What are square-root sign errors?

The magnitude uses the sum of squares; negative components still contribute positively.

Find the magnitude of pmatrix 2 \\ -3 \\ 6 pmatrix. [2 marks]

SEAB Original-style practice: The points A and B have position vectors a = pmatrix 1 \\ 2 \\ -1 pmatrix and b = pmatrix 5 \\ 0 \\ 3 pmatrix. Find AB, its magnitude, and a unit vector in its direction.

AB = b - a = pmatrix 4 \\ -2 \\ 4 pmatrix. Magnitude: |AB| = sqrt(4^2 + (-2)^2 + 4^2) = sqrt(16 + 4 + 16) = sqrt(36) = 6. Unit vector: 16pmatrix 4 \\ -2 \\ 4 pmatrix = pmatrix 2/3 \\ -1/3 \\ 2/3 pmatrix. Markers reward the subtraction b - a, the correct magnitude, and dividing by the magnitude for the unit vector.

SEAB Original-style practice: The point P divides AB in the ratio 2 : 3. Given a = pmatrix 0 \\ 4 pmatrix and b = pmatrix 10 \\ -1 pmatrix, find the position vector of P.

By the ratio theorem, the point dividing AB in ratio λ : μ has position vector μ a + λ bλ + μ. Here λ : μ = 2 : 3, so p = 3a + 2b5 = 15(pmatrix 0 \\ 12 pmatrix + pmatrix 20 \\ -2 pmatrix) = 15pmatrix 20 \\ 10 pmatrix = pmatrix 4 \\ 2 pmatrix. Markers reward the correct ratio-theorem weighting (the larger weight on the nearer point), the substitution, and the final position vector.

SingaporeMathsSyllabus dot point

How do we represent and manipulate vectors in two and three dimensions?

Represent vectors in component and position form, add and scale them, find magnitudes and unit vectors, and use the ratio theorem for points dividing a line segment

A focused answer to the H2 Mathematics outcome on vectors. Component and position vectors, addition and scalar multiplication, magnitude and unit vectors, collinearity, and the ratio theorem for a dividing point.

Generated by Claude Opus 4.89 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
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What this dot point is asking

SEAB wants you to represent vectors in component and position form, perform vector addition and scalar multiplication, compute magnitudes and unit vectors, test for collinearity and parallelism, and apply the ratio theorem to find the position vector of a point dividing a line segment.

The answer

Vectors and their components

A vector has magnitude and direction. In three dimensions a vector is written in component form $\begin{pmatrix} a \\ b \\ c \end{pmatrix} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$ . A position vector $\overrightarrow{OP}$ locates a point $P$ relative to the origin.

Addition, scaling and the displacement vector

Vectors add componentwise, and scalar multiplication scales each component. The displacement from $A$ to $B$ is

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

the position vector of the destination minus that of the start.

Magnitude and unit vectors

The magnitude (length) of $\begin{pmatrix} a \\ b \\ c \end{pmatrix}$ is $\sqrt{a^2 + b^2 + c^2}$ . A unit vector in the same direction is the vector divided by its magnitude; it has length $1$ and captures pure direction.

Parallelism, collinearity and the ratio theorem

Two vectors are parallel if one is a scalar multiple of the other. Three points are collinear if the displacement vectors between them are parallel and share a point. The ratio theorem gives the point $P$ dividing $AB$ in ratio $\lambda : \mu$ as

\mathbf{p} = \frac{\mu \mathbf{a} + \lambda \mathbf{b}}{\lambda + \mu}.

The midpoint is the special case $\dfrac{\mathbf{a} + \mathbf{b}}{2}$ .

Worked example

Points $A$ , $B$ , $C$ have position vectors $\mathbf{a} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$ , $\mathbf{b} = \begin{pmatrix} 3 \\ 2 \\ 4 \end{pmatrix}$ , $\mathbf{c} = \begin{pmatrix} 6 \\ 5 \\ 7 \end{pmatrix}$ . Show that $A$ , $B$ , $C$ are collinear and find the ratio $AB : BC$ .

Step 1: Find the displacement vectors

\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix}, \qquad \overrightarrow{BC} = \mathbf{c} - \mathbf{b} = \begin{pmatrix} 3 \\ 3 \\ 3 \end{pmatrix}

Step 2: Test for parallelism

$\overrightarrow{BC} = \dfrac{3}{2}\overrightarrow{AB}$ , so the two displacements are parallel.

Step 3: Conclude collinearity

They share the point $B$ and are parallel, so $A$ , $B$ , $C$ lie on one straight line.

Step 4: Find the ratio

The lengths are in the same ratio as the scalar multiples: $|\overrightarrow{AB}| : |\overrightarrow{BC}| = 2 : 3$ . Hence $AB : BC = 2 : 3$ .

Examples in context

Example 1. Navigation displacement. A drone moves from $\begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}$ to $\begin{pmatrix} 5 \\ 5 \\ 12 \end{pmatrix}$ (units in metres). Its displacement is $\begin{pmatrix} 3 \\ 4 \\ 12 \end{pmatrix}$ , of magnitude $\sqrt{9 + 16 + 144} = 13$ m, the straight-line distance flown.

Example 2. Centre of mass of two points. Two equal masses at $\mathbf{a}$ and $\mathbf{b}$ have centre of mass at the midpoint $\dfrac{\mathbf{a} + \mathbf{b}}{2}$ , the ratio theorem with $1 : 1$ , which is why the balance point sits exactly halfway.

Try this

Q1. Find the magnitude of $\begin{pmatrix} 2 \\ -3 \\ 6 \end{pmatrix}$ . [2 marks]

Cue. $\sqrt{4 + 9 + 36} = \sqrt{49} = 7$ .

Q2. Find the midpoint of $A(1, 4, -2)$ and $B(3, 0, 6)$ . [2 marks]

Cue. $\dfrac{1}{2}\begin{pmatrix} 4 \\ 4 \\ 4 \end{pmatrix} = (2, 2, 2)$ .

Q3. State the condition for two vectors to be parallel. [1 mark]

Cue. One is a scalar multiple of the other.

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.

Original4 marksThe points

A

and

B

have position vectors

\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}

and

\mathbf{b} = \begin{pmatrix} 5 \\ 0 \\ 3 \end{pmatrix}

. Find

\overrightarrow{AB}

, its magnitude, and a unit vector in its direction.

Show worked answer →

$\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 4 \\ -2 \\ 4 \end{pmatrix}$ .

Magnitude: $|\overrightarrow{AB}| = \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$ .

Unit vector: $\dfrac{1}{6}\begin{pmatrix} 4 \\ -2 \\ 4 \end{pmatrix} = \begin{pmatrix} 2/3 \\ -1/3 \\ 2/3 \end{pmatrix}$ .

Markers reward the subtraction $\mathbf{b} - \mathbf{a}$ , the correct magnitude, and dividing by the magnitude for the unit vector.

Original4 marksThe point

P

divides

AB

in the ratio

2 : 3

. Given

\mathbf{a} = \begin{pmatrix} 0 \\ 4 \end{pmatrix}

and

\mathbf{b} = \begin{pmatrix} 10 \\ -1 \end{pmatrix}

, find the position vector of

P

Show worked answer →

By the ratio theorem, the point dividing $AB$ in ratio $\lambda : \mu$ has position vector $\dfrac{\mu \mathbf{a} + \lambda \mathbf{b}}{\lambda + \mu}$ .

Here $\lambda : \mu = 2 : 3$ , so $\mathbf{p} = \dfrac{3\mathbf{a} + 2\mathbf{b}}{5} = \dfrac{1}{5}\left(\begin{pmatrix} 0 \\ 12 \end{pmatrix} + \begin{pmatrix} 20 \\ -2 \end{pmatrix}\right) = \dfrac{1}{5}\begin{pmatrix} 20 \\ 10 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$ .

Markers reward the correct ratio-theorem weighting (the larger weight on the nearer point), the substitution, and the final position vector.

What this dot point is asking

The answer

Vectors and their components

Addition, scaling and the displacement vector

Magnitude and unit vectors

Parallelism, collinearity and the ratio theorem

Examples in context

Try this

Exam-style practice questions

Related dot points