Find the magnitude of v = 6i - 8j. [2 marks]

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

Points P and Q have position vectors i + j and 5i + 9j. Find the position vector of the midpoint. [2 marks]

CCEA CCEA 2020-style practice: Vectors are given by a = 3i - 4j and b = 5i + 2j. Find a + 2b and its magnitude.

a + 2b = (3i - 4j) + 2(5i + 2j) = (3 + 10)i + (-4 + 4)j = 13i + 0j. So a + 2b = 13i. Its magnitude is sqrt(13^2 + 0^2) = 13. Markers reward scaling b by 2, adding components correctly, and the magnitude using Pythagoras.

CCEA CCEA 2019-style practice: The points A and B have position vectors a = 2i + j and b = 6i + 4j. Find the vector AB, its magnitude, and the position vector of the midpoint of AB.

The displacement is AB = b - a = (6 - 2)i + (4 - 1)j = 4i + 3j. Its magnitude is |AB| = sqrt(4^2 + 3^2) = sqrt(25) = 5. The midpoint has position vector (1)/(2)(a + b) = (1)/(2)(8i + 5j) = 4i + 2.5j. Markers reward AB = b - a, the magnitude, and the midpoint as the mean of the position vectors.

Northern IrelandMathsSyllabus dot point

How do vectors represent quantities with magnitude and direction in two dimensions?

Two-dimensional vectors in component and unit-vector form, the magnitude and direction of a vector, addition and scalar multiplication, position vectors, and using vectors in geometric problems.

A CCEA A-Level Mathematics answer on two-dimensional vectors in component and unit-vector form, finding magnitude and direction, vector addition and scalar multiplication, position vectors and displacement, and applying vectors to geometric problems.

Generated by Claude Opus 4.812 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
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Examples in context
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What this dot point is asking

CCEA wants you to represent two-dimensional vectors in component and unit-vector form, find a vector's magnitude and direction, add and scale vectors, work with position vectors and displacements, and use vectors to solve geometric problems such as finding midpoints or proving points are collinear. Vectors connect coordinate geometry to the motion problems in mechanics.

The answer

Component and unit-vector form

Magnitude and direction

Addition and scalar multiplication

To add vectors, add components: $(a\mathbf{i} + b\mathbf{j}) + (c\mathbf{i} + d\mathbf{j}) = (a + c)\mathbf{i} + (b + d)\mathbf{j}$ . To scale by a number $k$ , multiply each component: $k(a\mathbf{i} + b\mathbf{j}) = ka\mathbf{i} + kb\mathbf{j}$ . Geometrically, addition is the tip-to-tail (triangle) rule, and scaling stretches or reverses the vector.

Position vectors and geometry

Worked example: proving collinearity

Show that three points are collinear

The points $A$ , $B$ and $C$ have position vectors $\mathbf{a} = \mathbf{i} + 2\mathbf{j}$ , $\mathbf{b} = 3\mathbf{i} + 5\mathbf{j}$ and $\mathbf{c} = 7\mathbf{i} + 11\mathbf{j}$ . Show that $A$ , $B$ and $C$ are collinear.

step Find two displacement vectors

\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = 2\mathbf{i} + 3\mathbf{j}, \qquad \overrightarrow{AC} = \mathbf{c} - \mathbf{a} = 6\mathbf{i} + 9\mathbf{j}.

step Test for a scalar multiple

\overrightarrow{AC} = 6\mathbf{i} + 9\mathbf{j} = 3(2\mathbf{i} + 3\mathbf{j}) = 3\,\overrightarrow{AB}.

step Conclude

Because $\overrightarrow{AC}$ is a scalar multiple of $\overrightarrow{AB}$ , the two displacements are parallel and share the point $A$ , so $A$ , $B$ and $C$ lie on the same straight line. (Indeed $C$ is three times as far from $A$ as $B$ is, in the same direction.)

Examples in context

Example 1. Resultant displacement. A walker goes $4\,\text{km}$ east then $3\,\text{km}$ north. The resultant is $4\mathbf{i} + 3\mathbf{j}$ , of magnitude $5\,\text{km}$ at $\tan^{-1}(3/4) \approx 37^\circ$ north of east. Adding vectors tip to tail gives the straight-line displacement, the heart of navigation.

Example 2. Forces in equilibrium. Three forces in equilibrium add to the zero vector, so their $\mathbf{i}$ -components sum to zero and their $\mathbf{j}$ -components sum to zero. This component method is exactly how forces are resolved in the mechanics unit.

Try this

Q1. Find the magnitude of $\mathbf{v} = 6\mathbf{i} - 8\mathbf{j}$ . [2 marks]

Cue. $\sqrt{36 + 64} = 10$ .

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

Cue. $(6 - 1)\mathbf{i} + (3 + 3)\mathbf{j} = 5\mathbf{i} + 6\mathbf{j}$ .

Q3. Points $P$ and $Q$ have position vectors $\mathbf{i} + \mathbf{j}$ and $5\mathbf{i} + 9\mathbf{j}$ . Find the position vector of the midpoint. [2 marks]

Cue. $\frac{1}{2}(6\mathbf{i} + 10\mathbf{j}) = 3\mathbf{i} + 5\mathbf{j}$ .

Exam-style practice questions

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

CCEA 20205 marksVectors are given by

\mathbf{a} = 3\mathbf{i} - 4\mathbf{j}

and

\mathbf{b} = 5\mathbf{i} + 2\mathbf{j}

. Find

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

and its magnitude.

Show worked answer →

$\mathbf{a} + 2\mathbf{b} = (3\mathbf{i} - 4\mathbf{j}) + 2(5\mathbf{i} + 2\mathbf{j}) = (3 + 10)\mathbf{i} + (-4 + 4)\mathbf{j} = 13\mathbf{i} + 0\mathbf{j}.$

So $\mathbf{a} + 2\mathbf{b} = 13\mathbf{i}$ .

Its magnitude is $\sqrt{13^2 + 0^2} = 13$ .

Markers reward scaling $\mathbf{b}$ by $2$ , adding components correctly, and the magnitude using Pythagoras.

CCEA 20195 marksThe points

A

and

B

have position vectors

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

and

\mathbf{b} = 6\mathbf{i} + 4\mathbf{j}

. Find the vector

\overrightarrow{AB}

, its magnitude, and the position vector of the midpoint of

AB

Show worked answer →

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

Its magnitude is $|\overrightarrow{AB}| = \sqrt{4^2 + 3^2} = \sqrt{25} = 5$ .

The midpoint has position vector $\frac{1}{2}(\mathbf{a} + \mathbf{b}) = \frac{1}{2}(8\mathbf{i} + 5\mathbf{j}) = 4\mathbf{i} + 2.5\mathbf{j}$ .

Markers reward $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ , the magnitude, and the midpoint as the mean of the position vectors.

Related dot points

Sources & how we know this

CCEA GCE Mathematics specification — CCEA (2018)