Cambridge International AS A Level Mathematics Pure Mathematics 2 and 3 second edition

Points 10

In three dimensions, a point has three coordinates, usually called x, y and z.

z

2 This point is

1 (3, 4, 1).
–1
10.2 Vectors in three dimensions
–3 –2 –1 1 O 1 2 3 4y
2 –1 P

3

x

▲ Figure 10.5

The axes are conventionally arranged as shown in Figure 10.5, where the
point P is (3, 4, 1). Even on correctly drawn three-dimensional grids, it is
often hard to see the relationship between the points, lines and planes, so it is
seldom worth your while trying to plot points accurately.

The unit vectors i, j and k are used to describe vectors in three dimensions.

Equal vectors

The statement that two vectors a and b are equal means two things.

» The direction of a is the same as the direction of b.
» The magnitude of a is the same as the magnitude of b.
If the vectors are given in component form, each component of a equals the
corresponding component of b.

Position vectors

Saying the vector a is given by 3i + 4j + k tells you the components of the
vector, or equivalently its magnitude and direction. It does not tell you where
the vector is situated; indeed it could be anywhere.

All of the lines in Figure 10.6 represent the vector a.

a a
a a

k j
i

▲ Figure 10.6 237

10 There is, however, one special case which is an exception to the rule, that of

238 a vector which starts at the origin.This is called a position vector.Thus
⎛ 3⎞
the line joining the origin to the point P(3, 4, 1) is the position vector ⎜ 4⎟
⎜⎝ 1 ⎠⎟
or 3i + 4j + k.

Another way of expressing this is to say that the point P(3, 4, 1) has the

10 VECTORS ⎛ 3⎞
position vector ⎜ 4⎟ .

⎝⎜ 1 ⎠⎟

Displacement vectors

Vectors can also be used to represent displacement. For example, start at

A and walk 5 km east and then 12 km north to point B can be represented

by the vector AB = ⎛ 5⎞ . Note that magnitude of displacement is not the
⎜⎝ 12⎟⎠

same as distance – the distance travelled is 5 km + 12 km = 17 km; whereas the

magnitude of displacement is 52 + 122 = 13 km.

Example 10.2 Points L, M and N have coordinates (4, 3), (−2, −1) and (2, 2).
(i) Write down, in component form, the position vector of L and the vector

M⎯→N.
(ii) What do your answers to part (i) tell you about the lines OL and MN?

Solution 43⎠⎞ .

(i) The position vector of L is O⎯→L = ⎛
The vector M⎯→N is also ⎛ 4⎞ ⎝
⎝ 3⎠ (see Figure 10.7).

y

4

3L

2N

1

–2 –1 O 1 2 3 4x

M –1

▲ Figure 10.7

(ii) Since O⎯→L = M⎯→N, lines OL and MN are parallel and equal in length.

Note 10

A line joining two points, like MN in Figure 10.7, is often called a line
segment, meaning that it is just that particular part of the infinite straight

line that passes through those two points.

The vector M⎯→N is an example of a displacement vector. Its length represents 10.2 Vectors in three dimensions
the magnitude of the displacement when you move from M to N.

The length of a vector

In two dimensions, the use of Pythagoras’ theorem leads to the result that a

vector a1i + a2 j has length | a | given by

|a|= a12 + a 2 .
2

❯ Show that the length of the three-dimensional vector a1i + a2 j + a3k ?
is given by CP

|a|= a12 + a 2 + a 2 .
2 3

Example 10.3 Find the magnitude of the vector a = ⎜⎜⎛⎝−253⎞⎟⎟⎠.

Solution

| a | = 22 + (−5)2 + 32
= 4 + 25 + 9
= 38
= 6.16 (to 2 d.p.)

Exercise 10A 1 Express the following vectors in component form.

(i) y (ii) y

3 a x 3 x
2 1 23 4 2
1
b
–2 –1 0 1
–1
–2 –1 0 1 2 3
–1

239

10 (iii) y (iv) y

240 3 3
2c 2d
1 1

0 1234 x 0 1234 x

10 VECTORS 2 Draw diagrams to show each of these vectors and find the
magnitude and direction.

(i) 2i + 3j (ii) ⎛ 3⎞ (iii) ⎛ –4⎞
⎝ –2⎠ ⎝ –4⎠

(iv) −i + 2j (v) 3i − 4j

3 Find the magnitude of these vectors.

⎛ 1⎞ ⎛ 4⎞ (iii) 2i + 4j + 2k
(i) ⎜ –2⎟ (ii) ⎜ 0⎟

⎜⎝ 3⎟⎠ ⎝⎜ −2⎠⎟

(iv) i + j − 3k ⎛ 6⎞ (vi) i − 2k
(v) ⎜ –2⎟

⎜⎝ −3⎟⎠

4 Write, in component form, the vectors represented by the line
segments joining the following points.

(i) (2, 3) to (4, 1) (ii) (4, 0) to (6, 0)

(iii) (0, 0) to (0, −4) (iv) (0, −4) to (0, 0)

(v) (0, 0, 0) to (0, 0, 5) (vi) (0,0,0) to (−1,−2,3)

(vii) (−1, −2 , 3) to (0, 0, 0) (viii) (0, 2, 0) to (4, 0, 4)

(ix) (1, 2, 3) to (3, 2, 1) (x) (4,−5,0) to (−4,5,1)

PS 5 The points A, B and C have coordinates (2, 3), (0, 4) and (−2, 1).

(i) Write down the position vectors of A and C.

(ii) Write down the vectors of the line segments joining AB
and CB.

(iii) What do your answers to parts (i) and (ii) tell you about

(a) AB and OC

(b) CB and OA?

(iv) Describe the quadrilateral OABC.

10.3 Vector calculations

Multiplying a vector by a scalar

When a vector is multiplied by a number (a scalar) its length is
altered but its direction remains the same.

The vector 2a in Figure 10.8 is twice as long as the vector a but in the same 10
direction.

a 2a

▲ Figure 10.8 10.3 Vector calculations

When the vector is in component form, each component is multiplied by the
number. For example:

2 ×(3i − 5j + k) = 6i − 10j + 2k
⎛ 3⎞ ⎛ 6⎞

2 × ⎜⎜⎝–51⎠⎟⎟ = ⎜⎝⎜–120⎠⎟⎟ .

The negative of a vector

In Figure 10.9 the vector −a has the same length as the vector a but the
opposite direction.

a –a

▲ Figure 10.9

When a is given in component form, the components of −a are the same as
those for a but with their signs reversed. So

⎛ 23⎞ ⎛–23⎞
–⎜ 0⎟ ⎜ 0⎟
⎝⎜–11⎟⎠ = ⎝⎜ +11⎠⎟

Adding vectors

When vectors are given in component form, they can be added component
by component.This process can be seen geometrically by drawing them on
graph paper, as in Example 10.4 on the next page.

241

10 Example 10.4 Add the vectors 2i − 3j and 3i + 5j.

Solution
2i − 3j + 3i + 5j = 5i + 2j

10 VECTORS 5i + 2j

2i 3i + 5j 5j

–3j

2i – 3j

3i

▲ Figure 10.10

The sum of two (or more) vectors is called the resultant and is usually
indicated by being marked with two arrowheads.
Adding vectors is like adding the legs of a journey to find its overall outcome
(see Figure 10.11).

resultant

leg 1

leg 3

leg 2

▲ Figure 10.11

When vectors are given in magnitude−direction form, you can find their
resultant by making a scale drawing, as in Figure 10.11. If, however, you
need to calculate their resultant, it is usually easiest to convert the vectors
into component form, add component by component, and then convert the
answer back to magnitude−direction form.

Subtracting vectors

Subtracting one vector from another is the same as adding the negative of the
vector.

242

Example 10.5 Two vectors a and b are given by 10
a = 2i + 3j b = −i + 2j.
10.3 Vector calculations
(i) Find a − b.
(ii) Draw diagrams showing a, b, a − b.

Solution
(i) a − b = (2i + 3j) − (−i + 2j)

= 3i + j
(ii)

b a –b
a
j
i a + (–b) = a – b

▲ Figure 10.12

When you find the vector y Q(3, 5)
represented by the line segment 6
joining two points, you are in
effect subtracting their position 5
vectors. If, for example,

P is the point (2, 1) and Q is the 4
14 ⎠⎞ , 3
point (3, 5), P⎯→Q is ⎛ as 2 )41)
⎝
Figure 10.13 shows.

You find this by saying
P⎯→Q = P⎯→O + O⎯→Q = −p + q.

In this case, this gives 1
P(2, 1)
P⎯→Q = ⎛ 2⎞ ⎛ 3⎞ ⎛1⎞
– ⎝ 1⎠ + ⎝ 5⎠ = ⎝ 4⎠ 0 12
▲ Figure 10.13
as expected. 3 4 5x
243
This is an important result:
P⎯→Q = q − p

where p and q are the position vectors of P and Q.

10 Geometrical figures

It is often useful to be able to express lines in a geometrical figure in terms of
given vectors.

ACTIVITY 10.1

10 VECTORS The diagram shows a cuboid OABCDEFG. P, Q, R, S and T are the
midpoints of the edges they lie on.The origin is at O and the axes lie
along OA, OC and OD, as shown in Figure 10.14.

O⎯→A = ⎛ 6⎞ O⎯→C = ⎛ 0⎞ , O⎯→D = ⎛ 0⎞
⎜ 0⎟ , ⎜ 5⎟ ⎜ 0⎟
⎜⎝ 0⎠⎟ ⎜⎝ 0⎠⎟ ⎝⎜ 4⎟⎠

GS F

TR

DE

zC B
y Q

O xP A
▲ Figure 10.14

(i) Name the points with the following coordinates.

(a) (6, 5, 4) (b) (0, 5, 0) (c) (6, 2.5, 0)

(d) (0, 2.5, 4) (e) (3, 5, 4)

(ii) Use the letters in the diagram to give displacements which are equal

to the following vectors. Give all possible answers; some of them have

more than one.

⎛ 6⎞ ⎛ 6⎞ ⎛ 0⎞ ⎛ −6⎞ ⎛ −3 ⎞
(a) ⎜ 5⎟ (b) ⎜ 0⎟ (c) ⎜ 5⎟ (d) ⎜ −5⎟ (e) ⎜ 2.5⎟

⎝⎜ 4⎟⎠ ⎜⎝ 4⎠⎟ ⎜⎝ 4⎠⎟ ⎜⎝ 4⎟⎠ ⎜⎝ 4 ⎠⎟

Example 10.6 Figure 10.15 shows a hexagonal prism.

G H

r

B C
q

p I

AD
J

FE
▲ Figure 10.15
244

The hexagonal cross-section is regular and consequently A⎯→D = 2B⎯→C. 10

A⎯→B = p, B⎯→C = q and B⎯→G = r. Express the following in terms of p, q and r.

(i) A⎯→C (ii) A⎯→D (iii) H⎯→I (iv) →IJ

(v) E⎯→F (vi) B⎯→E (vii) A⎯→H (viii) F⎯→I

Solution Bq C 10.3 Vector calculations
(i) A⎯→C = A⎯→B + B⎯→C
p p+q
=p+q A C
(ii) A⎯→D = 2B⎯→C = 2q
(iii) H⎯→I = C⎯→D p+q

Since A⎯→C + C⎯→D = A⎯→D A 2q D
B C
p + q + C⎯→D = 2q
C⎯→D = q − p D

So H⎯→I = q − p E
(iv) →IJ = D⎯→E
▲ Figure 10.16
= −A⎯→B
C⎯→H = B⎯→G
= −p
(v) E⎯→F = −B⎯→C F⎯→E = B⎯→C, E→⎯ J = B⎯→G, →JI = A⎯→B

= −q

(vi) B⎯→E = B⎯→C + C⎯→D + D⎯→E

= q + (q − p) + −p

= 2q − 2p
Notice that B⎯→E = 2C⎯→D.
(vii) A⎯→H = A⎯→B + B⎯→C + C⎯→H

=p+q+r

(viii) F→I = F⎯→E + E→J + →JI

=q+r+p

Unit vectors

A unit vector is a vector with a magnitude of 1, like i and j.To find the
unit vector in the same direction as a given vector, divide that vector by its
magnitude.

245

10 Thus the vector 3i + 5j (in Figure 10.17) has magnitude 32 + 52 = 34, and
3 5
246 so the vector 34 i + 34 j is a unit vector. It has magnitude 1.

The unit vector in the direction of vector a is written as â and read as ‘a hat’.

y

10 VECTORS 5j

4j

3j This is the unit vector
3i + 5j
3 i + 5 j
2j 34 34

j

O i 2i 3i 4i x
▲ Figure 10.17

Example 10.7 Relative to an origin O, the position vectors of the points A, B and C are

given by ⎛ −2⎞ 0⎞ ⎛ −2⎞
⎜ 3⎟, 1⎟ ⎜ 3⎟.
O⎯→A = O⎯→B = ⎛ and O⎯→C = ⎝⎜ 1⎟⎠
⎜
⎝⎜ −2⎠⎟ ⎝⎜ −3⎠⎟
(i) Find the unit vector in the direction A⎯→B.

(ii) Find the perimeter of triangle ABC.

Solution

For convenience call O⎯→A = a, O⎯→B = b and O⎯→C = c.

(i) A⎯→B = b − a = ⎛ 0⎞ ⎛ −2⎞ = ⎛ 2⎞
⎜ 1⎟ − ⎜ 3⎟ ⎜ −2⎟
⎜⎝ −3⎠⎟ ⎝⎜ −2⎟⎠ ⎝⎜ −1⎟⎠

To find the unit vector in the direction A⎯→B, you need to divide A⎯→B by

its magnitude.

| |A⎯→B = 22 + (−2)2 + (−1)2 This is the
magnitude of A⎯→B.
=9

=3

A⎯→B ⎛ 2⎞ ⎛ 2 ⎞
⎜ −2⎟ ⎜ 3
⎜⎝ −1⎟⎠ ⎜ ⎟
So the unit vector in the direction is 1 = ⎝ − 2 ⎟
3 − 3 ⎠
1
3

(ii) The perimeter of the triangle is given by | A⎯→B | + | A⎯→C | + | B⎯→C |. 10

A⎯→C = c − a = ⎛ −2⎞ ⎛ −2⎞ = ⎛ 0⎞
⎜ 3⎟ − ⎜ 3⎟ ⎜ 0⎟
⎝⎜ 1⎠⎟ ⎝⎜ −2⎟⎠ ⎜⎝ 3⎟⎠

| |⇒ A⎯→C = 02 + 02 + 32

=3 10.3 Vector calculations

B⎯→C = c − b = ⎛ −2⎞ ⎛ 0⎞ = ⎛ −2⎞
⎜ 3⎟ −⎜ 1⎟ ⎜ 2⎟
⎝⎜ 1⎠⎟ ⎝⎜ −3⎠⎟ ⎝⎜ 4⎠⎟

| |⇒ B⎯→C = (−2)2 + 22 + 42

= 24

Perimeter of ABC = | A⎯→B | + | A⎯→C | + | B⎯→C |

= 3 + 3 + 24

= 10.9

Example 10.8 Figure 10.18 shows triangle AOB. A

C is a point on AB and divides it in the C
ratio 2 : 3. a
Find O⎯→C in terms of the vectors a and b.

B

b

O
▲ Figure 10.18

Solution When you divide AB in the ratio 2 : 3 then

O⎯→C = O⎯→A + 2 A⎯→B AC is 2 2 3 = 2 of AB and CB is
5 + 5

O⎯→A = a and A⎯B→ = b − a 2 3 3 = 3 of AB.
+ 5
O⎯→C
= a + 2 (b − a)
5

= a + 2 b − 2 a
5 5

= 3 a + 2 b
5 5

The above example made use of the ratio theorem. ?
❯ Prove that when C divides AB in the ratio s : t then CP

O⎯→C = s t ta + s s t b 247
+ +

10 Exercise 10B 1 Simplify the following.
CP
248 CP (i) ⎛ 2⎞ + ⎛ 4⎞ (ii) ⎛ 2⎞ + ⎛ –1⎞
CP ⎝ 3⎠ ⎝ 5⎠ ⎝ –1⎠ ⎝ 2⎠

(iii) ⎛ 3⎞ + ⎛ –3⎞ (iv) 3 ⎛ 2⎞ + 2 ⎛ 1⎞
⎝ 4⎠ ⎝ –4⎠ ⎝ 1⎠ ⎝ –2⎠

10 VECTORS (v) 6(3i − 2j) − 9(2i − j)

2 The vectors p, q and r are given by

p = 3i + 2j + k q = 2i + 2j + 2k r = −3i − j − 2k.

Find, in component form, the following vectors.

(i) p + q + r (ii) p − q (iii) p + r

(iv) 3(p − q) + 2(p + r) (v) 4p − 3q + 2r

3 In the diagram, PQRS is a parallelogram and P⎯→Q = a, P⎯→S = b.

(i) Write, in terms of a and b, Q R

the following vectors.
(a) Q⎯→R (b) P⎯→R

(c) Q⎯→S a

(ii) The midpoint of PR is M. Find
(a) P⎯→M (b) Q⎯→M.
P bS
(iii) Explain why this shows you that the B
C
diagonals of a parallelogram bisect each other. M

4 In the diagram,ABCD is a kite. D
AC and BD meet at M.
A⎯→B = i + j and A⎯→D = i − 2j

(i) Use the facts that the diagonals j A
i
of a kite meet at right angles

and that M is the midpoint of

AC to find,in terms of i and j,
(a) A⎯→M (b) A⎯→C

(c) B⎯→C (d) C⎯→D.

(ii) Verify that | A⎯→B | = | B⎯→C | and | A⎯→D | = | C⎯→D |.

5 In the diagram,ABC is a triangle. A M
L, M and N are the midpoints of N
the sides BC, CA and AB.
A⎯→B = p and A⎯→C = q.

(i) Find, in terms of p and q,
B⎯→C, M⎯→N, L⎯→M and L⎯→N.

(ii) Explain how your results B L C
from part (i) show you that

the sides of triangle LMN are parallel to those of triangle ABC,

and half their lengths.

6 Find unit vectors in the same directions as the following vectors. 10
⎛ 2⎞ ⎛ –2⎞
(i) ⎝ 3⎠ (ii) 3i + 4j (iii) ⎝ –2⎠ (iv) 5i − 12j

7 Find unit vectors in the same direction as the following vectors.

⎛ 1⎞ (ii) 2i – 2j + k (iii) 3i – 4k
(i) ⎜ 2⎟
10.3 Vector calculations
⎜⎝ 3⎠⎟

⎛ −2⎞ ⎛ 4⎞
(iv) ⎜ 4⎟ (v) 5i – 3j + 2k (vi) ⎜ 0⎟

⎜⎝ −3⎠⎟ ⎝⎜ 0⎟⎠

8 Relative to an origin O, the position vectors of the points A, B and C are

given by

O⎯→A = ⎛ 2⎞ O⎯→B = ⎛ −2⎞ and O⎯→C = ⎛ −1⎞
⎜ 1⎟ , ⎜ 4⎟ ⎜ 2⎟ .
⎜⎝ 3⎠⎟ ⎜⎝ 3⎠⎟ ⎜⎝ 1⎠⎟

Find the perimeter of triangle ABC.

9 Relative to an origin O, the position vectors of the points P and Q are
given by

O⎯→P = 3i + j + 4k and O⎯→Q = i + xj − 2k.

Find the values of x for which the magnitude of PQ is 7.
CP 10 In the cuboid, O⎯→A = p, O⎯→E = q,O⎯→G = r.

(i) Express the following vectors BC
MD
in terms of p,q and r. A
(a) ⎯G→F (b) ⎯C→F

(c) O⎯→B (d) O⎯→D

(e) O⎯→C p E F
q r G

(ii) The point M divides O

AD in the ratio 3 : 2.
Find O⎯M→ in terms of p, q and r.

(iii) Use vectors to prove that OC and BG bisect each other.

CP 11 Relative to an origin O, the position vectors of the points A, B, C and D
are given by

O⎯→A = ⎛ 3 ⎞ O⎯→B = ⎛ 5 ⎞ , O⎯→C = ⎛ 8 ⎞ O⎯→D = ⎛ 6⎞
⎜ 1 ⎟, ⎜ 5 ⎟ ⎜ 2 ⎟, and ⎜ −2 ⎟
⎜⎝ 5 ⎟⎠ ⎜ 13 ⎟ ⎝⎜ 7 ⎠⎟ ⎜⎝ −1 ⎟⎠
⎝ ⎠

Use vectors to prove that ABCD is a parallelogram.

249

10 12 Relative to an origin O, the position vectors of the points A and B are
given by
250
O⎯→A = ⎛ 4⎞ and O⎯→B = ⎛ 3⎞
⎜ 1⎟ ⎜ 2⎟ .
⎝⎜ −2⎟⎠ ⎜⎝ –4⎟⎠

(i) Given that C is the point such that A⎯→C = 2A⎯→B, find the unit vector
in the direction of O⎯→C.
10 VECTORS ⎛ 1⎞
The position vector of the point D is given by O⎯→D = ⎜ 4⎟ , where k is a
⎜⎝ k⎟⎠

constant,and it is given that O⎯→D = mO⎯→A + nO⎯→B,where m and n are

constants.

(ii) Find the values of m, n and k.
Cambridge International AS & A Level Mathematics
9709 Paper 1 Q9 June 2007

10.4 The angle between two vectors

❯ As you work through the proof in this section, make a list of all the ?
results that you are assuming.
CP

To find the angle θ between the two vectors A B
O⎯→A = a = a1i + a2 j (a1, a2) (b1, b2)

and y x
O⎯→B = b = b1i + b2 j

start by applying the cosine rule to
triangle OAB in Figure 10.19.

cos θ = OA2 + OB2 – AB2 a b
2OA × OB

In this, OA, OB and AB are the θ
lengths of the vectors O⎯→A, O⎯→B and
A⎯→B, and so O
▲ Figure 10.19

OA = | a | = a21 + a22 and OB = | b | = b21 + b22 .
The vector A⎯→B = b − a = (b1i + b2 j) − (a1i + a2 j)

= (b1 − a1)i + (b2 − a2)j
and so its length is given by

AB = | b − a | = (b1 – a1)2 + (b2 – a2)2.

Substituting for OA, OB and AB in the cosine rule gives 10
cos θ = (a12 + a22) + (b21 + b22) – [(b1 – a1)2 + (b2 – a2)2]
2 a21 + a22 × b21 + b22

( )=
a21 + a22 + b21 + b22 – b21 – 2a1b1 + a21 + b22 – 2a2b2 + a22
2 |a||b|

This simplifies to 2a1b1 + 2a2b2 a1b1 + a2b2 10.4 The angle between two vectors
2 |a||b| |a||b|
cos θ = =

The expression on the top line, a1b1 + a2b2, is called the scalar product
(or dot product) of the vectors a and b and is written a .b.Thus

cos θ = |aa|.|bb|.

This result is usually written in the form

a .b = | a | | b | cos θ.

The next example shows you how to use it to find the angle between two
vectors given numerically.

Example 10.9 Find the angle between the vectors ⎛ 3⎞ and ⎛ 5⎞ .
⎝ 4⎠ ⎝ –12⎠

Solution

Let a = ⎛ 3⎞ ⇒ | a | = 32 + 42 = 5
⎝ 4⎠ ⇒ | b | = 52 + (–12)2 = 13.

and b = ⎛ 5⎞
⎝ –12⎠

The scalar product

⎝⎛⎜43⎞⎟⎠ ⋅ ⎜⎛⎝–152⎟⎠⎞ = 3 ×5 + 4 ×(−12)

= 15 − 48
= −33

Substituting in a .b = | a | | b | cos θ gives

−33 = 5 ×13 ×cos θ

cos θ = –33
65
⇒ θ = 120.5°

Perpendicular vectors 251

Since cos 90° = 0, it follows that if vectors a and b are perpendicular then
a .b = 0.

Conversely, if the scalar product of two non-zero vectors is zero, they are
perpendicular.

10 Example 10.10 Show that the vectors a = ⎛ 2⎞ and b = ⎛ 6⎞ are perpendicular.
⎝ 4⎠ ⎝ –3⎠

10 VECTORS Solution

The scalar product of the vectors is
a . b = ⎛⎜⎝42⎟⎞⎠ . ⎝⎛⎜–36⎞⎠⎟
= 2 ×6 + 4 ×(−3)
= 12 − 12 = 0

Therefore the vectors are perpendicular.

Further points concerning the scalar product

» You will notice that the scalar product of two vectors is an ordinary
number. It has size but no direction and so is a scalar, rather than a vector.
It is for this reason that it is called the scalar product.There is another way
of multiplying vectors that gives a vector as the answer; it is called the
vector product.This is beyond the scope of this book.

» The scalar product is calculated in the same way for three-dimensional
vectors. For example:

⎛ 2⎞ ⎛ 5⎞
⎜ 3⎟ . ⎜ 6⎟ = 2 × 5 + 3 × 6 + 4 × 7 = 56
⎝⎜ 4⎟⎠ ⎝⎜ 7⎟⎠

In general

⎛ a1 ⎞ ⎛ b1 ⎞
⎜ a2 ⎟ ⎜ b2 ⎟ a1b1 + a2b2 + a3b3
⎝⎜ a3 ⎟⎠ . ⎝⎜ b3 ⎠⎟ =

» The scalar product of two vectors is commutative. It has the same value
whichever of them is on the left-hand side or right-hand side.Thus
a .b = b .a, as in the following example.

⎛ 2⎞ . ⎛ 6⎞ = 2× 6+3× 7 = 33
⎝ 3⎠ ⎝ 7⎠

⎛ 6⎞ . ⎛ 2⎞ = 6× 2+7× 3= 33.
⎝ 7⎠ ⎝ 3⎠

❯ How would you prove this result? ?
CP

252

The angle between two vectors in three dimensions 10

The angle θ between the vectors a = a1i + a2 j and b = b1i + b2 j in two
dimensions is given by

cos θ = a1b1 + a2b2 = a.b
a12 + a22 × b12 + b22 |a||b|

where a .b is the scalar product of a and b.This result was proved by using 10.4 The angle between two vectors
the cosine rule on page 250.

❯ Show that the angle between the three-dimensional vectors ?
CP

a = a1i + a2 j + a3k and b = b1i + b2 j + b3k

is also given by

cos θ = a.b
|a||b|

but that the scalar product a .b is now

a .b = a1b1 + a2b2 + a3b3.

Example 10.11 When working in two dimensions you found the angle between two lines by
using the scalar product.As you have just proved, this method can be extended
into three dimensions, and its use is shown in the following example.

The points P, Q and R are (1, 0, −1), (2, 4, 1) and (3, 5, 6). Find ∠QPR.

Solution
The angle between P⎯→Q and P⎯→R is given by θ in

cosθ = ⎯PQ→ . ⎯PR→
|⎯PQ→||⎯PR→|

In this

P⎯Q→ = ⎛2⎞ – ⎛ 1⎞ = ⎛ 1⎞ | P⎯→Q | = 12 + 42 + 22 = 21
⎜⎝⎜41⎠⎟⎟ ⎝⎜⎜–01⎟⎠⎟ ⎜⎜⎝42⎠⎟⎟

Similarly

P⎯→R== ⎛3⎞ ⎛ 1⎞ ⎛2⎞ | P⎯→R | = 22 + 52 + 72 = 78
⎝⎜⎜65⎠⎟⎟ ⎜ –01⎟⎟⎠ ⎜⎜⎝75 ⎟
– ⎜⎝ = ⎟⎠

➜

253

10 Therefore (3, 5, 6) R

P⎯→Q. . P⎯→R = ⎛ 1⎞ ⎛2⎞
⎜⎜⎝42⎠⎟⎟ .⎝⎜⎜75⎟⎟⎠

10 VECTORS = 1 ×2 + 4 ×5 + 2 ×7 ) )2
= 36
5
7

Substituting gives

cosθ = 36 (2, 4, 1) Q
21 × 78

⇒ θ = 27.2° ) )1

θ 4
P (1, 0, –1) 2

▲ Figure 10.20

You must be careful to find the correct angle.To find ∠QPR
(see Figure 10.21), you need the scalar product P⎯→Q . P⎯→R, as in the
example above. If you take Q⎯→P . P⎯→R, you will obtain ∠Q´PR, which is
(180° − ∠QPR).
R

Q
θ
P
Q'

▲ Figure 10.21

Exercise 10C 1 Find the angles between these vectors.

(i) 2i + 3j and 4i + j (ii) 2i − j and i + 2j

(iii) ⎛ –1⎞ and ⎛ –1⎞ (iv) 4i + j and i + j
⎝ –1⎠ ⎝ –2⎠

(v) ⎛⎝⎜32⎠⎟⎞ and ⎝⎜⎛ –46⎟⎠⎞ (vi) ⎛ 3⎞ and ⎛ –6⎞
⎝ –1⎠ ⎝ 2⎠

2 The points A, B and C have coordinates (3, 2), (6, 3) and (5, 6),
respectively.
(i) Write down the vectors A⎯→B and B⎯→C.

(ii) Show that the angle ABC is 90°.

(iii) Show that | A⎯→B | = | B⎯→C |.

(iv) The figure ABCD is a square.

254 Find the coordinates of the point D.

3 Find the angles between these pairs of vectors. 10

⎛ 2⎞ ⎛ 2⎞ ⎛ 1⎞ ⎛ 3⎞
(i) ⎜1⎟ and ⎜ –1⎟ (ii) ⎜ –1⎟ and ⎜1⎟

⎝⎜ 3⎠⎟ ⎜⎝ 4⎠⎟ ⎝⎜ 0⎠⎟ ⎜⎝ 5⎟⎠

(iii) 3i + 2j − 2k and −4i − j + 3k

PS 4 The diagram shows a room with rectangular walls,floor and ceiling. 10.4 The angle between two vectors

A string has been stretched in a straight line between the corners A and G.

z G F
(0, 0, 3) D string B

E

y spider
C

(0, 4, 0)

O (0, 0, 0) Ax
(5, 0, 0)

The corner O is taken as the origin.A is (5, 0, 0), C is (0, 4, 0) and
D is (0, 0, 3), where the lengths are in metres.

A spider walks up the string, starting at A.

(i) Write down the coordinates of G.

(ii) Find the vector A⎯→G and the distance the spider walks along the
string from A to G.

(iii) Find the angle of elevation of the spider’s journey along the string.

5 In the diagram, OABCDEFG is a cube in which each side has length 6.
Unit vectors i, j and k are parallel to O⎯→A, O⎯→C and O⎯→D respectively.The
that A⎯→P = A⎯→B
point P is such 1 and the point Q is the midpoint of DF.
3

G F
Q

DE

kC B

j
P

Oi A

(i) Express each of the vectors O⎯→Q and P⎯→Q in terms of i, j and k.

(ii) Find the angle OQP.

Cambridge International AS & A Level Mathematics 255
9709 Paper 12 Q6 November 2009

10 6 Relative to an origin O, the position vectors of points A and B are
2i + j + 2k and 3i − 2j + pk respectively.
256
10 VECTORS (i) Find the value of p for which OA and OB are perpendicular.

(ii) In the case where p = 6, use a scalar product to find angle AOB,
correct to the nearest degree.

(iii) Express the vector A⎯→B in terms of p and hence find the values of p
for which the length of AB is 3.5 units.
Cambridge International AS & A Level Mathematics
9709 Paper 1 Q10 June 2008

7 Relative to an origin O, the position vectors of the points A and B are
given by

O⎯→A = 2i − 8j + 4k and O⎯→B = 7i + 2j − k.

(i) Find the value of O⎯→A . O⎯→B and hence state whether angle AOB is
acute, obtuse or a right angle.

(ii) The point X is such that A⎯→X = 25A⎯→B. Find the unit vector in the
direction of OX.
Cambridge International AS & A Level Mathematics
9709 Paper 1 Q6 June 2009

8 Relative to an origin O, the position vectors of the points A and B are
given by

O⎯→A = 2i + 3j − k and O⎯→B = 4i − 3j + 2k.

(i) Use a scalar product to find angle AOB, correct to the nearest
degree.

(ii) Find the unit vector in the direction of A⎯→B.

(iii) The point C is such that O⎯→C = 6j + pk, where p is a constant.
Given that the lengths of A⎯→B and A⎯→C are equal, find the possible
values of p.
Cambridge International AS & A Level Mathematics

9709 Paper 1 Q11 June 2005

9 Relative to an origin O, the position vectors of the points P and Q are

given by ⎛ −2⎞ ⎛ 2⎞
⎜ 3⎟ ⎜1⎟,
O⎯→P = ⎜⎝ 1⎠⎟ and O⎯→Q = ⎝⎜ q ⎟⎠ where q is a constant.

(i) In the case where q = 3, use a scalar product to show that
cos POQ = 71.

(ii) Find the values of q for which the length of P⎯→Q is 6 units.
Cambridge International AS & A Level Mathematics
9709 Paper 1 Q4 November 2005

10 The diagram shows a semicircular prism with a horizontal rectangular 10
base ABCD.The vertical ends AED and BFC are semicircles of radius
6 cm.The length of the prism is 20 cm.The midpoint of AD is the origin
O, the midpoint of BC is M and the midpoint of DC is N.The points E
and F are the highest points of the semicircular ends of the prism.The
point P lies on EF such that EP = 8 cm.

F

P BM 10.4 The angle between two vectors
8 cm
E N
20 cm
C

k

A j

6cm O i D

Unit vectors i, j and k are parallel to OD, OM and OE respectively.
(i) Express each of the vectors P⎯→A and P⎯→N in terms of i, j and k.

(ii) Use a scalar product to calculate angle APN.
Cambridge International AS & A Level Mathematics
9709 Paper 1 Q4 November 2008

11 The diagram shows the roof of a house.The base of the roof, OABC, is
rectangular and horizontal with OA = CB = 14 m and OC = AB = 8 m.
The top of the roof DE is 5 m above the base and DE = 6 m.The sloping
edges OD, CD,AE and BE are all equal in length.

D 6m E
B

8m

C A

k 14 m
j
i
O

Unit vectors i and j are parallel to OA and OC respectively and the unit
vector k is vertically upwards.

(i) Express the vector O⎯→D in terms of i, j and k, and find its
magnitude.

(ii) Use a scalar product to find angle DOB.
Cambridge International AS & A Level Mathematics
9709 Paper 1 Q8 June 2006

257

10 12 The diagram shows a cube OABCDEFG in which the length of each
side is 4 units.The unit vectors i, j and k are parallel to O⎯→A, O⎯→C and O⎯→D
258
respectively.The midpoints of OA and DG are P and Q respectively and
R is the centre of the square face ABFE.

GF

Q

10 VECTORS DE
R

C B
k

j

O iP A

(i) Express each of the vectors P⎯→R and P⎯→Q in terms of i, j and k.

(ii) Use a scalar product to find angle QPR.

(iii) Find the perimeter of triangle PQR, giving your answer correct to
1 decimal place.
Cambridge International AS & A Level Mathematics

9709 Paper 1 Q10 November 2007

10.5 The vector equation of a line

The vector joining two points

In Figure 10.22, start by looking at two points A(2, −1) and B(4, 3); that is the

points with position vectors O⎯→A = ⎛ 2⎞ and O⎯→B = ⎛ 4⎞ , alternatively 2i − j
⎜⎝ −1⎠⎟ ⎜⎝ 3⎠⎟

and 4i + 3j.

y

3 B(4, 3)

2N

1M

–1 O 1 2 3 4 5x
–1 A(2, –1)
–2

▲ Figure 10.22

The vector joining A to B is A⎯→B and this is given by 10

A⎯→B = A⎯→O + O⎯→B
= −O⎯→A + O⎯→B
= O⎯→B − O⎯→A

= ⎛ 4⎞ − ⎛ 2⎞ = ⎛ 2⎞ 10.5 The vector equation of a line
⎝⎜ 3⎟⎠ ⎝⎜ –1⎟⎠ ⎜⎝ 4⎟⎠

Since A⎯→B = ⎛ 2⎞ , then it follows that the length of AB is given by
⎝⎜ 4⎠⎟

⏐A⎯→B⏐ = 22 + 4 2

= 20

You can find the position vectors of points along AB as follows.
The midpoint, M, has position vector O⎯M→, given by

O⎯M→ = O⎯→A + 21A⎯→B

= ⎛ 2⎞ + 1 ⎛2⎞
⎜⎝–1⎟⎠ 2 ⎜⎝4⎟⎠

= ⎛ 3⎞
⎜⎝ 1 ⎟⎠

In the same way, the position vector of the point N, three-quarters of the
distance from A to B, is given by

O⎯→N = ⎛ 2⎞ + 3 ⎛ 2⎞
⎝⎜ –1⎠⎟ 4 ⎜⎝ 4⎟⎠

= ⎛ 3 1 ⎞
2
⎝⎜ 2 ⎟⎠

and it is possible to find the position vector of any other point of subdivision
of the line AB in the same way.

A point P has position vector O⎯→P = O⎯→A + λA⎯→B where λ is a fraction. ?
❯ Show that this can be expressed as CP

O⎯→P = (1 − λ)O⎯→A + λO⎯→B.

259

10 The vector equation of a line

It is now a small step to go from finding the position vector of any point on
the line AB to finding the vector form of the equation of the line AB.To take
this step, you will find it helpful to carry out the following activity.

10 VECTORS ACTIVITY 10.2

The position vectors of a set of points are given by

r = ⎛ 2⎞ + λ ⎛ 2⎞ λ is the Greek
⎜⎝ –1⎠⎟ ⎜⎝ 4⎠⎟ letter ‘lamda’.

where λ is a parameter which may take any value.
(i) Show that λ = 2 corresponds to the point with position
6⎞
vector ⎛ 7⎟⎠ .
⎜⎝

(ii) Find the position vectors of points corresponding to values
of λ of −2, −1, 0, 21, 43, 1, 3.

(iii) Mark all your points on a sheet of squared paper and show that

when they are joined up they give the line AB in Figure 10.23.

(iv) State what values of λ correspond to the points A, B, M and N.

(v) What can you say about the position of the point if

(a) 0 < λ < 1?

(b) λ > 1?

(c) λ < 0?

This activity should have convinced you that

r = ⎛ 2⎞ + λ ⎛ 2⎞ The number λ is called a parameter
⎝⎜ –1⎟⎠ ⎝⎜ 4⎟⎠ and it can take any value. Of course,
you can use other letters for the

parameter such as µ, s and t.

is the equation of the line passing through (2, −1) and (4, 3), written in vector
form.

260

You may find it helpful to think of this in these terms. 10

( )2 Move to the point A with r = ⎛ 2⎞ + λ ⎛ 2⎞ ( )3 Move λ steps of 2 (i.e.
⎜⎝ –1⎠⎟ ⎜⎝ 4⎠⎟ 4
2 in the direction A⎯→B). λ need
position vector −1 .

not be a whole number and

may be negative.

and then 10.5 The vector equation of a line
y

3 B(4, 3)

1 Start at the origin. 2
1

–1 O 1 2 3 4 5x

–1 A(2, –1)

–2

▲ Figure 10.23

You should also have noticed that when:

λ= 0 the point corresponds to the point A

λ= 1 the point corresponds to the point B

0ϽλϽ1 the point lies between A and B

λϾ1 the point lies beyond B

λϽ0 the point lies beyond A.

The vector form of the equation is not unique; there are many (in fact
infinitely many) different ways in which the equation of any particular line
may be expressed.There are two reasons for this: direction and location.

Direction 2⎞
4⎠⎟
The direction of the line in the example is ⎛ .That means that for every
⎜⎝

2 units along (in the i direction), the line goes up 4 units (in the j direction).
This is equivalent to stating that for every 1 unit along, the line goes up 2
units, corresponding to the equation

r= ⎛ 2⎞ + ⎛1 ⎞
⎜⎝–1⎠⎟ λ ⎝⎜2⎠⎟ .

261

10 The only difference is that the two equations have different values of λ for

262 particular points. In the first equation, point B, with position vector ⎛ 4⎞ ,
⎝⎜ 3⎠⎟

corresponds to a value of λ of 1. In the second equation, the value of λ
for B is 2.
⎛ 2⎞ 1⎞ ⎛1⎞ 3⎞
The direction ⎜⎝ 4⎠⎟ is the same as ⎛ 2⎠⎟ , or as any multiple of ⎝⎜ 2⎟⎠ such as ⎛ 6⎠⎟ ,
⎝⎜ ⎜⎝
10 VECTORS
⎛ –5⎞ or ⎛ 100.5⎞ . Any of these could be used in the vector equation of the line.
⎝⎜ –10⎠⎟ ⎝⎜ 201 ⎠⎟

Location

In the equation

r = ⎛ 2⎞ + λ ⎛ 2⎞
⎜⎝ –1 ⎠⎟ ⎝⎜ 4⎟⎠

⎛ 2⎞ is the position vector of the point A on the line, and represents the
⎜⎝ –1⎠⎟

point at which the line was joined. However, this could have been any other

point on the line, such as M(3, 1), B(4, 3), etc. Consequently

r = ⎛ 3⎞ + λ ⎛ 2⎞
⎜⎝ 1 ⎟⎠ ⎝⎜ 4⎠⎟

and ⎛ 4⎞ 2⎞
⎜⎝ 3⎟⎠ 4⎠⎟
r = + λ ⎛
⎝⎜

are also equations of the same line, and there are infinitely many other
possibilities, one corresponding to each point on the line.

Notes

1 It is usual to refer to any valid vector form of the equation as the vector
equation of the line even though it is not unique.

2 It is often a good idea to give the direction vector in its simplest integer
form:
for example, replacing ⎛2⎞ with ⎛1 ⎞ .
⎟
⎜⎟ ⎜
⎝⎜2⎟⎠
⎜⎝4 ⎟⎠

The general vector form of the equation of a line

If A and B are points with position a and b, then the equation
r = O⎯→A + λ A⎯→B

may be written as r = a + λ(b − a)

which implies r = (1 − λ)a + λb.

This is the general vector form of the equation of the line joining two points.

ACTIVITY 10.3 10

Plot the following lines on the same sheet of squared paper.When you 263
have done so, explain why certain among them are the same as each other,
others are parallel to each other, and others are in different directions.

(i) r = ⎛ 2⎞ + λ ⎛1 ⎞ (ii) r = ⎛ 2⎞ + λ ⎛ –1⎞ 10.5 The vector equation of a line
⎜⎝–1⎟⎠ ⎜⎝2⎠⎟ ⎜⎝–1⎠⎟ ⎜⎝ 2⎟⎠

(iii) r = ⎛0⎞ + λ ⎛1 ⎞ (iv) r = ⎛ 1⎞ + λ ⎛ 3⎞
⎝⎜2⎟⎠ ⎝⎜2⎠⎟ ⎜⎝ –3⎠⎟ ⎝⎜ 6⎠⎟

(v) r = ⎛ 4⎞ + λ ⎛ 1⎞
⎝⎜ 3⎟⎠ ⎜⎝ –2⎟⎠

The same methods can be used to find the vector equation of a line in three
dimensions, as shown in the next example.

Example 10.12 The coordinates of A and B are (–2, 4, 1) and (2, 1, 3) respectively.

(i) Find the vector equation of the line AB.
(ii) Does the point P(6, –2, 7) lie on the line AB?
(iii) The point N lies on the line AB.

Given that 3⏐A⎯→N⏐=⏐N⎯→B⏐ find the coordinates of N.

Solution ⎛ −2⎞ O⎯→B ⎛ 2⎞
(i) a = O⎯→A 41⎠⎟ and b 31⎟⎠
= ⎜ = = ⎜
⎝ ⎝
There are other ways of
A⎯→B = b − a = ⎛2⎞ − ⎛− 2⎞ = ⎛ 4⎞ writing this equation, for
⎝⎜⎜13⎟⎟⎠ ⎜⎜⎝ 41⎠⎟⎟ ⎜⎜⎝− 23⎟⎠⎟ example

The vector equation of a line can r = ⎝⎛⎜231⎟⎠⎞ + λ ⎛ 423⎞⎟⎠ or
be written as ⎝⎜−

r = O⎯→A + λA⎯→B r = ⎝⎜⎛−−761⎟⎞⎠ + λ ⎛ 423⎠⎟⎞
⎝⎜−
⎛−2⎞ ⎛ 4⎞
⇒ r = ⎜⎜⎝ 41⎠⎟⎟ + λ ⎝⎜⎜−23⎟⎟⎠ but they are all equivalent
to each other.

(ii) If P lies on the line AB then for some value of λ

⎛x⎞ ⎛ 6⎞ ⎛− 2⎞ ⎛ 4⎞
⎝⎜⎜zy⎟⎟⎠ = ⎜⎝⎜− 72⎟⎠⎟ = ⎜⎜⎝ 41⎟⎟⎠ + λ ⎝⎜⎜− 23⎠⎟⎟

Find the value of λ for the x-coordinate. ➜
x : 6 = −2 + 4λ ⇒ λ = 2

10 Then check whether this value of λ gives a y-coordinate of −2 and
a z-coordinate of 7.

y : −2 = 4 − 3 ×2 z : 7 ≠ 1 + 2 ×2

So the point P(6, –2, 7) does not lie on the line.

(iii) Since 3⏐A⎯→N⏐=⏐N⎯→B⏐, N must lie 1 of the way along the line AB so
1 4
the value of λ is 4 .
10 VECTORS
O⎯→N = O⎯→A + 1 A⎯→B
4

O⎯→N ⎛−2⎞ ⎛ 4⎞ ⎛ −1⎞
⎜ 41⎟⎠⎟ 1 ⎜⎝⎜−23⎟⎟⎠ ⎜⎜⎝31 1 ⎟
= ⎜ + 4 = 4 ⎟
⎝ 1 ⎠
2

So the coordinates of N are (–1, 3.25, 1.5).

Exercise 10D 1 For each of these pairs of points,A and B, write down:
(a) the vector A⎯→B
(b) ⏐A⎯→B⏐
(c) the position vector of the midpoint of AB.

(i) A is (2, 3), B is (4, 11).

(ii) A is (4, 3), B is (0, 0).

(iii) A is (−2, −1), B is (4, 7).

(iv) A is (−3, 4), B is (3, −4).

(v) A is (−10, −8), B is (−5, 4).

2 Find the equation of each of these lines in vector form.
(i) Joining (2, 1) to (4, 5)

(ii) Joining (3, 5) to (0, 8)

(iii) Joining (−6, −6) to (4, 4)

(iv) Through (5, 3) in the same direction as i + j

(v) Through (2, 1) parallel to 6i + 3j
–1⎞
(vi) Through (0, 0) parallel to ⎛ 4⎠⎟
(vii) Joining (0, 0) to (−2, 8) ⎝⎜

(viii) Joining (3, −12) to (−1, 4)

264

3 Find the equation of each of these lines in vector form. 10

(i) Through (2, 4, −1) in the direction ⎛ 3⎞ 265
46⎟⎠
⎜
⎝

(ii) Through (1, 0, −1) in the direction ⎛ 1⎞
⎜ 00⎟⎠
⎝ 10.6 The intersection of two lines

(iii) Through (1, 0, 4) and (6, 3, −2)

(iv) Through (0, 0, 1) and (2, 1, 4)

(v) Through (1, 2, 3) and (−2, −4, −6)

4 Determine whether the given point P lies on the line in each of the
following cases.

⎛ 1⎞ ⎛ 2⎞
(i) P(5, 1, 4) and the line r = ⎜⎝⎜43⎟⎠⎟ + λ ⎜⎜⎝−01⎠⎟⎟

⎛ 1⎞ ⎛ 2⎞
(ii) P(−1, 5, 1) and the line r = ⎝⎜⎜43⎠⎟⎟ + λ ⎜⎝⎜− 23⎠⎟⎟

(iii) P(−5, 3, 12) and the line r = ⎛ 1⎞ + ⎛−2⎞
⎜⎜⎝−20⎟⎟⎠ λ ⎝⎜⎜ 51⎟⎠⎟

⎛ 1⎞ ⎛ 4⎞
(iv) P(9, 0, −6) and the line r = ⎝⎜⎜02⎟⎟⎠ + λ ⎝⎜⎜−−21⎠⎟⎟

⎛ 1⎞ ⎛2⎞
(v) P(−9, −2, −17) and the line r = ⎝⎜⎜−23⎟⎠⎟ + λ ⎝⎜⎜31⎠⎟⎟

PS 5 The coordinates of three points are A(−1, −2, 1), B( −3, 4, −5) and

C(0, −2, 4).
(i) Find a vector equation of the line AB.

(ii) Find the coordinates of the midpoint M of AB.

(iii) The point N lies on BC.
Given that 2⏐B⎯→N⏐=⏐N⎯→C⏐, find the equation of the line MN.

10.6 The intersection of two lines

Hold a pen and a pencil to represent two distinct straight lines as follows:

» hold them to represent parallel lines;
» hold them to represent intersecting lines;
» hold them to represent lines which are not parallel and which do not

intersect (even if you extend them).

10 In three-dimensional space two or more straight lines which are not parallel
and which do not meet are known as skew lines. In a plane two distinct
lines are either parallel or intersecting, but in three dimensions there are
three possibilities: the lines may be parallel, or intersecting, or skew.The next
example illustrates a method of finding whether two lines meet, and, if they
do meet, the coordinates of the point of intersection.

10 VECTORS Example 10.13 Find the position vector of the point where the following lines intersect.

r = ⎛2⎞ + λ ⎛1 ⎞ and r = ⎛6⎞ + µ ⎛ 1⎞
⎝⎜3⎠⎟ ⎜⎝2⎠⎟ ⎝⎜1 ⎠⎟ ⎝⎜–3⎠⎟

Note here that different letters are used for the parameters in the two

equations to avoid confusion.

Solution

When the lines intersect, the position vector is the same for each of them.

r = ⎛x⎞ = ⎛2⎞ + ⎛1 ⎞ = ⎛6⎞ + ⎛ 1⎞
⎝⎜y ⎟⎠ ⎝⎜3⎠⎟ λ ⎜⎝2⎟⎠ ⎜⎝1 ⎟⎠ µ ⎜⎝− 3⎟⎠
This gives two simultaneous equations for λ and µ.

x: 2 + λ = 6 + µ ⇒ λ − µ = 4

y : 3 + 2λ = 1 − 3µ ⇒ 2λ + 3µ = −2

Solving these gives λ = 2 and µ = −2. Substituting in either equation gives

r = ⎛ 4⎞
⎝⎜ 7⎟⎠

which is the position vector of the point of intersection.

Example 10.14 Find the coordinates of the point y
of intersection of the lines joining 6 A(1, 6)
A(1, 6) to B(4, 0), and C(1, 1) to
D(5, 3).

3 D(5, 3)

1 C(1, 1) B(4, 0) x
O1 4

▲ Figure 10.24
266

Solution ⎛ 4⎞ ⎛1⎞ ⎛ 3⎞ 10
⎜⎝ 0⎠⎟ ⎜⎝ 6⎠⎟ ⎜⎝ –6⎟⎠
A⎯→B = − = 267

and so the vector equation of line AB is

r = O⎯→A + λA⎯→B

r = ⎛1 ⎞ + λ ⎛ 3⎞ 10.6 The intersection of two lines
⎝⎜6⎟⎠ ⎝⎜–6⎟⎠

C⎯→D = ⎛ 5⎞ − ⎛ 1⎞ = ⎛ 4⎞
⎝⎜ 3⎟⎠ ⎜⎝ 1⎠⎟ ⎜⎝ 2⎟⎠

and so the vector equation of line CD is

r = O⎯→C + µ C⎯→D

r = ⎛1⎞ + ⎛4⎞
⎝⎜1⎟⎠ µ ⎜⎝2⎠⎟

The intersection of these lines is at

r = ⎛1 ⎞ ⎛ 3⎞ = ⎛1⎞ + ⎛4⎞
⎜⎝6⎠⎟ + λ ⎜⎝− 6⎠⎟ ⎝⎜1⎟⎠ µ ⎜⎝2⎟⎠

x : 1 + 3λ = 1 + 4µ ⇒ 3λ − 4µ = 0 ᭺1
᭺2
y : 6 − 6λ = 1 + 2µ ⇒ 6λ + 2µ = 5

Solve ᭺1 and ᭺2 simultaneously:

᭺1 : 3λ − 4µ = 0
᭺2 × 2: 12λ + 4µ = 10
Add: 15λ = 10

⇒ λ = 2
3

Substitute λ = 2 in the equation for AB:
3

⇒ r= ⎛1⎞ + 2 ⎛ 3⎞
⎝⎜ 6⎠⎟ 3 ⎜⎝ –6⎠⎟

⇒ r = ⎛ 23⎠⎞⎟ . The point of intersection has coordinates (3, 2).
⎝⎜

Note

Alternatively, you could have found µ = 1 and substituted in the equation
2
for CD.

10 In three dimensions, lines may be parallel, they may intersect or they may be
skew.

Example 10.15 Determine whether each pair of lines are parallel, intersect or are skew.

⎛ 1⎞ ⎛−3⎞ ⎛ 1⎞ ⎛ 6⎞
(i) r = ⎜⎝⎜−21⎟⎟⎠ + λ ⎜⎝⎜ 21⎟⎠⎟ and r = ⎝⎜⎜−23⎟⎟⎠ + µ ⎝⎜⎜−−42⎟⎟⎠

10 VECTORS (ii) r = ⎛ 1⎞ + ⎛ 2⎞ and r = ⎛ 4⎞ + ⎛−1⎞
⎜⎜⎝−21⎟⎟⎠ λ ⎜⎝⎜−43⎟⎟⎠ ⎜⎜⎝−−25⎟⎠⎟ µ ⎝⎜⎜ 21⎟⎟⎠

Solution ⎛ −3⎞ ⎛ 6⎞
⎜ 21⎠⎟ ⎜ −−42⎠⎟
(i) The vecto−rs2 ⎝ and ⎝ are in the same direction as

⎛ 6⎞ ⎛ −3⎞ Note the lines are different as one
−−42⎠⎟ = −2 ⎜ 21⎟⎠
⎜ ⎝ line passes through (1, –2, 1) and
⎝ the other through (1, 3, –2).

So the lines are parallel.

(ii) These lines are not parallel, so either they intersect or they are skew.
If the two lines intersect then there is a point (x, y, z) that lies on
both lines.

⎛x⎞ ⎛ 1⎞ ⎛ 2⎞ ⎛x⎞ ⎛ 4⎞ ⎛− 1⎞
⎝⎜⎜zy⎟⎟⎠ = ⎝⎜⎜−21⎠⎟⎟ + λ ⎝⎜⎜− 43⎟⎠⎟ and ⎝⎜⎜zy⎟⎠⎟ = ⎜⎝⎜−− 25⎟⎠⎟ + µ ⎜⎜⎝ 21⎟⎟⎠

This gives three simultaneous equations for λ and µ.

x : 1 + 2λ = 4 − µ ⇒ 2λ + µ = 3 ᭺1

y : 2 − 3λ = −2 + 2µ ⇒ 3λ + 2µ = 4 ᭺2

z : −1 + 4λ = −5 + µ ⇒ 4λ − µ = −4 ᭺3

Now solve any two of the three equations above simultaneously.

Using ᭺1 and ᭺2 :

{ } { }2λ + µ = 3
3λ + 2µ = 4
⇒ 4λ + 2µ = 6 ⇒ λ = 2, µ = −1
3λ + 2µ = 4

If these solutions satisfy the previously unused equation (equation ᭺3
here) then the lines meet, and you can substitute the value of λ (or

µ) into equations ᭺1 , ᭺2 and ᭺3 to find the coordinates of the point of
intersection.

If these solutions do not satisfy equation ᭺3 then the lines are skew.

4λ − µ = −4 ᭺3

268

When λ = 2 and µ = −1 10

4λ − µ = 9 ≠ −4

As there are no values for λ and µ that satisfy all three equations, the
lines do not meet and so are skew; you have already seen that they are
not parallel.

Note 10.6 The intersection of two lines

If the equation of the second line was

⎛ x⎞ ⎛ 4⎞ ⎛ − 1⎞
⎜⎜⎝ yz⎠⎟⎟ = ⎜⎜⎝ − 28⎠⎟⎟ + µ ⎜⎝⎜ 12⎟⎟⎠
then the values of λ = 2 and µ = –1 would produce the same point for both

lines:

⎛ x⎞ ⎛ 1⎞ ⎛ 2⎞ ⎛ 5⎞
⎜⎝⎜ yz⎠⎟⎟ = ⎜⎜⎝ − 12⎠⎟⎟ + 2 ⎜⎜⎝ − 43⎟⎟⎠ = ⎝⎜⎜ − 47⎟⎟⎠

⎛ x⎞ ⎛ 4⎞ ⎛ − 1⎞ ⎛ 5⎞
and ⎜⎜⎝ zy ⎟⎟⎠ = ⎝⎜⎜ − 82⎟⎠⎟ − 1⎜⎜⎝ 12⎠⎟⎟ = ⎜⎝⎜ − 47⎠⎟⎟ .

So the lines would intersect at (5, –4, 7).

Exercise 10E 1 Find the position vector of the point of intersection of each of these
pairs of lines.

(i) r = ⎛2⎞ + λ ⎛1 ⎞ and r = ⎛3⎞ + µ ⎛1⎞
⎜⎝1 ⎟⎠ ⎝⎜0⎟⎠ ⎜⎝0⎟⎠ ⎝⎜1⎟⎠

(ii) r = ⎛ 2⎞ + ⎛1⎞ and r ⎛1⎞
⎝⎜–1⎟⎠ λ ⎝⎜2⎠⎟ = µ ⎝⎜1⎠⎟

(iii) r = ⎛0⎞ + ⎛–2⎞ and r = ⎛ 0⎞ + ⎛1⎞
⎜⎝5⎟⎠ λ ⎝⎜–2⎟⎠ ⎜⎝–7⎟⎠ µ ⎝⎜2⎟⎠

(iv) r = ⎛−2⎞ + λ ⎛−1⎞ and r = ⎛1 ⎞ + µ ⎛ 2⎞
⎝⎜–3⎠⎟ ⎜⎝ 3⎠⎟ ⎜⎝3⎟⎠ ⎜⎝–1⎟⎠

(v) r = ⎛2⎞ + λ ⎛ 1⎞ and r = ⎛5⎞ + µ ⎛1⎞
⎜⎝7⎠⎟ ⎜ –1⎠⎟ ⎝⎜1 ⎟⎠ ⎝⎜2⎟⎠
⎝

269

10 2 Decide whether each of these pairs of lines intersect, are parallel or
are skew. If the lines intersect, find the coordinates of the point of
intersection.

⎛ 1⎞ ⎛1⎞ ⎛9⎞ ⎛ 2⎞
(i) r = ⎜⎜⎝−−61⎟⎠⎟ + λ ⎜⎜⎝23⎠⎟⎟ and r = ⎜⎝⎜72⎠⎟⎟ + µ ⎜⎜⎝−31⎠⎟⎟

10 VECTORS ⎛ 1⎞ ⎛ 6⎞ and ⎛−5⎞ ⎛ 2⎞
(ii) r = ⎜⎝⎜−60⎟⎠⎟ + λ ⎜⎝⎜−−93⎟⎠⎟ r = ⎜⎝⎜ 03⎟⎠⎟ + µ ⎜⎜⎝−−31⎟⎠⎟

⎛ 6⎞ ⎛ 1⎞ and ⎛ 1⎞ ⎛ 1⎞
(iii) r = ⎜⎜⎝−42⎠⎟⎟ + λ ⎜⎜⎝−25⎟⎟⎠ r = ⎝⎜⎜−147⎟⎟⎠ + µ ⎝⎜⎜−21⎟⎠⎟

⎛−1⎞ ⎛2⎞ ⎛−4⎞ ⎛ 5⎞
(iv) r = ⎝⎜⎜ 42⎠⎟⎟ + λ ⎜⎝⎜03⎟⎟⎠ and r = ⎜⎝⎜ 46⎟⎟⎠ + µ ⎝⎜⎜−21⎟⎟⎠

⎛ 0⎞ ⎛ 5⎞ ⎛ 2⎞ ⎛ 4⎞
(v) r = ⎜⎜⎝−41⎠⎟⎟ + λ ⎝⎜⎜−33⎟⎟⎠ and r = ⎜⎜⎝−51⎟⎟⎠ + µ ⎜⎝⎜−23⎟⎟⎠

⎛ 9⎞ ⎛ 1⎞ and ⎛ 1⎞ ⎛ 1⎞
(vi) r = ⎜⎝⎜−43⎟⎟⎠ + λ ⎜⎝⎜−23⎟⎟⎠ r = ⎝⎜⎜−45⎟⎟⎠ + µ ⎝⎜⎜−21⎠⎟⎟

⎛2⎞ ⎛ 1⎞ ⎛− 1⎞ ⎛ 1⎞
(vii) r = ⎜⎝⎜31⎟⎠⎟ + λ ⎜⎝⎜− 21⎟⎟⎠ and r = ⎜⎝⎜−−31⎟⎟⎠ + µ ⎜⎝⎜23⎟⎟⎠

PS 3 In this question the origin is taken to be at a harbour and the unit

vectors i and j to have lengths of 1 km in the directions E and N.

A cargo vessel leaves the harbour and its position vector t hours later is
given by

r1 = 12ti + 16tj.
A fishing boat is trawling nearby and its position at time t is given by

r2 = (10 − 3t)i + (8 + 4t)j.
(i) How far apart are the two boats when the cargo vessel leaves

harbour?

(ii) How fast is each boat travelling?

(iii) What happens to the boats?

270

4 The points A(1, 0), B(7, 2) and C(13, 7) are the vertices of a triangle. 10
The midpoints of the sides BC, CA and AB are L, M and N.
(i) Write down the position vectors of L, M and N. 10.6 The intersection of two lines
(ii) Find the vector equations of the lines AL, BM and CN.
(iii) Find the intersections of these pairs of lines.
(a) AL and BM (b) BM and CN
(iv) What do you notice?

⎛ −4⎞ ⎛ 2⎞ ⎛ 4⎞ ⎛ 2⎞
5 The line r = ⎝⎜⎜−142⎠⎟⎟ + q⎜⎜⎝−1101⎠⎟⎟ meets r = ⎝⎜⎜−−1165⎟⎠⎟ + s⎝⎜⎜−−35⎟⎟⎠ at A and meets

⎛ −1⎞ ⎛ 1⎞
r = ⎜⎜⎝−−293⎟⎠⎟ + t ⎝⎜⎜81⎟⎠⎟ at B.

Find the coordinates of A and the length of AB.

PS 6 To support a tree damaged in a gale a tree surgeon attaches wire guys to

four of the branches (see the diagram). He joins (2, 0, 3) to (−1, 2, 6) and
(0, 3, 5) to (−2, −2, 4). Do the guys, assumed straight, meet?

⎛ −7⎞ ⎛ 4⎞ ⎛ 3⎞ ⎛ 2⎞
7 Show that the three lines r = −244⎟⎠ + q −47⎟⎠ , r = −1105⎠⎟ + s −21⎠⎟ and
⎜ ⎜ ⎜ ⎜
⎝ ⎝ ⎝ ⎝

⎛ −3⎞ ⎛ 8⎞ t
r=⎜ 66⎟⎠ + −23⎟⎠ form a triangle and find the lengths of its sides.
⎝ ⎜
⎝

271

10 10.7 The angle between two lines

Earlier in this chapter, you learnt that the angle, θ, between two

vectors a = ⎛ aaa132 ⎞ and b = ⎛ bbb132 ⎞ can be found
⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ b

10 VECTORS using the formula

cos θ = a.b a
|a||b| θ

where a . b is the scalar product ▲ Figure 10.25
and a . b = a1b1 + a2b2 + a3b3.

The angle between two lines is the same as the angle between their direction
vectors.

Example 10.16 Verify that the lines

⎛ 1 ⎞ ⎛ −9 ⎞ ⎛ 1 ⎞ ⎛ 2 ⎞
⎜ 2 ⎟ ⎜ −2 ⎟ ⎜ 2 ⎟ ⎜ ⎟
r = ⎜⎜⎝ 3 ⎟⎟⎠ + λ ⎝⎜⎜ ⎟⎠⎟ and r = ⎜⎜⎝ 3 ⎟⎠⎟ + µ ⎜⎝⎜ −3 ⎠⎟⎟ are perpendicular.
4 3

Solution

When two lines are perpendicular, the angle between their direction
vectors is 90°.
Since cos 90° = 0 then a . b = 0.
So if the scalar product of two non-zero vector lines is zero then the lines are
perpendicular.

⎛−9⎞ ⎛ 2⎞
The direction vectors of these two lines are ⎜⎜⎝−42⎟⎟⎠ and ⎜⎜⎝−33⎠⎟⎟ .

⎛ −9⎞ ⎛ 2⎞
⎜ −2⎟ . ⎜ −3⎟ = (−9) × 2 + (−2) × (−3) + 4 × 3
⎝ 4⎠ ⎝ 3⎠

= (−18) + 6 + 12
=0

Therefore, the two lines are perpendicular.

272

Even if two lines do not meet, it is still possible to specify the angle between 10
them.The lines l and m shown in Figure 10.26 do not meet; they are skew.

l

θ

m' 10.8 The perpendicular distance from a point to a line

m

▲ Figure 10.26

The angle between them is that between their directions; it is shown
in Figure 10.26 as the angle θ between the lines l and m', where m' is a
translation of the line m to a position where it does intersect the line l.

Example 10.17 Find the angle between the skew lines

r = ⎛ 401⎞⎟⎠⎟⎟ + λ ⎛ 2 ⎞ and r = ⎛ 2 ⎞ + µ ⎛ 3 ⎞ .
⎜ ⎜⎜⎝⎜−− 1 ⎟ ⎜ ⎟ ⎜ 0 ⎟
⎜⎜⎝ 1 ⎠⎟⎟ ⎝⎜⎜ −1 ⎟⎟⎠ ⎜⎜⎝ 1 ⎟⎠⎟
3

Solution

The angle between the lines is the angle between their direction vectors

⎛ 2⎞ ⎛ 3⎞

⎜ ––11⎟⎟⎠⎟ and ⎜ 10⎟⎠⎟⎟ .
⎝⎜⎜ ⎜⎜⎝

Using cos θ = a.b
|a||b|

cos θ = 2 × 3 + (–1) × 0 + (–1) × 1
22 + (−1)2 + (−1)2 × 32 + 02 + 12

cos θ = 5
6 × 10

⇒ θ = 49.8°

10.8 The perpendicular distance from
a point to a line

The scalar product is also useful when determining the distance between a
point and a line.

273

10 Example 10.18 Find the shortest distance from point P(11, −5, −3) to the line l with equation

r ⎛1⎞ ⎛ − 3 ⎞ .
= ⎜ 5 ⎟ + λ ⎜ 1 ⎟
⎝⎜ 0 ⎟⎠ ⎝⎜ 4 ⎠⎟

10 VECTORS Solution
The shortest distance from P to the line l is ⏐N⎯→P⏐ where N is a point on the
line l and PN is perpendicular to the line l.

P

N
l

▲ Figure 10.27

You need to find the coordinates of N and then you can find ⏐N⎯→P⏐.

N lies on the line l. Let the value of λ at N be t.
So, relative to the origin O

O⎯→N = ⎛ 1⎞ ⎛ − 3⎞ ⎛1 − 3t⎞
05⎠⎟ + t⎜ 41⎟⎠ = 5+ t
⎜ ⎝ ⎜ 4t ⎟
⎝ ⎝ ⎠

and N⎯→P = O⎯→P − O⎯→N

⎛ 11⎞ ⎛1 − 3t⎞ t
−−53⎟⎠ − 5+
= ⎜ ⎜ 4t ⎟
⎝ ⎝ ⎠

⎛ 10 + 3t ⎞
= −−310−−4tt ⎠⎟
⎜
⎝ When two vectors are
perpendicular, their
As N⎯→P is perpendicular to the line l, scalar product is 0.

N⎯→P . ⎛ −3⎞ = 0 The direction of
⎜ 41⎟⎠ the line l
⎝

N⎯→P . ⎛ −3⎞ ⎛ 10 + 3t⎞ ⎛−3⎞
41⎟⎠ ⎝⎜⎜ −10 − t ⎠⎟⎟ ⎜⎜⎝ 41⎠⎟⎟
⎜ = −3 − 4t •
⎝

= (10 + 3t) ×(−3) + (−10 − t) ×1 + (−3 − 4t) ×4
= −30 − 9t − 10 − t − 12 − 16t
= −52 − 26t

274

The scalar product is 0, so 10

−52 − 26t = 0 ⇒ t = −2
Substituting t = −2 into O⎯→N and N⎯→P gives

O⎯→N ⎛ 1⎞ ⎛ − 3⎞ ⎛ 7⎞
05⎟⎠ − 2⎜ 41⎟⎠ = 83⎠⎟
= ⎜ ⎝ ⎜ −
⎝ ⎝
10.8 The perpendicular distance from a point to a line
N⎯→P ⎛ 10 + 3 × (−2)⎞ ⎛ 4⎞
and −−31−04−×(−(−2)2)⎠⎟ = −85⎠⎟
= ⎜ ⎜
⎝ ⎝

So ⏐N⎯→P⏐ = 4 2 + (−8)2 + 52

= 105
= 10.25 units

Exercise 10F ⎛ 1⎞ ⎛ 0⎞ ⎛ 001⎠⎟⎞
⎜ 00⎠⎟ ⎜
Remember i = ⎝ , j= ⎜ 01⎠⎟ and k = ⎝ .
⎝

In questions 1 to 5, find the angle between each pair of lines.

r = ⎛ 2⎞ s ⎛ 1⎞ and r = ⎛ 6⎞ ⎛ 2⎞
1 31⎟⎠ + 40⎟⎠ 140⎠⎟ + t 11⎠⎟
⎜ ⎜ ⎜ ⎜
⎝ ⎝ ⎝ ⎝

s ⎛ 4⎞ ⎛ 7⎞ ⎛ 1⎞
2 r= 41⎟⎠ and r = ⎜⎝⎜−03⎠⎟⎟ + t ⎝⎜⎜−21⎟⎟⎠
⎜
⎝

⎛ 4⎞ s ⎛ 3⎞ ⎛ 5⎞ ⎛ 2⎞
3 r = −21⎟⎠ + −47⎟⎠ and r = 01⎟⎠ + t 85⎟⎠
⎜ ⎜ ⎜ ⎜ −
⎝ ⎝ ⎝ ⎝

4 r = 2i + 3j + 4k + s(i + j − k) and r = t(i − k)

5 r = i − 2j − k + s(2i + 3j + 2k) and r = 2i + j + tk

6 For each point P and line l find

(a) the coordinates of the point N on the line such that PN is

perpendicular to the line

(b) the distance PN.

⎛ 0⎞ ⎛ −1⎞
(i) P(–2, 11, 5) and r = −23⎠⎟ + t⎜ 25⎠⎟
⎜ ⎝
⎝

275

10 (ii) P(7, –1, 6) and ⎛ 2⎞ ⎛ 1⎞
r = 13⎟⎠ + t − 42⎠⎟
276 ⎜ ⎜
⎝ ⎝

(iii) P(8, 4, –1) and ⎛ 1⎞ ⎛ −1⎞ t
r = −53⎟⎠ + −02⎟⎠
⎜ ⎜
⎝ ⎝

10 VECTORS 7 Find the perpendicular distance of the point P(–7, –2, 13) to the line

⎛ 1⎞ ⎛ 1⎞
r = ⎜⎜⎝25⎟⎠⎟ + λ ⎜⎜⎝− 43⎟⎟⎠ .

8 Find the distance of the point C(0, 6, 0) to the line joining the points
A(–4, 2, –3) and B(–2, 0, 1).

M 9 The diagram shows an extension to a house. Its base and walls are

rectangular and the end of its roof, EPF, is sloping, as illustrated.

E Q (2, 5, 4) G (4, 5, 3)
(0, 0, 3) H B (4, 5, 0)

(2, 1, 4)
P

F
C

(0, 0, 0) O A

(i) Write down the coordinates of A and F.

(ii) Find, using vector methods, the angles FPQ and EPF.

The owner decorates the room with two streamers which are pulled
taut. One goes from O to G, the other from A to H. She says that they
touch each other and that they are perpendicular to each other.
(iii) Is she right?

10 The points A and B have position vectors, relative to the origin O, given by 10
O⎯→A = i + 2j + 3k and O⎯→B = 2i + j + 3k.
10.8 The perpendicular distance from a point to a line
The line l has vector equation

r = (1 − 2t)i + (5 + t)j + (2 − t)k.

(i) Show that l does not intersect the line passing through A and B.

(ii) The point P lies on l and is such that angle PAB is equal to 60°.
Given that the position vector of P is (1 − 2t)i + (5 + t)j + (2 − t)k,
show that 3t2 + 7t + 2 = 0. Hence find the only possible position
vector of P.
Cambridge International AS & A Level Mathematics
9709 Paper 3 Q10 June 2008

11 With respect to the origin O, the position vectors of two points A and B
are given by O⎯→A = i + 2 j + 2k and O⎯→B = 3i + 4 j.The point P lies on
the line through A and B, and ⎯A→P = λ ⎯A→B .
(i) Show that O⎯→P = (1 + 2λ)i + (2 + 2λ) j + (2 − 2λ)k.

(ii) By equating expressions for cos AOP and cos BOP in terms of λ,
find the value of λ for which OP bisects the angle AOB.

(iii) When λ has this value, verify that AP : PB = OA : OB.
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q7 November 2011

12 The equations of two straight lines are
r = i + 4 j − 2k + λ(i + 3k) and r = ai + 2 j − 2k + µ(i + 2 j + 3ak),
where a is a constant.

(i) Show that the lines intersect for all values of a.

(ii) Given that the point of intersection is at a distance of 9 units from
the origin, find the possible values of a.
Cambridge International AS & A Level Mathematics
9709 Paper 33 Q7 November 2014

13 The straight line l1 passes through the points (0, 1, 5) and (2, −2, 1).The
straight line l2 has equation r = 7i + j + k + µ(i + 2 j + 5k) .

(i) Show that the lines l1 and l2 are skew.
(ii) Find the acute angle between the direction of the line l2 and the

direction of the x-axis.
Cambridge International AS & A Level Mathematics
9709 Paper 31 Q6 June 2015

277

10 KEY POINTS

278 1 A vector quantity has magnitude and direction.

2 A scalar quantity has magnitude only. tah,eofroarsm⎯O→OA⎯.→A.They

3 Vectors are typeset in bold, a or OA, or in are
handwritten either in the underlined form

10 VECTORS 4 The length (or modulus or magnitude) of the vector a is written as a
or as |a|.

5 Unit vectors in the x, y and z directions are denoted by i, j and k,
respectively.

6 A vector may be specified in

● magnitude−direction form: (r, θ ) (in two dimensions)

( )● x
component form: x i + y j or y (in two dimensions)

xi yj zk or ⎛ x⎞ (in three dimensions).
+ + zy⎟⎠
⎜
⎝

7 The position vector O⎯→P of a point P is the vector joining the origin
to P.

8 The vector A⎯→B is b − a, where a and b are the position vectors of A
and B.

9 The angle between two vectors, a and b, is given by θ in

cosθ = a⋅b

|a||b|

where a . b = a1b1 + a2b2 (in two dimensions)
= a1b1 + a2b2 + a3b3 (in three dimensions).

10 The position vector O⎯→P of a point P is the vector joining the origin
to P.

11 The vector A⎯→B is b – a, where a and b are the position vectors of A
and B.

12 The vector r often denotes the position vector of a general point.

13 The vector equation of the line through A with direction vector u is
given by

r = a + λu.
14 The vector equation of the line through points A and B is given by

r = O⎯→A + λA⎯→B
= a + λ(b − a)
= (1 − λ)a + λb.

15 The vector equation of the line through (a1, a2, a3) in the direction 10

⎛u1 ⎞ ⎜⎜⎝⎛aaa123 ⎞ ⎛⎝⎜⎜uuu132 ⎞
⎜ u ⎟ is r = ⎟ ⎟ .
⎝⎜ u 2 ⎟⎠ ⎟ + λ ⎟
3 ⎠ ⎠

LEARNING OUTCOMES 10.8 The perpendicular distance from a point to a line

Now you have finished this chapter, you should be able to
■ understand the terms vector and scalar
■ understand vectors in two and three dimensions, and express them

■ using i, j and k vectors
■ using column vectors
■ using O⎯→A or a notation
■ understand equal vectors, position vectors and displacement vectors
■ understand the link between the coordinates of a point and its position
vector
■ multiply a vector by a scalar
■ add and subtract vectors
■ find a unit vector in the direction of a given vector
■ understand that vectors are parallel when one is a scalar multiple of the
other
■ find the vector equation of a line
■ calculate the scalar product between two vectors and use it to find the
angle between
■ two vectors
■ two lines
■ determine whether two lines are
■ parallel
■ intersect
■ skew
■ use vectors in geometry problems.

279

P3

11 Complex numbers

11 COMPLEX NUMBERS …that wonder
of analysis, that
portent of the
ideal world,
that amphibian
between being
and not-being,
which we call
the imaginary
root of negative
unity.
Gottfried
Wilhelm Leibniz
(1646–1716)

Real numbers
Rational numbers
Integers

Natural numbers

▲ Figure 11.1

❯ What is the meaning of each of the terms shown in Figure 11.1? ?

❯ Suggest two numbers that could be placed in each part of the
diagram.

280

11.1 Extending the number system 11

The number system we use today has taken thousands of years to develop.
To classify the different types of numbers used in mathematics the following
letter symbols are used:

‫ ގ‬Natural numbers ? 11.1 Extending the number system
‫ ޚ‬Integers
‫ ޑ‬Rational numbers
‫ ޑ‬Irrational numbers
‫ ޒ‬Real numbers

❯ Why is there no set shown on the diagram for irrational numbers?

You may have noticed that some of these sets of numbers fit within the other
sets.This can be seen in Figure 11.1.

ACTIVITY 11.1

On a copy of Figure 11.1 write the following numbers in the correct

positions. .
0. 3
7 5 –13 227 –5 3.1415 π 0.33
109

What are complex numbers?

ACTIVITY 11.2

Solve each of these equations and decide which set of numbers the roots
belong to in each case.

(i) x + 7 = 9
(ii) 7x = 9
(iii) x2 = 9
(iv) x + 10 = 9
(v) x2 + 7x = 0

Now think about the equation x2 + 9 = 0. Writing this quadratic

You could rewrite it as x2 = –9. However,since equation as x2 + 0x + 9 = 0
the square of every real number is positive or zero,
there is no real number with a square of –9.This and calculating the
is an example of a quadratic equation which,up discriminant for this
to now,you would have classified as having no quadratic gives
real roots.
b2 – 4ac = –36 which is

less than zero.

281

11 11 COMPLEX NUMBERS The existence of such equations was recognised for hundreds of years, in
the same way that Greek mathematicians had accepted that x + 10 = 9 had
282 no solution; the concept of a negative number had yet to be developed.
The number system has expanded as mathematicians increased the range of
mathematical problems they wanted to tackle.

You can solve the equation x2 + 9 = 0 by extending the number system to
include a new number, i (sometimes written as j).This has the property that
i2 = −1 and it follows the usual laws of algebra. i is called an imaginary
number.

The square root of any negative number can be expressed in terms of i. For
example, the solution of the equation x2 = −9 is x = ± −9. This can be
written as ± 9 × −1 which simplifies to ±3i.

11.2 Working with complex numbers

Faced with the problem of wanting the square root of a negative number, we
make the following Bold Hypothesis.

The real number system can be extended by including a new number,denoted by i,
which combines with itself and the real numbers according to the usual laws of algebra,
but which has the additional property that i2 = −1.

The original notation for i was ι, the Greek letter iota.The letter j is also
commonly used instead of i.

The first thing to note is that we do not need further symbols for other
square roots. For example, since −196 = 196 × (−1) = 142 × i2, we see that
−196 has two square roots, ±14i.The following example uses this idea to
solve a quadratic equation with no real roots.

Example 11.1 Solve the equation z2 − 6z + 58 = 0, We use the letter z for the variable
and check the roots. here because we want to keep x
and y to stand for real numbers.

Solution

Using the quadratic formula:

z= 6± 62 − 4 × 58
2

= 6± −196
2

= 6 ± 14i
2

= 3 ± 7i

To check: 11

z = 3 + 7i ⇒ z2 − 6z + 58 = (3 + 7i)2 − 6(3 + 7i) + 58

= 9 + 42i + 49i2 − 18 − 42i + 58

Notice that here = 9 + 42i − 49 − 18 − 42i + 58
= 0 i2 = −1
0 means 0 + 0i.

ACTIVITY 11.3 11.2 Working with complex numbers

Check the other root, z = 3 − 7i.

Notice in particular A number z of the form x + iy, where x and y are real, is called a complex
that the imaginary number. x is called the real part of the complex number, denoted by Re(z),
part is real! and y is called the imaginary part, denoted by Im(z). So if, for example,
z = 3 − 7i then Re(z) = 3 and Im(z) = −7.

In Example 11.1 you did some simple calculations with complex numbers.
The general methods for addition, subtraction and multiplication are similarly
straightforward.
» Addition: add the real parts and add the imaginary parts.

(x + iy) + (u + iv) = (x + u) + i(y + v)
» Subtraction: subtract the real parts and subtract the imaginary parts.

(x + iy) − (u + iv) = (x − u) + i(y − v)

» Multiplication: multiply out the brackets in the usual way and simplify,
remembering that i2 = −1.

(x + iy)(u + iv) = xu + ixv + iyu + i2yv ?
= (xu − yv) + i(xv + yu)

Division of complex numbers is dealt with later in the chapter.

❯ What are the values of i3, i4, i5?
❯ Explain how you would work out the value of in for any positive

integer value of n.

Complex conjugates

The complex number x − iy is called the complex conjugate, or just the
conjugate, of x + iy. Simarly x + iy is the complex conjugate of x − iy. x + iy
and x − iy are a conjugate pair.The complex conjugate of z is denoted by z*.
If a polynomial equation, such as a quadratic, has real coefficients, then any
complex roots will be conjugate pairs.This is the case in Example 11.1. If,
however, the coefficients are not all real, this is no longer the case.

283

11 You can solve quadratic equations with complex coefficients in the same
way as an ordinary quadratic, either by completing the square or by using the
quadratic formula.This is shown in the next example.

Example 11.2 Solve z2 − 4iz − 13 = 0.

11 COMPLEX NUMBERS Solution
Substitute a = 1, b = −4i and c = −13 into the quadratic formula.

z = −b ± b 2 − 4ac
2a

= 4i ± (−4i)2 − 4 × 1 × (−13)
2

= 4i ± −16 + 52
2

= 4i ± 36
2

= 4i ± 6
2

= 2i ± 3

So the roots are 3 + 2i and –3 + 2i.

ACTIVITY 11.4

(i) Let z = 3 + 5i and w = 1 − 2i.
Find the following.

(a) z + z* (b) w + w* (c) zz* (d) ww*

What do you notice about your answers?

(ii) Let z = x + iy.
Show that z + z* and zz* are real for any values of x and y.

Equality of complex numbers

Two complex numbers z = x + i y and w = u + i v are equal if both x = u
and y = v. If y ≠ u or y ≠ v, or both, then z and w are not equal.

You may feel that this is obvious, but it is interesting to compare this situation
with the equality of rational numbers.

284

The rational numbers x and u are equal if x= u and y= v. 11
y v

?

❯ Is it possible for the rational numbers x and u to be equal if u ≠ x
and v ≠ y? y v

For two complex numbers to be equal, the real parts must be equal and the 11.2 Working with complex numbers
imaginary parts must be equal. Using this result is described as equating real
and imaginary parts, as shown in the following example.

Example 11.3 The complex numbers z1 and z2 are given by : z1 = (3 − a) + (2b − 4)i
and z2 = (7b − 4) + (3a − 2)i

(i) Given that z1 and z2 are equal, find the values of a and b.
(ii) Check your answer by substituting your values for a and b into the

expressions above.

Solution

(i) (3 − a) + (2b − 4)i = (7 − 2b) + (3a − 2)i Equating real and
imaginary parts leads
Equating real parts: 3 − a = 7b − 4 to two equations.

Equating imaginary parts: 2b − 4 = 3a − 2

7b + a = 7 Simplifying the
2b − 3a = 2 equations.

Solving simultaneously gives b = 1 and a = 0.

(ii) Substituting a = 0 and b = 1 gives z1 = 3 − 2i and z2 = 3 − 2i
so z1 and z2 are indeed equal.

Exercise 11A 1 Express the following in the form x + iy.

(i) (8 + 6i) + (6 + 4i) (ii) (9 − 3i) + (−4 + 5i)

(iii) (2 + 7i) − (5 + 3i) (iv) (5 − i) − (6 − 2i)

(v) 3(4 + 6i) + 9(1 − 2i) (vi) 3i(7 − 4i)

(vii) (9 + 2i)(1 + 3i) (viii) (4 − i)(3 + 2i)

(ix) (7 + 3i)2 (x) (8 + 6i)(8 − 6i)

(xi) (1 + 2i)(3 − 4i)(5 + 6i) (xii) (3 + 2i)3

285

11 11 COMPLEX NUMBERS 2 Solve each of the following equations, and check the roots in each case.
(i) z2 + 2z + 2 = 0
286 (ii) z2 − 2z + 5 = 0
(iii) z2 − 4z + 13 = 0
(iv) z2 + 6z + 34 = 0
(v) 4z2 − 4z + 17 = 0
(vi) z2 + 4z + 6 = 0

3 Solve each of the following equations.
(i) z2 − 4iz − 4 = 0
(ii) z2 − 2iz + 15 = 0
(iii) z2 − 2iz − 2 = 0
(iv) z2 + 6iz − 13 = 0
(v) z2 + 8iz − 17 = 0
(vi) z2 + iz + 6 = 0

4 Given that z = 2 + 3i and w = 6 − 4i, find the following.

(i) Re(z)

(ii) Im(w)
(iii) z*
(iv) w*
(v) z* + w*
(vi) z* − w*
(vii) Im(z + z*)
(viii) Re(w − w*)
(ix) zz* − ww*
(x) (z3)*
(xi) (z*)3
(xii) zw* − z*w

CP 5 Let z = x + iy.
Show that (z*)* = z.

CP 6 Let z1 = x1 + iy1 and z2 = x2 + iy2.
Show that (z1 + z2)* = z1* + z2*.

CP 7 Given that the complex numbers
z1 = a2 + (3 + 2b)i
z2 = (5a − 4) +b2i

are equal, find the possible values of a and b.

Hence list the possible pairs of complex numbers z1 and z2.