What is sign slips when subtracting?

-4 - 3 is -7, not -1. Subtracting moves further negative.

SingaporeMathsSyllabus dot point

What is a vector, and how do we add vectors and multiply them by a number?

Represent a vector as a column vector, add and subtract vectors, and multiply a vector by a scalar

A focused answer to the N(A)-Level Mathematics outcome on vectors. Column vector notation, adding and subtracting vectors, scalar multiples, and finding the magnitude of a vector.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

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

SEAB wants you to write a vector as a column vector, add and subtract vectors, and multiply a vector by a number (a scalar). A vector carries both size and direction, which makes it perfect for describing movements on a grid. This is a gentle introduction, kept to two dimensions.

The answer

What a vector is

A vector has both a magnitude (size) and a direction. A movement of " $3$ right and $2$ up" is a vector. We write it as a column vector, with the across-component on top and the up-component below:

\mathbf{a} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}

The top number is the change in $x$ and the bottom number is the change in $y$ . A negative top means moving left; a negative bottom means moving down.

Adding and subtracting vectors

To add or subtract vectors, combine the top components together and the bottom components together:

\begin{pmatrix} 3 \\ 2 \end{pmatrix} + \begin{pmatrix} 1 \\ 4 \end{pmatrix} = \begin{pmatrix} 4 \\ 6 \end{pmatrix}, \qquad \begin{pmatrix} 5 \\ 7 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}

Adding vectors is like making one journey and then another: the result is the single vector from start to finish.

Multiplying by a scalar

Multiplying a vector by a number multiplies each component by that number:

2 \begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}

The new vector points the same way but is twice as long. A negative scalar reverses the direction as well as scaling.

The magnitude of a vector

The magnitude (length) of a vector is found with Pythagoras, just like the distance between two points:

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

So the vector $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ has magnitude $\sqrt{9 + 16} = 5$ .

Worked example

Given $\mathbf{p} = \begin{pmatrix} 4 \\ -2 \end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix} 1 \\ 3 \end{pmatrix}$ , find $2\mathbf{p} - \mathbf{q}$ and its magnitude.

Step 1: Multiply p by the scalar

2\mathbf{p} = 2 \begin{pmatrix} 4 \\ -2 \end{pmatrix} = \begin{pmatrix} 8 \\ -4 \end{pmatrix}.

Step 2: Subtract q

2\mathbf{p} - \mathbf{q} = \begin{pmatrix} 8 \\ -4 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 8 - 1 \\ -4 - 3 \end{pmatrix} = \begin{pmatrix} 7 \\ -7 \end{pmatrix}.

Step 3: Find the magnitude

\left| \begin{pmatrix} 7 \\ -7 \end{pmatrix} \right| = \sqrt{7^2 + (-7)^2} = \sqrt{49 + 49} = \sqrt{98} \approx 9.90.

So $2\mathbf{p} - \mathbf{q} = \begin{pmatrix} 7 \\ -7 \end{pmatrix}$ with magnitude about $9.90$ .

Examples in context

Example 1. A movement on a map. Walking $5$ east then $3$ north is the vector $\begin{pmatrix} 5 \\ 3 \end{pmatrix}$ . If you then walk $\begin{pmatrix} 2 \\ -1 \end{pmatrix}$ , your total displacement is the sum $\begin{pmatrix} 7 \\ 2 \end{pmatrix}$ . Adding vectors gives the single straight-line journey from start to finish.

Example 2. Equal and parallel vectors. The vector $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and the vector $\begin{pmatrix} 4 \\ 2 \end{pmatrix}$ point the same way, because the second is $2$ times the first. A scalar multiple always produces a parallel vector, which is how we test whether two vectors are parallel.

Try this

Cue. Find $\begin{pmatrix} 2 \\ 5 \end{pmatrix} + \begin{pmatrix} 3 \\ -1 \end{pmatrix}$ . Combine components: $\begin{pmatrix} 5 \\ 4 \end{pmatrix}$ .
Cue. Find $4 \begin{pmatrix} 1 \\ -2 \end{pmatrix}$ . Scale both: $\begin{pmatrix} 4 \\ -8 \end{pmatrix}$ .
Cue. Find the magnitude of $\begin{pmatrix} 5 \\ 12 \end{pmatrix}$ . Compute $\sqrt{25 + 144} = \sqrt{169} = 13$ .

Exam-style practice questions

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

Original3 marksGiven

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

and

\mathbf{b} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}

, find

\mathbf{a} + \mathbf{b}

and

\mathbf{a} - \mathbf{b}

Show worked answer →

Add and subtract the matching components separately.

$\mathbf{a} + \mathbf{b} = \begin{pmatrix} 3 + (-1) \\ 1 + 4 \end{pmatrix} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}$ .

$\mathbf{a} - \mathbf{b} = \begin{pmatrix} 3 - (-1) \\ 1 - 4 \end{pmatrix} = \begin{pmatrix} 4 \\ -3 \end{pmatrix}$ .

What markers reward: combining the top components together and the bottom components together, and care with the signs (especially subtracting the negative $-1$ ). Mixing the rows is the usual error.

Original3 marksA vector is

\mathbf{v} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}

. Find (a) the vector

3\mathbf{v}

and (b) the magnitude of

\mathbf{v}