SCHOLAR Study Guide

CfE Higher MathematicsAssessment Practice 2:Vectors

Authored by:Margaret Ferguson

Reviewed by:Jillian Hornby

Previously authored by:Jane S Paterson

Dorothy A Watson

Heriot-Watt University

Edinburgh EH14 4AS, United Kingdom.

First published 2014 by Heriot-Watt University.

This edition published in 2017 by Heriot-Watt University SCHOLAR.

Copyright © 2017 SCHOLAR Forum.

Members of the SCHOLAR Forum may reproduce this publication in whole or in part for educationalpurposes within their establishment providing that no profit accrues at any stage, Any other use of thematerials is governed by the general copyright statement that follows.

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmittedin any form or by any means, without written permission from the publisher.

Heriot-Watt University accepts no responsibility or liability whatsoever with regard to the informationcontained in this study guide.

Distributed by the SCHOLAR Forum.

SCHOLAR Study Guide Assessment Practice: CfE Higher Mathematics

1. CfE Higher Mathematics Course Code: C747 76

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

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

Grateful acknowledgement is made for permission to use the following material in the SCHOLARprogramme:

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

The Scottish Government for financial support.

The content of this Study Guide is aligned to the Scottish Qualifications Authority (SQA) curriculum.

All brand names, product names, logos and related devices are used for identification purposes only and aretrademarks, registered trademarks or service marks of their respective holders.

1

Topic 7

Vectors

Contents7.1 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

7.2 Assessment practice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2 TOPIC 7. VECTORS

Learning objective

You should be able to:

• work with vectors in 2 and 3 dimensions;

• identify and use the components of a position vector;

• use and interpret unit vector form;

• use and interpret a zero vector;

• use and identify parallel vectors;

• determine collinearity of vectors;

• determine the coordinates of a point that divides a line internally;

• calculate the scalar product;

• determine the angle between two vectors;

• identify perpendicular vectors.

By the end of this topic, you should have identified your strengths and areas for furtherrevision.

© HERIOT-WATT UNIVERSITY

TOPIC 7. VECTORS 3

7.1 Learning points

Read through the learning points before you attempt the assessments and go back to the coursematerials if you need to revise further.

Vector definitions

• 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 lowercase letter in bold.

• Displacement is the shortest distance from A to B.

• A vector journey is a description of its displacement.

• i =

⎛⎜⎝

1

0

0

⎞⎟⎠ , j =

⎛⎜⎝

0

1

0

⎞⎟⎠ and k =

⎛⎜⎝

0

0

1

⎞⎟⎠ are unit vectors.

• A unit vector has magnitude 1.

• The zero vector has components

(0

0

)in 2D or

⎛⎜⎝

0

0

0

⎞⎟⎠ in 3D and is written as 0.

• Points which are collinear lie on the same straight line.

• Vectors are parallel if they have the same direction and one is a scalar multiple of the othere.g. e = 1

2 f tells us that vectors e and f are parallel and that the components of e are halfthose of f .

• The scalar product is not a vector it is a scalar.

Vector calculations

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

e.g. in

(x

y

)2D or in

⎛⎜⎝

x

y

z

⎞⎟⎠ 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 Pythagoras' Theoreme.g.

u =

(x

y

), |u | =

√x2 + y2

v =

⎛⎜⎝

x

y

z

⎞⎟⎠ , |v | =

√x2 + y2 + z2

© HERIOT-WATT UNIVERSITY

4 TOPIC 7. VECTORS

• The components of a position vector are the same as the coordinates of the point.

e.g. A(1,2,3) then a =

⎛⎜⎝

1

2

3

⎞⎟⎠.

• −−→AB = b − a

• P, Q and R are collinear if you can show that:−−→PQ = k

−−→QR (i.e.

−−→QR and

−−→PQ are parallel)

andQ is a common point.

• If the point P divides AB in the ratio m:n then:p = n

m + na + mm + nb by the section formula.

• Alternatively if P is mm + n of the way from A to B then:

−→AP = m

m + n

−−→AB

• The scalar product in component form is defined as:

for a =

⎛⎜⎝

x1

y1

z1

⎞⎟⎠ and b =

⎛⎜⎝

x2

y2

z2

⎞⎟⎠

a · b = x1x2 + y1y2 + z1z2

• The scalar product for an angle is defined as:a · b = |a||b|cosθ,wherea and b project outwards from the vertex of the angle θand0 ≤ θ ≤ 180◦

cos θ = a·b| a ||b |

Properties of the scalar product

• a · (b + c) = a · b + a · c• a · b = b · a• a · a = |a|2

• Vectors are perpendicular if a · b = 0; θ = 90◦

© HERIOT-WATT UNIVERSITY

TOPIC 7. VECTORS 5

7.2 Assessment practice

Make sure that you have read through the learning points and completed some revision beforeattempting these questions.

Go onlineAssessment practice: Vectors

SQA Past Paper: 2002 Paper 1

Q1: The point Q divides the line joining P (-1,-1,0) to R(5,2,-3) in the ratio 2:1.What are the coordinates of Q?

(3 marks)

SQA Past Paper: 2003 Paper 1

Q2: Vectors u and v are defined by u = 3i + 2j and v = 2i − 3j + 4kAre vectors u and v perpendicular?

(2 marks)

SQA Objective Question

Q3: The vector u is given by u = 14 i + pk where p < 0.

If u is a unit vector, what is the value of p as an exact value in its simplest form?

(3 marks)

SQA Objective Question

Q4: The point N divides the line LM in the ratio 3:2. L has coordinates (-1,1,0) and−−→LM =⎛

⎜⎝2

1

5

⎞⎟⎠.

What are the coordinates of N?

(3 marks)

SQA Objective Question

Q5: The vectors

⎛⎜⎝

1

2

4

⎞⎟⎠ and

⎛⎜⎝

−5

2

z

⎞⎟⎠ are perpendicular.

What is the value of z?

(2 marks)

© HERIOT-WATT UNIVERSITY

6 TOPIC 7. VECTORS

SQA Objective Question

Q6: Vectors u = 2i − 4j − 8k and v = 5i + pj − 20k are parallel.What is the value of p?

(2 marks)

SQA Past Paper: 2003 Paper 1

Q7: A and B are the points (-1,-3,2) and (2,-1,1) respectively. B and C are the points oftrisection of AD, that is AB = BC = CD.

What are the coordinates of D?

(3 marks)

SQA Past Paper: 2003 Paper 2

Q8: The diagram shows vectors a and b. |a| = 5 and |b| = 4 and a.(a + b) = 36.

What is the size of the acute angle θ, between a and b?

(4 marks)

SQA Past Paper: 2004 Paper 1

A, B and C have coordinates(-3,4,7), (-1,8,3) and (0,10,1).

Q9: What are the components of−−→AB?

(1 mark)

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

Q10: What are the components of−−→BC?

(1 mark)

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

© HERIOT-WATT UNIVERSITY

TOPIC 7. VECTORS 7

Q11: Why are A, B and C collinear?

(1 mark)

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

Q12: In what ratio does B divide AC?

(1 mark)

SQA Past Paper: 2001 Paper 1

Road makers look along the tops of a set of T-rods to ensure that straight sections of roadare being created. Relative to suitable axes the top left corners of the T-rods are the pointsA(-8,-10,-2), B(-2,-1,1) and C(6,11,5).

Q13: Has the section of road ABC been built in a straight line?

(3 marks)

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

Q14:

A further T-rod is placed such that D has coordinates (x,-4,4) and DB is perpendicular to AB.

What is the value of x?

(3 marks)

© HERIOT-WATT UNIVERSITY

8 TOPIC 7. VECTORS

SQA Past Paper: 2001 Paper 2

Q15: A box in the shape of a cuboid is designed with circles of different sizes on each face.The diagram shows three of the circles, where the origin represents one of the corners of thecuboid.

The centres of the circles are A(6,0,7), B(0,5,6) and C(4,5,0).What is the size of angle ABC?

(7 marks)

SQA Past Paper: 2002 Paper 2

The diagram shows a square-based pyramid of height 8 units. Square OABC has side of 6units.

The coordinates of A and D are (6,0,0) and (3,3,8). C lies on the y-axis.

Q16: What are the coordinates of B?

(1 mark)

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

Q17: What are the components of−−→DA?

(1 mark)

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

© HERIOT-WATT UNIVERSITY

TOPIC 7. VECTORS 9

Q18: What are the components of−−→DB?

(1 mark)

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

Q19: What is the size of the angle ADB?

(4 marks)

© HERIOT-WATT UNIVERSITY

10 ANSWERS: UNIT 2 TOPIC 2

Answers to questions and activities

Topic 2: Vectors

Assessment practice: Vectors (page 5)

Q1:

Hints:

•

−−→PQ =

2

3

−→PR

q − p =2

3(r − p)

3q − 3p = 2r − 2p

3q = 2r + p

3q = 2

⎛⎜⎝

5

2

−3

⎞⎟⎠ +

⎛⎜⎝

−1

−1

0

⎞⎟⎠

• Use your answer to find the coordinates of Q.

Steps:

• What fraction of−→PR is

−−→PQ? 2

3

• Express−−→PQ as a fraction of

−→PR and solve for position vector q.

Answer: Q(3,1,-2)

Q2:

Hints:

• u.v = 3 × 2 + 2 × (−3) + 0 × 4 = 0

• since u.v = 0, cos α = 0 and α = 90◦

Steps:

• What is the value of u.v? 0

• Use your answer to determine the angle between the 2 vectors.

Answer: Yes

Q3:

Hints:

• u =

⎛⎜⎝

14

p

0

⎞⎟⎠

• |u| =

√(14

)2+ p2 + 02 and since u is a unit vector |u| = 1

• so,√(

14

)2+ p2 + 02 = 1

© HERIOT-WATT UNIVERSITY

ANSWERS: UNIT 2 TOPIC 2 11

•

1

16+ p2 + 0 = 12

p2 = 1 − 1

16

p2 =15

16

Steps:

• What is a unit vector? One whose magnitude is 1.

• Find p given that you know the magnitude of u.

Answer:√154

Q4: Hints:

•

−−→LN =

3

5

−−→LM

n − l =3

5

⎛⎜⎝

2

1

5

⎞⎟⎠

n −

⎛⎜⎝

−1

1

0

⎞⎟⎠ =

⎛⎜⎝

6535

3

⎞⎟⎠

Steps:

• What fraction of−−→LM is

−−→LN? 3

5

• Use your answer to write an equation for−−→LN and solve for n.

Answer: N ( 15 ,85 ,3)

Q5:

Steps:

• What is the value of the scalar product for perpendicular vectors? 0

• Use the scalar product and solve for z.

Answer: 14

Q6:

Steps:

• u =

⎛⎜⎝

2

−4

−8

⎞⎟⎠ = 2

⎛⎜⎝

1

−2

−4

⎞⎟⎠ and v =

⎛⎜⎝

5

p

−20

⎞⎟⎠ = 5

⎛⎜⎝

1

−2

−4

⎞⎟⎠

Answer: -10

© HERIOT-WATT UNIVERSITY

12 ANSWERS: UNIT 2 TOPIC 2

Q7:

Hints:

•

3−−→AB =

−−→AD

3 (b − a) = d − a

3b − 3a = d − a

3b − 2a = d

3

⎛⎜⎝

2

−1

1

⎞⎟⎠ − 2

⎛⎜⎝

−1

−3

2

⎞⎟⎠ = d

Answer: D(8,3,-1)

Q8:

Hints:

•cos θ =

a.b

|a| |b|cos θ =

11

5 × 4

Steps:

• What is the value of a.a? 25Hints:

a.a = |a|2 cos 0◦

= 5 2 × 1

solve to find a.a• What is the value of a.b? 11

Hints:a. (a + b) = 36

a.a + a.b = 36use your answer from the previous step to find the value of a.b

• Use the scalar product for an angle.

Answer: 56·6◦

Q9:

⎛⎜⎝

2

4

−4

⎞⎟⎠

Q10:

⎛⎜⎝

1

2

−2

⎞⎟⎠

Q11:−−→AB = 2

−−→BC so AB and BC are parallel but B is a common point so A, B and C are

collinear.

© HERIOT-WATT UNIVERSITY

ANSWERS: UNIT 2 TOPIC 2 13

Q12: 2:1

Q13:

Steps:

• What are the components of−−→AB?

⎛⎜⎝

6

9

3

⎞⎟⎠ = 3

⎛⎜⎝

2

3

1

⎞⎟⎠

• What are the components of−−→BC?

⎛⎜⎝

8

12

6

⎞⎟⎠ = 4

⎛⎜⎝

2

3

1

⎞⎟⎠

• Are A, B and C collinear? yesHints: If A, B and C are collinear (i.e. in a straight line) then you must state why

e.g.−−→AB = 3

⎛⎜⎝

2

3

1

⎞⎟⎠ and

−−→BC = 4

⎛⎜⎝

2

3

1

⎞⎟⎠ they are parallel but B is a common point so A,

B and C are collinear.

Answer: Yes

Q14:

Hints:

• To use the scalar product, vectors must project outwards from B.• Perpendicular vectors are at right angles to each other.

• The scalar product for an angle is cos θ = a.b|a||b| and since cos 90◦ = 0 it follows that

a.b = 0

• Find−−→BA.

−−→BD = 0 and solve for x.

Steps:

• What are the components of−−→BD?

⎛⎜⎝

x + 2

−3

3

⎞⎟⎠

• What are the components of−−→BA?

⎛⎜⎝

−6

−9

−3

⎞⎟⎠

• What is an expression for−−→BA.

−−→BD? −6x − 12 + 27 − 9 = − 6x + 6

• What is−−→BA.

−−→BD equal to for perpendicular gradients? 0

• Use your answer to solve for x.

Answer: 1

© HERIOT-WATT UNIVERSITY

14 ANSWERS: UNIT 2 TOPIC 2

Q15:

Steps:

• What are the components of−−→BA?

⎛⎜⎝

6

−5

1

⎞⎟⎠

• What are the components of−−→BC?

⎛⎜⎝

4

0

−6

⎞⎟⎠

• What is∣∣∣−−→BA

∣∣∣? √62

• What is∣∣∣−−→BC

∣∣∣? √52

• What is−−→BA.

−−→BC? 18

• Use the scalar product for an angle.

Answer: 71·5◦

Q16: (6,6,0)

Q17:

⎛⎜⎝

3

−3

8

⎞⎟⎠

Q18:

⎛⎜⎝

3

−3

8

⎞⎟⎠

Q19:

Steps:

• What is∣∣∣−−→DA

∣∣∣? √82

• What is∣∣∣−−→DB

∣∣∣? √82

• What is−−→DA.

−−→DB? 64

• Find the angle by using the scalar product.

Answer: 38·7◦

© HERIOT-WATT UNIVERSITY