What are representing vectors?

A vector in two dimensions can be written as a column pmatrix x \\ y pmatrix or in unit-vector form xi + yj, where i and j point along the x- and y-axes. Two vectors are equal if they have the same components, regardless of where they are drawn.

Find the magnitude of v = 5i - 12j. [2 marks]

Given a = 3i + j and b = i - 2j, find 2a - b. [2 marks]

M is the midpoint of AB where a = 4i and b = 2i + 6j. Find the position vector of M. [2 marks]

WJEC WJEC AS style-style practice: The points A and B have position vectors a = 2i + 3j and b = 6i - j. Find AB and its magnitude.

The vector from A to B is found by subtracting the start position from the end position. AB = b - a = (6 - 2)i + (-1 - 3)j = 4i - 4j. Its magnitude is |AB| = sqrt(4^2 + (-4)^2) = sqrt(16 + 16) = sqrt(32) = 4sqrt(2). Markers reward computing b - a (not a - b), correct components, and using Pythagoras for the magnitude. Reversing the subtraction gives the opposite vector, which still has the same magnitude but the wrong direction.

WJEC WJEC AS style-style practice: The point P divides the segment AB in the ratio 1:3, where A has position vector a and B has position vector b. Express the position vector of P in terms of a and b.

Moving from A towards B, the point P is a fraction of the way along AB set by the ratio. The ratio 1:3 means P is 11+3 = 14 of the way from A to B. OP = a + 14(b - a) = 34a + 14b. Markers reward identifying the fraction 14 from the ratio, building the position vector as start plus a fraction of the displacement, and simplifying. A common error is using 13 instead of 14 by misreading the ratio.

WalesMathsSyllabus dot point

How do we represent and combine vectors in two dimensions, and use position vectors in geometry?

Vectors in two dimensions, magnitude and direction, addition and scalar multiplication, position vectors, and dividing a line segment in a given ratio.

A focused answer to WJEC AS Unit 1 vectors, covering two-dimensional vectors, magnitude and direction, vector addition and scalar multiplication, position vectors, and dividing a line segment in a given ratio.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

WJEC wants you to represent two-dimensional vectors in component or $\mathbf{i}$ , $\mathbf{j}$ form, find their magnitude and direction, perform addition and scalar multiplication, work with position vectors, and find the point that divides a segment in a given ratio. Vectors underpin the mechanics unit (forces and velocities are vectors) and extend to three dimensions in the A2 course.

The answer

Representing vectors

A vector in two dimensions can be written as a column $\begin{pmatrix} x \\ y \end{pmatrix}$ or in unit-vector form $x\mathbf{i} + y\mathbf{j}$ , where $\mathbf{i}$ and $\mathbf{j}$ point along the $x$ - and $y$ -axes. Two vectors are equal if they have the same components, regardless of where they are drawn.

Magnitude, direction and arithmetic

A unit vector has magnitude $1$ ; divide any vector by its magnitude to get the unit vector in the same direction.

Position vectors and displacement

Dividing a segment in a ratio

To find the point $P$ that divides $AB$ in the ratio $m:n$ , start at $A$ and move the fraction $\dfrac{m}{m+n}$ of the way along $\overrightarrow{AB}$ .

Find the point dividing a segment in a ratio

$A$ has position vector $\mathbf{a} = \mathbf{i} + 2\mathbf{j}$ and $B$ has $\mathbf{b} = 7\mathbf{i} + 8\mathbf{j}$ . Find the position vector of $P$ dividing $AB$ in the ratio $2:1$ .

step Find the fraction along $AB$

The ratio $2:1$ means $P$ is $\dfrac{2}{2+1} = \dfrac{2}{3}$ of the way from $A$ to $B$ .

step Find the displacement

$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = 6\mathbf{i} + 6\mathbf{j}$ .

step Build the position vector

$\overrightarrow{OP} = \mathbf{a} + \dfrac{2}{3}(6\mathbf{i} + 6\mathbf{j}) = (\mathbf{i} + 2\mathbf{j}) + (4\mathbf{i} + 4\mathbf{j})$ .

step Simplify

$\overrightarrow{OP} = 5\mathbf{i} + 6\mathbf{j}$ , so $P$ is at $(5, 6)$ .

Examples in context

Example 1. Collinearity. Points $A(1, 1)$ , $B(3, 4)$ and $C(7, 10)$ are tested for being collinear. $\overrightarrow{AB} = 2\mathbf{i} + 3\mathbf{j}$ and $\overrightarrow{AC} = 6\mathbf{i} + 9\mathbf{j} = 3\overrightarrow{AB}$ . Since $\overrightarrow{AC}$ is a scalar multiple of $\overrightarrow{AB}$ , the three points lie on a straight line. The parallel test settles collinearity instantly.

Example 2. Resultant displacement. A walker goes $3\,\text{km}$ east then $4\,\text{km}$ north, a resultant of $3\mathbf{i} + 4\mathbf{j}$ . The magnitude is $\sqrt{9 + 16} = 5\,\text{km}$ and the bearing is $\tan^{-1}(3/4) = 36.9^{\circ}$ east of north. Vectors add the two legs into one direct displacement, the same idea used for forces in mechanics.

Try this

Q1. Find the magnitude of $\mathbf{v} = 5\mathbf{i} - 12\mathbf{j}$ . [2 marks]

Cue. $|\mathbf{v}| = \sqrt{25 + 144} = \sqrt{169} = 13$ .

Q2. Given $\mathbf{a} = 3\mathbf{i} + \mathbf{j}$ and $\mathbf{b} = \mathbf{i} - 2\mathbf{j}$ , find $2\mathbf{a} - \mathbf{b}$ . [2 marks]

Cue. $2\mathbf{a} = 6\mathbf{i} + 2\mathbf{j}$ , subtract $\mathbf{b}$ : $(6-1)\mathbf{i} + (2+2)\mathbf{j} = 5\mathbf{i} + 4\mathbf{j}$ .

Q3. $M$ is the midpoint of $AB$ where $\mathbf{a} = 4\mathbf{i}$ and $\mathbf{b} = 2\mathbf{i} + 6\mathbf{j}$ . Find the position vector of $M$ . [2 marks]

Cue. Midpoint is $\tfrac{1}{2}(\mathbf{a} + \mathbf{b}) = \tfrac{1}{2}(6\mathbf{i} + 6\mathbf{j}) = 3\mathbf{i} + 3\mathbf{j}$ .

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC AS style4 marksThe points

A

and

B

have position vectors

\mathbf{a} = 2\mathbf{i} + 3\mathbf{j}

and

\mathbf{b} = 6\mathbf{i} - \mathbf{j}

. Find

\overrightarrow{AB}

and its magnitude.

Show worked answer →

The vector from $A$ to $B$ is found by subtracting the start position from the end position.

$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (6 - 2)\mathbf{i} + (-1 - 3)\mathbf{j} = 4\mathbf{i} - 4\mathbf{j}$ .

Its magnitude is $|\overrightarrow{AB}| = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$ .

Markers reward computing $\mathbf{b} - \mathbf{a}$ (not $\mathbf{a} - \mathbf{b}$ ), correct components, and using Pythagoras for the magnitude. Reversing the subtraction gives the opposite vector, which still has the same magnitude but the wrong direction.

WJEC AS style4 marksThe point

P

divides the segment

AB

in the ratio

1:3

, where

A

has position vector

\mathbf{a}

and

B

has position vector

\mathbf{b}

. Express the position vector of

P

in terms of

\mathbf{a}

and

\mathbf{b}

Show worked answer →

Moving from $A$ towards $B$ , the point $P$ is a fraction of the way along $\overrightarrow{AB}$ set by the ratio.

The ratio $1:3$ means $P$ is $\dfrac{1}{1+3} = \dfrac{1}{4}$ of the way from $A$ to $B$ .

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

Markers reward identifying the fraction $\dfrac{1}{4}$ from the ratio, building the position vector as start plus a fraction of the displacement, and simplifying. A common error is using $\dfrac{1}{3}$ instead of $\dfrac{1}{4}$ by misreading the ratio.

Related dot points

Sources & how we know this

WJEC GCE AS/A Level Mathematics specification (from 2017) — WJEC (2017)