1MA1 · G24–G25Geometry & MeasuresHigherGrades 5–9

Vectors

A vector has both size and direction — a displacement like "3 right and 2 up". Vectors can be added, subtracted and multiplied by numbers, which turns geometry into algebra: to show two lines are parallel, or three points lie on a straight line, you write each line as a vector and compare.

50 min Transformations & Vectors

The big picture

The central idea is the route. To get from one point to another, you can travel along any path of known vectors, and the total displacement is their sum. Going backwards along a vector changes its sign. Once you can write routes, vector geometry questions become careful bookkeeping.

Two vectors are parallel when one is a multiple of the other. Three points are collinear when the vectors between them are parallel and share a point. Edexcel Higher vector proofs, typically 4 or 5 marks, are built entirely on these two facts.

What you'll learn

Add, subtract and multiply column vectors, and find their magnitude
Write vectors for routes through geometric shapes
Use ratios to find vectors to points dividing a line
Prove that lines are parallel or points are collinear

Column vectors

$\begin{pmatrix} 3 \\ -1 \end{pmatrix}$ means 3 across and 1 down. Vectors are written in bold, $\mathbf{a}$ , or with an arrow, $\overrightarrow{AB}$ .

Add and subtract component by component; multiply every component by a scalar.

The magnitude (length) of $\begin{pmatrix} x \\ y \end{pmatrix}$ is $\sqrt{x^{2} + y^{2}}$ , by Pythagoras.

Example 1

\mathbf{a} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}

and

\mathbf{b} = \begin{pmatrix} -2 \\ 4 \end{pmatrix}

. Find

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

and its magnitude.

2\mathbf{a} = \begin{pmatrix} 6 \\ -2 \end{pmatrix}

2\mathbf{a} - \mathbf{b} = \begin{pmatrix} 8 \\ -6 \end{pmatrix}

3Magnitude

= \sqrt{64 + 36}

Answer

\begin{pmatrix} 8 \\ -6 \end{pmatrix}

, magnitude

10

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

$\mathbf{a} = \begin{pmatrix} 4 \\ 1 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}$ . Find $2\mathbf{a} - \mathbf{b}$ .

Your answer

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

Show worked solution

Solution

1 $2\mathbf{a} = \begin{pmatrix} 8 \\ 2 \end{pmatrix}$ .
2 $\begin{pmatrix} 8 - (-1) \\ 2 - 3 \end{pmatrix} = \begin{pmatrix} 9 \\ -1 \end{pmatrix}$ .

Answer: $\begin{pmatrix} 9 \\ -1 \end{pmatrix}$

Routes through shapes

$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$ : travel from $A$ to $C$ via $B$ .

Reversing direction changes the sign: $\overrightarrow{BA} = -\overrightarrow{AB}$ .

In a parallelogram $OABC$ with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$ , opposite sides are equal vectors, so $\overrightarrow{CB} = \mathbf{a}$ and $\overrightarrow{AB} = \mathbf{c}$ .

Example 1

\overrightarrow{OA} = \mathbf{a}

and

\overrightarrow{OB} = \mathbf{b}

M

is the midpoint of

AB

. Find

\overrightarrow{OM}

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

\overrightarrow{OM} = \overrightarrow{OA} + \tfrac12\overrightarrow{AB} = \mathbf{a} + \tfrac12(\mathbf{b} - \mathbf{a})

Answer

\overrightarrow{OM} = \tfrac12(\mathbf{a} + \mathbf{b})

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

$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ . Write $\overrightarrow{AB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .

Your answer

Type it like b - a.

Show worked solution

Solution

1 $\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB}$ .
2 $= -\mathbf{a} + \mathbf{b}$ .

Answer: $\mathbf{b} - \mathbf{a}$

Tip — Always go through known vectors. To find $\overrightarrow{AB}$ , route back to $O$ first: $\overrightarrow{AO} + \overrightarrow{OB}$ .

Dividing a line in a ratio

If $P$ lies on $AB$ with $AP : PB = m : n$ , then $\overrightarrow{AP} = \dfrac{m}{m + n}\overrightarrow{AB}$ .

So $\overrightarrow{OP} = \overrightarrow{OA} + \dfrac{m}{m + n}\overrightarrow{AB}$ .

Example 1With

\overrightarrow{OA} = \mathbf{a}

and

\overrightarrow{OB} = \mathbf{b}

P

lies on

AB

with

AP : PB = 1 : 2

. Find

\overrightarrow{OP}

\overrightarrow{AP} = \tfrac13\overrightarrow{AB} = \tfrac13(\mathbf{b} - \mathbf{a})

\overrightarrow{OP} = \mathbf{a} + \tfrac13\mathbf{b} - \tfrac13\mathbf{a}

Answer

\overrightarrow{OP} = \tfrac23\mathbf{a} + \tfrac13\mathbf{b}

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

With $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ , $P$ lies on $AB$ with $AP : PB = 1 : 3$ . Find $\overrightarrow{OP}$ .

Your answer

Type it like 3a/4 + b/4.

Show worked solution

Solution

1 $\overrightarrow{OP} = \mathbf{a} + \frac{1}{4}(\mathbf{b} - \mathbf{a})$ .
2 $= \frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}$ .

Answer: $\frac{3}{4}\mathbf{a} + \frac{1}{4}\mathbf{b}$

Why this works

A ratio of $1 : 2$ splits the line into 3 equal parts, so $P$ is one third of the way along. The denominator is always the total number of parts.

Parallel vectors and collinear points

Vectors are parallel if one is a scalar multiple of the other: $\begin{pmatrix} 6 \\ -4 \end{pmatrix} = 2\begin{pmatrix} 3 \\ -2 \end{pmatrix}$ .

Points $A$ , $B$ and $C$ are collinear if $\overrightarrow{AB}$ is a multiple of $\overrightarrow{BC}$ (or $\overrightarrow{AC}$ ) — parallel vectors sharing the point $B$ must lie on one line.

A complete proof states both facts: the multiple, and the common point.

Example 1

\overrightarrow{OA} = \mathbf{a}

\overrightarrow{OB} = \mathbf{b}

C

is such that

\overrightarrow{OC} = 3\mathbf{b} - 2\mathbf{a}

. Prove that

A

B

and

C

are collinear.

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

\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = 2\mathbf{b} - 2\mathbf{a} = 2(\mathbf{b} - \mathbf{a})

3So

\overrightarrow{BC} = 2\overrightarrow{AB}

: the vectors are parallel.

Answer

\overrightarrow{BC}

is a multiple of

\overrightarrow{AB}

and they share point

B

, so

A

B

and

C

are collinear.

Tip — Factorise vector expressions — $2\mathbf{b} - 2\mathbf{a} = 2(\mathbf{b} - \mathbf{a})$ — so the multiple is visible. That factorised line is usually a mark.

Think like an examiner

Write each route as a sum of known vectors.
Simplify and factorise expressions to show multiples.
For collinearity, state the multiple and the common point.

Vectors

\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}

Vector between points.

\left|\begin{pmatrix} x \\ y \end{pmatrix}\right| = \sqrt{x^{2} + y^{2}}

Magnitude.

\overrightarrow{AP} = \tfrac{m}{m + n}\overrightarrow{AB}

Point dividing AB in m : n.

Watch out for these

Writing

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

when

\overrightarrow{OA} = \mathbf{a}

\overrightarrow{OB} = \mathbf{b}

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

(end minus start).

Using

\tfrac12

for a ratio

1 : 2

1 : 2

means

\tfrac13

of the way.

Claiming collinearity from parallel vectors alone.

The vectors must also share a point.

Adding magnitudes instead of vectors.

Add components, then find the magnitude.

Stretch yourself

$OABC$ is a parallelogram with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$ . $M$ is the midpoint of $AB$ , and $N$ lies on $OB$ with $ON : NB = 2 : 1$ . Prove that $C$ , $N$ and $M$ are collinear.

Hint — Find $\overrightarrow{OB}$ , $\overrightarrow{ON}$ and $\overrightarrow{OM}$ first. Then compare $\overrightarrow{CN}$ with $\overrightarrow{NM}$ .

Reveal solution

Solution

\overrightarrow{OB} = \mathbf{a} + \mathbf{c}

, so

\overrightarrow{ON} = \tfrac23(\mathbf{a} + \mathbf{c})

\overrightarrow{AB} = \mathbf{c}

, so

\overrightarrow{OM} = \mathbf{a} + \tfrac12\mathbf{c}

. Then

\overrightarrow{CN} = \overrightarrow{ON} - \overrightarrow{OC} = \tfrac23\mathbf{a} - \tfrac13\mathbf{c} = \tfrac13(2\mathbf{a} - \mathbf{c})

, and

\overrightarrow{NM} = \overrightarrow{OM} - \overrightarrow{ON} = \tfrac13\mathbf{a} - \tfrac16\mathbf{c} = \tfrac16(2\mathbf{a} - \mathbf{c})

. So

\overrightarrow{CN} = 2\overrightarrow{NM}

: they are parallel and share point

N

, so

C

N

and

M

are collinear.

\blacksquare

Questions students ask

What is a scalar?

An ordinary number, with size but no direction. Multiplying a vector by a scalar changes its length (and reverses it if negative).

Are equal vectors in the same place?

Not necessarily. Equal vectors have the same size and direction, wherever they are drawn — like opposite sides of a parallelogram.

How do I show two lines are parallel?

Show one vector is a scalar multiple of the other.

Key takeaways

Add and scale column vectors component by component.
Magnitude is $\sqrt{x^{2} + y^{2}}$ .
Routes: sum known vectors; reversing changes the sign.
A point dividing $AB$ in $m : n$ is $\tfrac{m}{m + n}$ of the way along.
Parallel: scalar multiples. Collinear: parallel and sharing a point.

Exam-style questions

Practise it the way it is examined

3 original questions · 7 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 52 marks

Work out the magnitude of the vector $\begin{pmatrix} 5 \\ -12 \end{pmatrix}$ .[2]
Show mark scheme
- M1 $\sqrt{25 + 144}$
- A1 $13$
Answer: $13$

Question 2Grade 72 marks

$OABC$ is a parallelogram with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$ . $M$ is the midpoint of $BC$ .

Find $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and $\mathbf{c}$ .[2]
Show mark scheme
- M1 $\overrightarrow{OB} = \mathbf{a} + \mathbf{c}$ and $\overrightarrow{BC} = -\mathbf{a}$
- A1 $\frac{1}{2}\mathbf{a} + \mathbf{c}$
Answer: $\frac{1}{2}\mathbf{a} + \mathbf{c}$

Question 3Grade 93 marks

$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ . $P$ lies on $AB$ with $AP : PB = 2 : 1$ . $X$ is the point with $\overrightarrow{OX} = \mathbf{a} + 2\mathbf{b}$ .

Prove that $O$ , $P$ and $X$ lie on a straight line.[3]
Show mark scheme
- M1 $\overrightarrow{OP} = \mathbf{a} + \frac{2}{3}(\mathbf{b} - \mathbf{a}) = \frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{b}$
- M1 $\overrightarrow{OX} = 3\overrightarrow{OP}$
- C1Parallel with the common point $O$ , so collinear

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