Class XII Chapter 11 – Three Dimensional Geometry Mathscbseocean.weebly.com/uploads/2/8/1/5/28152469/cmn1… · · 2017-04-18Class XII Chapter 11 – Three Dimensional Geometry

Class XII Chapter 11 – Three Dimensional Geometry Maths

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Exercise 11.1

Question 1:

If a line makes angles 90°, 135°, 45° with x, y and z-axes respectively, find its direction

cosines.

Answer

Let direction cosines of the line be l, m, and n.

Therefore, the direction cosines of the line are

Question 2:

Find the direction cosines of a line which makes equal angles with the coordinate axes.

Answer

Let the direction cosines of the line make an angle α with each of the coordinate axes.

∴ l = cos α, m = cos α, n = cos α


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Thus, the direction cosines of the line, which is equally inclined to the coordinate axes,

are

Question 3:

If a line has the direction ratios −18, 12, −4, then what are its direction cosines?

Answer

If a line has direction ratios of −18, 12, and −4, then its direction cosines are

Thus, the direction cosines are .

Question 4:

Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.

Answer

The given points are A (2, 3, 4), B (− 1, − 2, 1), and C (5, 8, 7).

It is known that the direction ratios of line joining the points, (x1, y1, z1) and (x2, y2, z2),

are given by, x2 − x1, y2 − y1, and z2 − z1.

The direction ratios of AB are (−1 − 2), (−2 − 3), and (1 − 4) i.e., −3, −5, and −3.

The direction ratios of BC are (5 − (− 1)), (8 − (− 2)), and (7 − 1) i.e., 6, 10, and 6.

It can be seen that the direction ratios of BC are −2 times that of AB i.e., they are

proportional.

Therefore, AB is parallel to BC. Since point B is common to both AB and BC, points A, B,

and C are collinear.


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Question 5:

Find the direction cosines of the sides of the triangle whose vertices are (3, 5, − 4), (−

1, 1, 2) and (− 5, − 5, − 2)

Answer

The vertices of ∆ABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2).

The direction ratios of side AB are (−1 − 3), (1 − 5), and (2 − (−4)) i.e., −4, −4, and 6.

Therefore, the direction cosines of AB are

The direction ratios of BC are (−5 − (−1)), (−5 − 1), and (−2 − 2) i.e., −4, −6, and −4.

Therefore, the direction cosines of BC are

The direction ratios of CA are (−5 − 3), (−5 − 5), and (−2 − (−4)) i.e., −8, −10, and 2.

Therefore, the direction cosines of AC are


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Exercise 11.2

Question 1:

Show that the three lines with direction cosines

are mutually perpendicular.

Answer

Two lines with direction cosines, l1, m1, n1 and l2, m2, n2, are perpendicular to each

other, if l1l2 + m1m2 + n1n2 = 0

(i) For the lines with direction cosines, and , we obtain

Therefore, the lines are perpendicular.

(ii) For the lines with direction cosines, and , we obtain


(iii) For the lines with direction cosines, and , we obtain


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Thus, all the lines are mutually perpendicular.

Question 2:

Show that the line through the points (1, −1, 2) (3, 4, −2) is perpendicular to the line

through the points (0, 3, 2) and (3, 5, 6).

Answer

Let AB be the line joining the points, (1, −1, 2) and (3, 4, − 2), and CD be the line

joining the points, (0, 3, 2) and (3, 5, 6).

The direction ratios, a1, b1, c1, of AB are (3 − 1), (4 − (−1)), and (−2 − 2) i.e., 2, 5, and

−4.

The direction ratios, a2, b2, c2, of CD are (3 − 0), (5 − 3), and (6 −2) i.e., 3, 2, and 4.

AB and CD will be perpendicular to each other, if a1a2 + b1b2+ c1c2 = 0

a1a2 + b1b2+ c1c2 = 2 × 3 + 5 × 2 + (− 4) × 4

= 6 + 10 − 16

= 0

Therefore, AB and CD are perpendicular to each other.

Question 3:

Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the

points (−1, −2, 1), (1, 2, 5).

Answer

Let AB be the line through the points, (4, 7, 8) and (2, 3, 4), and CD be the line through

the points, (−1, −2, 1) and (1, 2, 5).

The directions ratios, a1, b1, c1, of AB are (2 − 4), (3 − 7), and (4 − 8) i.e., −2, −4, and

−4.

The direction ratios, a2, b2, c2, of CD are (1 − (−1)), (2 − (−2)), and (5 − 1) i.e., 2, 4,

and 4.


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AB will be parallel to CD, if

Thus, AB is parallel to CD.

Question 4:

Find the equation of the line which passes through the point (1, 2, 3) and is parallel to

the vector .

Answer

It is given that the line passes through the point A (1, 2, 3). Therefore, the position

vector through A is

It is known that the line which passes through point A and parallel to is given by

is a constant.

This is the required equation of the line.

Question 5:

Find the equation of the line in vector and in Cartesian form that passes through the

point with position vector and is in the direction .

Answer

It is given that the line passes through the point with position vector


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It is known that a line through a point with position vector and parallel to is given by

the equation,

This is the required equation of the line in vector form.

Eliminating λ, we obtain the Cartesian form equation as

This is the required equation of the given line in Cartesian form.

Question 6:

Find the Cartesian equation of the line which passes through the point

(−2, 4, −5) and parallel to the line given by

Answer

It is given that the line passes through the point (−2, 4, −5) and is parallel to

The direction ratios of the line, , are 3, 5, and 6.

The required line is parallel to

Therefore, its direction ratios are 3k, 5k, and 6k, where k ≠ 0

It is known that the equation of the line through the point (x1, y1, z1) and with direction

ratios, a, b, c, is given by


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Therefore the equation of the required line is

Question 7:

The Cartesian equation of a line is . Write its vector form.

Answer

The Cartesian equation of the line is

The given line passes through the point (5, −4, 6). The position vector of this point is

Also, the direction ratios of the given line are 3, 7, and 2.

This means that the line is in the direction of vector,

It is known that the line through position vector and in the direction of the vector is

given by the equation,

This is the required equation of the given line in vector form.

Question 8:

Find the vector and the Cartesian equations of the lines that pass through the origin and

(5, −2, 3).

Answer

The required line passes through the origin. Therefore, its position vector is given by,

The direction ratios of the line through origin and (5, −2, 3) are

(5 − 0) = 5, (−2 − 0) = −2, (3 − 0) = 3


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The line is parallel to the vector given by the equation,

The equation of the line in vector form through a point with position vector and parallel

to is,

The equation of the line through the point (x1, y1, z1) and direction ratios a, b, c is given

by,

Therefore, the equation of the required line in the Cartesian form is

Question 9:

Find the vector and the Cartesian equations of the line that passes through the points (3,

−2, −5), (3, −2, 6).

Answer

Let the line passing through the points, P (3, −2, −5) and Q (3, −2, 6), be PQ.

Since PQ passes through P (3, −2, −5), its position vector is given by,

The direction ratios of PQ are given by,

(3 − 3) = 0, (−2 + 2) = 0, (6 + 5) = 11

The equation of the vector in the direction of PQ is

The equation of PQ in vector form is given by,

The equation of PQ in Cartesian form is


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i.e.,

Question 10:

Find the angle between the following pairs of lines:

(i)

(ii) and

Answer

(i) Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by,

The given lines are parallel to the vectors, and ,

respectively.


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(ii) The given lines are parallel to the vectors, and ,

respectively.

Question 11:

Find the angle between the following pairs of lines:

(i)

(ii)

i. Answer


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ii. Let and be the vectors parallel to the pair of lines,

, respectively.

and

The angle, Q, between the given pair of lines is given by the relation,

(ii) Let be the vectors parallel to the given pair of lines, and

, respectively.


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If Q is the angle between the given pair of lines, then

Question 12:

Find the values of p so the line and

are at right angles.

Answer

The given equations can be written in the standard form as

and

The direction ratios of the lines are −3, , 2 and respectively.

Two lines with direction ratios, a1, b1, c1 and a2, b2, c2, are perpendicular to each other, if

a1a2 + b1 b2 + c1c2 = 0


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Thus, the value of p is .

Question 13:

Show that the lines and are perpendicular to each other.

Answer

The equations of the given lines are and

The direction ratios of the given lines are 7, −5, 1 and 1, 2, 3 respectively.

Two lines with direction ratios, a1, b1, c1 and a2, b2, c2, are perpendicular to each other, if

a1a2 + b1 b2 + c1c2 = 0

∴ 7 × 1 + (−5) × 2 + 1 × 3

= 7 − 10 + 3

= 0

Therefore, the given lines are perpendicular to each other.

Question 14:

Find the shortest distance between the lines


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Answer

The equations of the given lines are

It is known that the shortest distance between the lines, and , is

given by,

Comparing the given equations, we obtain

Substituting all the values in equation (1), we obtain


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Therefore, the shortest distance between the two lines is units.

Question 15:

Find the shortest distance between the lines and

Answer

The given lines are and

It is known that the shortest distance between the two lines,

, is given by,

Comparing the given equations, we obtain


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Since distance is always non-negative, the distance between the given lines is

units.

Question 16:

Find the shortest distance between the lines whose vector equations are

Answer

The given lines are and


given by,


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Comparing the given equations with and , we obtain


Therefore, the shortest distance between the two given lines is units.

Question 17:

Find the shortest distance between the lines whose vector equations are

Answer

The given lines are


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given by,

For the given equations,


Therefore, the shortest distance between the lines is units.


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Exercise 11.3

Question 1:

In each of the following cases, determine the direction cosines of the normal to the plane

and the distance from the origin.

(a)z = 2 (b)

(c) (d)5y + 8 = 0

Answer

(a) The equation of the plane is z = 2 or 0x + 0y + z = 2 … (1)

The direction ratios of normal are 0, 0, and 1.

∴

Dividing both sides of equation (1) by 1, we obtain

This is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to

the plane and d is the distance of the perpendicular drawn from the origin.

Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the

origin is 2 units.

(b) x + y + z = 1 … (1)


∴

Dividing both sides of equation (1) by , we obtain


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This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of

normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal are and the distance of

normal from the origin is units.

(c) 2x + 3y − z = 5 … (1)

The direction ratios of normal are 2, 3, and −1.




Therefore, the direction cosines of the normal to the plane are and

the distance of normal from the origin is units.

(d) 5y + 8 = 0

⇒ 0x − 5y + 0z = 8 … (1)

The direction ratios of normal are 0, −5, and 0.



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Therefore, the direction cosines of the normal to the plane are 0, −1, and 0 and the

distance of normal from the origin is units.

Question 2:

Find the vector equation of a plane which is at a distance of 7 units from the origin and

normal to the vector .

Answer

The normal vector is,

It is known that the equation of the plane with position vector is given by,

This is the vector equation of the required plane.

Question 3:

Find the Cartesian equation of the following planes:

(a) (b)

(c)

Answer

(a) It is given that equation of the plane is

For any arbitrary point P (x, y, z) on the plane, position vector is given by,


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Substituting the value of in equation (1), we obtain

This is the Cartesian equation of the plane.

(b)



This is the Cartesian equation of the plane.

(c)



This is the Cartesian equation of the given plane.

Question 4:

In the following cases, find the coordinates of the foot of the perpendicular drawn from

the origin.

(a) (b)

(c) (d)

Answer

(a) Let the coordinates of the foot of perpendicular P from the origin to the plane be

(x1, y1, z1).


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2x + 3y + 4z − 12 = 0

⇒ 2x + 3y + 4z = 12 … (1)





The coordinates of the foot of the perpendicular are given by

(ld, md, nd).

Therefore, the coordinates of the foot of the perpendicular are

(b) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1,

y1, z1).

⇒ … (1)

The direction ratios of the normal are 0, 3, and 4.



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(ld, md, nd).


(c) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1,

y1, z1).

… (1)

The direction ratios of the normal are 1, 1, and 1.





(ld, md, nd).


(d) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1,

y1, z1).

⇒ 0x − 5y + 0z = 8 … (1)

The direction ratios of the normal are 0, −5, and 0.


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(ld, md, nd).


Question 5:

Find the vector and Cartesian equation of the planes

(a) that passes through the point (1, 0, −2) and the normal to the plane is .

(b) that passes through the point (1, 4, 6) and the normal vector to the plane is

.

Answer

(a) The position vector of point (1, 0, −2) is

The normal vector perpendicular to the plane is

The vector equation of the plane is given by,

is the position vector of any point P (x, y, z) in the plane.

Therefore, equation (1) becomes


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This is the Cartesian equation of the required plane.

(b) The position vector of the point (1, 4, 6) is

The normal vector perpendicular to the plane is

The vector equation of the plane is given by,

is the position vector of any point P (x, y, z) in the plane.



Question 6:

Find the equations of the planes that passes through three points.

(a) (1, 1, −1), (6, 4, −5), (−4, −2, 3)

(b) (1, 1, 0), (1, 2, 1), (−2, 2, −1)

Answer

(a) The given points are A (1, 1, −1), B (6, 4, −5), and C (−4, −2, 3).


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Since A, B, C are collinear points, there will be infinite number of planes passing through

the given points.

(b) The given points are A (1, 1, 0), B (1, 2, 1), and C (−2, 2, −1).

Therefore, a plane will pass through the points A, B, and C.

It is known that the equation of the plane through the points, , and

, is


Question 7:

Find the intercepts cut off by the plane

Answer


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It is known that the equation of a plane in intercept form is , where a, b, c

are the intercepts cut off by the plane at x, y, and z axes respectively.

Therefore, for the given equation,

Thus, the intercepts cut off by the plane are .

Question 8:

Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

Answer

The equation of the plane ZOX is

y = 0

Any plane parallel to it is of the form, y = a

Since the y-intercept of the plane is 3,

∴ a = 3

Thus, the equation of the required plane is y = 3

Question 9:

Find the equation of the plane through the intersection of the planes

and and the point (2, 2, 1)

Answer


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The equation of any plane through the intersection of the planes,

3x − y + 2z − 4 = 0 and x + y + z − 2 = 0, is

The plane passes through the point (2, 2, 1). Therefore, this point will satisfy equation

(1).

Substituting in equation (1), we obtain

This is the required equation of the plane.

Question 10:

Find the vector equation of the plane passing through the intersection of the planes

and through the point (2, 1, 3)

Answer

The equations of the planes are

The equation of any plane through the intersection of the planes given in equations (1)

and (2) is given by,

, where


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The plane passes through the point (2, 1, 3). Therefore, its position vector is given by,




Question 11:

Find the equation of the plane through the line of intersection of the planes

and which is perpendicular to the plane

Answer

The equation of the plane through the intersection of the planes, and

, is

The direction ratios, a1, b1, c1, of this plane are (2λ + 1), (3λ + 1), and (4λ + 1).

The plane in equation (1) is perpendicular to


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Its direction ratios, a2, b2, c2, are 1, −1, and 1.

Since the planes are perpendicular,



Question 12:

Find the angle between the planes whose vector equations are

and .

Answer

The equations of the given planes are and

It is known that if and are normal to the planes, and , then the

angle between them, Q, is given by,

Here,


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Substituting the value of , in equation (1), we obtain

Question 13:

In the following cases, determine whether the given planes are parallel or perpendicular,

and in case they are neither, find the angles between them.

(a)

(b)

(c)

(d)

(e)

Answer

The direction ratios of normal to the plane, , are a1, b1, c1 and

.

The angle between L1 and L2 is given by,

(a) The equations of the planes are 7x + 5y + 6z + 30 = 0 and

3x − y − 10z + 4 = 0

Here, a1 = 7, b1 =5, c1 = 6


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Therefore, the given planes are not perpendicular.

It can be seen that,

Therefore, the given planes are not parallel.

The angle between them is given by,

(b) The equations of the planes are and

Here, and

Thus, the given planes are perpendicular to each other.

(c) The equations of the given planes are and

Here, and

Thus, the given planes are not perpendicular to each other.


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∴

Thus, the given planes are parallel to each other.

(d) The equations of the planes are and

Here, and

∴

Thus, the given lines are parallel to each other.

(e) The equations of the given planes are and

Here, and

Therefore, the given lines are not perpendicular to each other.

∴

Therefore, the given lines are not parallel to each other.

The angle between the planes is given by,


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Question 14:

In the following cases, find the distance of each of the given points from the

corresponding given plane.

Point Plane

(a) (0, 0, 0)

(b) (3, −2, 1)

(c) (2, 3, −5)

(d) (−6, 0, 0)

Answer

It is known that the distance between a point, p(x1, y1, z1), and a plane, Ax + By + Cz =

D, is given by,

(a) The given point is (0, 0, 0) and the plane is

(b) The given point is (3, − 2, 1) and the plane is

∴

(c) The given point is (2, 3, −5) and the plane is

(d) The given point is (−6, 0, 0) and the plane is


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Miscellaneous Solutions

Question 1:

Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line

determined by the points (3, 5, −1), (4, 3, −1).

Answer

Let OA be the line joining the origin, O (0, 0, 0), and the point, A (2, 1, 1).

Also, let BC be the line joining the points, B (3, 5, −1) and C (4, 3, −1).

The direction ratios of OA are 2, 1, and 1 and of BC are (4 − 3) = 1, (3 − 5) = −2, and

(−1 + 1) = 0

OA is perpendicular to BC, if a1a2 + b1b2 + c1c2 = 0

∴ a1a2 + b1b2 + c1c2 = 2 × 1 + 1 (−2) + 1 ×0 = 2 − 2 = 0

Thus, OA is perpendicular to BC.

Question 2:

If l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually perpendicular lines,

show that the direction cosines of the line perpendicular to both of these are m1n2 −

m2n1, n1l2 − n2l1, l1m2 − l2m1.

Answer

It is given that l1, m1, n1 and l2, m2, n2 are the direction cosines of two mutually

perpendicular lines. Therefore,

Let l, m, n be the direction cosines of the line which is perpendicular to the line with

direction cosines l1, m1, n1 and l2, m2, n2.


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l, m, n are the direction cosines of the line.

∴l2 + m2 + n2 = 1 … (5)

It is known that,

∴

Substituting the values from equations (5) and (6) in equation (4), we obtain

Thus, the direction cosines of the required line are


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Question 3:

Find the angle between the lines whose direction ratios are a, b, c and b − c,

c − a, a − b.

Answer

The angle Q between the lines with direction cosines, a, b, c and b − c, c − a,

a − b, is given by,

Thus, the angle between the lines is 90°.

Question 4:

Find the equation of a line parallel to x-axis and passing through the origin.

Answer

The line parallel to x-axis and passing through the origin is x-axis itself.

Let A be a point on x-axis. Therefore, the coordinates of A are given by (a, 0, 0), where

a ∈ R.

Direction ratios of OA are (a − 0) = a, 0, 0

The equation of OA is given by,

Thus, the equation of line parallel to x-axis and passing through origin is


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Question 5:

If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9,

2) respectively, then find the angle between the lines AB and CD.

Answer

The coordinates of A, B, C, and D are (1, 2, 3), (4, 5, 7), (−4, 3, −6), and

(2, 9, 2) respectively.

The direction ratios of AB are (4 − 1) = 3, (5 − 2) = 3, and (7 − 3) = 4

The direction ratios of CD are (2 −(− 4)) = 6, (9 − 3) = 6, and (2 −(−6)) = 8


Therefore, AB is parallel to CD.

Thus, the angle between AB and CD is either 0° or 180°.

sQuestion 6:

If the lines and are perpendicular, find the value

of k.

Answer

The direction of ratios of the lines, and , are −3,

2k, 2 and 3k, 1, −5 respectively.

It is known that two lines with direction ratios, a1, b1, c1 and a2, b2, c2, are

perpendicular, if a1a2 + b1b2 + c1c2 = 0


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Therefore, for , the given lines are perpendicular to each other.

Question 7:

Find the vector equation of the plane passing through (1, 2, 3) and perpendicular to the

plane

Answer

The position vector of the point (1, 2, 3) is

The direction ratios of the normal to the plane, , are 1, 2, and −5

and the normal vector is

The equation of a line passing through a point and perpendicular to the given plane is

given by,

Question 8:

Find the equation of the plane passing through (a, b, c) and parallel to the plane

Answer

Any plane parallel to the plane, , is of the form

The plane passes through the point (a, b, c). Therefore, the position vector of this

point is



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Question 9:

Find the shortest distance between lines

and .

Answer

The given lines are

It is known that the shortest distance between two lines, and , is

given by

Comparing to equations (1) and (2), we obtain


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Therefore, the shortest distance between the two given lines is 9 units.

Question 10:

Find the coordinates of the point where the line through (5, 1, 6) and

(3, 4, 1) crosses the YZ-plane

Answer

It is known that the equation of the line passing through the points, (x1, y1, z1) and (x2,

y2, z2), is

The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by,

Any point on the line is of the form (5 − 2k, 3k + 1, 6 −5k).

The equation of YZ-plane is x = 0

Since the line passes through YZ-plane,

5 − 2k = 0


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Therefore, the required point is .

Question 11:

Find the coordinates of the point where the line through (5, 1, 6) and

(3, 4, 1) crosses the ZX − plane.

Answer

It is known that the equation of the line passing through the points, (x1, y1, z1) and (x2,

y2, z2), is

The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by,

Any point on the line is of the form (5 − 2k, 3k + 1, 6 −5k).

Since the line passes through ZX-plane,

Therefore, the required point is .


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Question 12:

Find the coordinates of the point where the line through (3, −4, −5) and (2, − 3, 1)

crosses the plane 2x + y + z = 7).

Answer

It is known that the equation of the line through the points, (x1, y1, z1) and (x2, y2, z2), is

Since the line passes through the points, (3, −4, −5) and (2, −3, 1), its equation is

given by,

Therefore, any point on the line is of the form (3 − k, k − 4, 6k − 5).

This point lies on the plane, 2x + y + z = 7

∴ 2 (3 − k) + (k − 4) + (6k − 5) = 7

Hence, the coordinates of the required point are (3 − 2, 2 − 4, 6 × 2 − 5) i.e.,

(1, −2, 7).

Question 13:

Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to

each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

Answer

The equation of the plane passing through the point (−1, 3, 2) is

a (x + 1) + b (y − 3) + c (z − 2) = 0 … (1)

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


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It is known that two planes, and , are

perpendicular, if

Plane (1) is perpendicular to the plane, x + 2y + 3z = 5

Also, plane (1) is perpendicular to the plane, 3x + 3y + z = 0

From equations (2) and (3), we obtain

Substituting the values of a, b, and c in equation (1), we obtain


Question 14:

If the points (1, 1, p) and (−3, 0, 1) be equidistant from the plane

, then find the value of p.

Answer

The position vector through the point (1, 1, p) is

Similarly, the position vector through the point (−3, 0, 1) is


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The equation of the given plane is

It is known that the perpendicular distance between a point whose position vector is

and the plane, is given by,

Here, and d

Therefore, the distance between the point (1, 1, p) and the given plane is

Similarly, the distance between the point (−3, 0, 1) and the given plane is

It is given that the distance between the required plane and the points, (1, 1, p) and

(−3, 0, 1), is equal.

∴ D1 = D2


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Question 15:

Find the equation of the plane passing through the line of intersection of the planes

and and parallel to x-axis.

Answer

The given planes are

The equation of any plane passing through the line of intersection of these planes is

Its direction ratios are (2λ + 1), (3λ + 1), and (1 − λ).

The required plane is parallel to x-axis. Therefore, its normal is perpendicular to x-axis.

The direction ratios of x-axis are 1, 0, and 0.



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Therefore, its Cartesian equation is y − 3z + 6 = 0

This is the equation of the required plane.

Question 16:

If O be the origin and the coordinates of P be (1, 2, −3), then find the equation of the

plane passing through P and perpendicular to OP.

Answer

The coordinates of the points, O and P, are (0, 0, 0) and (1, 2, −3) respectively.

Therefore, the direction ratios of OP are (1 − 0) = 1, (2 − 0) = 2, and (−3 − 0) = −3

It is known that the equation of the plane passing through the point (x1, y1 z1) is

where, a, b, and c are the direction ratios of normal.

Here, the direction ratios of normal are 1, 2, and −3 and the point P is (1, 2, −3).

Thus, the equation of the required plane is

Question 17:

Find the equation of the plane which contains the line of intersection of the planes

, and which is perpendicular to the plane

.

Answer

The equations of the given planes are

The equation of the plane passing through the line intersection of the plane given in

equation (1) and equation (2) is


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The plane in equation (3) is perpendicular to the plane,



The Cartesian equation of this plane can be obtained by substituting in

equation (3).

Question 18:

Find the distance of the point (−1, −5, −10) from the point of intersection of the line

and the plane .

Answer

The equation of the given line is

The equation of the given plane is

Substituting the value of from equation (1) in equation (2), we obtain


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Substituting this value in equation (1), we obtain the equation of the line as

This means that the position vector of the point of intersection of the line and the plane

is

This shows that the point of intersection of the given line and plane is given by the

coordinates, (2, −1, 2). The point is (−1, −5, −10).

The distance d between the points, (2, −1, 2) and (−1, −5, −10), is

Question 19:

Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes

and .

Answer

Let the required line be parallel to vector given by,

The position vector of the point (1, 2, 3) is

The equation of line passing through (1, 2, 3) and parallel to is given by,

The equations of the given planes are


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The line in equation (1) and plane in equation (2) are parallel. Therefore, the normal to

the plane of equation (2) and the given line are perpendicular.


Therefore, the direction ratios of are −3, 5, and 4.


This is the equation of the required line.

Question 20:

Find the vector equation of the line passing through the point (1, 2, − 4) and

perpendicular to the two lines:

Answer

Let the required line be parallel to the vector given by,


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The position vector of the point (1, 2, − 4) is

The equation of the line passing through (1, 2, −4) and parallel to vector is

The equations of the lines are

Line (1) and line (2) are perpendicular to each other.

Also, line (1) and line (3) are perpendicular to each other.


∴Direction ratios of are 2, 3, and 6.


This is the equation of the required line.


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Question 21:

Prove that if a plane has the intercepts a, b, c and is at a distance of P units from the

origin, then

Answer

The equation of a plane having intercepts a, b, c with x, y, and z axes respectively is

given by,

The distance (p) of the plane from the origin is given by,

Question 22:

Distance between the two planes: and is

(A)2 units (B)4 units (C)8 units

(D)

Answer



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It can be seen that the given planes are parallel.

It is known that the distance between two parallel planes, ax + by + cz = d1 and ax + by

+ cz = d2, is given by,

Thus, the distance between the lines is units.

Hence, the correct answer is D.

Question 23:

The planes: 2x − y + 4z = 5 and 5x − 2.5y + 10z = 6 are

(A) Perpendicular (B) Parallel (C) intersect y-axis

(C) passes through

Answer


2x − y + 4z = 5 … (1)

5x − 2.5y + 10z = 6 … (2)



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∴

Therefore, the given planes are parallel.

Hence, the correct answer is B.

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53

3D GEOMETRY

INTRODUCTION

Summary

1. Distance formula: Distance between two points A(𝑥1, 𝑦1, 𝑧1) and B (𝑥2, 𝑦2, 𝑧2)is

AB=√(𝑥2 − 𝑥1)2 + (𝑦2 − 𝑦1)2 + (𝑧2 − 𝑧1)2

2. Section formula: Coordinates of a point P, which divides the line segment joining two

given points

A(𝑥1, 𝑦1, 𝑧1) and B(𝑥2, 𝑦2, 𝑧2) in the ratio m:n

(i). internally,

are P(𝑚𝑥2+𝑛𝑥1

𝑚+𝑛,𝑚𝑦2+𝑛𝑦1

𝑚+𝑛,𝑚𝑧2+𝑛𝑧1

𝑚+𝑛) ,

the Coordinates of a point Q divides the line segment joining two given points in the

ratio m:n

(ii). externally are Q(𝑚𝑥2−𝑛𝑥1

𝑚−𝑛,𝑚𝑦2−𝑛𝑦1

𝑚−𝑛,𝑚𝑧2−𝑛𝑧1

𝑚−𝑛)

(iii).coordinate of mid-point are R(𝑥2+𝑥1

2,𝑦2+𝑦1

2,𝑧2+𝑧1

2)

3. Direction cosines of a line :

(i).The direction of a line OP is determined by the angles 𝛼, 𝛽, 𝛾 which makes with OX,

OY,OZ respectively. These angles are called the direction angles and their cosines are

called the direction cosines.

(ii).Direction cosines of a line are denoted by l, m, n; l =cos 𝛼 ,m= cos 𝛽, 𝑛 = cos 𝛾

(iii).Sum of the squares of direction cosines of a line is always 1.

l2+m2+n2=1 i.e cos2𝛼 + cos2𝛽 +cos2𝛾 =1

4. Direction ratio of a line :(i)Numbers proportional to the direction cosines of a line

are called direction ratios of a line .If a ,b ,c, are , direction ratios of a line, then𝑙

𝑎=

𝑚

𝑏=

𝑛

𝑐.

(ii). If a ,b ,c, are , direction ratios of a line , then the direction cosines are

±𝑎

√𝑎2+𝑏2+𝑐2 , ±

𝑏

√𝑎2+𝑏2+𝑐2±

𝑐

√𝑎2+𝑏2+𝑐2

(𝑖𝑖𝑖). Direction ratio of a line AB passing through the points A(x1,y1,z1)and B (x2,y2,z2)

are𝑥2 − 𝑥1 , 𝑦2 − 𝑦1 , 𝑧2 − 𝑧1

5. STRAIGHT LINE:. (i). Vector equation of a Line passing through a point 𝑎 and along

the direction�� , :𝑟 =𝑎 + 𝜇�� ,

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54

(ii).Cartesian equation of a Line:𝑥−𝑥1

𝑎=

𝑦−𝑦1

𝑏=

𝑧−𝑧1

𝑐. Where (x1,y1,z1)is the given point

and its direction ratios are a,b,c.

6. (i). Vector equation of a Line passing through two points, with position vectors

𝑎 𝑎𝑛𝑑 �� 𝑟 =𝑎 + 𝝁(�� - 𝑎 )

(ii). ).Cartesian equation of a Line:𝑥−𝑥1

𝑥2−𝑥1=

𝑦−𝑦1

𝑦2−𝑦1=

𝑧−𝑧1

𝑧2−𝑧1, two points are (x1,y1) and

(x2,y2).

7. ANGLE between two lines (i). Vector equations: 𝑟 =𝑎1 + 𝝀𝑏1 and 𝑟 =𝑎2 + 𝝁𝑏2

,

(ii) ).Cartesian equations: If lines are 𝑥−𝑥1

𝑎1=

𝑦−𝑦1

𝑏1=

𝑧−𝑧1

𝑐1 ,

𝑥−𝑥2

𝑎2=

𝑦−𝑦2

𝑏2=

𝑧−𝑧2

𝑐2

cos 𝜃 =𝑏1 . 𝑏2

|𝑏1 |. |𝑏2 |

(iii). If two lines are perpendicular, then �� 1.�� 2 = 0, i.e. 𝑎1𝑎2 + 𝑏1𝑏2 + 𝑐1𝑐2 = 0

(iv) . If two lines are parallel, then �� 1 = 𝑡 �� 2, where t is a scalar. OR �� 1 × �� 2 =0, OR 𝑎1

𝑎2=

𝑏1

𝑏2=

𝑐1

𝑐2

(v).If 𝜃 𝑖𝑠 𝑡ℎ𝑒 angle between two lines with direction cosines ,l1,m1,n1 and l2,m2,n2 then

(a).cos 𝜃 =l1l2+m1m2+n1n2 (b). if the lines are parallel, then 𝑙1

𝑙2=

𝑚1

𝑚2=

𝑛1

𝑛2

(c). If the lines are perpendicular, then l1l2+m1m2+n1n2=0

8.(a).Shortest distance between two skew- lines:

(i).Vector equations: 𝑟 =𝑎1 + 𝝀𝑏1 , and::𝑟 =𝑎2 + 𝝁𝑏2

,

d=|(𝑎2 −𝑎1 ).(𝑏1 ×𝑏2 )

|𝑏1 ×𝑏2 ||.

If shortest distance is zero, then lines intersect and line intersects in space if they are

coplanar. Hence if above lines are coplanar

If(𝑎2 − 𝑎1 ). (𝑏1 × 𝑏2 ) = 0

(ii). Cartesian equations: 𝑥−𝑥1

𝑎1=

𝑦−𝑦1

𝑏1=

𝑧−𝑧1

𝑐1,𝑥−𝑥2

𝑎2=

𝑦−𝑦2

𝑏2=

𝑧−𝑧2

𝑐2

9.If shortest distance is zero, then lines intersect and line intersects in space if they are

coplanar. Hence if above lines are coplanar

|

𝑥2 − 𝑥1 𝑦2 − 𝑦1 𝑧2 − 𝑧1

𝑎1 𝑏1 𝑐1

𝑎2 𝑏2 𝑐2

| = 0

(b).Shortest distance between two parallel lines: If two lines are parallel, then they are

coplanar.

Let the lines be : 𝑟 =𝑎1 + 𝝀�� , and:𝑟 =𝑎2 + 𝝁�� ,

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55

D=|�� ×(𝑎2 −𝑎1 )

|�� ||

10.General equation of a plane in vector form :- It is given by 𝑟 . �� + 𝑑 = 0 , �� is a vector

normal to plane.

11.General equation of a plane in Cartesian form :- 𝒂𝒙 + 𝒃𝒚 + 𝒄𝒛 + 𝒅 = 𝟎 , Where a,b,c are

directionratios of normal to the plane.

12.General equation of a plane passing through a point :- if position vector of given point is

𝑎 then equation is given by (𝑟 − 𝑎 ). �� = 0,�� is a vector perpendicular tothe plane.

13.General equation of a plane passing through a point :- if coordinates of point are(𝑥, 𝑦, 𝑧)

then equation is𝑎(𝑥 − 𝑥1) + 𝑏(𝑦 − 𝑦1 ) + 𝑐(𝑧 − 𝑧1) = 0, a,b,care direction ratios of a line

perpendicular to the plane.

14.Intercept form of equation of a plane :-General equation of a plane which cuts off

intercepts a, b and c on x-axis, y-axis, z-axis respectively is 𝑥

𝑎+

𝑦

𝑏+

𝑧

𝑐= 1.

15. Equation of a plane in normal form:- 𝑟 . �� = p,where �� is a unit vector along perpendicular

from origin and ‛p’ is distance of plane from origin.p is always positive.

16.Equation of a plane in normal form :- It is given by 𝑙𝑥 + 𝑚𝑦 + 𝑛𝑧 = 𝑝, where 𝑙, 𝑚, 𝑛 are

direction cosines of perpendicular from origin and ‛p’ is distance of plane from origin. p is

always positive.

17.Equation of a plane passing through three non-collinear points :- If 𝑎 , �� , 𝑐 are the position

vectors of three non-collinear points, then equation of a plane through three points is given by –

(𝑟 − 𝑎 ). {(�� − 𝑎 ) × (𝑐 − 𝑎 )} = 0.

18. Equation of a plane passing through three non-collinear points(Cartesian system) :- If

plane passing through points (𝒙𝟏, 𝒚𝟏, 𝒛𝟏 ) , (𝒙𝟐, 𝒚𝟐, 𝒛𝟐) and (𝒙𝟑, 𝒚𝟑, 𝒛𝟑 ) then equation is-

|

(𝑥 − 𝑥1) (𝑦 − 𝑦1) (𝑧 − 𝑧1)

(𝑥2 − 𝑥1) (𝑦2 − 𝑦1) (𝑧2 − 𝑧1)(𝑥3 − 𝑥1) (𝑦3 − 𝑦1) (𝑧3 − 𝑧1)

| = 0

19.If 𝜽 is angle between two planes �� . 𝒏𝟏 + 𝒅𝟏 = 𝟎 and �� . 𝒏𝟐 + 𝒅𝟐 = 𝟎thencos 𝜃 = 𝑛1 .𝑛2

|𝑛1 ||𝑛2 |

(i) If planes are perpendicular, then 𝑛1 . 𝑛2 =0

(ii) If planes are parallel, then 𝑛1 × 𝑛2 = 0 or 𝑛1 = 𝝀𝑛2 , 𝝀is a scalar.

20. If 𝜽 is angle between two planes 𝒂𝟏 𝒙 + 𝒃𝟏𝒚 + 𝒄𝟏𝒛 + 𝒅𝟏 = 𝟎 𝒂𝒏𝒅 𝒂𝟐𝒙 + 𝒃𝟐𝒚 + 𝒄𝟐𝒛 +

𝒅𝟐 = 𝟎

Then 𝐜𝐨𝐬 𝜽 =𝒂𝟏𝒂𝟐+𝒃𝟏𝒃𝟐+𝒄𝟏𝒄𝟐

√(𝒂𝟏𝟐+𝒃𝟏

𝟐+𝒄𝟏𝟐)( 𝒂𝟐

𝟐+𝒃𝟐𝟐+𝒄𝟐

𝟐 )

(i) If planes are perpendicular ,then 𝑎1𝑎2 + 𝑏1𝑏2 + 𝑐1𝑐2 = 0

(ii) If planes are parallel , then 𝑎1

𝑎2=

𝑏1

𝑏2=

𝑐1

𝑐2

21. If 𝜽 is angle between line �� = �� + 𝝀�� and the plane �� . �� + 𝒅 = 𝟎 ,then 𝐬𝐢𝐧𝜽 =

�� .��

|�� |.|�� |

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(i)If line is parallel to plane ,then �� . �� =0 and

(ii)If line is perpendicular to plane , then �� × �� = 0 or �� = 𝑡�� ,t is a scalar.

22 . If 𝜽 is angle between line 𝒙−𝒙𝟏

𝒂𝟏=

𝒚−𝒚𝟏

𝒃𝟏=

𝒛−𝒛𝟏

𝒄𝟏 and the plane 𝒂𝒙 + 𝒃𝒚 + 𝒄𝒛 + 𝒅 = 𝟎

,then

𝐬𝐢𝐧 𝜽 =𝒂𝒂𝟏 + 𝒃𝒃𝟏 + 𝒄𝒄𝟏

√(𝒂𝟏𝟐 + 𝒃𝟏

𝟐 + 𝒄𝟏𝟐)( 𝒂𝟐 + 𝒃𝟐 + 𝒄𝟐 )

(i)If line is parallel to the plane ,then 𝑎𝑎1 + 𝑏𝑏1 + 𝑐𝑐1 = 0

(ii)If line is perpendicular to the plane, then 𝑎

𝑎1=

𝑏

𝑏1=

𝑐

𝑐1

23. General equation of a plane parallel to the plane𝑟 . �� + 𝑑 = 0 𝑖𝑠𝑟 . �� + 𝜆 = 0, where 𝜆 is a

constant and can be calculated from given condition.

24. General equation of a plane parallel to the plane ax+by+cz+d=0 is ax+by+cz+ 𝜆 = 0, where

𝜆 is a constant and can be calculated from given condition.

25. General equation of a plane (vector form) passing through the line of the intersection of

planes

𝑟 . 𝑛1 + 𝑑1 = 0 and 𝑟 . 𝑛2 + 𝜆𝑑2 = 0 is 𝑟 . (�� 1 + 𝜆�� 2) + (𝑑1 + 𝜆𝑑2) = 0 , where 𝜆 is a

constant and can be calculated from given condition.

26. General equation of a plane(Cartesian form) passing through the line of the intersection of

planes a1x+b1y+c1z+d1=0 and a2x+b2y+c2z+d2=0 is( a1x+b1y+c1z+d1)+𝜆 (a2x+b2y+c2z+d2 )=0 ,

where 𝜆 is a constant and can be calculated from given condition.

27. Distance of a plane(vector form) 𝑟 . �� + 𝑑 = 0

, from a point with position vector 𝑎 , is|�� .�� +𝑑

|�� ||.

28. Distance of a plane(Cartesian form) ax+by+cz+d=0, , from a point (x1,y1,z1)

is|𝑎𝑥1+𝑏𝑦1+𝑐𝑧1+𝑑

√𝑎2+𝑏2+𝑐2|.

FLOW CHART

(a) To find shortest distance between two skew lines

1 Find 𝑎 1, 𝑎 2, �� 1&�� 2

2 Find 𝑎 2 − 𝑎 1

3 Find 𝑏 1 × �� 2

4 Find |�� 1 × �� 2|

5 (𝑏1 × 𝑏2

). (𝑎2 − 𝑎1 )

6 Distance = |(�� 1×�� 2).(�� 2−�� 1)

|�� 1×�� 2||

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57

(b) To find Coordinates of foot of perpendicular from origin to the plane:

1. Direction ratio of any line perpendicular to the given plane.

2. Equation of the line through origin and perpendicular to the given plane.

3. Coordinates of any point(P) on the line.

4. Find the value of λ, by putting the coordinates of P in the plane.

5. Find required coordinates of foot of perpendicular

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58

(c) To find coordinates of image of a point in the plane

1. Direction ratios of any line (PP’) perpendicular to the given plane.

2. Equation of the line PP’ through a given point (P) and perpendicular the given plane.

3. Coordinates of any point on the plane (say P’)

4. Coordinates of midpoint (M) of PP’

5. Find the value of r ,as M will satisfy the plane

6. Then find coordinates of P’ Images of point P

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59

QUESTIONS ON 3-D.

LEVEL-1

6) The equation of a line is given by4−𝑥

2=

𝑦+3

3=

𝑧+2

6 . Write the direction cosines of a line

parallel to given line. Ans:<𝟐

𝟕,𝟑

𝟕,𝟔

𝟕>

7) Find the equation of the plane passing through the point (-1,3, 2) and perpendicular to the

planes x +2y +3z = 5 and3x +3y + z = 0.

8) Find the co-ordinates of the point where the line through (3, 4, 1) and (5, 1, 6) crosses the xy-

plane .

9) Find the shortest distance between the following pair of lines :

𝑥 − 1

2=

𝑦 − 2

3=

𝑧 − 3

4;

𝑥 − 2

3=

𝑦 − 4

4=

𝑧 − 5

5 𝑨𝒏𝒔:

1

√6

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60

LEVEL-2

7. Find the Coordinate of foot of the perpendicular from origin to the plane 3x+4y-5z=7

8. Q 1: Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the

origin, then 2 2 2 2

1 1 1 1

a b c p

9. Find the distance of the point (−1,−5,−10) from the point of intersection of the line

𝑟 = (2𝑖 − 𝑗 + 2��) + 𝜆(3 𝑖 + 4𝑗 + 2�� ) and the plane 𝑟 . (𝑖 − 𝑗 + ��) =

5. 𝐴𝑁𝑆: 13 𝑢𝑛𝑖𝑡𝑠.

10. Find the distance between the point P(6,5,9) and the plane determined by the points

A(3, −1,2),B(5,2,4) and C(−1,−1,6). ANS:6

√34units.

11. Find the equation of the plane passing through the points (3,4,1), (0,1,0) and is parallel to

line 𝑥+3

2=

𝑦−3

7=

𝑧−2

5Ans:𝟖𝒙 − 𝟏𝟑𝒚 + 𝟏𝟓𝒛 + 𝟏𝟑 = 𝟎

12. Find the values of p so that the lines :1−𝑥

3=

7𝑦−14

2𝑝=

𝑧−3

2 and

7−7𝑥

3𝑝=

𝑦−5

1=

6−𝑧

5 are at

right angles Ans: 𝟕𝟎

𝟏𝟏

13. Find the shortest distance between the lines 𝑥 + 1 = 2𝑦 = −12𝑧 and 𝑥 = 𝑦 + 2 = 6𝑧 −

6 Ans: 2

14. Find the shortest distance between the following pair of lines : 𝑥−1

2=

𝑦+1

3= 𝑧 ;

𝑥+1

5=

𝑦−2

1, 𝑧 = 2Ans:

𝟑

√𝟓𝟗

LEVEL-3

10. Find the equation of the plane which is perpendicular to the plane 08635 zyx

and which contains the line of intersection of the planes

0520432 zyxandzyx .

0173501551: zyxAns

11. Find the equation of a plane which is at a distance of 7 units from the origin and which is

normal to the vector .653

kji Ans.:- 0707)653.(

kjir

12. Find a unit vector perpendicular to the plane of the triangle ABC, where the coordinates

of its vertices are A (3, −1,2), B (1, −1,−3) and C (4, −3,1).

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61

Ans: 𝟏

√𝟏𝟔𝟓(𝟏𝟎�� + 𝟕𝒋 − 𝟒��)

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

1=

𝑦−1

2=

𝑧−2

3.

14. Find the image of the point (1,3,4) in the plane 𝑥 − 𝑦 + 𝑧 = 5.

15. Find the vector and the Cartesian equations of the plane passing through the intersection

of the planes 𝑟 . (𝑖 + 𝑗 + ��) = 6 and 𝑟 . (2𝑖 + 3𝑗 + 4��) = −5 and the point(1, 1, 1).

Ans: 𝟐𝟎𝒙 + 𝟐𝟑𝒚 + 𝟐𝟔𝒛 − 𝟔𝟗 = 𝟎�� . (𝟐𝟎�� + 𝟐𝟑𝒋 + 𝟐𝟔��) = 𝟔𝟗

16. Find the shortest distance between the lines whose vector equations are:

𝑟 = (1 − 𝑡)𝑖 + (𝑡 − 2)𝑗 + (3 − 2𝑡)�� 𝑎𝑛𝑑 𝑟 = (𝑠 + 1)𝑖 + (2𝑠 − 1)𝑗 − (2𝑠 +

1)��Ans:𝟖√𝟐𝟗

𝟐𝟗

17. . Find the distance of the point (1, −2,3)𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 plane x-y+z=5 measured ‖ to the

line𝑥+1

2=

𝑦+3

3=

𝑧+1

−6. ANS: other point (

9

7, −

11

7,15

7) , 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 1 𝑢𝑛𝑖𝑡

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Class XII Chapter 11 – Three Dimensional Geometry Mathscbseocean.weebly.com/uploads/2/8/1/5/28152469/cmn1… · · 2017-04-18Class XII Chapter 11 – Three Dimensional Geometry