What are position vectors?

The position vector of a point A relative to an origin O is OA, often written a. It locates the point as a displacement from the origin, and every point has its own position vector.

What is not following a connected route?

Each step in a route must share a point with the next, joining nose to tail.

Given OA = a and OB = b, write AB in terms of a and b. [1 mark]

If PQ = 2XY, what can you say about lines PQ and XY? [1 mark]

M is the midpoint of AB with position vectors a and b. Write OM. [2 marks]

SEAB Original-style practice: Points A and B have position vectors a and b relative to an origin O. The point M is the midpoint of AB. Express OM in terms of a and b, showing your reasoning.

The vector from A to B is AB = b - a (end minus start). Since M is the midpoint, AM = 12AB = 12(b - a). Then OM = OA + AM = a + 12(b - a) = 12a + 12b = 12(a + b). Markers reward AB = b - a, the halving to reach M, and the route to give OM = (1)/(2)(a + b).

SingaporeMathsSyllabus dot point

How do position vectors and the route method let us prove geometric facts?

Use position vectors and the relationship between points to express vectors, and apply parallel and ratio properties in geometry

A focused answer to the O-Level E-Maths outcome on vector geometry. Position vectors, expressing one vector via a route through others, the parallel condition, and using vectors to prove collinearity and ratios.

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
Examples in context
Try this

What this dot point is asking

SEAB wants you to use position vectors, to express a vector between points by following a route through other known vectors, and to apply the parallel and ratio properties of vectors to prove geometric results such as collinearity. This is the reasoning side of vectors, building on the component arithmetic.

The answer

Position vectors

The position vector of a point $A$ relative to an origin $O$ is $\overrightarrow{OA}$ , often written $\mathbf{a}$ . It locates the point as a displacement from the origin, and every point has its own position vector.

Expressing a vector between two points

The vector from $A$ to $B$ is the end position vector minus the start:

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

This end-minus-start rule is the workhorse of vector geometry. To find any vector, follow a route through the origin or through known points and add the steps.

The parallel condition

Two vectors are parallel when one is a scalar multiple of the other:

\overrightarrow{PQ} = k\,\overrightarrow{RS} \implies PQ \parallel RS

The scalar $k$ also gives the ratio of their lengths. This is how vectors prove that lines are parallel.

Collinearity and ratios

Three points are collinear (on one straight line) if the vector between two of them is a scalar multiple of the vector between another pair sharing a point. Ratios such as $\overrightarrow{PR} = 3\overrightarrow{RQ}$ fix where a point divides a segment, giving $PR : RQ = 3 : 1$ .

Worked example

Points $A$ , $B$ and $C$ have position vectors $\mathbf{a}$ , $\mathbf{b}$ and $\mathbf{c}$ . The point $D$ is such that $\overrightarrow{AD} = \overrightarrow{BC}$ . Find the position vector of $D$ in terms of $\mathbf{a}$ , $\mathbf{b}$ and $\mathbf{c}$ .

Step 1: Express the vector BC

By end minus start, $\overrightarrow{BC} = \mathbf{c} - \mathbf{b}$ .

Step 2: Use the given equality

Since $\overrightarrow{AD} = \overrightarrow{BC}$ , we have $\overrightarrow{AD} = \mathbf{c} - \mathbf{b}$ .

Step 3: Follow the route from O to D

\overrightarrow{OD} = \overrightarrow{OA} + \overrightarrow{AD} = \mathbf{a} + (\mathbf{c} - \mathbf{b})

Step 4: State the position vector

\overrightarrow{OD} = \mathbf{a} - \mathbf{b} + \mathbf{c}

This makes $ABCD$ a parallelogram, since $\overrightarrow{AD} = \overrightarrow{BC}$ .

Examples in context

Example 1. Proving a parallelogram. Showing that $\overrightarrow{AB} = \overrightarrow{DC}$ proves opposite sides are equal and parallel, so the quadrilateral $ABCD$ is a parallelogram. Vectors give a clean proof without coordinates.

Example 2. Dividing a line in a ratio. A point dividing $AB$ in the ratio $2 : 1$ has position vector $\mathbf{a} + \frac{2}{3}(\mathbf{b} - \mathbf{a})$ , found by travelling two thirds of the way from $A$ to $B$ . Ratio reasoning with vectors locates such points exactly.

Try this

Q1. Given $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$ , write $\overrightarrow{AB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ . [1 mark]

Cue. End minus start: $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ .

Q2. If $\overrightarrow{PQ} = 2\overrightarrow{XY}$ , what can you say about lines $PQ$ and $XY$ ? [1 mark]

Cue. They are parallel, and $PQ$ is twice the length of $XY$ .

Q3. $M$ is the midpoint of $AB$ with position vectors $\mathbf{a}$ and $\mathbf{b}$ . Write $\overrightarrow{OM}$ . [2 marks]

Cue. $\overrightarrow{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$ .

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 marksPoints

A

and

B

have position vectors

\mathbf{a}

and

\mathbf{b}

relative to an origin

O

. The point

M

is the midpoint of

AB

. Express

\overrightarrow{OM}

in terms of

\mathbf{a}

and

\mathbf{b}

, showing your reasoning.

Show worked answer →

The vector from $A$ to $B$ is $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ (end minus start).

Since $M$ is the midpoint, $\overrightarrow{AM} = \dfrac{1}{2}\overrightarrow{AB} = \dfrac{1}{2}(\mathbf{b} - \mathbf{a})$ .

Then $\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \dfrac{1}{2}(\mathbf{b} - \mathbf{a}) = \dfrac{1}{2}\mathbf{a} + \dfrac{1}{2}\mathbf{b} = \dfrac{1}{2}(\mathbf{a} + \mathbf{b})$ .

Markers reward $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ , the halving to reach $M$ , and the route to give $\overrightarrow{OM} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$ .

Original4 marksGiven

\overrightarrow{OP} = \mathbf{p}

and

\overrightarrow{OQ} = \mathbf{q}

, the point

R

lies on

PQ

so that

\overrightarrow{PR} = 3\overrightarrow{RQ}

. Show that

\overrightarrow{PQ}

and

\overrightarrow{PR}

are parallel and state the ratio

PR : RQ