G24, G25Geometry & MeasuresFoundation & HigherGrades 4–9

Vectors

Q: How do I write vectors by hand?

Underline the letter: a. Bold type is only used in print.

Q: What is the magnitude of a vector?

Its length, found with Pythagoras: | pmatrix 3 4 pmatrix | = 3^2 + 4^2 = 5.

Q: Is AB the same as BA?

No. They have the same length but opposite directions: BA = - AB.

A vector describes a movement with both a size and a direction. It can be written as a column vector such as $\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ , as a bold letter $\mathbf{a}$ , or as $\overrightarrow{AB}$ . Vector arithmetic is simple, and it powers the hardest geometry proofs on the Higher paper.

50 min AQA GCSE Maths

The big picture

A vector is a journey, not a place. $\overrightarrow{AB}$ means "the journey from $A$ to $B$ " — and any route from $A$ to $B$ gives the same total. That is why you can find an unknown vector by going the long way round through points you know.

Two vectors are parallel when one is a multiple of the other. Showing that $\overrightarrow{PQ} = 3\overrightarrow{RS}$ proves the lines are parallel; showing $\overrightarrow{PQ} = k\overrightarrow{QR}$ proves $P$ , $Q$ and $R$ lie on a straight line. Those two facts are the core of every vector proof.

What you'll learn

Use vector notation and column vectors
Add, subtract and multiply vectors by a scalar
Use vectors to describe routes around a shape
Prove results about parallel lines and collinear points using vectors (Higher)

Column vectors and arithmetic

In $\begin{pmatrix} x \\ y \end{pmatrix}$ , the top number is the movement right (negative means left) and the bottom is the movement up (negative means down).

Add and subtract component by component. Multiplying by a scalar multiplies both components: $3\begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ -3 \end{pmatrix}$ . The vector $-\mathbf{a}$ has the same length as $\mathbf{a}$ but the opposite direction.

Example 1

\mathbf{a} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}

and

\mathbf{b} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}

. Find

2\mathbf{a} - 3\mathbf{b}

2\mathbf{a} = \begin{pmatrix} 8 \\ 2 \end{pmatrix}

and

3\mathbf{b} = \begin{pmatrix} -6 \\ 9 \end{pmatrix}

2Subtract:

\begin{pmatrix} 8 - (-6) \\ 2 - 9 \end{pmatrix}

Answer

\begin{pmatrix} 14 \\ -7 \end{pmatrix}

Your turnSame method, new numbers — try it before moving on.

$\mathbf{p} = \begin{pmatrix} 5 \\ -2 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}$ . Find $\mathbf{p} + 2\mathbf{q}$ .

Your answer

Type the column vector as (top, bottom), e.g. (3, -2).

Show worked solution

Solution

1 $2\mathbf{q} = \begin{pmatrix} 2 \\ 8 \end{pmatrix}$ .
2 $\mathbf{p} + 2\mathbf{q} = \begin{pmatrix} 7 \\ 6 \end{pmatrix}$ .

Answer: $\begin{pmatrix} 7 \\ 6 \end{pmatrix}$

Example 2

A

(1, 5)

and

B

(6, 2)

. Write

\overrightarrow{AB}

as a column vector.

1From

A

B

x

changes by

6 - 1 = 5

;

y

changes by

2 - 5 = -3

Answer

\overrightarrow{AB} = \begin{pmatrix} 5 \\ -3 \end{pmatrix}

Your turnSame method, new numbers — try it before moving on.

$P$ is $(-2, 3)$ and $Q$ is $(4, -1)$ . Write $\overrightarrow{QP}$ as a column vector.

Your answer

Type the column vector as (top, bottom), e.g. (3, -2).

Show hint

$\overrightarrow{QP}$ starts at $Q$ .

Show worked solution

Solution

1From $Q$ to $P$ : $-2 - 4 = -6$ and $3 - (-1) = 4$ .

Answer: $\begin{pmatrix} -6 \\ 4 \end{pmatrix}$

Vectors around shapes

To find an unknown vector, travel along known vectors from the start to the end. Going along a vector backwards changes its sign: $\overrightarrow{BA} = -\overrightarrow{AB}$ .

For a point part-way along a line, use fractions of the vector: if $M$ is the midpoint of $AB$ , then $\overrightarrow{AM} = \frac{1}{2}\overrightarrow{AB}$ .

\overrightarrow{OA} = \mathbf{a}

\overrightarrow{OB} = \mathbf{b}

and

M

is the midpoint of

AB

Example 1In the diagram, find

\overrightarrow{AB}

and

\overrightarrow{OM}

in terms of

\mathbf{a}

and

\mathbf{b}

\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b}

\overrightarrow{OM} = \overrightarrow{OA} + \frac{1}{2}\overrightarrow{AB} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a})

3Simplify:

\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}

Answer

\overrightarrow{AB} = \mathbf{b} - \mathbf{a}

;

\overrightarrow{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{b})

Example 2

OACB

is a parallelogram with

\overrightarrow{OA} = \mathbf{a}

and

\overrightarrow{OB} = \mathbf{b}

N

is on

AC

with

AN : NC = 1 : 3

. Find

\overrightarrow{ON}

\overrightarrow{AC} = \overrightarrow{OB} = \mathbf{b}

(opposite sides of a parallelogram).

\overrightarrow{AN} = \frac{1}{4}\mathbf{b}

\overrightarrow{ON} = \mathbf{a} + \frac{1}{4}\mathbf{b}

Answer

\overrightarrow{ON} = \mathbf{a} + \frac{1}{4}\mathbf{b}

Tip — A ratio $1 : 3$ splits a line into 4 parts, so $AN$ is $\frac{1}{4}$ of $AC$ — not $\frac{1}{3}$ .

Parallel vectors and proofs (Higher)

Vectors are parallel if one is a scalar multiple of the other: $\begin{pmatrix} 6 \\ -9 \end{pmatrix} = 3\begin{pmatrix} 2 \\ -3 \end{pmatrix}$ , and $4\mathbf{a} - 2\mathbf{b} = 2(2\mathbf{a} - \mathbf{b})$ .

To prove three points $P$ , $Q$ , $R$ are on a straight line (collinear), show $\overrightarrow{PQ}$ is a multiple of $\overrightarrow{QR}$ (they are parallel) $\textbf{and}$ state that they share the point $Q$ .

Example 1

\overrightarrow{PQ} = 2\mathbf{a} + 6\mathbf{b}

and

\overrightarrow{QR} = \mathbf{a} + 3\mathbf{b}

. Prove that

P

Q

and

R

are collinear.

\overrightarrow{PQ} = 2(\mathbf{a} + 3\mathbf{b}) = 2\overrightarrow{QR}

2So

PQ

is parallel to

QR

3They share the point

Q

, so

P

Q

and

R

lie on a straight line.

Answer

Collinear, since

\overrightarrow{PQ} = 2\overrightarrow{QR}

with common point

Q

Your turnSame method, new numbers — try it before moving on.

Which value of $k$ makes $\begin{pmatrix} k \\ 10 \end{pmatrix}$ parallel to $\begin{pmatrix} 3 \\ 5 \end{pmatrix}$ ?

Your answer

Show worked solution

Solution

1 $10$ is $2 \times 5$ , so the vector must be $2\begin{pmatrix} 3 \\ 5 \end{pmatrix}$ .
2 $k = 2 \times 3 = 6$ .

Answer: $k = 6$

Why the common point matters

Parallel alone is not enough for collinearity: two train tracks are parallel but never meet. The shared point is what forces the two segments onto the same line — which is why the examiner wants it stated.

Think like an examiner

Write the route you are taking, e.g. $\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM}$ , before substituting — it earns method marks.
Simplify fully: collect the $\mathbf{a}$ terms and the $\mathbf{b}$ terms.
In proofs, a final sentence using the words "parallel" and "common point" is needed for the last mark.

Vector facts

\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = \mathbf{b} - \mathbf{a}

Travel via known points.

\overrightarrow{BA} = -\overrightarrow{AB}

Reverse direction, change sign.

\mathbf{u} = k\mathbf{v} \Rightarrow \mathbf{u} \parallel \mathbf{v}

Parallel vectors.

Watch out for these

\overrightarrow{AB} = \mathbf{a} - \mathbf{b}

when

\overrightarrow{OA} = \mathbf{a}

\overrightarrow{OB} = \mathbf{b}

\overrightarrow{AB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}

Writing a column vector as coordinates

(3, -2)

in working.

Use

\begin{pmatrix} 3 \\ -2 \end{pmatrix}

For

AN : NC = 1 : 3

, using

\overrightarrow{AN} = \frac{1}{3}\overrightarrow{AC}

There are 4 parts:

\frac{1}{4}\overrightarrow{AC}

Concluding collinear from parallel vectors alone.

Also state the common point.

Stretch yourself

$OAB$ is a triangle with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ . $P$ is the midpoint of $OA$ and $Q$ is the midpoint of $OB$ . Prove that $PQ$ is parallel to $AB$ and half its length.

Hint — Find $\overrightarrow{PQ}$ via $O$ , and compare it with $\overrightarrow{AB}$ .

Reveal solution

Solution

\overrightarrow{PQ} = \overrightarrow{PO} + \overrightarrow{OQ} = -\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} = \frac{1}{2}(\mathbf{b} - \mathbf{a})

. And

\overrightarrow{AB} = \mathbf{b} - \mathbf{a}

. So

\overrightarrow{PQ} = \frac{1}{2}\overrightarrow{AB}

PQ

is parallel to

AB

and half its length. (This is the midpoint theorem.)

\blacksquare

Questions students ask

How do I write vectors by hand?

Underline the letter:

\underline{a}

. Bold type is only used in print.

What is the magnitude of a vector?

Its length, found with Pythagoras:

\left|\begin{pmatrix} 3 \\ 4 \end{pmatrix}\right| = \sqrt{3^2 + 4^2} = 5

\overrightarrow{AB}

the same as

\overrightarrow{BA}

No. They have the same length but opposite directions:

\overrightarrow{BA} = -\overrightarrow{AB}

Key takeaways

Column vectors: top is right/left, bottom is up/down.
Add, subtract and scale component by component.
Find unknown vectors by travelling via known points; reversing changes the sign.
Use fractions of vectors for points dividing lines in a ratio.
Parallel means a scalar multiple; collinear also needs a common point.

Exam-style questions

Practise it the way it is examined

3 original questions · 9 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 42 marks

$\mathbf{a} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -2 \\ 5 \end{pmatrix}$ .

Work out $3\mathbf{a} + \mathbf{b}$ as a column vector.[2]
Show mark scheme
- M1 $3\mathbf{a} = \begin{pmatrix} 9 \\ -3 \end{pmatrix}$
- A1 $\begin{pmatrix} 7 \\ 2 \end{pmatrix}$
Answer: $\begin{pmatrix} 7 \\ 2 \end{pmatrix}$

Question 2Grade 63 marks

$ABCDEF$ is a regular hexagon with centre $O$ . $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ .

Write $\overrightarrow{AB}$ and $\overrightarrow{AC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .[3]
Show mark scheme
- B1 $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$
- M1Uses $\overrightarrow{BC} = \overrightarrow{AO} = -\mathbf{a}$ (the hexagon is made of equilateral triangles)
- A1 $\overrightarrow{AC} = \mathbf{b} - 2\mathbf{a}$

Question 3Grade 84 marks

$OAB$ is a triangle. $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ . $M$ is the midpoint of $AB$ . $N$ is the point on $OB$ such that $ON : NB = 2 : 1$ . $X$ is the point such that $\overrightarrow{OX} = 2\mathbf{a}$ .

Prove that $X$ , $M$ and $N$ lie on a straight line.[4]
Show mark scheme
- M1 $\overrightarrow{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$ and $\overrightarrow{ON} = \frac{2}{3}\mathbf{b}$
- M1 $\overrightarrow{XM} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} - 2\mathbf{a} = -\frac{3}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}$
- M1 $\overrightarrow{MN} = \frac{2}{3}\mathbf{b} - \frac{1}{2}\mathbf{a} - \frac{1}{2}\mathbf{b} = -\frac{1}{2}\mathbf{a} + \frac{1}{6}\mathbf{b}$
- A1 $\overrightarrow{XM} = 3\overrightarrow{MN}$ , so $XM$ is parallel to $MN$ , and they share the point $M$ — so $X$ , $M$ and $N$ are collinear

Examiner note

Write both vectors in simplified form first; the multiple (here 3) is then easy to spot. Without the statement about the common point $M$ , the final mark is lost.

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

$\begin{pmatrix} 2 \\ 5 \end{pmatrix} + \begin{pmatrix} 3 \\ -1 \end{pmatrix} =$
- $\begin{pmatrix} 5 \\ 4 \end{pmatrix}$
- $\begin{pmatrix} 5 \\ 6 \end{pmatrix}$
- $\begin{pmatrix} -1 \\ 6 \end{pmatrix}$
- $\begin{pmatrix} 6 \\ -5 \end{pmatrix}$
Show answer and explanation
Answer: $\begin{pmatrix} 5 \\ 4 \end{pmatrix}$
Add each component.
$4\begin{pmatrix} -1 \\ 2 \end{pmatrix} =$
- $\begin{pmatrix} -4 \\ 8 \end{pmatrix}$
- $\begin{pmatrix} 3 \\ 6 \end{pmatrix}$
- $\begin{pmatrix} -4 \\ 2 \end{pmatrix}$
- $\begin{pmatrix} 4 \\ -8 \end{pmatrix}$
Show answer and explanation
Answer: $\begin{pmatrix} -4 \\ 8 \end{pmatrix}$
Multiply both components.
If $\overrightarrow{AB} = \mathbf{p}$ , then $\overrightarrow{BA} =$
- $-\mathbf{p}$
- $\mathbf{p}$
- $2\mathbf{p}$
- $\frac{1}{\mathbf{p}}$
Show answer and explanation
Answer: $-\mathbf{p}$
Opposite direction.
$\overrightarrow{OA} = \mathbf{a}$ , $\overrightarrow{OB} = \mathbf{b}$ . $\overrightarrow{AB} =$
- $\mathbf{b} - \mathbf{a}$
- $\mathbf{a} - \mathbf{b}$
- $\mathbf{a} + \mathbf{b}$
- $\frac{1}{2}(\mathbf{a} + \mathbf{b})$
Show answer and explanation
Answer: $\mathbf{b} - \mathbf{a}$
Go from $A$ back to $O$ , then to $B$ .
Which vector is parallel to $2\mathbf{a} - 3\mathbf{b}$ ?
- $6\mathbf{a} - 9\mathbf{b}$
- $3\mathbf{a} - 2\mathbf{b}$
- $2\mathbf{a} + 3\mathbf{b}$
- $4\mathbf{a} - 3\mathbf{b}$
Show answer and explanation
Answer: $6\mathbf{a} - 9\mathbf{b}$
$3(2\mathbf{a} - 3\mathbf{b})$ .
Magnitude of $\begin{pmatrix} 6 \\ 8 \end{pmatrix}$ ?
- $10$
- $14$
- $100$
- $2$
Show answer and explanation
Answer: $10$
$\sqrt{36 + 64}$ .
$M$ is the midpoint of $AB$ ; $\overrightarrow{OA} = \mathbf{a}$ , $\overrightarrow{OB} = \mathbf{b}$ . $\overrightarrow{OM} =$
- $\frac{1}{2}(\mathbf{a} + \mathbf{b})$
- $\frac{1}{2}(\mathbf{b} - \mathbf{a})$
- $\mathbf{a} + \mathbf{b}$
- $2(\mathbf{a} + \mathbf{b})$
Show answer and explanation
Answer: $\frac{1}{2}(\mathbf{a} + \mathbf{b})$
$\mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a})$ .
To prove $P$ , $Q$ , $R$ are collinear, you show…
- $\overrightarrow{PQ} = k\overrightarrow{QR}$ and they share $Q$
- $\overrightarrow{PQ} = \overrightarrow{QR}$ only
- the vectors have equal magnitudes
- $\overrightarrow{PQ}$ is perpendicular to $\overrightarrow{QR}$
Show answer and explanation
Answer: $\overrightarrow{PQ} = k\overrightarrow{QR}$ and they share $Q$
Parallel plus a common point.

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