Edexcel - Kumarmaths · Mathematics 4 Vectors Edited by: K V Kumaran . kumarmaths.weebly.com 2 Vectors By the end of this unit you should be able to find: a unit vector in the direction

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Edexcel Core

Mathematics 4 Vectors

Edited by: K V Kumaran


Vectors

By the end of this unit you should be able to find:

a unit vector in the direction of a.

the distance between two points (x1, y1, z1) and (x2, y2, z2) by

d2 = (x1 – x2)2 + (y1 – y2)2 + (z1 – z2)2

the vector equation of a line in the form r = a + μb

the point of intersection of two lines

By the end of this unit you should also know that :

If a and b are two vectors with an angle θ between them then:

cos. baba

Where a.b is the scalar product of a and b.

If a and b are two non zero vectors then if a.b = 0 then the two vectors are

perpendicular.


1. Position vectors:

The coordinates of the point, A (2, -1, 4) the position vector of A is

𝑂𝐴⃗⃗⃗⃗ ⃗ = 2𝑖 − 𝑗 + 4𝑘 As the column vector form is(2

−14

).

2. Direction vectors:

A vectors between two points, A(2, -1, 4) and B(-3, 0, 2), the direction

vector of 𝐴𝐵⃗⃗⃗⃗ ⃗= (the position vector of B) – (the position vector of A),

𝐴𝐵⃗⃗⃗⃗ ⃗ = −5𝑖 + 𝑗 − 2𝑘 As the column vector form is(−5+1−2

).

3. Magnitude of vectors:

The length of the vector, the magnitude of the vector of AB is

|𝐴𝐵| = √(−5)2 + (1)2 + (−2)2 = √30

4. Parallel vectors:

Two vectors are parallel if they are multiples of each other.

The vector AB is parallel to a vector CD, 𝐶𝐷⃗⃗⃗⃗ ⃗ = λ *𝐴𝐵⃗⃗⃗⃗ ⃗,

𝐶𝐷⃗⃗⃗⃗ ⃗ = 𝜆(−5𝑖 + 𝑗 − 2𝑘)

5. Vector equation of a straight line:

r = position vector + 𝜆 direction vector

A point on the line either the direction vector between two points on the

line or the vector parallel to the line

a) The vector equation of the straight line, l1, passes through the

points A and B is

r = (+2−1+4

)+ 𝜆 (−5+1−2

)

Any coordinates of the point lies on the line can be given as

(2 − 5𝜆, −1 + 𝜆, 4 − 2𝜆)


b) The vector equation of the straight line, l2, passes through the

points B and parallel to the vector −3𝑖 + 2𝑗 − 𝑘 is

r = (−30

+2)+ µ (

−3+2−1

)

6. Interception between the two vector equations:

Equate the x and y components of both lines and solve for 𝜆 and µ. Put

these values back into the original lines and compare the z coordinates, if

they are the same then they intercept, if different then skew.

7. The angle between two vectors: Scaler product – “dot product”

a= x1i + y1j+ z1k, b=x2i +y2j +z2k, a.b=x1x2 +y1y2 +z1z2, the angle, α, between

the vectors a and b is, 𝑐𝑜𝑠𝛼 =𝑎.𝑏

|𝑎||𝑏|,

the two vectors are at perpendicular ⇔ the dot product is zero.

8. The angle between two straight lines:

Dot product between the direction victors, the angle between the line l1

and l2 is 𝑐𝑜𝑠𝛼 =−5∗−3+1∗2−2∗−1

√30∗ √12

9. Finding a point on the line which is closest to an another given point

Draw a sketch, let the point on the line closest to the point, C, to be P. So

the coordinates of P can be found in terms of𝜆, we require 𝐶𝑃⃗⃗⃗⃗ ⃗ to be

perpendicular to the line if it’s the closest point.

Thus 𝐶𝑃⃗⃗⃗⃗ ⃗. 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑣𝑒𝑐𝑡𝑜𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 = 0. , find the value of 𝜆 and

the coordinates of the point P.


Example 1

The points A and B have coordinates (-3, -3, -3) and (0, 0, 2) respectively. The line

l1 which passes through A has equation

3 0

r 3 t 1

3 0.6

Show that AB is perpendicular to l1

Exam papers have only recently started to use this type of notation for vector

questions but the same principles apply.

To find AB we simply need to subtract the two position vectors

AB =

3

3

5

The second bracket in the equation of the line is the direction component of the

line. The direction component is used time and again in vectors questions. If the

line is perpendicular to AB then the scalar product of the two should equal zero.

3

3

5

.

0

1

0.6

= 3 × 0 + 3 × 1 – 5 × 0.6 = 0 as required.

This is a very regularly asked question and it should be a good opportunity to pick

up some marks.


Example 2

The points A and B have coordinates (2, 3, -3) and (1, -2, -2) respectively. The line

l1 which passes through A has equation

2 1

r 3 t 1

3 4

The line l2 which passes through B has equation

1 2

r 2 s 3

2 8.5

Show that the lines l1 and l2 intersect, and find the coordinates of the point of

intersection.

Another standard question as all you have to do is equate the i,j,k components

and solve the simultaneous equations.

Therefore:

2 + t = 1 + 2s (1)

3 – t = -2 – 3s (2)

-3 – 4t = -2 - 8.5s (3)


Adding equations (1) and (2) gives:

5 = -1 – s s = -6 hence t = -13

Substituting the value of t into l1 gives:

(-11, 16, 49) as the point of intersection.

Check that substituting the value of s into l2 gives the same answer.

The next example involves more concepts but hopefully it is manageable.

Example 3

Relative to a fixed point O, the points A and B have position vectors

i + 2j -7k and 8i + 16j +7k respectively.

a) Find the vector AB .

b) Find the cosine of OAB

c) Show that for all values of µ, the point P with position vector

µi + 2µj + (2µ -9)k lies on the line through A and B.

d) Find the value of µ for which OP is perpendicular to AB.

e) Hence find the foot of the perpendicular from O to AB.

a) Once again subtract the two position vectors.

AB = 7i + 14j +14k

b) Find the cosine of OAB

The use of a diagram may help here.


We can use the scalar product to find the angle between two vectors BUT both of

the vectors must be pointing towards the angle or away from it. Therefore in this

case we will use AO

Scalar Product a.b = |a||b|cosθ

Where a = AO and b = AB

AO = -i - 2j +7k AB = 7i + 14j +14k

a.b = -7 -28 + 98 = 63 |a| = √54 |b| = √441

Therefore

a.bcos

a b

9 3cos

7 3 621 54

6cos

42

410

5421

63

.cos

cos


c) Show that for all values of µ, the point P with position vector

µi + 2µj + (2µ -9)k lies on the line through A and B.

When µ = 1 the point P has position vector i + 2j -7k (ie at A)

When µ = 8 the point P has position vector 8i + 16j +7k (ie at B)

Therefore for any other value of µ the point P will lie on the line through A and B

d) Find the value of µ for which OP is perpendicular to AB

The scalar product of OP and AB is equal to zero at the point where the two

vectors are perpendicular.

(7i + 14j +14k) . (µi + 2µj + (2µ -9)k) = 0

7µ + 28µ + 28µ - 126 = 0

µ = 2

e) Hence find the foot of the perpendicular from O to AB.

We have just shown in part (d) that when µ = 2 OP is perpendicular to AB and

that P is on the line between A and B.

Therefore we only need to substitute µ = 2 into the position vector for P to find

the foot of the perpendicular.

µi + 2µj + (2µ -9)k

2i + 4j -5k


Example 4

The line l1 has equation

7 1

r 10 t 1

14 1

where t is a parameter. The point A has coordinates (2, 5, a), where a is a

constant. The point B has coordinates (b, 12, 12), where b is a constant. The

points A and B lie on the line l1.

a) Find the values of a and b.

Given that the point O is the origin, and that the point P lies on l1 such that OP is

perpendicular to l1,

b) find the coordinates of P.

c) Hence find the distance OP, giving your answer in surd form.

a) Find the values of a and b

The x coord of A is 2 therefore:

7 + t = 2 t = -5

So the z coord must be:

14 - - 5 = 19 a = 19

The z coord of B is 12 therefore:

14 - t = 12 t = 2

So the x coord must be:

7 + 2 = 9 b = 9


b) find the coordinates of P

The scalar product of OP and the direction component of the line l1 will equal zero

if the two are perpendicular. Therefore:

OP = (7 + t, 10 + t, 14 – t) Direction component = (1, 1, -1)

(7 + t, 10 + t, 14 – t) . (1, 1, -1) = 0

7 + t + 10 + t – 14 + t = 0

t = -1

Substituting this value of t into the equation of the line gives the coordinates of P

as:

(6, 9, 15)

c) Hence find the distance OP, giving your answer in surd form.

Using Pythagoras in three dimensions:

OP = √(62 + 92 + 152) = √342


Past paper questions vectors

1. The line l1 has vector equation

r =

2

1

3

+

4

1

1

and the line l2 has vector equation

r =

2

4

0

+

0

1

1

,

where and are parameters.

The lines l1 and l2 intersect at the point B and the acute angle between l1 and l2 is .

(a) Find the coordinates of B.

(4)

(b) Find the value of cos , giving your answer as a simplified fraction.

(4)

The point A, which lies on l1, has position vector a = 3i + j + 2k.

The point C, which lies on l2, has position vector c = 5i – j – 2k.

The point D is such that ABCD is a parallelogram.

(c) Show that AB = BC .

(3)

(d) Find the position vector of the point D.

(2)

(C4, June 2005 Q7)


r = 8i + 12j + 14k + (i + j – k),

where is a parameter.

The point A has coordinates (4, 8, a), where a is a constant. The point B has coordinates (b, 13, 13), where

b is a constant. Points A and B lie on the line l1.

(a) Find the values of a and b.

(3)

Given that the point O is the origin, and that the point P lies on l1 such that OP is perpendicular to l1,

(b) find the coordinates of P. (5)

(b) Hence find the distance OP, giving your answer as a simplified surd.

(2)

(C4, Jan 2006 Q6)


3. The point A, with coordinates (0, a, b) lies on the line l1, which has equation

r = 6i + 19j – k + (i + 4j – 2k).

(a) Find the values of a and b.

(3)

The point P lies on l1 and is such that OP is perpendicular to l1, where O is the origin.

(b) Find the position vector of point P.

(6)

Given that B has coordinates (5, 15, 1),

(c) show that the points A, P and B are collinear and find the ratio AP : PB.

(4)

(C4, June 2006 Q5)

4. The point A has position vector a = 2i + 2j + k and the point B has position vector b = i + j – 4k, relative to

an origin O.

(a) Find the position vector of the point C, with position vector c, given by c = a + b.

(1)

(b) Show that OACB is a rectangle, and find its exact area.

(6)

The diagonals of the rectangle, AB and OC, meet at the point D.

(c) Write down the position vector of the point D.

(1)

(d) Find the size of the angle ADC.

(6)

(C4, Jan 2007Q7)

5. The line l1 has equation r =

1

0

1

+

0

1

1

.

The line l2 has equation r =

6

3

1

+

1

1

2

.

(a) Show that l1 and l2 do not meet.

(4)

The point A is on l1 where = 1, and the point B is on l2 where = 2.

(b) Find the cosine of the acute angle between AB and l1.

(6)

(C4, June 2007 Q5)


6. The points A and B have position vectors 2i + 6j – k and 3i + 4j + k respectively.

The line 1l passes through the points A and B.

(a) Find the vector AB .

(2)

(b) Find a vector equation for the line 1l .

(2)

A second line 2l passes through the origin and is parallel to the vector i + k. The line

1l meets the line 2l

at the point C.

(c) Find the acute angle between 1l and

2l .

(3)

(d) Find the position vector of the point C.

(4)

(C4, Jan 2008 Q6)

7. With respect to a fixed origin O, the lines l1 and l2 are given by the equations

l1 : r = (–9i + 10k) + λ(2i + j – k)

l2 : r = (3i + j + 17k) + μ(3i – j + 5k)

where λ and μ are scalar parameters.

(a) Show that l1 and l2 meet and find the position vector of their point of intersection.

(6)

(b) Show that l1 and l2 are perpendicular to each other.

(2)

The point A has position vector 5i + 7j + 3k.

(c) Show that A lies on l1.

(1)

The point B is the image of A after reflection in the line l2.

(d) Find the position vector of B.

(3)

(C4, June 2008 Q6)


8. With respect to a fixed origin O the lines l1 and l2 are given by the equations

l1 : r =

17

2

11

+

4

1

2

l2 : r =

p

11

5

+ μ

2

2

q

where λ and μ are parameters and p and q are constants. Given that l1 and l2 are perpendicular,

(a) show that q = –3. (2)

Given further that l1 and l2 intersect, find

(b) the value of p, (6)

(c) the coordinates of the point of intersection. (2)

The point A lies on l1 and has position vector

13

3

9

. The point C lies on l2.

Given that a circle, with centre C, cuts the line l1 at the points A and B,

(d) find the position vector of B. (3)

(C4, Jan2009 Q4)

9. Relative to a fixed origin O, the point A has position vector (8i + 13j – 2k), the point B has position vector

(10i + 14j – 4k), and the point C has position vector (9i + 9j + 6k).

The line l passes through the points A and B.

(a) Find a vector equation for the line l.

(3)

(b) Find CB .

(2)

(c) Find the size of the acute angle between the line segment CB and the line l, giving your answer in

degrees to 1 decimal place.

(3)

(d) Find the shortest distance from the point C to the line l.

(3)

The point X lies on l. Given that the vector CX is perpendicular to l,

(e) find the area of the triangle CXB, giving your answer to 3 significant figures.

(3)

(C4, June 2009 Q7)



r =

1

4

6

+ λ

3

1

4

and the line l2 has vector equation

r =

1

4

6

+

1

4

3

where λ and μ are parameters.

The lines l1 and l2 intersect at the point A and the acute angle between l1 and l2 is θ.

(a) Write down the coordinates of A. (1)

(b) Find the value of cos θ. (3)

The point X lies on l1 where λ = 4.

(c) Find the coordinates of X. (1)

(d) Find the vector AX . (2)

(e) Hence, or otherwise, show that AX = 4√26. (2)

The point Y lies on l2. Given that the vector YX is perpendicular to l1,

(f) find the length of AY, giving your answer to 3 significant figures. (3)

(C4, Jan 2010 Q4)

11. The line 1l has equation r =

1

2

1

4

3

2

, where λ is a scalar parameter.

The line 2l has equation r =

2

0

5

3

9

0

, where is a scalar parameter.

Given that 1l and

2l meet at the point C, find

(a) the coordinates of C. (3)

The point A is the point on 1l where λ = 0 and the point B is the point on 2l where μ = –1.

(b) Find the size of the angle ACB. Give your answer in degrees to 2 decimal places.

(4)

(c) Hence, or otherwise, find the area of the triangle ABC.

(5)

(C4, June 2010 Q7)


12. Relative to a fixed origin O, the point A has position vector i − 3j + 2k and the point B has position vector

−2i + 2j − k. The points A and B lie on a straight line l.

(a) Find AB .

(2)

(b) Find a vector equation of l.

(2)

The point C has position vector 2i + pj − 4k with respect to O, where p is a constant.

Given that AC is perpendicular to l, find

(c) the value of p,

(4)

(d) the distance AC.

(2)

(C4, Jan2011 Q4)


l1: r =

2

3

6

+

3

2

1

, l2: r =

3

15

5

+ μ

1

3

2

,

where μ and are scalar parameters.

(a) Show that l1 and l2 meet and find the position vector of their point of intersection A.

(6)

(b) Find, to the nearest 0.1°, the acute angle between l1 and l2.

(3)

The point B has position vector

1

1

5

.

(c) Show that B lies on l1.

(1)

(d) Find the shortest distance from B to the line l2, giving your answer to 3 significant figures.

(4)

(C4, June 2011 Q6)


14. Relative to a fixed origin O, the point A has position vector (2i – j + 5k),

the point B has position vector (5i + 2j + 10k),

and the point D has position vector (–i + j + 4k).



(2)

(b) Find a vector equation for the line l.

(2)

(c) Show that the size of the angle BAD is 109°, to the nearest degree.

(4)

The points A, B and D, together with a point C, are the vertices of the parallelogram ABCD, where AB =

DC .

(d) Find the position vector of C.

(2)

(e) Find the area of the parallelogram ABCD, giving your answer to 3 significant figures.

(3)

(f) Find the shortest distance from the point D to the line l, giving your answer to 3 significant figures.

(2)

(C4, Jan2012 Q7)

15. Relative to a fixed origin O, the point A has position vector (10i + 2j + 3k), and the point B has position

vector (8i + 3j + 4k).



(2)

(b) Find a vector equation for the line l.

(2)

The point C has position vector (3i + 12j + 3k) .

The point P lies on l. Given that the vector CP is perpendicular to l,

(c) find the position vector of the point P.

(6)

(C4, June 2012 Q8)



l1 : r = (9i + 13j – 3k) + (i + 4j – 2k)

l2 : r = (2i – j + k) + (2i + j + k)

where and are scalar parameters.

(a) Given that l1 and l2 meet, find the position vector of their point of intersection.

(5)

(b) Find the acute angle between l1 and l2, giving your answer in degrees to 1 decimal place.

(3)

Given that the point A has position vector 4i + 16j – 3k and that the point P lies on l1 such that AP is

perpendicular to l1,

(c) find the exact coordinates of P.

(5)

(C4, Jan2013 Q7)

17. With respect to a fixed origin O, the line l has equation

13 2

8 2

1 1

r , where λ is a scalar parameter.

The point A lies on l and has coordinates (3, – 2, 6).

The point P has position vector (–pi + 2pk) relative to O, where p is a constant.

Given that vector PA is perpendicular to l,

(a) find the value of p.

(4)

Given also that B is a point on l such that <BPA = 45°,

(b) find the coordinates of the two possible positions of B.

(5)

(C4, June 2013 Q8)


18. Relative to a fixed origin O, the point A has position vector 21i – 17j + 6k and the point B has position

vector 25i – 14j + 18k.

The line l has vector equation

6

10 1

a

b c

r

where a, b and c are constants and λ is a parameter.

Given that the point A lies on the line l,

(a) find the value of a. (3)

Given also that the vector AB is perpendicular to l,

(b) find the values of b and c, (5)

(c) find the distance AB. (2)

The image of the point B after reflection in the line l is the point B´.

(d) Find the position vector of the point B´. (2)

(C4, June 2013_R Q6)

19. Relative to a fixed origin O, the point A has position vector

2

4

7

and the point B has position vector

1

3

8

.

The line l1 passes through the points A and B.

(a) Find the vector AB . (2)

(b) Hence find a vector equation for the line l1. (1)

The point P has position vector

0

2

3

.

Given that angle PBA is θ,

(c) show that 1

cos3

(3)

The line l2 passes through the point P and is parallel to the line l1.

(d) Find a vector equation for the line l2. (2)

The points C and D both lie on the line l2.

Given that AB = PC = DP and the x coordinate of C is positive,

(e) find the coordinates of C and the coordinates of D. (3)

(f) find the exact area of the trapezium ABCD, giving your answer as a simplified surd. (4)

(C4, June 2014 Q8)


20. With respect to a fixed origin, the point A with position vector i + 2j + 3k lies on the line l1 with equation

1 0

2 2

3 1

r , where λ is a scalar parameter,

and the point B with position vector 4i + pj + 3k, where p is a constant, lies on the line l2 with equation

7 3

0 5

7 4

r , where μ is a scalar parameter.

(a) Find the value of the constant p.

(1)

(b) Show that l1 and l2 intersect and find the position vector of their point of

intersection, C.

(4)

(c) Find the size of the angle ACB, giving your answer in degrees to 3 significant figures.

(3)

(d) Find the area of the triangle ABC, giving your answer to 3 significant figures.

(2)

(C4, June 2014_R Q6)


l1: r =

p

3

5

+

3

1

0

, l2: r =

2

5

8

+

5

4

3

,

where and are scalar parameters and p is a constant.

The lines l1 and l2 intersect at the point A.

(a) Find the coordinates of A.

(2)

(b) Find the value of the constant p.

(3)

(c) Find the acute angle between l1 and l2, giving your answer in degrees to 2 decimal places.

(3)

The point B lies on l2 where = 1.

(d) Find the shortest distance from the point B to the line l1, giving your answer to 3 significant figures.

(3)

(C4, June 2015 Q4)


22. With respect to a fixed origin O, the line l1 is given by the equation

r =

8 5

1 4

3 3

,

where μ is a scalar parameter.

The point A lies on l1 where μ = 1.

(a) Find the coordinates of A.

(1)

The point P has position vector

1

5

2

.

The line l2 passes through the point P and is parallel to the line l1.

(b) Write down a vector equation for the line l2.

(2)

(c) Find the exact value of the distance AP.

Give your answer in the form k 2 , where k is a constant to be determined.

(2)

The acute angle between AP and l2 is θ.

(d) Find the value of cos θ.

(3)

A point E lies on the line l2.

Given that AP = PE,

(e) find the area of triangle APE,

(2)

(f) find the coordinates of the two possible positions of E.

(5)

(C4, June 2016 Q8)



l1 : r =

4

28

4

æ

è

ççç

ö

ø

÷÷÷

+ 𝜆

-1

-5

1

æ

è

ççç

ö

ø

÷÷÷

, l2 : r =

5

3

1

æ

è

ççç

ö

ø

÷÷÷

+ μ

3

0

-4

æ

è

ççç

ö

ø

÷÷÷

where 𝜆 and μ are scalar parameters.

The lines l1 and l2 intersect at the point X.

(a) Find the coordinates of the point X.

(3)

(b) Find the size of the acute angle between l1 and l2, giving your answer in degrees to

2 decimal places.

(3)


2

18

6

æ

è

ççç

ö

ø

÷÷÷

(c) Find the distance AX, giving your answer as a surd in its simplest form.

(2)

The point Y lies on l2. Given that the vector YA

is perpendicular to the line l1

(d) find the distance YA, giving your answer to one decimal place.

(2)

The point B lies on l1 where | AX

| = 2| AB

|.

(e) Find the two possible position vectors of B.

(3)

(C4, June 2017 Q6)


l1 : r = (i + 5j + 5k) + λ(2i + j – k)

l2 : r = (2j + 12k) + μ(3i – j + 5k)


(a) Show that l1 and l2 meet and find the position vector of their point of intersection. (6)

(b) Show that l1 and l2 are perpendicular to each other. (2)

The point A, with position vector 5i + 7j + 3k, lies on l1.

The point B is the image of A after reflection in the line l2.

(c) Find the position vector of B. (3)

(C34, IAL Jan 2014 Q10)


25. Relative to a fixed origin O, the line l has vector equation

1 2

4 1

6 1

r

where λ is a scalar parameter.

Points A and B lie on the line l, where A has coordinates (1, a, 5)

and B has coordinates (b, –1, 3).

(a) Find the value of the constant a and the value of the constant b. (3)

(b) Find the vector AB .

(2) The point C has coordinates (4, –3, 2).

(c) Show that the size of the angle CAB is 30°. (3)

(d) Find the exact area of the triangle CAB, giving your answer in the form k√3, where k is a constant to

be determined. (2)

The point D lies on the line l so that the area of the triangle CAD is twice the area of the triangle CAB.

(e) Find the coordinates of the two possible positions of D. (4)

(C34, IAL June 2014 Q14)

26. With respect to a fixed origin O the lines l1 and l2 are given by the equations

l1 :

14 2

6 1

13 4

r l2 : 7 2

4 1

p q

r

where λ and 𝜇 are scalar parameters and p and q are constants.

Given that l1 and l2 are perpendicular,

(a) show that q = 3

(2)

Given further that l1 and l2 intersect at point X,

find

(b) the value of p,

(5)

(c) the coordinates of X.

(2)


6

2

3

.

Given that point B also lies on l1 and that AB = 2AX

(d) find the two possible position vectors of B. (3)

(C34, IAL Jan 2015 Q11)


27. (i) Relative to a fixed origin O, the line l1 is given by the equation

l1:

5 2

1 3

6 1

r where λ is a scalar parameter.

The point P lies on l1. Given that

OP

is perpendicular to l1, calculate the

coordinates of P.

(5)

(ii) Relative to a fixed origin O, the line l2 is given by the equation

l2:

4 5

3 3

12 4

r where 𝜇 is a scalar parameter.

The point A does not lie on l2. Given that the vector

OA

is parallel to the line l2

and

OA

= 2 units, calculate the possible position vectors of the point A.

(5)

(C34, IAL June 2015 Q12)


1

12 5

: 4 4

5 2

l

r = 2

2 0

: 2 6

0 3

l

r =


(a) Show that l1 and l2 meet, and find the position vector of their point of intersection A.

(6)

(b) Find, to the nearest 0.1°, the acute angle between l1 and l2

(3)

The point B has position vector

7

0

3

(c) Show that B lies on l1

(1)

(d) Find the shortest distance from B to the line l2, giving your answer to 3 significant

figures.

(4)

(C34, IAL Jan 2016 Q12)



1

7 1

: 4 1

9 4

l

r

2

6 5

: 7 4

3

l

b

r

where 𝜆 and μ are scalar parameters and b is a constant.

Given that l1 and l2 meet at the point X,

(a) show that b = –3 and find the coordinates of X. (5)

The point A lies on l1 and has coordinates (6, 3, 5)

The point B lies on l2 and has coordinates (14, 9, –9)

(b) Show that angle AXB = arccos 1

10

(4)

(c) Using the result obtained in part (b), find the exact area of triangle AXB.

Write your answer in the form p q where p and q are integers to be determined.

(3)

(C34, IAL June 2016 Q11)

30. ABCD is a parallelogram with AB parallel to DC and AD parallel to BC.

The position vectors of A, B, C, and D relative to a fixed origin O are a, b, c and d

respectively.

Given that

a = i + j – 2k, b = 3i – j + 6k, c = – i + 3j + 6k

(a) find the position vector d,

(3)

(b) find the angle between the sides AB and BC of the parallelogram,

(4)

(c) find the area of the parallelogram ABCD.

(2)

The point E lies on the line through the points C and D, so that D is the midpoint of CE.

(d) Use your answer to part (c) to find the area of the trapezium ABCE.

(2)

(C34, IAL Jan 2017 Q14)


31.

Figure 2 shows a sketch of a triangle ABC.

Given AB ®

= 2i + 3j – 2k and AC ®

= 5i – 6j + k,

(a) find the size of angle CAB, giving your answer in degrees to 2 decimal places,

(3)

(b) find the area of triangle ABC, giving your answer to 2 decimal places.

(2)

Using your answer to part (b), or otherwise,

(c) find the shortest distance from A to BC, giving your answer to 2 decimal places.

(3)

(C34, IAL June 2017 Q9)

Leave

blank

28

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E IN

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IS A

REA

D

O N

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RIT

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REA

D

O N

OT W

RIT

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9.

A

C

B

Figure 2

Figure 2 shows a sketch of a triangle ABC.

Given AB = 2i + 3j – 2k and AC = 5i – 6j + k,

(a) find the size of angle CAB, giving your answer in degrees to 2 decimal places,

(3)

(b) find the area of triangle ABC, giving your answer to 2 decimal places.

(2)

Using your answer to part (b), or otherwise,

(c) find the shortest distance from A to BC, giving your answer to 2 decimal places.

(3)

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Edexcel - Kumarmaths · Mathematics 4 Vectors Edited by: K V Kumaran . kumarmaths.weebly.com 2 Vectors By the end of this unit you should be able to find: a unit vector in the direction