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9780521740494c15.xml CUAU033-EVANS September 13, 2008 9:27

C H A P T E R

15Vectors

ObjectivesTo understand the concept of vector

To apply basic operations to vectors

To understand the zero vector

To use the unit vectors i and j to represent vectors in two dimensions

To use the fact that, if a and b are parallel, then a = kb for a real value k. The

converse of this also holds.

To use the unit vectors i, j and k to represent vectors in three dimensions

15.1 Introduction to vectorsFor experiments in science or engineering some of the things which are measured are

completely determined by their magnitude. For example, mass, length or time are determined

by a number and an appropriate unit of measurement.

e.g., length: 30 cm is the length of the page of a particular book

time: 10 s is the time for one athlete to run 100 m

More is required to describe velocity, displacement or force. The direction must be recorded

as well as the magnitude.

e.g., displacement: 30 km in a direction north

velocity: 60 km/h in a direction south east

Quantities that have direction as well as magnitude are represented by arrows that point in

the direction of the action and whose lengths give the magnitude of the quantity in terms of a

suitably chosen unit.

Arrows with the same length and direction are regarded as equivalent. These arrows are

directed line segments and the sets of equivalent segments are called vectors.

390

SAMPLE

Cambridge University Press • Uncorrected Sample Pages • 978-0-521-61252-4 2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

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Chapter 15 — Vectors 391

The five directed line segments shown all have the same

magnitude and direction.

y

A

B D

C

O

P

xF

EG

H

A directed line segment from a point A to a point B is

denoted by−→AB.

For simplicity of language this is also called vector−→AB,

i.e., the set of equivalent segments can be named through

one member of that set.

Note:−→AB = −→

CD = −→OP = −→

EF = −→GH

In Chapter 8, vectors were introduced in the context of

translations (in two dimensions). A column of numbers

was introduced to represent the translation and it was

called a vector.

This is consistent with the approach here as the column

of numbers corresponds to a set of equivalent directed

line segments.

The column

[3

2

]corresponds to the directed line segment

which goes 3 across and 2 up.

x

y

B

2 units

3 unitsA

0

This notation will be used to represent a directed line segment in the first section of this

chapter. The notation is widely used to represent vectors but not to a large extent in Victorian

schools and so the notation, although useful, will be abandoned in the latter sections of the

chapter.

Vectors are often denoted by a single bold face roman letter. The vector from A to B can be

denoted by−→AB or by a single v. That is, v = −→

AB.

When a vector is handwritten the notation is v∼.

Addition of vectorsTwo vectors u and v can be added geometrically by drawing

a line segment representing u from A to B and then a line

segment from B to C representing v.

The sum u + v is the vector from A to C.

That is, u + v = −→AC .

C

v

B

u

A

u + v

C

v

B

u

u + v

u

D

v

A

The same result is achieved if the order is reversed. This is

represented in the following diagram

i.e. u + v = −→AC

and u + v = v + uSAMPLE


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392 Essential Advanced General Mathematics

The addition can also be achieved with the column

vector notation.

e.g., if u =[

4

1

]and v =

[−1

3

]

then u + v =[

4

1

]+

[−1

3

]=

[3

4

]

3

v

u

1

4

4

1

3u + v

Scalar multiplicationMultiplication by a real number (scalar) changes the length of

the vector.

u

2u

u12

e.g., 2u = u + u and1

2u + 1

2u = u

2u is twice the length of u and1

2u is half the length of u.

The vector ku, k ∈ R+, with u �= 0, has the same direction as u, but its length is

multiplied by a factor of k.

When a vector is multiplied by −2 the vector’s direction is reversed

and its length doubled.u

–2u

When a vector is multiplied by −1 the vector’s direction is reversed

and the length remains the same.

If u =[

3

2

], −u =

[−3

−2

], 2u =

[6

4

]and −2u =

[−6

−4

]

If u = −→AB then −u = − −→

AB = −→BA.

The directed line segment − −→AB goes from B to A.

Zero vectorThe zero vector is denoted by 0 and represents a line segment of zero length. The zero vector

has no direction.

Subtraction of vectorsIn order to subtract v from u, add −v to u.

For example

uv

–v

uu – v

v

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Example 1

Draw the directed line segment defined by

[3

−2

].

Solution[3

−2

]is the vector ‘3 across to the right and 2 down’

Note: Here the vector starts at (1, 1) and goes to

(4, −1). It can start at any point.

x

y

A

B

4320

–1

1

1

Example 2

The vector u is defined by the directed line segment from (2, 6) to (3, 1).

If u =[

a

b

], find a and b.

Solution

The vector u =[

3

1

− 2

− 6

]=

[1

−5

]

Hence a = 1 and b = −5x

y

A

0

B (3, 1)

(2, 6)

Example 3

If the vector u =[

3

−1

]and the vector v =

[−2

2

], find 2u + 3v.

Solution

2u + 3v = 2 ×[

3

−1

]+ 3 ×

[−2

2

]

=[

6

−2

]+

[−6

6

]

=[

0

4

]

Polygons of vectorsFor two vectors

−→AB and

−→BC,

−→AB + −→

BC = −→AC

B

A

CSAM

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For a polygon ABCDEF,−→AB + −→

BC + −→CD + −→

DE + −→EF + −→

FA = 0 B C

F E

DA

Example 4

Illustrate the vector sum−→AB + −→

BC + −→CD, where A, B, C and D are points in the plane.

Solution

−→AB + −→

BC + −→CD = −→

ADB

C

A D

Parallel vectorsThe non-zero vectors u and v are said to be parallel if there exists k ∈ R\{0} such that u = kv.

If u =[

−2

3

]and v =

[−6

9

]then vector u is parallel to v as v = 3u.

Position vectorsA point O, the origin, can be used as a starting point for a vector to indicate the position of a

point in space relative to that point. In this chapter, O is the origin for a cartesian plane (three

dimensional work is considered briefly).

For a point A the position vector is−→OA.

Linear combinations of non-parallel vectorsIf a and b are non-zero, non-parallel vectors, then

ma + nb = pa + qb

implies m = p and n = q

An argument is as follows:

ma − pa = qb − nb

∴ (m − p)a = (q − n)b

∴ a = q − n

m − pb or b = m − p

q − na

but a and b are not parallel and not zero

∴ q = n and m = p

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Example 5

Let A, B and C be the vertices of a triangle.

Let D be the midpoint of BC.

Let−→AB = a and

−→BC = b.

Find the following in terms of a and b.

a−→BD b

−→DC c

−→AC d

−→AD e

−→CA

BD

C

A

Solution

a−→BD = 1

2

−→BC = 1

2b (same direction and half the length)

b−→

DC = −→BD = 1

2b (equivalent vectors)

c−→AC = −→

AB + −→BC = a + b

d−→AD = −→

AB + −→BD = a + 1

2b

e−→CA = − −→

AC = −(a + b) (−→CA + −→

AC = 0)

Example 6

In the figure,−→

DC = kp where k ∈ R\{0}.a Express p in terms of k, q and r.

b Express−→FE in terms of k and p to show FE is parallel to DC.

c If−→FE = 4

−→AB, find the value of k.

Ap

B

rq

D C

r

F

q

E

Solution

a p = −→AB = −→

AD + −→DC + −→

CB

= q + kp − r

∴ (1 − k)p = q − r

and hence p = 1

1 − k(q − r)

b−→FE = −2q + p + 2r

= 2(r − q) + p

But r − q = kp − p = (k − 1)p . . . from a

∴ −→FE = 2kp − 2p + p

= (2k − 1)p

c If−→FE = 4

−→AB

(2k − 1)p = 4p

2k − 1 = 4

k = 5

2

Exercise 15A

1 On the same graph, draw the arrows which represent the following vectors.Example 1

a

[1

5

]b

[0

−2

]c

[−1

−2

]d

[−4

3

]

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2 The vector u is defined by the directed line segment from (1, 5) to (6, 6). If u =[

a

b

],Example 2

find a and b.

3 The vector v is defined by the directed line segment from (−1, 5) to (2, −10). If

v =[

a

b

], find a and b.

4 A = (1, −2), B = (3, 0), C = (2, −3) and O is the origin. Express the following vectors

in the form

[a

b

].

a−→OA b

−→AB c

−→BC d

−→CO e

−→CB

5 A = (2, −3), B = (4, 0), C = (1, −4) and O is the origin. Sketch the following vectors.Example 4

a−→OA b

−→AB c

−→BC d

−→CO e

−→CB

6 On graph paper, sketch the vectors joining the following pairs of points in the direction

indicated.

a (0, 0) → (2, 1) b (3, 4) → (0, 0) c (1, 3) → (3, 4)

d (2, 4) → (4, 3) e (−2, 2) → (5, −1) f (−1, −3) → (3, 0)

7 Identify vectors from 6 which are parallel to each other.

8 a Plot the points A (−1, 0) , B (1, 4) , C (4, 3) , D (2, −1) on a set of coordinate axes.

b Sketch the vectors−→AB,

−→BC,

−→AD, and

−→DC.

c Show that

i−→AB = −→

DC ii−→BC = −→

AD

d Describe the shape of the quadrilateral ABCD.

9 Let a =[

1

2

], b =

[1

−3

]and c =

[−2

1

].Example 3

a Find

i a + b ii 2c − a iii a + b − c

b Show that a + b is parallel to c.

10 Find the values of m and n such that m

[3

−3

]+ n

[2

4

]=

[−19

61

]

AM

B

CN

D

11 In the figure A, B, C, D are the vertices of a parallelogram.

−→AB = a and

−→AD = b

M, N are the midpoints of AB and DC respectively.

a Express the following in terms of a and b.

i−→

MD ii−→

MN

b Find the relationship between−→

MN and−→AD.

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12 The figure represents the triangle ABC with−→AB = a

and−→AC = b.

M, N are midpoints of−→AB and

−→AC respectively.

a Express−→CB and

−→MN in terms of a and b.

b Hence describe the relation between the

two vectors (or directed line segments).

AM

C

N

B

13 The figure represents the regular hexagon ABCDEF

Example 5

with vectors−→AF = a and

−→AB = b.

Express the following vectors in terms of a and b.

a−→

CD b−→ED c

−→BE d

−→FC

e−→FA f

−→FB g

−→FE

B

A F

E

DC

14 In parallelogram ABCD,−→AB = a and

−→BC = b. Express each of the following vectors in

Example 6

terms of a and b.

a−→

DC b−→DA c

−→AC d

−→CA e

−→BD

15 In triangle OAB,−→OA = a and

−→OB = b. P is a point on AB such that

−→AP = 2

−→PB and Q is a

point such that−→OP = 3

−→PQ. Express each of the following in terms of a and b.

a−→BA b

−→PB c

−→OP d

−→PQ e

−→BQ

16 PQRS is a quadrilateral in which−→PQ = u,

−→QR = v,

−→RS = w . Express the following

vectors in terms of u, v and w.

a−→PR b

−→QS c

−→PS

17 OABC is a parallelogram.−→OA = u,

−→OC = v. M is the midpoint of AB.

a Express−→OB and

−→OM in terms of u and v.

b Express−→

CM in terms of u and v.

c If P is a point on−→

CM and−→CP = 2

3

−→CM , express

−→CP in terms of u and v.

d Find−→OP and hence show that P lies on

−→OB.

e Find the ratio OP : PB.

x

y B(5, 7)

A(2, 3) X

O

15.2 Components of vectorsThe vector

−→AB illustrated opposite can be described

by the column vector

[3

4

]. From the diagram it is

possible to see that−→AB can be expressed as the sum

of two vectors−→AX and

−→XB,

i.e.,−→AB = −→

AX + −→XB.

SAMPLE


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In column vector notation[3

4

]=

[3

0

]+

[0

4

]

This suggests the introduction of two important vectors.

Let i be the vector of unit length with direction

the positive direction of the x axis.

x

y

O

j

i

Ox

y

u yj

xi

Let j be the vector of unit length with direction

the positive direction of the y axis.

Note that using the column notation, i =[

1

0

]and j =

[0

1

].

For the example above,−→AX = 3i and

−→XB = 4j

Therefore−→AB = 3i + 4j.

It is possible to describe any two-dimensional vectors in this way.

For a vector u =[

x

y

], u = x i + yj. It is said that u is

the sum of two components xi and yj.

The magnitude of vector u = x i + yj is denoted

by |u| and |u| =√

x2 + y2

Operations with vectors now look more like basic algebra.

(x i + yj) + (mi + nj) = (x + m) i + (y + n) j

k (x i + yj) = kx i + kyj

For a = x i + yj and b = mi + nj,

a = b if x = m and y = n

Example 7

a Find−→AB if

−→OA = 3i and

−→OB = 2i − j

b Find |2i − 3j|.

Solution

a−→AB = − −→

OA + −→OB

= −3i + (2i − j)

= −i − j

b |2i − 3j| = √4 + 9 = √

13

Example 8

A, B are points on the cartesian plane such that−→OA = 2i + j and

−→OB = i − 3j. Find−→

AB and | −→AB|.

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Solution−→AB = −→

AO + −→OB

= − −→OA + −→

OB

∴ −→AB = −(2i + j) + i − 3j

= −i − 4j

∴ | −→AB| = √

1 + 16 = √17

Unit vectorsA unit vector is a vector of length one unit (i and j are unit vectors).

The unit vector in the direction of a is a (pronounced ‘a hat’).

|a| = 1 so |a| a = a

a = a

|a| or1

|a|a

Example 9

Let a = 3i + 4j.

Find |a|, the magnitude of a, and hence the unit vector in the direction of a.

Solution

a = 3i + 4j so |a| =√

x2 + y2

=√

32 + 42

|a| = 5

a = 1

|a|a

So a = 1

5(3i + 4j)

Exercise 15B

1 A, B are points on the cartesian plane such that−→OA = i + 2j and

−→OB = 3i − 5j. Find

−→AB.Example 7

2 OAPB is a rectangle in which the vector−→OA = 5i and the vector

−→OB = 6j.

Express the following vectors in terms of i and j.

a−→OP b

−→AB c

−→BA

3 Determine the magnitude of the following vectors.Example 8

a 5i b −2j c 3i + 4j d −5i + 12j

4 The vectors u and v are given by u = 7i + 8j and v = 2i − 4j.

a Find |u − v|.b Find constants x and y such that xu + yv = 44j.

SAMPLE


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5 In the triangle OAB,−→OA = 10i and

−→OB = 4i + 5j. If M is the midpoint of AB, find

−→OM in

terms of i and j.

6 In the rectangle OPAQ,−→OP = 2i and

−→OQ = j. M is the point on OP such that OM = 1

5OP.

N is the point on MQ such that MN = 1

6MQ.

a Find the following vectors in terms of i and j.

i−→

OM ii−→

MQ iii−→MN iv

−→ON v

−→OA

b i Hence show that N is on the diagonal OA.

ii State the ratio of the lengths ON : NA.

7 The position vectors of A and B are given by−→OA =

[1

3

]and

−→OB =

[5

−1

].

Find the distance between A and B.

8 Find the pronumerals in the following equations.

a 2i + 3j = 2 (li + kj) b (x − 1) i + yj = 5i + (x − 4) j

c (x + y) i + (x − y) j = 6i d k (i + j) = 3i − 2j + l (2i − j)

9 Let A = (2, 3) and B = (5, 1). Find

a−→AB b | −→

AB|

10 Let OA = 3i, OB = i + 4j and OC = −3i + j. Find

a−→AB b

−→AC c | −→

BC|

11 Let A = (5, 1) , B = (0, 4) and C = (−1, 0). Find

a D such that−→AB = −→

CD b F such that−→AF = −→

BC c G such that−→AB = 2

−→GC

12 Let a = i + 4j and b = −2i + 2j. A, B and C are points such that−→AO = a,

−→OB = b,−→

BC = 2a and O is the origin. Find the coordinates of A, B and C.

13 A, B, C and D are the vertices of a parallelogram and O is the origin.

A = (2, −1), B = (−5, 4) and C = (1, 7).

a Find

i−→OA ii

−→OB iii

−→OC iv

−→BC v

−→AD

b Hence find the coordinates of D.

x

y

O

RQ

P

14 The diagram shows a parallelogram OPQR.

The points P and Q have coordinates (12, 5) and (18, 13)

respectively. Find

a−→OP and

−→PQ b | −→

RQ | and | −→OR|

15 A(1, 6), B(3, 1), C(13, 5) are the vertices of a triangle ABC

a Find

i | −→AB| ii | −→

BC| iii | −→CA|

b Hence show that ABC is a right-angled triangle.

SAMPLE


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16 A(4, 4), B(3, 1) and C(7, 3) are the vertices of the triangle ABC.

a Find the vector

i−→AB ii

−→BC iii

−→CA

b Find

i | −→AB| ii | −→

BC| iii | −→CA|

c Hence show that triangle ABC is an isosceles right-angled triangle.

17 A(−3, 2), B(0, 7) are points on the cartesian plane. O is the origin. M is the

midpoint of AB.

a Find

i−→OA ii

−→OB iii

−→BA iv

−→BM

b Hence find the coordinates of M. (Hint:−→

OM = −→OB + −→

BM)

18 Find the unit vector in the direction of each of the following vectors.Example 9

a a = 3i + 4j b b = 3i − j c c = −i + j

d d = i − j e e = 1

2i + 1

3j f f = 6i − 4j

y

x

(x, y, z)

z

y

z

k

i

j

11

1

xz

A

yj

zk

Bx

xiO y

15.3 Vectors in three dimensionsPoints in three dimensions (3-D) are represented in axes

as shown. Vectors in 3-D are of the form

a =

x

y

z

= x i + yj + zk where

i =

1

0

0

, j =

0

1

0

and k =

0

0

1

i, j and k are represented in the figure.

−→OA = x i + yj + zk

By Pythagoras’ theorem

OB2 = x2 + y2

and OA2 = OB2 + BA2

= x2 + y2 + z2

∴ | −→OA| =

√x2 + y2 + z2

Example 10

Let a = i + j − k and b = i + 7k. Find

a a + b b b − 3a c |a|

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Solution

a a + b = i + j − k + i + 7k = 2i + j + 6k

b b − 3a = i + 7k − 3(i + j − k) = −2i − 3j + 10k

c |a| = √3

Example 11

OABCDEFG is a cuboid.−→OA = 3j,

−→OC = k,

−→OD = i.

a Express each of the following in terms of i, j, k.

i−→

OE ii−→

OF iii−→

GF iv−→GB

b M, N are the midpoints of−→

OD and−→

GF respectively.

Find MN .D

G

C

O

F

B

A

ESolution

a i−→

OE = −→OA + −→

AE = 3j + i (−→AE = −→

OD)

ii−→

OF = −→OE + −→

EF = 3j + k + i (−→EF = −→

OC)

iii−→

GF = −→OA = 3j

iv−→GB = −→

DA = −→DO + −→

OA = −i + 3j

b−→

MN = −→MD + −→

DN = −→MD + −→

DG + −→GN

= 1

2i + k + 3

2j

∴ | −→MN | =

√1

4+ 1 + 9

4=

√14

2

Example 12

If a = 3i + 2j + 2k, find a.

Solution

|a| = √9 + 4 + 4 =

√17

∴ a = 1√17

(3i + 2j + 2k)

= 3√17

i + 2√17

j + 2√17

k

Exercise 15C

1 Let a = i + j + 2k, b = 2i − j + 3k, c = −i + k. FindExample 10

a a − b b 3b − 2a + c c |b| d |b + c| e 3(a − b) + 2c

2 If a = 3i + j − kExample 12

a find

i a ii −2a

b find the vector b in the direction of a such that |b| = 5.

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3 If a = i − j + 5k and b = 2i − j − 3k find the vector c in the direction of a such that

|c| = |b|.4 P and Q are points defined by the position vectors i + 2j − k and 2i − j − k respectively.

M is the midpoint of−→PQ. Find

a−→PQ b | −→

PQ| c−→

OM

G

D

C

O

F

B

E

A

5 OABCDEFG is a cuboid.

−→OA = 2j,

−→OC = 2k,

−→OD = i

Express the following vectors in terms of i, j and k.

a−→OB b

−→OE c

−→OG d

−→OF

e−→ED f

−→EG g

−→CE h

−→BD

Example 11

G

D

C

O

FB

E

A

6 OABCDEFG is a cuboid.

−→OA = 3j,

−→OC = 2k,

−→OD = i

M is such that−→

OM = 1

3

−→OE.

N is the midpoint of BF. Find

a−→

MN b | −→MN |

15.4 ApplicationsExample 13

Three points P, Q, and R have position vectors p, q, and k(2p + q) respectively, relative to a

fixed origin O. O, P and Q are not collinear. Find the value of k if

a−→QR is parallel to p b

−→PR is parallel to q c P, Q and R are collinear.

Solution

a−→QR = −→

QO + −→OR

= −q + k(2p + q)

= −q + 2kp + kq

If−→QR is parallel to p, there exists � ∈ R\{0} such that

(k − 1)q + 2kp = �p

∴ k − 1 = 0 and 2k = �

∴ k = 1

b−→PR = −→

PO + −→OR

= −p + k(2p + q)

= (2k − 1)p + kq

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If−→PR is parallel to q there exists m ∈ R\{0} such that

(2k − 1)p + kq = mq

∴ k = 1

2

c If P, Q and R are collinear there exists n ∈ R\{0} such that

n−→PQ = −→

QR

∴ n(−p + q) = (k − 1)q + 2kp

∴ k − 1 = n and 2k = −n

which implies 3k − 1 = 0

i.e., k = 1

3

Exercise 15D

1 In the diagram OR = 4

5OP,

−→OP = p,

−→OQ = q

and PS : SQ = 1 : 4.

a Express each of the following in terms of p and q.

i−→OR ii

−→RP iii

−→PQ

iv−→PS v

−→RS

b What can be said about line segments RS and OQ?

c What type of quadrilateral is ORSQ?

d The area of triangle PRS is 5 cm2. What is the area of ORSQ?

q

Q

O

p

P S

R

2 The position vectors of three points A, B and C relative to an origin O are a, b and ka

respectively. The point P lies on AB and is such that AP = 2PB. The point Q lies on BC and

is such that CQ = 6QB.

a Find in terms of a and b

i the position vector of P

ii the position vector of Q

b Given that OPQ is a straight line, find

i the value of k

ii the ratioOP

PQ

c The position vector of a point R is7

3a. Show that PR is parallel to BC.

3 The position vectors of two points A and B relative to an origin O are 3i + 3.5j andExample 13

6i − 1.5j respectively.

a i Given that−→

OD = 1

3

−→OB and

−→AE = 1

4

−→AB, write down the position vectors of

D and E.

ii Hence find | −→ED|.

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b Given that OE and AD intersect at X and that−→

OX = p−→

OE and−→XD = q

−→AD, find the

position vector of X in terms of

i p ii q

c Hence determine the values of p and q.

4 The position vectors of P, Q with reference to an origin O are p and q and M is the point on

PQ such that

�−→

PM = �−→

MQ

a Prove that the position vector of M is m where

m = � p + �q

� + �

b The vector p = ka and the vector q = lb where k and l are positive real numbers and a

and b are unit vectors.

i Prove that the position vector of any point on the internal bisector of ∠POQ has the

form � (a + b)

ii If M is the point where the internal bisector of ∠POQ meets PQ, show that

�

�= k

l

5 ORST is a parallelogram. U is the midpoint of RS and V is the midpoint of ST. Relative to

the origin O, r, s, t, u and v are the position vectors of R, S, T, U and V respectively.

a Express s in terms of r and t.

b Express v in terms of s and t.

c Hence or otherwise show that 4 (u + v) = 3 (r + s + t)

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Chapter summary

A vector is a set of equivalent directed line segments.

A directed line segment from a point A to a point B is denoted by−→AB.

In two dimensions, a vector can be represented by a column of numbers, e.g.

[2

3

]is the

vector ‘2 across’ and ‘3 up’.

The sum u + v can be shown diagrammatically

A

B

v

u

u + v

C

u + v = v + u

If u =[

a

b

]and v =

[c

d

]then u + v =

[a + c

b + d

]

The vector ku, k ∈ R+ and u �= 0, has the same direction as u but its length is multiplied by

a factor k.

The vector −v is in the opposite direction to v but it has the same length.

u − v = u + (−v)

Two non-zero vectors u and v are said to be parallel if there exists k ∈ R\{0} such that

u = kv.

For a point A, the position vector of A is−→OA where O is the origin.

Every vector u can be expressed as the sum of two vectors

x i and yj, where i is the unit vector in the positive direction

of the x axis and j is the unit vector in the positive direction

of the y axis.

xO

xi

yju

y

The magnitude of vector u = x i + yj is denoted by |u| and |u| =√

x2 + y2

The unit vector in the direction of a isa

|a| . This vector is denoted by a.

In three dimensions a vector u can be written as u = x i + yj + zk, where i, j and k are unit

vectors as shown.

y

z

x

(x, y, z)

y

k

i

j

x

z

If u = x i + yj + zk, |u| =√

x2 + y2 + z2

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Review


Multiple-choice questions

1 The vector v is defined by the directed line segment from (1, 1) to (3, 5). If v = ai + bj then

A a = 3 and b = 5 B a = −2 and b = −4 C a = 2 and b = 4

D a = 2 and b = 3 E a = 4 and b = 2

2 If vector−→AB = u and vector

−→AC = v then vector

−→CB is equal to

A u + v B v − u C u − v D u × v E v + u

3 If vector a =[

1

−2

]and vector b =

[2

3

]then a + b =

A

[1

−5

]B

[1

5

]C

[1

−1

]D

[1

1

]E

[3

1

]

4 If vector a =[

3

−2

]and vector b =

[−1

3

]then 2a − 3b =

A

[9

−13

]B

[9

7

]C

[9

−7

]D

[3

−13

]E

[3

5

]

5 PQRS is a parallelogram. If−→PQ = p and

−→QR = q, then in terms of p and q,

−→SQ equals

A p + q B p − q C q − p D 2q E 2p

6 |3i − 5j| =A 2 B

√34 C 34 D 8 E −16

7 If−→OA = 2i + 3j and

−→OB = i − 2j then

−→AB equals

A −i − 5j B −i + 5j C −i − j D −i + j E i + j

8 If−→OA = 2i + 3j and

−→OB = i − 2j then

∣∣ −→AB

∣∣ equals

A 6 B 26 C√

26 D√

24 E 36

9 If a = 2i + 3j then the unit vector parallel to a is

A 2i + 3j B1

13(2i + 3j) C

1√5

(2i + 3j)

D1√13

(2i + 3j) E√

13(2i + 3j)

10 If a = −3i + j + 3k then a is

A1

7(−3i + j + 3k) B

1√7

(−3i + j + 3k) C1√19

(−3i + j + 3k)

D1

19(−3i + j + 3k) E −3i + j + 3k

Short-answer questions (technology-free)

1 Given that a = 7i + 6j and b = 2i + x j, find the values of x for which

a a is parallel to b b a and b have the same magnitude.

2 ABCD is a parallelogram where−→OA = 2i − j,

−→AB = 3i + 4j and

−→AD = −2i + 5j. Find the

coordinates of the four vertices of the parallelogram.

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3 If a = 2i − 3j + k, b = 2i − 4j + 5k and c = −i − 4j + 2k, find the values of p and

q if a + pb + qc is parallel to the x axis.

4 The position vectors of P and Q are 2i − 2j + 4k and 3i − 7j + 12k respectively.

Find

a | −→PQ| b a unit vector parallel to

−→PQ.

5 The position vectors of A, B and C are 2j + 2k, 4i + 10j + 18k and x i + 14j + 26k

respectively. Find x if A, B and C are collinear.

6−→OA = 4i + 3j and C is a point on OA such that

∣∣ −→OC

∣∣ = 16

5.

a Find the unit vector in the direction of−→OA.

b Hence find−→

OC.

7 In the diagram, ST = 2TQ,−→PQ = a,

−→SR = 2a and

−→SP = b.

a Find each of the following in terms of a and b.R

Q

a

Pb S

T

2ai

−→SQ ii

−→TQ iii

−→RQ

iv−→PT v

−→TR

b Show that P, T and R are collinear.

8 If a = 5i − sj + 2k and b = t i + 2j + uk are equal vectors, find

a i s ii t iii u

b |a|9 The vector p has magnitude 7 units and bearing 050◦ and the vector q has magnitude

12 units and bearing 170◦. (These are compass bearings on the horizontal plane.) Draw a

diagram (not to scale) showing p, q and p + q. Calculate the magnitude of p + q.

10 If a = 5i + 2j + k and b = 3i − 2j + k, find

a a + 2b b |a| c a d a − b

11 O, A and B are the points (0, 0), (3, 4) and (4, –6) respectively.

a C is the point such that−→OA = −→

OC + −→OB. Find the coordinates of C.

b D is the point (1, 24) and−→

OD = h−→OA + k

−→OB. Find the values of h and k.

12 Given that p = 3i + 7j and q = 2i − 5j, find the values of m and n such that

mp + nq = 8i + 9j.

13 The points A, B and C have position vectors a, b and c relative to an origin O. Write down

an equation connecting a, b and c for each of the following cases.

a OABC is a parallelogram.

b B divides AC in the ratio 3 : 2, i.e., AB : BC = 3 : 2.

SAMPLE


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Review


Extended-response questions

1 Let

[1

0

]represent a displacement 1 km due east and

[0

1

]represent a displacement 1 km due north. The diagram shows

a circle of radius 25 km with a centre at O(0, 0). A lighthouse

entirely surrounded by sea is located at O.x

O

P

y

20

10

30

0

–10–10 0 10 20 30

–20

–20

–30

–30

The lighthouse is not visible from points outside the circle.

The ship is initially at P, 31 km west and 32 km south of the lighthouse.

a Write down the vector−→OP.

The ship is travelling parallel to vector u =[

4

3

]with speed 20 km/h. An hour after

leaving P the ship is at R.

b Show that the vector−→PR =

[16

12

], and hence find the vector

−→OR.

c Show that when the ship reaches R, the lighthouse first becomes visible.

2 Given that p = 3i + j and q = −2i + 4j find

a |p − q| b |p| − |q| c r, such that p + 2q + r = 0

3 Let a =

−2

1

2

, b =

11

7

3

, c =

7

9

7

and d =

26

12

2

a Find the value of the scalar k such that a + 2b − c = kd

b Find the scalars x and y such that xa + yb = d

c Use your answers to a and b to find scalars p, q and r (not all zero) such that

pa + qb − rc = 0

4 The quadrilateral PQRS is a parallelogram. The point P has coordinates (5, 8), the point R

has coordinates (32, 17) and the vector−→PQ is given by

−→PQ =

[20

−15

].

a Find the coordinates of Q, and write down the vector−→QR.

b Write down the vector−→RS , and show that the coordinates of S are (12, 32).

5 The diagram shows the path of a light beam from its source

at O in the direction of the vector r =[

3

1

]. At P the beam

is reflected by an adjustable mirror and meets the x axis at

M. The position of M varies, depending on the adjustment

of the mirror at P.O

P

M

r

θ

a Given that−→OP = 4r, find the coordinates of P.

b The point M has coordinates (k, 0). Find in terms of k an expression for the vector−→

PM .

c Find the magnitudes of vectors−→OP,

−→OM and

−→PM , and hence find the value of k for

which � is equal to 90◦.

d Find the value � for which M has coordinates (9, 0).

SAMPLE


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