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SCHOLAR Study Guide

National 5 Mathematics

Course MaterialsTopic 17: Vectors

Authored by:Margaret Ferguson

Reviewed by:Jillian Hornby

Previously authored by:Eddie Mullan

Heriot-Watt University

Edinburgh EH14 4AS, United Kingdom.

First published 2014 by Heriot-Watt University.

This edition published in 2016 by Heriot-Watt University SCHOLAR.

Copyright © 2016 SCHOLAR Forum.

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Distributed by the SCHOLAR Forum.

SCHOLAR Study Guide Course Materials Topic 17: National 5 Mathematics

1. National 5 Mathematics Course Code: C747 75

AcknowledgementsThanks are due to the members of Heriot-Watt University's SCHOLAR team who planned andcreated these materials, and to the many colleagues who reviewed the content.

We would like to acknowledge the assistance of the education authorities, colleges, teachersand students who contributed to the SCHOLAR programme and who evaluated these materials.

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The Scottish Qualifications Authority for permission to use Past Papers assessments.

The Scottish Government for financial support.

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1

Topic 1

Vectors

Contents

17.1 Introduction to vectors in 2D . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

17.2 Vector journeys . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

17.3 Coordinates in 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

17.4 Using vector components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

17.5 Magnitude of vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

17.6 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

17.7 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

2 TOPIC 1. VECTORS

Learning objectives

By the end of this topic, you should be able to:

• add, subtract and scale 2D vectors using directed line segments;

• describe a vector journey;

• determine the coordinates of a point on a 3D object;

• add, subtract and scale vectors using components;

• calculate the magnitude of a vector.

© HERIOT-WATT UNIVERSITY

TOPIC 1. VECTORS 3

1.1 Introduction to vectors in 2D

Key point

A vector is a quantity which has both direction and magnitude (or size).You will meet examples of vectors in Physics such as displacement, velocity,acceleration, force and electromagnetic fields.

Examples

1.

Problem:

If B is 10 km East of A then−−→AB is a vector with distance (or magnitude) 10 km and

direction East.

Solution:

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.

Problem:

PQ and MN are vectors. A vector can be represented by a directed line segment, thatis a line from one point to another which has a direction arrow on it.

The two lines MN and PQ represent vectors in two dimensions.−−→MN and

−−→PQ are

directed line segments and the arrowheads indicate the direction.

Solution:

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


4 TOPIC 1. VECTORS

Examples

1.

Problem:

u and v are vectors. Vectors are often represented using a lower case letter in bold.

Solution:

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2. Equal vectors

Problem:−−→FE and u have the same length (or magnitude) and are parallel with the same direction.So we can say that

−−→FE = u.

Solution:

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3. Multiplying a vector by a scalar

Problem:−−→PQ and v are parallel with the same direction but

−−→PQ is double the size of v. So we can

say that−−→PQ = 2v or v = 1

2

−−→PQ.


TOPIC 1. VECTORS 5

Solution:

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4. Reversing the direction of a vector

Problem:−−→KL and a have the same magnitude and are parallel but are pointing in oppositedirections. So we can say that

−−→KL = −a or a = −−−→

KL.

Solution:

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Adding and subtracting vectors

Go online

The diagram shows 2 directed line segments a and b.


6 TOPIC 1. VECTORS

Draw the resultant vector a + b.Vector b is added to vector a by joining it to the end of vector a.

Draw the resultant vector a + 2b.Here, vector b is doubled in size and joined to the end of vector a.

Draw the resultant vector a − b.Vector b is negative so its direction is reversed then joined to the end of vector a.


TOPIC 1. VECTORS 7

Draw the resultant vector b + a.Vector a is added to vector b by joining it to the end of vector b.

Draw the resultant vector b − 2a.Vector a is reversed and doubled in size then joined to the end of vector b.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Top tip

When drawing vectors use the grid squares to count along and up or down.

Vectors in 2D practice

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Q1: Given vector a what is −a?


8 TOPIC 1. VECTORS

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q2: Given vector a what is 3a?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q3: Given vectors a and b what is a + b?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q4: Given vectors a and b what is a − b?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


TOPIC 1. VECTORS 9

Q5: Given vectors a and b what is 2a + b?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q6: Given vectors a and b what is 3b − a?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q7: Given vectors a, b and c what is a + b + c?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


10 TOPIC 1. VECTORS

Q8: Given vectors a, b and c what is a + 2b − c?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.2 Vector journeys

Key point

A displacement is the shortest distance from one point to another and a VectorJourney is a displacement.

Examples

1.

Problem:

Express each of the following displacements in terms of vectors a and b.

a)−→AC

b)−−→CB

c)−→CA

d)−−→BC

e)−−→AB

Solution:

a)−→AC = a

b)−−→CB = b


TOPIC 1. VECTORS 11

c)−→CA is just

−→AC with its direction reversed so

−→CA = - a.

d)−−→BC is just

−−→CB with its direction reversed so

−−→BC = - b.

e)−−→AB is the shortest distance from A to B but the long way would be from A to C to Bwhich is

−→AC +

−−→CB so

−−→AB = a + b

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.

Problem:

In the diagram−−→AB = 2

−−→DC and

−−→BC = v.

Express the following displacements in terms of vectors u and v.

a)−−→AB

b)−−→CD

c)−−→AD

If−−→BC = 2

3

−−→BE.

Express the displacements in terms of vectors u and v.

d)−−→CE

e)−−→ED

Solution:

a) Since−−→AB = 2

−−→DC it follows that AB and DC have the same direction, so

−−→AB =

2−−→DC = 2u.

b)−−→CD is just

−−→DC with its direction reversed so

−−→CD = - u.

c)−−→AD =

−−→AB +

−−→BC +

−−→CD = 2u + v + ( - u) = u + v

d) Sometimes we may have to employ a little bit of algebra to find the solution.

If−−→BC = 2

3

−−→BE then v = 2

3

−−→BE.

If we change the subject of the formula to−−→BE we get 3

2v =−−→BE.

Now−−→BC = 2

3

−−→BE so it follows that

−−→CE = 1

3

−−→BE.

If we substitute 32v for

−−→BE we get

−−→CE = 1

3 × 32v = 3

6v = 12v.


12 TOPIC 1. VECTORS

e)−−→ED =

−−→EC +

−−→CD.

We know that−−→CE = 1

2v and−−→DC = u.

If we reverse the direction of−−→CE and

−−→DC we get

−−→ED = - 1

2v + ( - u) or−−→ED = - 1

2v − u.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Vector journeys exercise

Go online

Express each of the following displacements in terms of vectors a and b.

Q9:−→AC

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q10:−−→BC

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q11:−→CA

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q12:−−→CB

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q13:−−→AB

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q14:−−→BA

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


TOPIC 1. VECTORS 13

EFGH is a rhombus. Express each of the following displacements in terms of vectorse and f .

Q15:−−→EF

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q16:−−→HG

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q17:−−→EG

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q18:−−→HF

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

PQR is a triangle. In the diagram−→ST = 3

4

−−→RQ.

Express each of the following displacements in terms of vectors w and x.

Q19:−→RS

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q20:−→ST

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q21:−→TQ

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

If−→RS = 1

4

−→RP , express each of the following displacements in terms of vectors w and

x.

Q22:−→SP

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


14 TOPIC 1. VECTORS

Q23:−→PS

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q24:−→PT

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q25:−−→PQ

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.3 Coordinates in 3D

Two dimensional objects, for example a square, lie on a two dimensional plane. You arealready familiar with this concept and can plot points on a coordinate diagram with anx-axis and a y-axis.

Three dimensional objects, for example a cuboid, have length, breadth and height. Apoint in space can be described with three coordinates: an x-coordinate, y-coordinateand a z-coordinate. This requires 3 axes: an x-axis, a y-axis and a z-axis.

Coordinates in 3D space

Go online

If we think of the x-axis and the y-axis lying on a flat page then the third z-axis, wouldproject out of the page perpendicularly. The axes are all perpendicular or at right anglesto each other.


TOPIC 1. VECTORS 15

The point A lies on the x-axis and has coordinates (4,0,0).It might help to think of the journey from the origin as along 4, back 0 and up 0.

The point B lies on the y-axis and has coordinates (0,5,0).It might help to think of the journey from the origin as along 0, back 5 and up 0.

The point C lies on the z-axis and has coordinates (0,0,2).It might help to think of the journey from the origin as along 0, back 0 and up 2.


16 TOPIC 1. VECTORS

The point D lies in space and has coordinates (3,4,3).It might help to think of the journey from the origin as along 3, back 4 and up 3.

The point E lies in space and has coordinates (-2,0,0).It might help to think of the journey from the origin as left 3, back 0 and up 0.

The point F lies in space and has coordinates (0,-1,0).It might help to think of the journey from the origin as along 0, forward 1 and up 0.


TOPIC 1. VECTORS 17

The point G lies in space and has coordinates (0,0,-1).It might help to think of the journey from the origin as along 0, back 0 and down 1.

It is much easier to find the coordinates of a point when a 3D shape is placed on 3Daxes.OABCDEFG is a cuboid which sits on the xy-plane.

x

y

z

ABCDE is a square based pyramid with base length 4cm and height 6cm.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


18 TOPIC 1. VECTORS

Key point

Coordinates in 3D are defined by (x, y, z). They describe the journey from theorigin:

• x is the distance along;

• y is the distance back or forward;

• z is the distance up or down.

Example

Problem:

OABCDEFG is a cube of length 8 cm.Identify the coordinates of the vertices of the cube.

Solution:

O is the origin O(0,0,0).

OA is 8 cm and lies on the x-axis so A(8,0,0) because we go along 8 from the origin.

AB is 8 cm and is parallel to the y-axis so B(8,8,0) because we go along 8 to A thenback 8.

OC is 8 cm and lies on the y-axis so C(0,8,0) because we go along 0 and back 8.

OD is 8 cm and lies on the z-axis so D(0,0,8) because we go along 0, back 0 and up 8.

E is 8 cm above A so E(8,0,8) because we go along 8 to A, back 0 and up 8.

F is 8 cm above B so F (8,8,8) because we go along 8 to A, back 8 to B and up 8.

G is 8 cm above C so G(0,8,8) because we go along 0, back 8 to C and up 8.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


TOPIC 1. VECTORS 19

Examples

1.

Problem:

ABCDEFGH is a cuboid of length 10 cm, breadth 5 cm and height 4 cm.Identify the coordinates of the vertices of this cuboid if A(10,1,0).

Solution:

This cuboid has been moved back from the x-axis by 1 unit.

A is along 10, back 1 and up 0 and AB is 5 cm so B(10, 6, 0) along 10, back 1 + 5 = 6.

C is on the y-axis and is the same distance back as B so C(0,6,0).

D sits on the y-axis and is the same distance back as A so D(0,1,0).

E is 4 cm above A(10,1,0) so E(10,1,4).

F is 4 cm above B(10,6,0) so F (10,6,4).

G is 4 cm above C(0,6,0) so G(0,6,4).

H is 4 cm above D(0,1,0) so H(0,1,4).

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.

Problem:

ABCDEFGH is a cuboid of length 10 cm, breadth 5 cm and height 4 cm.Identify the coordinates of the vertices of this cuboid if A(8,0,0).


20 TOPIC 1. VECTORS

Solution:

In this example the cuboid has been moved along the x-axis to the left of the origin by 2units.

A is along 8, back 0 and up 0 and AB is 5 cm so B(8, 5, 0) along 8, back 5.

D sits on the negative side of the x-axis and AD is 10 cm so D(-2,0,0).

C is the same distance along as D and the same distance back as B so C(-2,5,0).

E is 4 cm above A so E(8,0,4).

F is 4 cm above B so F (8,5,4).

G is 4 cm above C so G(-2,5,4).

H is 4 cm above D so H(-2,0,4).

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Coordinates in 3D space exercise

Go online

Q26: OABCDEFG is a cuboid of length 6 cm, breadth 10 cm and height 12 cm.Identify the coordinates of the vertices of this cuboid where O is the origin.

What are the coordinates of:

a) A?

b) B?

c) C?

d) D?

e) E?

f) F?

g) G?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


TOPIC 1. VECTORS 21

Q27: ABCDEFGH is a cube of length 5 cm.Identify the coordinates of the vertices of the cube if D(0,0,2).


a) A?

b) B?

c) C?

d) E?

e) F?

f) G?

g) H?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q28: ABCDE is a square based pyramid with base length 10 cm and height 15 cm.Identify the coordinates of the vertices of the pyramid if A(-5,0,0).


a) B?

b) C?

c) D?

d) E?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


22 TOPIC 1. VECTORS

Q29: ABCDEFGH is a cube of length 10 cm.Identify the coordinates of the vertices of the cube if A(10,-1,0).


a) B?

b) C?

c) D?

d) E?

e) F?

f) G?

g) H?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.4 Using vector components

Key point

A vector is a quantity which has both direction and magnitude. The componentsof a vector describe how to get from one end of the vector to the other followingits direction.

Vector components in 2D

Go online

The components of a vector are written vertically in brackets. In the following graph−−→AB =

(4

3

).


TOPIC 1. VECTORS 23

a =

(4

3

)along 4 then up 3 describes the journey from A to B.

In this graph−−→PQ =

(−3

−6

).

u =

(−3

−6

)along 3 to the left then down 6 describes the journey from P to Q.

We can add the vectors u and a together and identify the components of the resultant

vector. In the following graph u + a =

(1

−3

).


24 TOPIC 1. VECTORS

Notice that if we add the components of u and a we get the same answer.

u + a =

(−3

−6

)+

(4

3

)=

(−3 + 4

−6 + 3

)=

(1

−3

)

We can subtract the vector u from a and identify the components of the resultant vector.

In this graph a − u =

(7

9

).

Notice that if we subtract the components of u from a we get the same answer.

a − u =

(4

3

)−(

−3

−6

)=

(4 − (−3)

3 − (−6)

)=

(7

9

)

We can scale a vector and identify the components of the resultant vector. Here we can

see a =

(4

3

)and 2a =

(8

6

).

Notice that if we multiply the components of a by 2 we get the same answer.

2a = 2 ×(

4

3

)=

(2 × 4

2 × 3

)=

(8

6

)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


TOPIC 1. VECTORS 25

Examples

1.

Problem:

Name the vector and state its components.

Solution:

The line segment has the direction from L to K.

The journey from L to K is along 7 then down 4.−−→LK =

(7

−4

). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.

Problem:

r =

(2

−1

), s =

(−5

8

)and t =

(4

3

)

State the components of the resultant vectors:

a) r + s

b) s − t

c) r + s + t

d) s + 2t

e) 3r − 2s + t

Solution:

a) r + s =

(2

−1

)+

(−5

8

)=

(−3

7

)

b) s − t =

(−5

8

)−(

4

3

)=

(−9

5

)

c) r + s + t =

(2

−1

)+

(−5

8

)+

(4

3

)=

(1

10

)

d) s + 2t =

(−5

8

)+ 2

(4

3

)=

(−5

8

)+

(8

6

)=

(3

14

)


26 TOPIC 1. VECTORS

e)

3r − 2s + t = 3

(2

−1

)− 2

(−5

8

)+

(4

3

)

=

(6

−3

)−(

−10

16

)+

(4

3

)

=

(20

−16

)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Vector components in 2D practice

Go online

Q30:

a) Name the vector shown.

b) State its components.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q31: Draw a representation of the vector e =

(4

−7

).

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Vector components in 3D

Go online

The components of a vector in 3D are harder to visualize but the process is just the

same as in 2D. The following graph shows−−→AB =

⎛⎜⎝

5

2

1

⎞⎟⎠.


TOPIC 1. VECTORS 27

a =

⎛⎜⎝

5

2

1

⎞⎟⎠ along 5, back 2 then up 1 describes the journey from A to B.

We can scale a vector and identify the components of the resultant vector.

Remember a =

⎛⎜⎝

5

2

1

⎞⎟⎠?

This vector can be scaled to 2a =

⎛⎜⎝

10

4

2

⎞⎟⎠.

Notice that if we multiply the components of a by 2 we get the same answer.

2a = 2 ×

⎛⎜⎝

5

2

1

⎞⎟⎠ =

⎛⎜⎝

2 × 5

2 × 2

2 × 1

⎞⎟⎠ =

⎛⎜⎝

10

4

2

⎞⎟⎠

Let’s add three dimensional vectors.

Two forces have components u =

⎛⎜⎝

−2

3

−1

⎞⎟⎠ and w =

⎛⎜⎝

4

7

−5

⎞⎟⎠. Calculate u + w.


28 TOPIC 1. VECTORS

So, u + w =

⎛⎜⎝

−2

3

−1

⎞⎟⎠ +

⎛⎜⎝

4

7

−5

⎞⎟⎠ =

⎛⎜⎝

2

10

−6

⎞⎟⎠.

This time let’s subtract three dimensional vectors.

Two forces have components m =

⎛⎜⎝

3

2

1

⎞⎟⎠ and n =

⎛⎜⎝

−6

7

−8

⎞⎟⎠. Calculate m - n.

So, m − n =

⎛⎜⎝

3

2

1

⎞⎟⎠ −

⎛⎜⎝

−6

7

−8

⎞⎟⎠ =

⎛⎜⎝

9

−5

9

⎞⎟⎠.

Now let us add, subtract and multiply three dimensional vectors.

Three forces have components a =

⎛⎜⎝

1

4

−3

⎞⎟⎠, b =

⎛⎜⎝

6

1

1

⎞⎟⎠ and c =

⎛⎜⎝

−2

5

2

⎞⎟⎠.

Calculate a + 2b - c.

a + 2b − c =

⎛⎜⎝

1

4

−3

⎞⎟⎠ + 2

⎛⎜⎝

6

1

1

⎞⎟⎠ −

⎛⎜⎝

−2

5

2

⎞⎟⎠ multiply the components of b by 2

=

⎛⎜⎝

1

4

−3

⎞⎟⎠ +

⎛⎜⎝

12

2

2

⎞⎟⎠ −

⎛⎜⎝

−2

5

2

⎞⎟⎠ calculate the resultant vector

=

⎛⎜⎝

15

1

−3

⎞⎟⎠

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Using vector components exercise

Go onlineQ32: a =

(6

5

), b =

(−4

9

)and c =

(−7

−4

)

a) What are the components of a + b?

b) What are the components of a − b?

c) What are the components of 2a − c?

d) What are the components of 4a + b − 2c?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


⎛⎜⎝

1

2

3

⎞⎟⎠ , b =

⎛⎜⎝

4

−1

1

⎞⎟⎠ and c =

⎛⎜⎝

−7

9

0

⎞⎟⎠

Calculate the components of the resultant vectors.


TOPIC 1. VECTORS 29

Q33: a + b

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q34: b − c

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q35: 3c

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q36: 3a − b

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q37: 2a − 3b + c

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.5 Magnitude of vectors

Key point

The magnitude of a vector or directed line segment is simply its size or length.We can apply the Theorem of Pythagoras to find the magnitude of vectors in 2Dand 3D.

The magnitude of a vector

Go onlineHere we have

−−→AB = a =

(4

2

).

We have formed a right-angled triangle. The notation for magnitude is to place the nameof the vector or directed line segment between vertical lines.

For example,∣∣∣−−→AB∣∣∣ = |a| =

√42 + 32 = 5

Calculate the magnitude of a + b.


30 TOPIC 1. VECTORS

a + b =

(6

3

)

|a + b| =√

62 + 32 =√45 or 6 · 7

We could also calculate the magnitude of a + b using their components.

a + b =

(4

3

)+

(2

−6

)=

(6

3

)

|a + b| =√

62 + 32 =√45 or 6 · 7

The Theorem of Pythagoras can be used to find magnitude of a vector in 3D. Here we

have−−→AB =

⎛⎜⎝

5

2

1

⎞⎟⎠.

If we apply Pythagoras’ Theorem twice we can find∣∣∣−−→AB∣∣∣.


TOPIC 1. VECTORS 31

The first application of the Theorem of Pythagoras.

AC =√52 + 22 =

√29

The second application of the Theorem of Pythagoras uses the length of AC.

AB =

√(√29)2

+ 12 =√30 or 5 · 5 hence

∣∣∣−−→AB∣∣∣ =√30 or 5 · 5

There is a quicker way . . .

−−→AB =

⎛⎜⎝

5

2

1

⎞⎟⎠ so,

∣∣∣−−→AB∣∣∣ =√52 + 22 + 12 =

√30 or 5 · 5

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Key point

The magnitude is calculated from the components using the Theorem ofPythagoras e.g.

u =

(x

y

), |u| =

√x2 + y2

and


32 TOPIC 1. VECTORS

Key point continued

v =

⎛⎜⎝

x

y

z

⎞⎟⎠, |v| =

√x2 + y2 + z2

Given a =

(1

9

)and b =

(−4

−5

)

Examples

1.

Problem:

Calculate: |a|Solution:

|a| =√12 + 92 =

√82 or 9.1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.

Problem:

Calculate:|b|Solution:

|b| =

√(−4)2 + (−5)2 =

√41 or 6.4

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3.

Problem:

Calculate:|a + b|Solution:

a + b =

(1

9

)+

(−4

−5

)=

(−3

4

)so |a + b| =

√(−3)2 + 42 = 5

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.

Problem:

Calculate:|a| + |b|Solution:

|a| + |b| = 9.1 + 6.4 = 15.5

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


TOPIC 1. VECTORS 33

Key point

|a + b| �= |a| + |b|

Given u =

⎛⎜⎝

1

2

−1

⎞⎟⎠ and v =

⎛⎜⎝

3

−5

1

⎞⎟⎠

Examples

1.

Problem:

Calculate: |u|Solution:

|u| =

√12 + 22 + (−1)2 =

√6 or 2.4

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.

Problem:

Calculate: |v|Solution:

|v| =√

32 + (−5)2 + 12 =√35 or 5.9

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3.

Problem:

Calculate: |u + v|Solution:

u + v =

⎛⎜⎝

1

2

−1

⎞⎟⎠ +

⎛⎜⎝

3

−5

1

⎞⎟⎠ =

⎛⎜⎝

4

−3

0

⎞⎟⎠

so

|u + v| =

√42 + (−3)2 + 02 = 5

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

4.

Problem:

Calculate: |u − v|Solution:

u − v =

⎛⎜⎝

1

2

−1

⎞⎟⎠ −

⎛⎜⎝

3

−5

1

⎞⎟⎠ =

⎛⎜⎝

−2

7

−2

⎞⎟⎠

so |u − v| =√

(−2)2 + 72 + (−2)2 =√57 or 7.5

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


34 TOPIC 1. VECTORS

Magnitude of vectors exercise

Go online

Magnitude of vectors in 2D


(4

3

), b =

(−1

2

)and c =

(−6

−3

)

Q38: Calculate the magnitude of |a|. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q39: Calculate the magnitude of |b|. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q40: Calculate the magnitude of |c|. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q41: Calculate the magnitude of |a + b|. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q42: Calculate the magnitude of |b − c|. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q43: Calculate the magnitude of |a − 2b + c|. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


Three forces have components u =

⎛⎜⎝

3

4

5

⎞⎟⎠ , v =

⎛⎜⎝

2

−2

3

⎞⎟⎠ and w =

⎛⎜⎝

−4

−3

−2

⎞⎟⎠

Q44: Calculate the magnitude of |u|. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q45: Calculate the magnitude of |v|. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q46: Calculate the magnitude of |w|. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q47: Calculate the magnitude of |u + w|. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q48: Calculate the magnitude of |v − u|. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q49: Calculate the magnitude of |2v − w + u|. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


TOPIC 1. VECTORS 35

1.6 Learning points• A vector is a quantity which has both direction and magnitude.

• The magnitude of a vector is its size or length.

• A directed line segment from A to B is defined as−−→AB.

• A vector or force can also be defined by a lower-case letter in bold.

• Vectors are equal if they have the same direction and magnitude.

• When a vector has its direction reversed it is negative e.g. a becomes -a.

• A vector can be multiplied by a scalar e.g. doubling a gives 2a.

• Vectors can be added by joining one to the end of another.

• Displacement is the shortest distance from A to B.

• A vector journey is a description of its displacement.

• The components of a vector describe the journey from A to B e.g.

(x

y

)in 2D or

⎛⎜⎝

x

y

z

⎞⎟⎠ in 3D.

• Arithmetic can be performed on the components (+ − × ÷) e.g.

(1

2

)+(

4

3

)=

(1 + 4

2 + 3

)=

(5

5

)

• The magnitude is calculated from the components using the Theorem of

Pythagoras e.g. u =

(x

y

), |u| =

√x2 + y2 and v =

⎛⎜⎝

x

y

z

⎞⎟⎠,

|v| =√

x2 + y2 + z2


36 TOPIC 1. VECTORS

1.7 End of topic test

End of topic 17 test

Go online

Naming Vectors in 2D

Q50: Name the vector a as a directed line segment.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q51: Name the vector b as a directed line segment.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q52: Name the vector c as a directed line segment.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Vectors in 2D

Q53: Draw the resultant vector of a + b.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q54: Draw the resultant vector of b − c.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q55: Draw the resultant vector of 2a − b + c.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


TOPIC 1. VECTORS 37

Vector Journeys

ABC is an isosceles triangle and M is the mid-point of BC.

Q56: Express−−→CB in terms of u and w.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q57: Express−−→CM in terms of u and w.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q58: Express−−→AM in terms of u and w.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3D coordinates

Q59: State the coordinates of D.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


38 TOPIC 1. VECTORS

ABCDEFGH is a cuboid and B(12,7,-5).

Q60: What are the coordinates of E?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q61: What are the coordinates of G?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q62: What are the coordinates of D?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q63: What are the coordinates of A?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

OABCDE is a isosceles triangular prism of length 15 cm, breadth 8 cm and height 6 cm.

Q64: What are the coordinates of A?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q65: What are the coordinates of D?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q66: What are the coordinates of B?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q67: What are the coordinates of C?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q68: What are the coordinates of E?

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


TOPIC 1. VECTORS 39

Vector components

Q69: Name the vector from the graph above.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q70: State the components of the vector from the graph above.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Using vector components in 2D

p =

(1

−1

), q =

(5

−3

)and r =

(−2

4

)

State the components of the resultant vectors

Q71: p + r

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q72: p − q

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q73: 2q + 3p

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q74: 1/2r − 2p + 3q

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Using vector components in 3D


⎛⎜⎝

6

8

3

⎞⎟⎠ , b =

⎛⎜⎝

−2

−7

1

⎞⎟⎠ and c =

⎛⎜⎝

−9

−5

−1

⎞⎟⎠

Calculate the components of the resultant vectors.

Q75: a + b

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


40 TOPIC 1. VECTORS

Q76: 2b − c

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q77: 2a − 3b + c

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


Three forces have components d =

(4

3

), e =

(−1

2

)and f =

(−6

−3

)

Q78: Calculate |f |. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q79: Calculate |e− f |. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q80: Calculate |d− 2e+ f |. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


Three forces have components u =

⎛⎜⎝

1

3

−2

⎞⎟⎠ , v =

⎛⎜⎝

−2

1

3

⎞⎟⎠ , w =

⎛⎜⎝

−5

4

−6

⎞⎟⎠

Q81: Calculate |w|. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q82: Calculate |u|+ |v|. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Q83: Calculate |2v −w|+ |u|. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .


ANSWERS: TOPIC 17 41

Answers to questions and activities

17 Vectors

Vectors in 2D practice (page 7)

Q1:

Vector −a is negative so its direction is reversed.

Q2:

Vector 3a is three times the size of vector a.

Q3:

The resultant vector a + b is found by adding vector b to vector a.To do this you join vector b to the end of vector a.Make sure the arrows are facing the same direction.


42 ANSWERS: TOPIC 17

Q4:

The resultant vector a − b is found by first reversing the direction of vector b so that itbecomes −b.Vector −b is then joined to the end of vector a, so that the arrows are facing the samedirection.

Q5:

The resultant vector 2a + b is found first by doubling the size of vector a to 2a.Then vector b is joined to the end of vector 2a.Make sure the arrows are facing the same direction.

Q6:

The resultant vector 3b − a is found first by changing the direction of vector a so that itbecomes −a.Vector b is then increased in size by a factor of 3 to become 3b.Vector −a is then joined to the end of vector 3b.Make sure the arrows are facing in the correct direction.



Q7:

Resultant vector a + b + c is found by joining vector b to the end of vector a.Vector c is then joined to the end of vector b.All the arrows must face the same direction.

Q8:

Resultant vector a + 2b − c is found by first changing the direction of vector c so thatit becomes −c.Vector b is the doubled in size to become 2b.The vectors are then joined together in order, with the arrows facing the correct direction.



Vector journeys exercise (page 12)

Q9: a

Q10: b

Q11: −a

Q12: −b

Q13: a − b

Q14: b − a

Q15: e

Q16: e

Q17: f + e

Q18: e − f

Q19: x

Q20: 3/4w

Q21:

Steps:

• −→TQ =

−→TS +

−→SR +

−−→RQ

• What is−→TS? − 3

4w

• What is−→SR? −x

• What is−−→RQ? w

• Use these answers to find−→TQ.

Answer: 1/4w − x

Q22:

Steps:

• What is−→RS? x

• −→RS = 1

4

−→RP change the subject of the formula to

−→RP .

−→RP = 4

−→RS

• What is−→RP? 4x

• −→SP =

−→RP − −→

RS

• Use these answers to find−→SP .

Answer: 3x

Q23: −3x



Q24:

Hint:

• −→PT =

−→PS +

−→ST

• You should have already found−→PS and

−→ST in the previous questions. Use your

answers to find−→PT .

Answer: 3/4w − 3x

Q25:

Hint:

• −−→PQ =

−→PT +

−→TQ

• You should have already found−→PT and

−→TQ in the previous questions. Use your

answers to find−−→PQ.

Answer: w − 4x

Coordinates in 3D space exercise (page 20)

Q26:

a) A(6,0,0)

b) B(6,10,0)

c) C(0,10,0)

d) D(0,0,12)

e) E(6,0,12)

f) F (6,10,12)

g) G(0,10,12)

Q27:

a) A(5,0,2)

b) B(5,5,2)

c) C(0,5,2)

d) E(5,0,7)

e) F (5,5,7)

f) G(0,5,7)

g) H(0,0,7)

Q28:

a) Step: The y-axis bisects the base of the pyramid. A sits on the x-axis at -5 andAB is 10 cm what is the x-coordinate of B? 5Answer: B(5,0,0)

b) C(5,10,0)



c) D(-5,10,0)

d) Hint: The y-axis bisects the base of the pyramid and E is vertically above thecentre of the square base.Answer: E(0,5,15)

Q29:

Steps:

• Part a) The cube has been moved forward from the origin by 1 unit so A is along10, forward 1. Since going back is positive it follows that going forward is negative.To get to B we go along 10, how many back do we go? 9

Answer:

a) B(10,9,0)

b) C(0,9,0)

c) D(0,-1,0)

d) E(10,-1,10)

e) F (10,9,10)

f) G(0,9,10)

g) H(0,-1,10)

Vector components in 2D practice (page 26)

Q30:

a)−−→BA

b)

(−6

2

)

Remember the direction shows the journey from B to A which is along 6 in the negativedirection then up 2.

Q31:

e =

(4

−7

)means go along 4 then down 7.



Using vector components exercise (page 28)

Q32:

a) a + b =

(2

14

)

b) a − b =

(10

−4

)

c) 2a − c =

(19

14

)

d) 4a + b − 2c =

(34

37

)

Q33: a + b =

⎛⎜⎝

5

1

4

⎞⎟⎠

Q34: b − c =

⎛⎜⎝

11

−10

1

⎞⎟⎠

Q35: 3c =

⎛⎜⎝

−21

27

0

⎞⎟⎠

Q36: 3a − b =

⎛⎜⎝

−1

7

8

⎞⎟⎠

Q37: 2a − 3b + c =

⎛⎜⎝

−17

16

3

⎞⎟⎠

Magnitude of vectors exercise (page 34)

Q38: 5

Q39: 2·2

Q40: 6·7

Q41:

Steps:

• What are the components of a + b?

(3

5

)



• Now find the magnitude of this resultant vector.

Answer: 5·8

Q42:

Steps:

• What are the components of b - c?

(5

5

)


Answer:7·1

Q43:

Steps:

• What are the components of a - 2b + c?

(0

−4

)


Answer: 4

Q44: 7·1

Q45: 4·1

Q46: 5·4

Q47:

Steps:

• What are the components of u + w?

⎛⎜⎝

−1

1

3

⎞⎟⎠


Answer: 3·3

Q48:

Steps:

• What are the components of v - u?

⎛⎜⎝

−1

−6

−2

⎞⎟⎠


Answer: 6·4

Q49:



Steps:

• What are the components of 2v - w + u?

⎛⎜⎝

11

3

13

⎞⎟⎠


Answer: 17·3

End of topic 17 test (page 36)

Q50:−−→AB

Q51:−−→HG

Q52:−−→QP

Q53:

Q54:



Q55:

Q56: w + u

Q57:

Steps:

• AM bisects CB. What fraction of CB is CM? 1/2

Answer: 1/2(w + u)

Q58:

Hint:

• −−→AM =

−→AC +

−−→CM

Answer: 1/2u − 1/2w

Q59: (-2,6,-5)

Q60: (12,0,0)

Q61: (0,7,0)

Q62: (0,0,-5)

Q63: (12,0,-5)

Q64: (15,0,0)

Q65: (0,8,0)

Q66: (15,8,0)

Q67: (15,4,6)

Q68: (0,4,6)

Q69:−−→HG



Q70:

(−5

−6

)

Q71: p + r =

(−1

3

)

Q72: p − q =

(−4

2

)

Q73: 2q + 3p =

(13

−9

)

Q74: 12r − 2p + 3q =

(−12

−5

)

Q75: a + b =

⎛⎜⎝

4

1

4

⎞⎟⎠

Q76: 2b − c =

⎛⎜⎝

5

−9

3

⎞⎟⎠

Q77: 2a − 3b + c =

⎛⎜⎝

9

32

2

⎞⎟⎠

Q78: |f | = 6 · 5 or√45

Q79:

Steps:

• What are the components of e− f? e− f =

(5

5

)


Answer: |e− f | = 7 · 1 or√50

Q80:

Steps:

• What are the components of d− 2e+ f? d− 2e+ f =

(0

−4

)


Answer: |d− 2e+ f | = √16 or 4



Q81: |w| = √77

Q82:

Hints:

• Find the magnitude of u, find the magnitude of v then add your answers.

Answer: |u|+ |v| = 7 · 5Q83:

Steps:

• What is |2v −w|? √149

• What is |u|? √14

• Add these answers together.

Answer: |2v −w|+ |u| = 15 · 9


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