B)

� �

� �a b

a b

−

2

(C)

� � � �

� �

a a b b

a b

−

2

(D) none

3. A, B, C & D are four points in a plane with pv's� � �a b c, , &

�d respectively such that

( ) ( ) ( ) ( )� � � � � � � �a d b c b d c a− − = − −. . = 0. Then for the triangle ABC, D is its:

(A) incentre (B) circumcentre (C) orthocentre (D) centroid

4. Vectors� �a b& make an angle θ =

2

3

π. If �a = 1,

�b = 2 then ( ) ( ){ }� � � �

a b x a b+ −3 32

=

(A) 225 (B) 250 (C) 275 (D) 300

5. Consider a tetrahedron with faces f1, f

2, f

3, f

4. Let

� � � �a a a a1 2 3 4, , , be the vectors whose magnitudes are

respectively equal to the areas of f1, f

2, f

3, f

4 & whose directions are perpendicular to these faces in the

outward direction. Then,

(A)� � � �a a a a1 2 3 4+ + + = 0 (B)

� � � �a a a a1 3 2 4+ = + (C)

� � � �a a a a1 2 3 4+ = + (D) none

6. For non−zero vectors� � �a b c, , ,

� � �a x b c. =

� � �a b c holds if and only if;

(A)� �a b. = 0,

� �b c. = 0 (B)

� �c a. = 0,

� �a b. = 0 (C)

� �a c. = 0,

� �b c. = 0 (D)

� �a b. =

� �b c. =

� �c a. = 0

7. If�a i j k= + +� � � ,

�b i j k= − +� � � ,

�c i j k= + −� � �2 , then the value of

� � � � � �

� � � � � �

� � � � � �

a a a b a c

b a b b b c

c a c b c c

. . .

. . .

. . .

=

(A) 2 (B) 4 (C) 16 (D) 64

8. If� � �a b c, & are any three vectors, then

� � � � � �a b c a b c× × = × ×( ) ( ) is true if:

(A)� �b c& are collinear (B)

� �a c& are collinear (C)

� �a b& are collinear (D) none

9. ( ) ( ) ( ) ( ) ( ) ( )� � �r i i r r j r k k rj r. . .× ×+ × + =

(A) 0 (B) �r (C) 2

�r (D) 3

�r

10. A point taken on each median of a triangle divides the median in the ratio 1: 3, reckoning from the vertex.Then the ratio of the area of the triangle with vertices at these points to that of the original triangle is:(A) 5: 13 (B) 25: 64 (C) 13: 32 (D) none

11. Given a parallelogram ABCD. If AB→

= a, AD→

= b & AC→

= c, then DB AB→ →

. has the value:

(A)3

2

2 2 2a b c+ −(B)

a b c2 2 23

2

+ −(C)

a b c2 2 23

2

− +(D) none

12. The points whose position vectors are p i q j r k� � �+ + ; q i r j p k� � �+ + & r i p j q k� � �+ + are collinear if:(A) p + q + r = 0 (B) p2 + q2 + r2 − pq − qr − rp = 0(C) p3 + q3 + r3 − 3 pqr = 0 (D) none of these

13. If� �p s& are not perpendicular to each other and

� � � �r x p qx p= &

� �r s. = 0, then

�r =

(A) � �p s. (B)

��

� ��

qq s

p sp−

.

.(C) �

� �

� ��

qq p

p sp+

.

. (D)

� �q p+ µ for all scalars µ

14. If a, b, c are pth, qth, rth terms of an H.P. and

u�

= (q − r) i�

+ (r − p) j�

+ (p − q) k�

, υ�

=c

k

b

j

a

i��

++ , then:

(A) u�

, υ�

are parallel vectors (B) u�

, υ�

are orthogonal vectors

(C) u�

. υ�

= 1 (D) u�

× υ�

= i�

+ j�

+ k�

15. If q,p��

are two noncollinear and nonzero vectors such that ( ) ( ) ( ) 0qbapacqpcb =−+−+×−��

,where a, b, c are the length of the sides of a triangle, then the triangle is(A) right angled (B) obtuse angled (C) equilateral (D) isoceles

16. If cos � � �α i j k+ + , � cos � �i j k+ +β & � � cos �i j k+ + γ (α ≠ β ≠ γ ≠ 2 n π) are coplanar then the value of

cos cos cosec ec ec2 2 2

2 2 2

α β γ+ +

=

(A) 1 (B) 2 (C) 3 (D) none of these

17. If� � � �r x b cx b= &

� �r a. = 0 where

�a i j k= + −2 3� � � ,

�b i j k= − +3� � � &

�c i j k= + +� � �3 , then

�r is equal to:

(A) 2 ( )� � �i j k− + (B) 2 ( )� � �i j k+ − (C) 2 ( )− + +� � �i j k (D) 2 ( )� � �i j k+ +

18. The value of [ ] [ ] [ ] [ ]� � � � � � � � � � � � � � � �d b c a d c a b d a b c d a b c+ + − is equal to:

(A) 0 (B) 2 [ ]� � � �a b c d (C) – 2 [ ]� � � �

a b c d (D) none of these

19. The three vectors � � , � � , � �i j j k k i+ + + taken two at a time form three planes. The three unit vectorsdrawn perpendicular to these three planes form a parallelopiped of volume:

(A) 1

3(B) 4 (C)

3 3

4(D)

4

3 3

20. For any four points P, Q, R, S, PQ RS Q R PS R P QS→ → → → → →

× − × + × is equal to 4 times the area of the

triangle:(A) PQR (B) QRS (C) PRS (D) PQS

21. If� � �a b c, , are three non

−

coplanar &

� � �p q r, , are reciprocal vectors, then:

�a mb n c→ → →

+ +

. �p mq n r

→ → →+ +

is equal to:

(A) �2 + m2 + n2 (B) � m + m n + n � (C) 0 (D) none of these

22. In a quadrilateral ABCD,AC

→ is the bisector of the AB A D

→ →∧

which is

2

3

π, 15| |AC

→ = 3| |AB

→

= 5| |A D→

then cos BA C D→ →∧

is:

(A) − 14

7 2(B) −

21

7 3(C)

2

7(D)

2 7

14

23. In the isosceles triangle ABC| |AB→

=| |BC→

= 8, a point E divides AB internally in the ratio 1: 3, then the

cosine of the angle between CE→

& CA→

is (where| |CA→

= 12)

(A) − 3 7

8(B)

3 8

17(C)

3 7

8(D)

− 3 8

17

24. If� � �p a b= −3 5 ;

� � �q a b= +3 ;

� � �r a b= + 4 ;

� � �s a b= − + are four vectors such that

sin ( )� �p q∧

= 1 and ( )� �r s∧

= 1 then cos ( )� �a b∧

is:

(A) − 19

5 43(B) 0 (C)

19

5 43(D) 1

25. If r,q,p��

be three mutually perpendicular vectors of the same magnitude. If a vector x� satisfies the

equation p�

× ( )( )pqx��

×− + q�

× ( )( )qrx��

×− + r�

( )( )rpx��

×− = 0�

, then x�

is given by [IIT - 1997]

(A) 2

1 ( )r2qp

��−+ (B)

2

1 ( )rqp

��++ (C)

3

1 ( )rqp��

++ (D) 3

1 ( )rqp2��

−+

26. Let� �a b& be two non−collinear unit vectors. If ( )� � � � �

u a a b b= − . &� � �v a x b= , then

�v is [IIT - 1999]

(A) �u (B)

� � �u u a+ . (C)

� � �u u b+ . (D) ( )� � � �

u u a b+ +.

27. If a�

, b�

, c�

are three non-zero, non coplanar vectors and 1b�

= b�

– 2|a|

a.b�

��

a�

, 2b�

= b�

+ 2|a|

a.b�

��

a�

,

1c�

= c�

– 2|a|

a.c�

��

a�

+ 2|c|

c.b�

��

1b�

, 2c�

= c�

– 2|a|

a.c�

��

a�

– 21

1

|b|

c.b�

��

1b�

, 3c�

= c�

– 2|c|

a.c�

��

a�

+ 2|c|

c.b�

��

1b�

,

4c�

= c�

– 2|c|

a.c�

��

a�

– 2|b|

c.b�

��

1b�

, then the set of orthogonal vectors is [IIT - 2005]

(A) )c,b,a( 31

��(B) )c,b,a( 21

��(C) )c,b,a( 11

��(D) )c,b,a( 22

��

28. Let A�

be vector parallel to line of intersection of planes P1 and P

2 through origin, P

1 is parallel to the vectors

2 i + 3 k and 4 j – 3 k and P2 is parallel to j – k and 3 i + 3 j , then the angle between vector

A�

and 2 i + j – 2 k is [IIT - 2006]

(A) 2

π(B)

4

π (C)

6

π(D)

4

3π

Part : (B) May have more than one options correct

29. If� � �a b c, , &

�d are linearly independent set of vectors & K a K b K c K d1 2 3 4

� � � �+ + + = 0 then:

(A) K1 + K

2 + K

3 + K

4 = 0 (B) K

1 + K

3 = K

2 + K

4 = 0 (C) K

1 + K

4 = K

2 + K

3 = 0 (D) none of these

30. Given three vectors� � �a b c, , such that they are non

−

zero, non

−

coplanar vectors, then which of the

following are coplanar.

(A) � � � � � �a b b c c a+ + +, , (B)

� � � � � �a b b c c a− + +, , (C)

� � � � � �a b b c c a+ − +, , (D)

� � � � � �a b b c c a+ − −, ,

31. Let�p i j a k= + +2 3� � � ,

�q b i j k= + −� � �5 &

�r i j k= + +� � �3 . If

� � �p q r, , are coplanar and

� �p q. = 20, a & b

have the values:(A) 1, 3 (B) 9, 7 (C) 5, 5 (D) 13, 9

32. If�z a i b j1 = +� � &

�z c i d j2 = +� � are two vectors in � & �i j system where

� �z z1 2= = r &

� �z z1 2. = 0

then�w a i c j1 = +� � &

�w b i d j2 = +� � satisfy:

(A)�w1 = r (B)

�w2 = r (C)

� �w w1 2. = 0 (D) none of these

33. If�a &�b are two non colinear unit vectors &

�a ,�b , x�a − y

�b form a triangle, then:

(A) x = − 1; y = 1 & �a + �b = 2 cos

� �a b

∧

2

(B) x = − 1; y = 1 & cos� �a b

∧

+ �a +

�b cos ( )� � �

a a b, − +

∧ = − 1

(C) �a + �b = − 2 cot

� �a b

∧

2

cos

� �a b

∧

2

& x = − 1, y = 1 (D) none

34. The value(s) of α ∈ [0, 2π] for which vector ( )�a i j k= + +� � sin �3 2α makes an obtuse angle with the

Z-axis and the vectors�b i j k= − +(tan ) � � sin �α

α2

2 and ( ) ( )�

c i j ec k= + −tan � tan � cos �α αα

32

are

orthogonal, is/are:(A) tan −1 3 (B) π − tan −1 2 (C) π + tan −1 3 (D) 2 π − tan −1 2

35. A parallelogram is constructed on the vectors�p &�q . A vector which coincides with the altitude of the

parallelogram & perpendicular to the side�p expressed in terms of the vectors

�p &�q is:

(A)�q −

( )

� �

p2

�p (B)

( )� � �

�p x q x p

p2(C)

� �

�q . p

p2

�p − �q (D)

( )� � �

�p x p x q

p2

36. Identify the statement(s) which is/are incorrect?

(A)�a x ( )[ ]� � �

a a b× × = ( ) ( )� � �a b a× 2

(B) If� � �a b c, , are non coplanar vectors and

� � � � � �v a v b v c. . .= = = 0 then

�v must be a null vector

(C) If�a and

�b l ie in a plane normal to the plane containing the vectors

�c and

�d

then ( )� �a b× x ( )� �

c d× = 0

(D) If� � �a b c, , and

� � �′ ′ ′a b c, , are reciprocal system of vectors then

� � � � � �a b b c c a. . .′ + ′ + ′ = 3

37. If a i j k= + +� � �2 4 ,�b i j k= − +2 3� � � ,

�c i j k= + −� � �4 4 , then the vector ( )� � �

a b c× × is orthogonal to:

�q . p

(A) �a (B)

�b (C)

�c (D)

� � �a b c+ +

38. If� � �a b c, , are non-zero, non-collinear vectors such that a vector

�p = a

b

cos ( )( )2π − ∧� � �a b c and a vector

�q = a

c cos ( )( )π − ∧� � �

a c b then p +�q is

(A) parallel to�a (B) perpendicular to

�a (C) coplanar with

� �b c& (D) none of these

39. Which of the following statement(s) is/are true?

(A) If� �n a. = 0,

� �n b. = 0 &

� �n c. = 0 for some non zero vector

�n , then [ ]� � �

a b c = 0

(B) there exist a vector having direction angles α = 30º & β = 45º(C) locus of point for which x = 3 & y = 4 is a line parallel to the z - axis whose distance from the

z axis is 5(D) the vertices of a regular tetrahedron are OABC where '

O

' is the origin. The vector

OA OB OC→ → →

+ + is perpendicular to the plane ABC.

40. In a ∆ ABC, let M be the mid point of segment AB and let D be the foot of the bisector of ∠ C. Then the

ratio Area CDM

Area ABC

∆∆

is:

(A)1

4 a b

a b

−+ (B)

1

2 a b

a b

−+ (C)

1

2 tan

A B−2

cot A B+

2(D)

1

4 cot

A B−2

tan A B+

2

41. The vectors� � �a b c, , are of the same length & pairwise form equal angles. If

�a i j= +� � &

�b j k= +� � , the

pv's of�c can be:

(A) (1, 0, 1) (B) − −

4

3

1

3

4

3, , (C)

1

3

4

3

1

3, ,−

(D) − −

1

3

4

3

1

3, ,

EXERCISE–81. Through the middle point M of the side AD of a parallelogram ABCD the straight line BM is drawn

cutting AC at R and CD produced at Q prove that QR = 2RB

2. Show that the perpendicular distance of the point c�

from the line joining b&a��

is,

ab

baaccb

��

��

−

×+×+×.

3. If�α = + +� � �i j k2 3 ;

�

β = − +2 � � �i j k ;�γ = + +3 2� � �i j k and ( )� � � � � �

α β γ α β γ× × = + +p q r then find the values

of p, q, r

4. If�a = 2 3� � �i j k− + ,

�b = � � �i j k− + 2 ,

�c = 2 � � �i j k+ − &

�d = 3 2� � �i j k− − , then find the value of

(�a x �b ) x (

�a x �c ).�d

5. Show that ( ) ( )( ) ( ) ( )( ) ( ) ( )( )� � � � � � � � � � � � � � �a q c p b b p c q a c p a q b× × × × = × × × × + × × × ×

6. It is given that��

� � ��

� � ��

� � �xbx c

a b cy

cxa

a b cz

a x b

a b c= = =

[ ];

[ ];

[ ] where

� � �a b c, , are non − coplanar vectors. Show

that� � �x y z, , also forms a non − coplanar system. Find the value of

� � � � � � � � �x a b y b c z c a.( ) .( ) .( )+ + + + + .

7. The median AD of a triangle ABC is bisected at E and BE is produced to meet the side AC in F. Provethat AF = (1/3) AC and EF = (1/4) BF.

8. Points X and Y are taken on the sides QR and RS, respectively of a parallelogram PQRS, so thatQX = 4XR and RY = 4YS. The line XY cuts the line PR at Z. Find the ratio PZ: ZR.

9. Forces� �P Q, act at O & have a resultant

�R . If any transversal cuts their line of action at A,B,C respectively,,

then show that P

OA

Q

OB

R

OC+ = .

10. In a tetrahedron, if two pairs of opposite edges are perpendicular, then show that the third pair ofopposite edges is also perpendicular & in this case the sum of the squares of two opposite edges isthe same for each pair. Also show that the segment joining the mid points of opposite edges bisect oneanother.

11. Use vectors to prove that the diagonals of a trapezium having equal non parallel sides are equal &conversely.

12. Given four non zero vectors� � � �a b c and d, , . The vectors

� � �a b c, & are coplanar but not collinear pair by

pair and vector�d is not coplanar with vectors

� � �a b c, & and ( ) ( ) , ( ) , ( )

� � � � � � � �a b b c d a d b∧

=∧

=∧

=∧

=π

α β3

,

prove that ( ) cos (cos cos )� �d c∧

= −−1 β α .

13. If � � �p q r, & are three non-coplanar vectors, prove that,

� �a b× =

[ ]1

� � � � � �q r r p p q× × ×, ,

� � �

� � � � � �

� � � � � �

p q r

p a q a r a

p b q b r b

. . .

. . .

14. Consider the non zero vectors� � � �a b c d, , & such that no three of which are coplanar then prove

that [ ] [ ] [ ] [ ]� � � � � � � � � � � � � � � �a b cd c a bd b a cd d a b c+ = + . Hence prove that

� � � �a b c d, , & represent the position vectors

of the vertices of a plane quadrilateral if and only if[ ] [ ][ ] [ ]

� � � � � �

� � � � � �

b cd a bd

a cd a b c

+

+= 1 .

15. Solve the following equation for the vector�p ; ( )� � � � � � �

pxa p b c bx c+ =. where� � �a b c, , are non zero non coplanar

vectors and�a is neither perpendicular to

�b nor to

�c , hence show that

[ ]� ��

� ��

pxaa b c

a cc+

.

is perpendicular

to� �b c− .

16. If 'c,'b,'a&c,b,a��

are reciprocal system of vectors then prove that:

(i) 1]'c'b'a[]cba[ =��

(ii) ]cba[

cba)'ax'c()'cx'b()'bx'a( ��

�� ++

=++ .

17. Let �A = 2i + k;

�B = i + j + k &

�C = 4i − 3j + 7k. Determine a vector

�R satisfying 0A.R&BxCBxR ==

��

18. For any two vectors� �u v& , prove that [IIT - 1998]

(a) ( . ) | | | | | |� � � � � �u v u v u v2 2 2 2+ × = & (b) ( | | )( | | ) ( . ) | ( )|1 1 12 2 2 2+ + = − + + + ×

� � � � � � � �u v u v u v u v

19. Let ABC and PQR be any two triangles in the same plane. Assume that the perpendiculars from thepoints A, B, C to the sides QR, RP, PQ respectively are concurrent. Using vector methods or otherwise,prove that the perpendiculars from P, Q, R to BC, CA, AB respectively are also concurrent. [IIT - 2000]

20. Find 3 – dimensional vectors 321 v,v,v��

satisfying

11 v.v��

= 4, 21 v.v��

= –2, 31 v.v��

= 6, 22 v.v��

= 2, 32 v.v��

= –5, 33 v.v��

= 29. [IIT - 2001]

21. If w,v,u be three non-coplanar unit vectors with angles between v&u is α, between w&v is βand between u&w is γ. If c,b,a

�� are the unit vectors along angle bisectors of α, β, γ respectively,,

then prove that, [ ]ac,cb,ba��

××× =16

1 [ ]2wvu sec2

α2

sec2

β2

sec2

γ2

. [IIT - 2003]

EXERCISE–71. B 2. B 3. C 4. D 5. A 6. D

7. C 8. B 9. A 10. B 11. A 12. B

13. B 14. B 15. C 16. B 17. C 18. A

19. D 20. B 21. A 22. C 23. C 24. C

25. B 26. A 27. B 28. D 29. ABC

30. BCD 31. AD 32. ABC 33. AB 34. BD

35. BD 36. ACD 37. AD 38. BC 39. ACD

40. BC 41. AD

EXERCISE–83. p = 0; q = 10; r = − 3

4. – 98 6. 3

20. i2v1 =�

, jiv2 ±−=�

, k4j2i3v3 ±±=�

are some

possible values

Assertion- Reason

Some questions (Assertion–Reason type) are given below. Each question contains Statement – 1 (Assertion)

and Statement – 2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is

correct. So select the correct choice :Choices are :

(A)Statement – 1 is True, Statement – 2 is True; Statement – 2 is a correct explanation for Statement – 1. (B)Statement – 1 is True, Statmnt – 2 is True; Statement – 2 is NOT a correct explanation for Statement – 1.

(C) Statement – 1 is True, Statement – 2 is False.

(D) Statement – 1 is False, Statement – 2 is True.

452. Let a, b, c be three non-coplanar vectors then ( b c).[(c a) (a b)]− − × − = 0

Statement 1: b c− can be expressed as linear combination of c a− and a b− .

Statement 2: Given non-coplanar vectors one vector can be expressed as a linear combination of other two. 453. A vector has components p and 1 with respect to a rectangular cartesian system. If the axes are rotated through an

angle α about the origin in the anticlockwise sense. Statement–1 : If the vector has component p + 2 and 1 with respect to the new system then p = –1

Statement–2 : Magnitude of vector original and new system remains same

454. Let | a | 4,=�

| b | 2=�

and angle between a�

and b�

is π/6

Statement–1 : 2(a b) 4× =�

�

Statement–2 : 2 2(a b) | a |× =�

� �

455. Statement–1 : b c c a a b 0 × × × =

� �

� � � �

Statement–2 : If a�

, b�

, r�

are linearly dependent vectors then they are coplanar.

456. Statement–1 : If a b a b+ = −� �

� �

then a�

is parallel to b�

.

Statement–2 : If a b+�

�

= a b−�

�

then a.b�

�

= 0.

457. Let r�

be a non-zero vector satisfying r.a r.b r.c 0= = =�

� � � � �

for given non–zero vectors a, b�

�

and c�

.

Statement–1 : a, b�

�

and c�

are coplanar vectors.

Statement–2 : r�

is perpendicular to the vectors a, b�

�

and c�

.

458. Let a�

and r�

be two non–collinear vectors.

Statement–1 : vector a�

× ( )a r×� �

is a vector in the plane of a�

and r�

, perpendicular to a�

.

Statement–2 : ( )a a b× ×�

� �

= 0�

, for any vector b�

.

459. Statement–1 : If three points P, Q, R have position vectors a�

, b�

, c�

respectively and

2a 3b 5c 0+ − =�

� �

, then the points P, Q, R must be collinear. Statement–2 : If for three

points A, B, C; AB AC= λ��

, then the points A, B, C must be collinear.

460. Statement–1 : Let a�

and b�

be two non collinear unit vectors. If ( )u a a.b b= −� �

� ��

and v a b= ×�

��

then | v | | u | .=� �

Statement–2 : The vector ( )1 ˆ ˆ ˆ2i 2 j k3

− + is makes an angle of 3

π with the vector

( )ˆ ˆ ˆ5i 4j 3k− + .

461. Statement-1: If u & v� �

are unit vectors inclined at an angle α and x�

is a unit vector bisecting the angle between

them, then u v

x

2cos2

+=

α

� �

�

Statement-2: If ∆ABC is an isosceles triangle with AB = AC = 1, then vector representing bisector of angle A is

given by AD��

= AB AC

2

+��

462. Statement-1: The direction ratios of line joining origin and point (x, y, z) must be x, y, z.

Statement-2: If P is a point (x, y, z) in space and OP = r, then direction cosines of OP are x y z

, ,r r r

.

463. Statement-1: If the vectors ˆ ˆ ˆ2i j k− + , ˆ ˆ î 2j 3k+ − and ˆ ˆ ˆ3i j 5k− λ + are coplanar, then |λ|2 is equal to 16.

Statement-2: The vectors a , b�

�

and c�

are coplanar iff a, (b c×�

� �

) = 0

464. Statement-1: A line L is perpendicular to the plane 3x – 4y + 5z = 10

Statement-2: Direction co-sines of L be < 3 4 1

, ,5 2 5 2 2

− >

465. Statement-1 : The points with position vectors a 2b 3c, 2a 3b c− + − + −� �

� � � �

, 4a 7b 7c− +�

� �

are collinear.

Statement-2: The position vectors a 2b 3c,− +�

� �

− 2 a 3b c,+ −�

� �

4a 7b 7c− +�

� �

are linearly dependent vectors.

466. Statement-1: If a,b, c�

� �

are three unit vectors such that 1

a (b c) b2

× × =� �

� �

then the angle between a & b�

�

is π/2

Statement-2: If 1

a (b c) b,2

× × =� �

� �

then a .b�

�

= 0.

467. Statement-1: If cosα, cosβ, cosγ are the direction cosine of any line segment, cos2α + cos

2β + cos2γ = 1.

Statement-2: If cosα, cosβ, cosγ are the direction cosine of line segment,

cos2α + cos2β + cos2γ = −1. 468. Statement-1: The direction cosines of one of the angular bisector of two intersecting lines having direction

cosines as l1 , m1, n1, & l2, m2, n2 is proportional to l1 + l2, m1+ m2, n1 + n2.

Statement-2: The angle between the two intersecting lines having direction cosines as l1, m1, n1 & l2, m2, n2 is

given by cosθ = l1 l2 + m1m2 + n1n2.

469. Statement-1: If a.b 0=�

�

⇒ a�

⊥ b�

Statement-2: a.b�

�

= 0 ⇒ either a 0=�

or b�

= 0 or a�

⊥ b�

470. Statement-1: A B B A× = ×� ��

Statement-2: A B | A || B |× =� ��

sinθ n , when θ is angle, when your fingers curls from A to B

471. Statement-1 : A vector ⊥r the plane of (1, -1, 0), (2, 1, -1) & (-1, 1, 2) is ˆ ˆ6i 6k+

Statement-2 : A B×� �

always gives a vector perpendicular to plane of A & B� �

472. Statement-1 : Angle between planes 1 1

r.n q=� � �

& 2 2

r.n q=� � �

.

(acute angle) is given by cosθ = 1 2n .n� �

Statement-2 : Angle between the planes in same as acute angle formed by their normals.

473. Statement-1: In ∆ABC, AB BC CA+ +��

= 0

Statement-2 : If OA a, OB b= =��

�

then AB a b= +��

�

474. Statement-1: a 3i pj 3k= + +� � �

�

and b 2i 3 j qk= + +� � � �

are parallel vectors it p = 9/2 and q = 2.

Statement-2 : If 1 2 3a a i a j a k= + +� � �

�

and 1 2 3b b i b j b k= + +

� � � �

are parallel 31 2

1 2 3

aa a

b b b= =

475. Statement-1: The direction ratios of line joining origin and point (x, y, z) must be x, y, z

Statement-2: If P is a point (x,y, z) in space and OP = r then directions cosines of OP are x y z

, ,r r r

476. Statement-1: The shortest distance between the skew lines r a b= + α�

� �

and r c d= + β�

� �

is | [a c bd] |

| b d |

−

×

��

� �

�

Statement-2: Two lines are skew lines if three axist no plane passing through them.

477. Statement-1: ˆ ˆ â i pj 2k= + +�

, ˆ ˆ ˆb 2i 3j qk= + +�

are parallel vectors of p = 3/2 and q = 4.

Statement-2: 1 2 3ˆ ˆ â a i a j a k= + +

�

and 1 2 3ˆ ˆ ˆb b i b j b k= + +

�

are parallel if 31 2

1 2 3

aa a

b b b= = .

478. Statement-1: If ˆ ˆ ˆ â 2i k,b 3j 4k= + = +�

�

and ˆ ˆc 8i 3j= −�

are coplanar then c 4a b= −�

� �

.

Statement-2: A set of vectors 1 2 na ,a ...a� � �

is said to be linearly independent if every relation of the form l1 1a�

+ l2

2a�

+ … + ln na�

= 0 implies that l1 = l2 = …. = ln = 0 (scalars).

479. Statement-1: The shortest distance between the skew lines r a b and r c d= + α = + β� �

� � � �

is (a c).(b d)

| b d |

− ×

×

� �

� �

� �

Statement-2: Two lines are skew lines if there exists no plane passing through them.

480. Statement-1: The curve which is tangent to a sphere at a given point is the equation of a plane. Statement-2: Infinite number of lines touch the sphere at a given point.

481. Statement-1: In ∆ABC AB BC CA O+ + =��

Statement-2: If OA a, OB b,= =��

�

then AB a b= +��

�

(∆ law of addition).

482. Statement-1: ˆ ˆ â i pj 2k= + +�

and ˆ ˆ ˆb 2i 3j qk= + +�

are parallel vectors if 3

P , q 42

= =

Statement-2: If 1 2 3ˆ ˆ â a i a j a k= + +

�

and 1 2 3ˆ ˆ ˆb b i b j b k= + +

�

are parallel then 31 2

1 2 3

aa a

b b b= = .

483. Statement-1: If 1 3 3 nˆ ˆ ˆ ˆ ˆ â 2i k a ,a , a .....a , b 3j 4k and c 8i 3j= + = + = −

�

� � � � � �

are coplanar then c 4a b= −�

� �

Statement-2: A set of vectors is said to be linearly independent if every relation of the form

1 2 2 n na a ..... a 0+ + + =� � �

l l l ⇒ l1 = l2 = ….. = ln = 0.

484. Statement-1: The shortest distance between the skew lines 1 1r a b+ α�

� �

= and 2 2r a b+ β�

� �

= is

1 2 2 1

1 2

[b b (a a )]

(b b )

−

×

� �

� �

� � Statement-2: Two lines are skew lines if there exists no plane passing through them.

485. Statement-1 : The value of expression ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ î( j k) j(k i) k(i j) 3× + × + × =

Statement-2 : ˆ ˆ ˆ ˆ ˆ î( j k) [i.j.k] 1× = =

486. Statement-1: A relation between the vectors r, a and b�

� �

is r a b× =�

� �

⇒ a b

ra.a

×=

�

�

�

� � Statement-2 : r.a 0=

� �

3-Dimension

487. The equation of two straight line are x 1 y 3 z 2

2 1 3

− + −= =

− and

x 2 y 1 z 3

1 3 2

− − += =

−

Statement–1 : The given lines are coplanar

Statement–2 : The equation 2x1 – y1 = 1, x1 + 3y1 = 4, 3x1 + 2y1 = 5 are consistent.

488. Statement–1 : The distance between the planes 4x – 5y + 3z = 5 and 4x – 5y + 3z + 2 = 0 is

3

5 2. Statement–2 The distance between ax + by + cz + d1 = 0 and ax + by + cz + d2

= 0 is 1 2

2 2 2

d d

a b c

−

+ +.

489. Given the line x 1 y 1 z 3

3 2 1

− + −= =

− and the plane π : x - 2y - z = 0

Statement-1: L lies in π Statement-2: L is parallel to π

490. The image of the point (1, b, 3) in the Statement-1: Line x y 1 z 2

1 2 3

− −= = will be (1, 0, 7)

Statement-2: Length of the perpendicular from the point A( α ) on the line r a tb,= +�

� �

is given by d =

| (a ) b |

| b |

− α ×

Answer

452. C 453. A 454. D 455. D 456. D 457. A 458. C

459. A 460. C 461. A 462. A 463. A 464. A 465. A

466. A 467. B 468. B 469. D 470. D 471. A 472. A

473. C 474. A 475. A 476. B 477. A 478. B 479. B

480. A 481. C 482. A 483. B 484. B 485. A 486. A

487. A 488. D 489. C 490. B

Que from Compt. Exams Co-ordinate Geometry of Three Dimensions

1. The direction cosines of a line segment AB are ,17/2− .17/2,17/3 − If 17=AB and the co-ordinates of A are (3, –

6, 10), then the co-ordinates of B are

(a) (1, –2, 4) (b) (2, 5, 8) (c) (–1, 3, –8) (d) (1, – 3, 8)

2. The projection of any line on co-ordinate axes be respectively 3, 4, 5 then its length is [MP PET 1995; RPET 2001]

(a) 12 (b) 50 (c) 25 (d) None of these

3. If centroid of the tetrahedron OABC , where CBA ,, are given by (a, 2, 3),(1, b, 2) and (2, 1, c) respectively be (1, 2, –1),

then distance of ),,( cbaP from origin is equal to

(a) 107 (b) 14 (c) 14/107 (d) None of these

4. If )1,0,0(),0,1,0( ≡≡ QP , then projection of PQ on the plane 3=++ zyx is [EAMCET 2002]

(a) 3 (b) 3 (c) 2 (d) 2

5. The points )4,9,3(),1,1,0(),1,5,4( CBA −− and )4,4,4(−D are

[Kurukshetra CEE 2002]

(a) Collinear (b) Coplanar (c) Non- coplanar (d) Non- Collinear and non-coplanar

6. The angle between two diagonals of a cube will be [MP PET 1996, 2000; RPET 2000, 02; UPSEAT 2004]

(a) 3/1sin 1− (b) 3/1cos 1− (c) Variable (d) None of these

7. The equations of the line passing through the point (1,2,–4) and perpendicular to the two lines 7

10

16

19

3

8 −=

−

+=

− zyx and

5

5

8

29

3

15

−

−=

−=

− zyx, will be [AI CBSE 1983]

(a) 6

4

3

2

2

1 +=

−=

− zyx (b)

8

4

3

2

2

1 +=

−=

−

− zyx (c)

8

4

2

2

3

1 +=

−=

− zyx (d) None of these

8. If three mutually perpendicular lines have direction cosines ),,(),,,( 222111 nmlnml and ),,( 333 nml , then the line having direction

cosines 321 lll ++ , 321 mmm ++ and 321 nnn ++ make an angle of ..... with each other

(a) °0 (b) °30 (c) °60 (d) °90

9. The straight lines whose direction cosines are given by 0,0 =++=++ hlmgnlfmncnbmal are perpendicular, if

(a) 0=++c

h

b

g

a

f (b) 0=++

h

c

g

b

f

a(c) chbgaf == (d)

h

c

g

b

f

a==

10. If the straight lines ,1 sx += ,3 sy λ−−= sz λ+= 1 and tztytx −=+== 2,1,2/ , with parameters s and t respectively,

are co-planar, then λ equals [AIEEE 2004]

(a) 0 (b) –1 (c) –1/2 (d) – 2

11. The co-ordinates of the foot of perpendicular drawn from point )3,0,1(P to the join of points )1,7,4(A and )3,5,3(B is [RPET 01]

(a) (5, 7, 1) (b)

3

17,

3

7,

3

5 (c)

3

7,

3

5,

3

2 (d)

3

7,

3

2,

3

5

12. If the lines 4

1

3

1

2

1 −=

+=

− zyx and

111

3 zkyx=

−=

− intersect, then k = [IIT Screening 2004]

(a) 9

2 (b)

2

9 (c) 0 (d) None of these

13. A square ABCD of diagonal 2a is folded along the diagonal AC so that the planes DAC and BAC are at right angle. The

shortest distance between DC and AB is [Kurukshetra CEE 1998]

(a) a2 (b) 3/2a (c) 5/2a (d) a)2/3(

14. A line with direction cosines proportional to 2,1, 2 meets each of the lines zayx =+= and zyax 22 ==+ . The co-ordinates

of each of the points of intersection are given by [AIEEE 2004]

(a) ),,2(),3,,2( aaaaaa (b) ),,(),3,2,3( aaaaaa (c) )2,,(),3,2,3( aaaaaa (d) ),,(),3,3,3( aaaaaa

15. The equation of the planes passing through the line of intersection of the planes 043 =−− zyx and 063 =++ yx whose

distance from the origin is 1, are

(a) 0322 =−−− zyx , 0322 =+−+ zyx (b) 0322 =−+− zyx , 0322 =+++ zyx

(c) 0322 =−−+ zyx , 0322 =+−− zyx (d) None of these

16. The co-ordinates of the points A and B are (2, 3, 4) (–2, 5,– 4) respectively. If a point P moves so that kPBPA =− 22 where k is a

constant, then the locus of P is

(a) A line (b) A plane (c) A sphere (d) None of these

17. The equation of the plane passing through the points )2,3,1( −− and perpendicular to planes 522 =++ zyx and

8233 =++ zyx , is [AISSE 1987]

(a) 08342 =−+− zyx (b) 08342 =+−− zyx (c) 08342 =+++ zyx (d) None of these

18. A variable plane at a constant distance p from origin meets the co-ordinates axes in CBA ,, . Through these points planes are

drawn parallel to co-ordinate planes. Then locus of the point of intersection is

(a) 2222

1111

pzyx=++ (b) 2222 pzyx =++ (c) pzyx =++ (d) p

zyx=++

111

19. P is a fixed point ),,( aaa on a line through the origin equally inclined to the axes, then any plane through P perpendicular to OP,

makes intercepts on the axes, the sum of whose reciprocals is equal to

(a) a (b) a2

3 (c)

2

3a (d) None of these

20. The equation of the plane through the intersection of the planes 0432 =−++ zyx , 01234 =+++ zyx and passing through

the origin will be [MP PET 1998]

(a) 0=++ zyx (b) 0111417 =++ zyx (c) 047 =++ zyx (d) 01417 =++ zyx

21. The d.r’s of normal to the plane through )0,1,0(),0,0,1( which makes an angle 4

π with plane 3=+ yx , are [AIEEE 2002]

(a) 1,2,1 (b) 1,1, 2 (c) 1, 1, 2 (d) 1,1,2

22. Two systems of rectangular axes have the same origin. If a plane cuts them at distance a, b, c and a', b', c' from the origin, then

(a) 0'

1

'

1

'

1111222222

=+++++cbacba

(b) 0'

1

'

1

'

1111222222

=−++−+cbacba

(c) 0'

1

'

1

'

1111222222

=−−+−−cbacba

(d) 0'

1

'

1

'

1111222222

=−−−++cbacba

[AIEEE 2003]

23. If 044 =−+ kzyx is the equation of the plane through the origin that contains the line ,43

1

2

1 zyx=

+=

−then k =

(a) 1 (b) 3 (c) 5 (d) 7 [MP PET 1992]

24. The distance of the point (1, –2, 3) from the plane 5=+− zyx measured parallel to the line ,632 −

==zyx

is

(a) 1 (b) 6/7 (c) 7/6 (d) None of these [AI CBSE 1984]

25. The distance of the point of intersection of the line 2

5

2

4

1

3 −=

−=

− zyxand the plane 17=++ zyx from the point (3, 4, 5)

is given by

(a) 3 (b) 3/2 (c) 3 (d) None of these

26. The lines δααδα +

−−=

−=

−

+− dazaydax and

γββγβ +

−−=

−=

−

+− cbzbycbx are coplanar and then equation to the plane in

which they lie, is

(a) 0=++ zyx (b) 0=+− zyx (c) 02 =+− zyx (d) 02 =−+ zyx

27. The line 4

5

3

4

2

3 −=

−=

− zyx lies in the plane 044 =−−+ dkzyx . The values of k and d are

(a) 4, 8 (b) –5, – 3 (c) 5, 3 (d) – 4, – 8

28. The value of k such that 21

2

1

4 kzyx −=

−=

− lies in the plane 742 =+− zyx , is [IIT Screening 2003]

(a) 7 (b) – 7 (c) No real value (d) 4

The shortest distance from the plane 3273412 =++ zyx to the sphere +++ 222 zyx 155624 =−− zyx is [AIEEE 2003]

(a) 26 (b) 13

411 (c) 13 (d) 39

29. The radius of the circle in which the sphere 019422222 =−−−+++ zyxzyx is cut by the plane 0722 =+++ zyx is

(a) 1 (b) 2 (c) 3 (d) 4 [AIEEE 2003]

30. The equation of motion of a rocket are: ,4,2 tytx −== tz 4= where the time 't' is given in seconds, and the co-ordinates of a

moving point in kilometers. What is the path of the rocket? At what distance will be the rocket be from the starting point 0(0, 0, 0)

in 10 seconds

(a) Straight line, 60 km (b) Straight line, 30 km (c) Parabola, 60 km (d) Ellipse, 60 km

31. The plane 0=+ mylx is rotated an angle α about its line of intersection with the plane 0=z , then the equation to the plane in

its new position is

(a) 0tan)( 22 =+±+ αmlzmylx

(b) 0tan)( 22 =+±− αmlzmylx

(c) 0cos)( 22 =+±+ αmlzmylx

(d) 0cos)( 22 =+±− αmlzmylx

32. The distance between two points P and Q is d and the length of their projections of PQ on the co-ordinate planes are 321 ,, ddd .

Then 223

22

21 kdddd =++ where ‘k’ is

(a) 1 (b) 5

(c) 3 (d) 2

33. If 1P and 2P are the lengths of the perpendiculars from the points (2,3,4) and (1,1,4) respectively from the plane

011263 =++− zyx , then 1P and 2P are the roots of the equation

(a) 07232 =+− PP (b) 016237

2 =+− PP

(c) 016172 =+− PP (d) 07162 =+− PP

34. The edge of a cube is of length ‘a’ then the shortest distance between the diagonal of a cube and an edge skew to it is

(a) 2a (b) a

(c) a/2 (d) 2/a

Que from Compt. Exams Co-ordinate Geometry of Three Dimensions

1 d 2 c 3 a 4 c 5 b

6 b 7 a 8 a 9 a 10 d

11 b 12 b 13 b 14 b 15 a

16 b 17 a 18 a 19 d 20 b

21 b 22 d 23 c 24 a 25 a

26 c 27 c 28 a 29 c 30 c

31 a 32 a 33 d 34 b 35 d

Que from Compt. Exams Vector Algebra

1. Three forces of magnitudes 1, 2, 3 dynes meet in a point and act along diagonals of three adjacent faces of a cube. The resultant force is [MNR 1987] (a) 114 dyne (b) 6 dyne (c) 5 dyne (d) None of these

2. The vectors b and c are in the direction of north-east and north-west respectively and |b|=|c|= 4. The magnitude and direction of the vector d = c – b, are [Roorkee 2000]

(a) ,24 towards north (b) 24 , towards west (c) 4, towards east (d) 4, towards south

3. If a, b and c are unit vectors, then 222 |||||| accbba −+−+− does not exceed[IIT Screening 2001]

(a) 4 (b) 9 (c) 8 (d) 6

4. The vectors kji 453 ++=AB and kji 255 +−=AC are the sides of a triangle ABC. The length of the median through

A is [UPSEAT 2004]

(a) 13 unit (b) 52 unit (c) 5 unit (d) 10 unit

5. Let the value of bap )12()4( ++++= yxyx and ,)132()22( baq −−++−= yxxy where a and b are non-collinear

vectors. If ,23 qp = then the value of x and y will be [RPET 1984; MNR 1984]

(a) – 1, 2 (b) 2, – 1 (c) 1, 2 (d) 2, 1\ 6. The points D, E, F divide BC, CA and AB of the triangle ABC in the ratio 1 : 4, 3 : 2 and 3 : 7 respectively and the point

K divides AB in the ratio 3:1 , then CKCFBEAD :)( ++ is equal to [MNR 1987]

(a) 1 : 1 (b) 2 : 5 (c) 5 : 2 (d) None of these 7. If two vertices of a triangle are ji − and kj + , then the third vertex can be [Roorkee 1995]

(a) ki + (b) kji −− 2 (c) ki − (d) ji −2

(e) All the above

8. If a of magnitude 50 is collinear with the vector ,2

1586

kjib −−= and makes an acute angle with the positive

direction of z-axis, then the vector a is equal to [Pb. CET 2004]

(a) kji 303224 +− (b) kji 303224 ++− (c) kji 151616 −− (d) kji 301612 −+−

9. If three non-zero vectors are ,321 kjia aaa ++= kjib 321 bbb ++= and .321 kjic ccc ++= If c is the unit vector

perpendicular to the vectors a and b and the angle between a and b is ,6

π then

2

321

321

321

ccc

bbb

aaa

is equal to

(a) 0 (b) 4

)()()(3 21

21

21 cba ΣΣΣ

(c) 1 (d) 4

)()( 21

21 ba ΣΣ

10. Let the unit vectors a and b be perpendicular and the unit vector c be inclined at an angle θ to both a and b. If ),( babac ×++= γβα then [Orissa JEE 2003]

(a) θγθβα 2cos,cos 2 === (b) θγθβα 2cos,cos 2 −===

(c) θγθβθα 2cos,sin,cos 2 === (d) None of these

11. The vector ba + bisects the angle between the vectors a and b, if (a) |||| ba = (b) |||| ba = or angle between a and b is zero

(c) |||| ba m= (d) None of these

12. The points DCBAO ,,,, are such that ,a=OA ,b=OB ba 32 +=OC and .2ba −=OD If |,|3|| ba = then the

angle between BD and AC is

(a) 3

π (b)

4

π (c)

6

π (d) None of these

13. If kjikji ++−=++= 2,32 BA and ,3 ji +=C then the value of t such that BtA + is at right angle to vector ,C is

[RPET 2002] (a) 2 (b) 4 (c) 5 (d) 6

14. Let jib 34 += and c be two vectors perpendicular to each other in the xy-plane. All vectors in the same plane having

projections 1 and 2 along b and c respectively, are given by [IIT 1987]

(a) jiji5

11

5

2,2 +− (b) jiji

5

11

5

2,2 +−+ (c) jiji

5

11

5

2,2 −−+ (d) jiji

5

11

5

2,2 +−−

15. Let kjibkjia −+=+−= 2,2 and kjic 2−+= be three vectors. A vector in the plane of b and c whose projection

on a is of magnitude 3/2 is

[IIT 1993; Pb. CET 2004] (a) kji 332 −+ (b) kji 332 ++ (c) kji 52 +−− (d) kji 52 ++

16. A vector a has components 2p and 1 with respect to a rectangular cartesian system. The system is rotated through a certain angle about the origin in the anti-clockwise sense. If a has components p+1 and 1 with respect to the new system, then [IIT 1984]

(a) 0=p (b) 1=p or 3

1− (c) 1−=p or

3

1 (d) 1=p or 1−

17. If kjiu −+= 22 and ,236 kjiv +−= then a unit vector perpendicular to both u and v is [MP PET 1987]

(a) kji 1810 −− (b)

−− kji

5

182

5

1

17

1(c) )18107(

473

1kji −− (d) None of these

18. If kjibkia ++=+= ,2 and .734 kjic +−= If bcbd ×=× and ,0. =ad then d will be [IIT 1990]

(a) kji 28 ++ (b) kji 28 +− (c) kji −+− 8 (d) kji 28 +−−

19. If abra λ+=× and ,3. =ra where kjia −+= 2 and ,2 kjib +−−= then r and λ are equal to

(a) 5

6,

3

2

6

7=+= λjir (b)

6

5,

3

2

6

7=+= λjir (c)

5

6,

3

2

7

6=+= λjir (d) None of these

20. Let the vectors a, b, c and d be such that 0)()( =××× dcba . Let 1P and 2P be planes determined by pair of vectors a,

b and c, d respectively. Then the angle between 1P and 2P is [IIT Screening 2000; MP PET 2004]

(a) o0 (b) 4

π (c)

3

π (d)

2

π

21. If 1., =++= bakjia and ,kjba −=× then =b [IIT Screening 2004]

(a) i (b) kji +− (c) kj −2 (d) i2

22. The position vectors of the vertices of a quadrilateral ABCD are cb,a, and d respectively. Area of the quadrilateral

formed by joining the middle points of its sides is [Roorkee 2000]

(a) ||4

1addbba ×+×+× (b) abdadccb ×+×+×+×

4

1

(c) addccbba ×+×+×+× 4

1 (d) bddccb ×+×+×

4

1

23. The moment about the point )6,4,2( −−M of the force represented in magnitude and position by AB where the points

A and B have the co-ordinates )3,2,1( − and )2,4,3( − respectively, is [MP PET 2000]

(a) 8i – 9j – 14k (b) 2i – 6j + 5k (c) – 3i + 2j – 3k (d) – 5i + 8j – 8k

2244.. If the vectors kjikji ++++ ba , and kji c++ )1( ≠≠≠ cba are coplanar, then the value of =−

+−

+− cba 1

1

1

1

1

1

[BIT Ranchi 1988; RPET 1987;IIIITT 11998877;; DDCCEE 22000011;; MMPP PPEETT 22000044;; OORRIISSSSAA JJEEEE 22000055]]

(a) – 1 (b) 2

1− (c)

2

1 (d) 1

25. If 0accbba =×+×+× )()()( γβα and at least one of the numbers βα , and γ is non-zero, then the vectors a, b and

c are (a) Perpendicular (b) Parallel (c) Coplanar (d) None of these

26. The volume of the tetrahedron, whose vertices are given by the vectors kjikji +−++− , and kji −+ with reference

to the fourth vertex as origin, is

(a) 3

5cubic unit (b)

3

2 cubic unit (c)

5

3cubic unit (d) None of these

27. Let .,, ikckjbjia −=−=−= If d is a unit vector such that ],ˆ[0ˆ. dcbda == then d is equal to [IIT 1995]

(a) 3

kji −+± (b)

3

kji ++± (c)

6

2kji −+± (d) k±

28. The value of 'a' so that the volume of parallelopiped formed by kjkji aa +++ , and ki +a becomes minimum is

[IIT Screening 2003]

(a) – 3 (b) 3 (c) 3

1 (d) 3

29. If b and c are any two non-collinear unit vectors and a is any vector, then =××

×++ )(

||

)(.).().( cb

cb

cbaccabba

[IIT 1996] (a) a (b) b (c) c (d) 0

30. If a, b, c are non-coplanar unit vectors such that 2

)(cb

cba+

=×× , then the angle between a and b is [IIT 1995]

(a) 4

π (b)

2

π (c)

4

3π (d) π

31. =××××××××× )]()()()()()[( baacaccbcbba

(a) 2][ cba (b) 3][ cba (c) 4][ cba (d) None of these

32. Unit vectors a, b and c are coplanar. A unit vector d is perpendicular to them. If kjidcba3

1

3

1

6

1)()( +−=××× and

the angle between a and b is o30 , then c is [Roorkee Qualifying 1998]

(a) 3

)22( kji +− (b)

3

)2( kji −+ (c)

3

)22( kji −+− (d)

3

)2( kji ++−

33. The radius of the circular section of the sphere 5|| =r by the plane 33)(. =++ kjir is [DCE 1999]

(a) 1 (b) 2 (c) 3 (d) 4

34. If x is parallel to y and z where kjix α++= 2 , kiy += α and jiz −= 5 , then α is equal to [J & K 2005]

(a) 5± (b) 6± (c) 7± (d) None of these

35. The vector c directed along the internal bisector of the angle between the vectors kjia 447 −−= and kjib 22 +−−=

with ,65|| =c is

(a) )27(3

5kji +− (b) )255(

3

5kji ++ (c) )27(

3

5kji ++ (d) )255(

3

5kji ++−

36. The distance of the point )32( kji ++B from the line which is passing through )224( kji ++A and which is parallel to

the vector kji 632 ++=C is [Roorkee 1993]

(a) 10 (b) 10 (c) 100 (d) None of these

37. Let a, b, c are three non-coplanar vectors such that ,,, 321 bacracbrcbar ++=−+=+−=

.432 cbar +−= If ,332211 rrrr λλλ ++= then

(a) 71 =λ (b) 331 =+ λλ (c) 4321 =++ λλλ (d) 223 =+ λλ

38. Let kjibkjia −+=++= 2,2 and a unit vector c be coplanar. If c is perpendicular to a, then c =

[IIT 1999; Pb. CET 2003; DCE 2005]

(a) )(2

1kj +− (b) )(

3

1kji −−− (c) )2(

5

1ji − (d) )(

3

1kji −−

39. Let p, q, r be three mutually perpendicular vectors of the same magnitude. If a vector x satisfies equation ,0}){(}){(}){( =×−×+×−×+×−× rpxrqrxqpqxp then x is given by [IIT 1997 Cancelled]

(a) )2(2

1rqp −+ (b) )(

2

1rqp ++ (c) )(

3

1rqp ++ (d) )2(

3

1rqp −+

40. The point of intersection of abar ×=× and babr ×=× , where jia += and kib −= 2 is [Orissa JEE 2004]

(a) kji −+3 (b) ki −3 (c) kji ++ 23 (d) None of these

Que from Compt. Exams

Vector Algebra

1 c 2 b 3 b 4 c 5 b

6 b 7 e 8 b 9 d 10 b

11 b 12 d 13 c 14 d 15 a,c

16 b 17 b 18 d 19 b 20 a

21 a 22 c 23 a 24 d 25 c

26 b 27 c 28 c 29 a 30 c

31 c 32 a,c 33 d 34 c 35 a

36 b 37 b,c 38 a 39 b 40 a

Documents

Maths Study Material - Three Dimentional Geometry