What is not stating the scalar multiple in a proof?

Showing one vector is k times another is what proves parallel or collinear.

CCEA CCEA 2021-style practice: In a diagram, OA = a and OB = b. M is the midpoint of AB. Find OM in terms of a and b. (Higher, non-calculator.)

Travel from O to M via A. First OA = a. Then AB = b - a, and M is the midpoint, so AM = 12(b - a). Add the route: OM = a + 12(b - a) = 12a + 12b = 12(a + b). Marks are for AB, for the half, for the route, and for the simplified answer. The midpoint of AB from the origin is always the average of the two position vectors.

Northern IrelandMathsSyllabus dot point

How do you use column vectors to add, subtract and scale, and prove geometric results?

Use column vector notation, add and subtract vectors, multiply by a scalar, find the magnitude of a vector, and use vectors to prove geometric facts such as collinearity (Higher tier).

A CCEA GCSE Mathematics Higher answer on vectors, covering column vector notation, addition subtraction and scalar multiplication, the magnitude of a vector, and using vectors to prove geometric results such as parallel lines and collinear points.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-14

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

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What this dot point is asking
Column vectors and notation
Adding, subtracting and scaling
Magnitude
Vectors between points and proof
Why this matters

What this dot point is asking

Vectors describe quantities with both size and direction, and they are a Higher-tier CCEA Geometry and Measures topic. You must use column vector notation, add and subtract vectors, multiply a vector by a scalar, find a vector's magnitude, and use vectors to prove geometric results such as that points are collinear or that lines are parallel. The proof questions, where you express a journey in terms of given vectors, are the higher-value marks and reward clear, route-based working.

Column vectors and notation

A vector has magnitude and direction, written as a column $\binom{x}{y}$ meaning $x$ to the right and $y$ up (negative values go left or down). A vector is often named by a bold letter such as $\mathbf{a}$ , or by its endpoints with an arrow, $\overrightarrow{AB}$ , meaning the vector from $A$ to $B$ . The reverse vector $\overrightarrow{BA}$ is $-\overrightarrow{AB}$ .

Adding, subtracting and scaling

Vector arithmetic works on each component separately.

Multiplying by a scalar stretches or shrinks the vector and, if the scalar is negative, reverses its direction. Two vectors are parallel exactly when one is a scalar multiple of the other.

The geometric picture behind addition is the triangle (or nose-to-tail) law: placing the second vector's tail at the first vector's head, the sum runs from the start of the first to the end of the second. Subtraction $\mathbf{a} - \mathbf{b}$ can be read as $\mathbf{a} + (-\mathbf{b})$ , adding the reverse of $\mathbf{b}$ . A position vector is the special case of the vector from the origin to a point, so the point $(4, 3)$ has position vector $\binom{4}{3}$ . These ideas turn a diagram of labelled points into algebra you can manipulate, which is the whole basis of the Higher-tier proof questions.

Magnitude

The magnitude of a vector is its length, found by Pythagoras from its components.

Vectors between points and proof

To find the vector from one point to another, travel along vectors you already know, in the right direction. From $A$ to $B$ via the origin $O$ , $\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\overrightarrow{OA} + \overrightarrow{OB} = \mathbf{b} - \mathbf{a}$ .

Prove two vectors are parallel

Points have $\overrightarrow{OP} = \mathbf{a} + 2\mathbf{b}$ and $\overrightarrow{OQ} = 3\mathbf{a} + 6\mathbf{b}$ . Show that $OP$ and $OQ$ are parallel.

step 1 Compare the two vectors

$\overrightarrow{OQ} = 3\mathbf{a} + 6\mathbf{b}$ and $\overrightarrow{OP} = \mathbf{a} + 2\mathbf{b}$ .

step 2 Factor out a scalar

$\overrightarrow{OQ} = 3(\mathbf{a} + 2\mathbf{b}) = 3\,\overrightarrow{OP}$ .

step 3 Interpret

Because $\overrightarrow{OQ}$ is a scalar multiple of $\overrightarrow{OP}$ , the two vectors are parallel.

step 4 Note the consequence

Since they share the point $O$ , the points $O$ , $P$ and $Q$ are also collinear (they lie on one straight line).

Why this matters

Vectors give a precise language for displacement and direction, used in physics, navigation and computer graphics, and they are a distinctive Higher-tier reasoning topic. The arithmetic builds on directed numbers, the magnitude uses Pythagoras, and the scalar-multiple test for parallel vectors connects to the gradient idea from straight-line graphs. CCEA rewards a clear route through the diagram and a stated conclusion, so write each step of the journey explicitly.

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 20203 marksGiven

\mathbf{a} = \binom{3}{-1}

and

\mathbf{b} = \binom{2}{4}

, find

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

as a column vector. (Higher, non-calculator.)

Show worked answer →

First scale: $2\mathbf{a} = 2\binom{3}{-1} = \binom{6}{-2}$ .

Then subtract $\mathbf{b}$ component by component: $\binom{6}{-2} - \binom{2}{4} = \binom{6 - 2}{-2 - 4} = \binom{4}{-6}$ .

Marks are for the scaling, for the component subtraction, and for the final vector $\binom{4}{-6}$ . Adding instead of subtracting, or mixing up the rows, are the common slips.

CCEA 20214 marksIn a diagram,

\overrightarrow{OA} = \mathbf{a}

and

\overrightarrow{OB} = \mathbf{b}

M

is the midpoint of

AB

. Find

\overrightarrow{OM}

in terms of

\mathbf{a}

and

\mathbf{b}

. (Higher, non-calculator.)

Show worked answer →

Travel from $O$ to $M$ via $A$ . First $\overrightarrow{OA} = \mathbf{a}$ .

Then $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ , and $M$ is the midpoint, so $\overrightarrow{AM} = \tfrac{1}{2}(\mathbf{b} - \mathbf{a})$ .

Add the route: $\overrightarrow{OM} = \mathbf{a} + \tfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \tfrac{1}{2}\mathbf{a} + \tfrac{1}{2}\mathbf{b} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b})$ .

Marks are for $\overrightarrow{AB}$ , for the half, for the route, and for the simplified answer. The midpoint of $AB$ from the origin is always the average of the two position vectors.

Related dot points

Sources & how we know this

CCEA GCSE Mathematics specification (2210) — CCEA (2017)