Post on 13-May-2018
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.
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SCHOLAR Study Guide Assessment Practice: CfE Higher Mathematics
1. CfE Higher Mathematics Course Code: C747 76
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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