National 5 Maths
Vectors

Course content

Adding or subtracting 2D vector pathways using directed line segments
Adding or subtracting 2D or 3D vectors using components
Calculating the magnitude of a 2D or 3D vector.

Magnitude

Magnitude means length.

The magnitude of a vector $v$ is written in print as $| v |$ and in handwriting as $| \underline{v} | .$

Similarly, the magnitude of $\vec{AB}$ is written as $| \vec{AB} | .$

If a vector is given in component form, we can use either 2D or 3D Pythagoras to find its magnitude.

Textbook page references

Zeta National 5+ Maths pp.234-249
TeeJay Maths Book N5 pp.141-152
Leckie National 5 Maths pp.312-327

Notation

$\vec{AB}$ represents the vector from point A to point B.
Lower case letters are also used to represent vectors.
- In handwriting, they are underlined, like this: $\underline{u}$
- In print, they are in bold, like this: $u$

Key idea

For any three points A, B and C, $\vec{AC} = \vec{AB} + \vec{BC}$ .
Think of this as going from A to B and then continuing from B to C. In other words: from A to C via B.
It's the same resultant journey whether we go via B or not. We start at A and end at C, so the vectors are equal. Vectors only care about the end-points, not the pathway.

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Example 1 (non-calculator)

In the diagram below, $u$ , $v$ and $w$ represent $\vec{AB}$ , $\vec{BC}$ and $\vec{AC}$ respectively.

Express $v$ in terms of $u$ and $w$ .

This simple vector pathways question uses the key idea above:

\begin{array}{rcl} \vec{AC} & = & \vec{AB} + \vec{BC} \\ \underline{w} & = & \underline{u} + \underline{v} \\ \underline{v} & = & \underline{w} - \underline{u} \end{array}

Yes, it's that simple! Questions like this are only worth one mark.

Example 2 (non-calculator)

PQRS is a trapezium with $\vec{RQ} = 2 \vec{SP}$ .

$\vec{PQ}$ and $\vec{RQ}$ represent vectors $u$ and $v$ respectively.

(a) Express $\vec{RP}$ in terms of $u$ and $v$ .
(b) Express $\vec{RS}$ in terms of $u$ and $v$ . Give your answer in its simplest form.

(a) Using the key idea at the top:

\begin{array}{rcl} \vec{RP} & = & \vec{RQ} + \vec{QP} \\ = & \vec{RQ} - \vec{PQ} \\ = & \underline{v} - \underline{u} \end{array}

Note: the reason for the minus sign is that $\vec{QP}$ and $\vec{PQ}$ are in opposite directions.

(b) We need to decide a route from R to S. We could go around the outside, via Q and then P. But let's "cut the corner" and just go via P:

\begin{array}{rcl} \vec{RS} & = & \vec{RP} + \vec{PS} \\ = & \vec{RP} - \frac{1}{2} \vec{RQ} (because \vec{RQ} = 2 \vec{SP}) \\ = & (\underline{v} - \underline{u}) - \frac{1}{2} \underline{v} (using part (a)) \\ = & \frac{1}{2} \underline{v} - \underline{u} \end{array}

Example 3 (non-calculator)

Given $u = (\begin{matrix} - 4 \\ 5 \end{matrix})$ and $v = (\begin{matrix} 3 \\ - 1 \end{matrix})$ , find:
  (a) $u + v$
  (b) $u - v$
  (c) $2 u - 3 v$
  (d) $\frac{1}{2} u + v$

These vectors are in "component form".

The top value is the $x$ -component: how far left (–) or right (+). The bottom value is the $y$ -component: how far up (+) or down (–).

The method here is just add or subtract the $x$ - and $y$ -components separately.

\begin{array}{rcl} (a) \underline{u} + \underline{v} & = & (\begin{array}{c} - 4 \\ 5 \end{array}) + (\begin{array}{c} 3 \\ - 1 \end{array}) \\ = & (\begin{array}{c} - 1 \\ 4 \end{array}) \\ (b) \underline{u} - \underline{v} & = & (\begin{array}{c} - 4 \\ 5 \end{array}) - (\begin{array}{c} 3 \\ - 1 \end{array}) \\ = & (\begin{array}{c} - 7 \\ 6 \end{array}) \\ (c) 2 \underline{u} - 3 \underline{v} & = & 2 (\begin{array}{c} - 4 \\ 5 \end{array}) - 3 (\begin{array}{c} 3 \\ - 1 \end{array}) \\ = & (\begin{array}{c} - 8 \\ 10 \end{array}) - (\begin{array}{c} 9 \\ - 3 \end{array}) \\ = & (\begin{array}{c} - 17 \\ 13 \end{array}) \\ (d) \frac{1}{2} \underline{u} + \underline{v} & = & \frac{1}{2} (\begin{array}{c} - 4 \\ 5 \end{array}) + (\begin{array}{c} 3 \\ - 1 \end{array}) \\ = & (\begin{array}{c} - 2 \\ \frac{5}{2} \end{array}) + (\begin{array}{c} 3 \\ - 1 \end{array}) \\ = & (\begin{array}{c} 1 \\ \frac{3}{2} \end{array}) \end{array}

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Example 4 (non-calculator)

Given $p = (\begin{matrix} 3 \\ - 2 \\ 1 \end{matrix})$ and $q = (\begin{matrix} 5 \\ 0 \\ - 3 \end{matrix})$ , find:
(a) $2 p - q$
(b) $\frac{1}{2} (p - q)$

This example includes $z$ -components: the third dimension.

\begin{array}{rcl} (a) 2 \underline{p} - \underline{q} & = & 2 (\begin{array}{c} 3 \\ - 2 \\ 1 \end{array}) - (\begin{array}{c} 5 \\ 0 \\ - 3 \end{array}) \\ = & (\begin{array}{c} 6 \\ - 4 \\ 2 \end{array}) - (\begin{array}{c} 5 \\ 0 \\ - 3 \end{array}) \\ = & (\begin{array}{c} 1 \\ - 4 \\ 5 \end{array}) \\ (b) \frac{1}{2} (\underline{p} - \underline{q}) & = & \frac{1}{2} [(\begin{array}{c} 3 \\ - 2 \\ 1 \end{array}) - (\begin{array}{c} 5 \\ 0 \\ - 3 \end{array})] \\ = & \frac{1}{2} (\begin{array}{c} - 2 \\ - 2 \\ 4 \end{array}) \\ = & (\begin{array}{c} - 1 \\ - 1 \\ 2 \end{array}) \end{array}

Example 5 (calculator)

Find $| v |$ , the magnitude of vector $v = (\begin{matrix} 45 \\ - 28 \end{matrix})$ .

The magnitude is just the length of the hypotenuse of the right-angled triangle made from 45 right and 28 down. So the method here is really just 2D Pythagoras.

\begin{array}{rcl} | \underline{v} | & = & \sqrt{45^{2} + (- 28)^{2}} \\ = & \sqrt{2025 + 784} \\ = & \sqrt{2809} \\ = & 53 \end{array}

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Example 6 (calculator)

Find $| r |$ , the magnitude of vector $r = (\begin{matrix} 8 \\ - 19 \\ - 4 \end{matrix})$ .

The method here is really just 3D Pythagoras.

\begin{array}{rcl} | \underline{r} | & = & \sqrt{8^{2} + (- 19)^{2} + (- 4)^{2}} \\ = & \sqrt{64 + 361 + 16} \\ = & \sqrt{441} \\ = & 21 \end{array}

Example 7 (non-calculator)

Find $| a - b |$ , where $a = (\begin{matrix} 5 \\ 6 \end{matrix})$ and $b = (\begin{matrix} 9 \\ 4 \end{matrix})$ . Express your answer as a surd in its simplest form.

We need to find $\underline{a} - \underline{b}$ in component form, and then calculate its magnitude.

\begin{array}{rcl} | \underline{a} - \underline{b} | & = & | (\begin{array}{c} 5 \\ 6 \end{array}) - (\begin{array}{c} 9 \\ 4 \end{array}) | \\ = & | (\begin{array}{c} - 4 \\ 2 \end{array}) | \\ = & \sqrt{(- 4)^{2} + 2^{2}} \\ = & \sqrt{16 + 4} \\ = & \sqrt{20} \\ = & \sqrt{4} \sqrt{5} \\ = & 2 \sqrt{5} \end{array}

Example 8 (non-calculator)

The diagram shows a rectangular-based pyramid, relative to the coordinate axes.

B is the point (8,6,0).

The apex D is directly above the centre of the base.

The height of the pyramid is 7 units.

(a) Express $\vec{AC}$ in component form.
(b) Express $\vec{AD}$ in component form.
(c) Determine the exact value of $| \vec{AD} | .$

This example links with the 3D Coordinates topic.

(a) $\vec{AC} = \vec{AO} + \vec{OC}$ so it involves moving 8 in the negative $x$ -direction and 6 in the positive $y$ -direction.

So $\vec{AC} = (\begin{matrix} - 8 \\ 6 \\ 0 \end{matrix})$ .

(b) The centre of the base is (4,3,0) so D is (4,3,7).

So the position vector $\vec{OD} = (\begin{matrix} 4 \\ 3 \\ 7 \end{matrix})$ .

$\begin{array}{rcl} \vec{AD} & = & \vec{AO} + \vec{OD} \\ = & (\begin{array}{c} - 8 \\ 0 \\ 0 \end{array}) + (\begin{array}{c} 4 \\ 3 \\ 7 \end{array}) \\ = & (\begin{array}{c} - 4 \\ 3 \\ 7 \end{array}) \end{array}$

\begin{array}{rcl} | \vec{AD} | & = & \sqrt{(- 4)^{2} + 3^{2} + 7^{2}} \\ = & \sqrt{16 + 9 + 49} \\ = & \sqrt{74} \end{array}

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Example 9 (calculator)

SQA National 5 Maths 2016 P2 Q3

The diagram below shows parallelogram ABCD.

$\vec{AB}$ represents vector $u$ and $\vec{BC}$ represents vector $v .$
Express $\vec{BD}$ in terms of $u$ and $v .$

This was a simple 1-mark question.

$\begin{array}{rcl} \vec{BD} & = & \vec{BA} + \vec{AD} \\ = & \vec{BA} + \vec{BC} \\ = & - \underline{u} + \underline{v} \end{array}$

Note that $v - u$ is also correct.

Example 10 (calculator)

SQA National 5 Maths 2018 P2 Q10

In the diagram below, $\vec{AB}$ and $\vec{EA}$ represent the vectors $u$ and $w$ respectively.

• $\vec{ED} = 2 \vec{AB}$
• $\vec{EA} = 2 \vec{DC}$
Express $\vec{BC}$ in terms of $u$ and $w .$
Give your answer in its simplest form.

This was a 2-mark question. We need to go the long way from B to C.

$\begin{array}{rcl} \vec{BC} & = & \vec{BA} + \vec{AE} + \vec{ED} + \vec{DC} \\ = & - \underline{u} - \underline{w} + 2 \underline{u} + \frac{1}{2} \underline{w} \\ = & \underline{u} - \frac{1}{2} \underline{w} \end{array}$

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Non-calculator papers and solutions

Calculator papers and solutions

Example 11 (non-calculator)

SQA National 5 Maths 2019 P1 Q10

In triangle PQR, $\vec{PR} = (\begin{matrix} 6 \\ - 4 \end{matrix})$ and $\vec{RQ} = (\begin{matrix} - 1 \\ 8 \end{matrix}) .$

(a) Express $\vec{PQ}$ in component form.

M is the midpoint of PR.
(b) Express $\vec{MQ}$ in component form.

(a) This part was only worth 1 mark. We will go from P to Q via R.

$\begin{array}{rcl} \vec{PQ} & = & \vec{PR} + \vec{RQ} \\ = & (\begin{array}{c} 6 \\ - 4 \end{array}) + (\begin{array}{c} - 1 \\ 8 \end{array}) \\ = & (\begin{array}{c} 5 \\ 4 \end{array}) \end{array}$

(b) In this 2-mark part, we could go via either P or R.

The answer below goes via R. You might like to try it the other way yourself.

$\begin{array}{rcl} \vec{MQ} & = & \vec{MR} + \vec{RQ} \\ = & \frac{1}{2} \vec{PR} + \vec{RQ} \\ = & \frac{1}{2} (\begin{array}{c} 6 \\ - 4 \end{array}) + (\begin{array}{c} - 1 \\ 8 \end{array}) \\ = & (\begin{array}{c} 3 \\ - 2 \end{array}) + (\begin{array}{c} - 1 \\ 8 \end{array}) \\ = & (\begin{array}{c} 2 \\ 6 \end{array}) \end{array}$

Example 12 (non-calculator)

SQA National 5 Maths 2021 P2 Q5

The vectors $u$ and $v$ are shown in the diagram below.

Find the resultant vector $u - v .$
Express your answer in component form.

This question was worth 2 marks.

First we need to read the components of $\underline{u}$ and $\underline{v}$ from the diagram.

$\underline{u}$ is 5 right and 2 up, so $\underline{u} = (\begin{matrix} 5 \\ 2 \end{matrix}) .$

$\underline{v}$ is 3 right and 4 down, so $\underline{v} = (\begin{matrix} 3 \\ - 4 \end{matrix}) .$

Now we just calculate $\underline{u} - \underline{v}$ in the usual way.

$\begin{array}{rcl} \underline{u} - \underline{v} & = & (\begin{array}{c} 5 \\ 2 \end{array}) - (\begin{array}{c} 3 \\ - 4 \end{array}) \\ = & (\begin{array}{c} 2 \\ 6 \end{array}) \end{array}$

Example 13 (calculator)

SQA National 5 Maths 2021 P2 Q17

The triangle ABC is shown below

$\vec{AB} = u$ and $\vec{AC} = t .$
G is the point such that CG = $\frac{1}{3}$ CB.
Express $\vec{AG}$ in terms of $u$ and $t .$
Give your answer in simplest form.

This vector pathways question was worth 3 marks, as it involves two steps before we can begin expressing the vector $\vec{AG}$ in terms of $u$ and $t .$

$\begin{array}{rcl} \vec{AG} & = & \vec{AC} + \vec{CG} \\ = & \vec{AC} + \frac{1}{3} \vec{CB} \\ = & \vec{AC} + \frac{1}{3} (\vec{CA} + \vec{AB}) \\ = & \underline{t} + \frac{1}{3} (- \underline{t} + \underline{u}) \\ = & \underline{t} - \frac{1}{3} \underline{t} + \frac{1}{3} \underline{u} \\ = & \frac{2}{3} \underline{t} + \frac{1}{3} \underline{u} \end{array}$

Example 14 (non-calculator)

SQA National 5 Maths 2024 P1 Q4

Given $a = (\begin{matrix} 3 \\ 4 \\ - 1 \end{matrix})$ and $b = (\begin{matrix} 5 \\ 3 \\ 2 \end{matrix}),$ find the resultant vector $3 a + b .$ Express your answer in component form.

$\begin{array}{rcl} 3 \underline{a} + \underline{b} & = & 3 (\begin{array}{c} 3 \\ 4 \\ - 1 \end{array}) + (\begin{array}{c} 5 \\ 3 \\ 2 \end{array}) \\ = & (\begin{array}{c} 9 \\ 12 \\ - 3 \end{array}) + (\begin{array}{c} 5 \\ 3 \\ 2 \end{array}) \\ = & (\begin{array}{c} 14 \\ 15 \\ - 1 \end{array}) \end{array}$

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Maths.scot worksheet

• Vectors worksheet
• Answer sheet
• See all National 5 Maths worksheets

Past paper questions

• All past paper questions by topic

Vector pathways:
• 2016 Paper 2 Q3
• 2017 Specimen Paper 1 Q11
• 2017 Paper 2 Q8
• 2018 Paper 2 Q10
• 2021 Paper 2 Q17
• 2024 Paper 2 Q14

Vector components:
• 2014 Paper 1 Q4
• 2015 Paper 2 Q5
• 2016 Paper 1 Q1
• 2018 Paper 1 Q4
• 2019 Paper 1 Q10
• 2021 Paper 2 Q5
• 2024 Paper 1 Q4

Magnitude:
• 2015 Paper 2 Q4
• 2017 Specimen Paper 1 Q3
• 2017 Paper 2 Q1
• 2018 Paper 2 Q3
• 2019 Paper 2 Q2
• 2021 Paper 1 Q1

Other great resources

Videos - Maths180.com

Videos - Mr Graham Maths
1. Introduction to vectors
2. Alternative routes
3. Magnitude of a vector
4. Scalar multiplication
5. Vector pathways

Video - Mr Hamilton Online
Resultant and magnitude

Videos - YouKenMaths
1. 2D vectors
2. Vector journeys: 1 of 4
3. Vector journeys: 2 of 4
4. Vector journeys: 3 of 4
5. Vector journeys: 4 of 4
6. Worked examples

Notes and videos - Mistercorzi
1. 2D vectors
2. 3D vectors

Theory guide - National 5 Maths

PowerPoint - MathsRevision.com

Detailed notes - BBC Bitesize
1. 2D vectors
2. Vector components
3. Magnitude of a vector

Notes - National5.com

Notes - D R Turnbull

Lesson notes - Maths 777
1. Vector arithmetic, resultant
2. Magnitude of a vector

Notes and examples - Maths Mutt

Practice questions - Maths Hunter

Worksheet - Cumnock Academy

Essential Skills worksheet
Vectors worksheet (Answers)

Exercises - Larkhall Academy
Pages 19-32 Ex 1-8

Click here to study the vectors notes on National5.com.

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