KBAT: Proving Geometry With Vectors

Q: How do I prove three points are collinear with vectors?

Form two vectors that start from the same point, say QP→\overrightarrow{QP}QP and QR→\overrightarrow{QR}QR. If you can write one as a scalar multiple of the other, QR→=k QP→\overrightarrow{QR}=k\,\overrightarrow{QP}QR=kQP, then the three points lie on one straight line. Here k=3k=3k=3, so PPP, QQQ and RRR are collinear.

Q: How do I find the position vector of a point that divides a segment?

Use OP→=OA→+mm+nAB→\overrightarrow{OP}=\overrightarrow{OA}+\tfrac{m}{m+n}\overrightarrow{AB}OP=OA+m+nmAB for AP:PB=m:nAP:PB=m:nAP:PB=m:n. For AP:PB=1:2AP:PB=1:2AP:PB=1:2 this gives OP→=a+13(b−a)=23a+13b\overrightarrow{OP}=\mathbf{a}+\tfrac{1}{3}(\mathbf{b}-\mathbf{a})=\tfrac{2}{3}\mathbf{a}+\tfrac{1}{3}\mathbf{b}OP=a+31(b−a)=32a+31b. A midpoint is the special case of ratio 1:11:11:1.

Q: Why can I compare the coefficients of a and b?

Because a\mathbf{a}a and b\mathbf{b}b are non-zero and not parallel, every vector has exactly one expression p a+q bp\,\mathbf{a}+q\,\mathbf{b}pa+qb. So if two expressions are equal, their a\mathbf{a}a-coefficients match and their b\mathbf{b}b-coefficients match. That uniqueness is what lets you solve for an unknown scalar or confirm a multiple.

Q: How is Add Math Paper 2 marked on vector proofs?

Paper 2 is 2 hours 30 minutes and 100 marks; marking is analytic, so method marks are awarded line by line. Correct position vectors, a clean scalar-multiple step and a stated collinearity conclusion earn marks even if the final ratio is simplified only at the end.

What makes this a KBAT question

A routine vectors question asks you to add two vectors or find a resultant, one step, one answer. A KBAT proof question in Vectors asks you to establish that something is true: that three points are collinear, that a line is parallel to another, or that a point divides a segment in a certain ratio.

That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: expressing a point as a position vector and applying a ratio are familiar skills, but here you must choose which vectors to build, decide what a proof of collinearity actually needs, and argue from the fact that two non-parallel vectors have unique coefficients. In Add Math this rewards students who treat vectors as a reasoning tool, not just a way to combine arrows.

One worked problem, in the style of Paper 2

This is an original question written in the style of SPM Paper 2. Try it yourself before reading the solution.

Q1[8 marks]

In a diagram, $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$ , where $\mathbf{a}$ and $\mathbf{b}$ are not parallel. The point $P$ lies on $AB$ such that $AP:PB=1:2$ .

The point $Q$ is the midpoint of $OB$ , and the point $R$ lies on $OA$ produced such that $\overrightarrow{OR}=2\mathbf{a}$ . (a) Express $\overrightarrow{OP}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .

(b) Show that the points $P$ , $Q$ and $R$ are collinear. (c) Find the ratio $QP:PR$ .

Show worked solution

Understand. Every point is measured from the single origin $O$ .

Because $\mathbf{a}$ and $\mathbf{b}$ are not parallel, each vector in the plane has exactly one expression of the form $p\,\mathbf{a}+q\,\mathbf{b}$ . Three points are collinear when two vectors joining them and sharing a common point are scalar multiples of each other.

Plan. (a) $P$ divides $AB$ internally in the ratio $1:2$ , so use $\overrightarrow{OP}=\overrightarrow{OA}+\tfrac{1}{3}\overrightarrow{AB}$ .

(b) Write $\overrightarrow{OP}$ , $\overrightarrow{OQ}$ and $\overrightarrow{OR}$ as position vectors, form $\overrightarrow{QP}$ and $\overrightarrow{QR}$ from the common point $Q$ , and show $\overrightarrow{QR}=k\,\overrightarrow{QP}$ . (c) Read the ratio from the value of $k$ .

Execute and check. (a) Since $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ and $AP:PB=1:2$ gives $\overrightarrow{AP}=\tfrac{1}{3}\overrightarrow{AB}$ :

\overrightarrow{OP}=\mathbf{a}+\tfrac{1}{3}(\mathbf{b}-\mathbf{a})=\tfrac{2}{3}\mathbf{a}+\tfrac{1}{3}\mathbf{b}

(b) The midpoint of $OB$ gives $\overrightarrow{OQ}=\tfrac{1}{2}\mathbf{b}$ , and $\overrightarrow{OR}=2\mathbf{a}$ is given. Form the two vectors from $Q$ :

\overrightarrow{QP}=\overrightarrow{OP}-\overrightarrow{OQ}=\tfrac{2}{3}\mathbf{a}+\tfrac{1}{3}\mathbf{b}-\tfrac{1}{2}\mathbf{b}=\tfrac{2}{3}\mathbf{a}-\tfrac{1}{6}\mathbf{b}

\overrightarrow{QR}=\overrightarrow{OR}-\overrightarrow{OQ}=2\mathbf{a}-\tfrac{1}{2}\mathbf{b}

Factor $\overrightarrow{QR}$ to compare it with $\overrightarrow{QP}$ :

\overrightarrow{QR}=2\mathbf{a}-\tfrac{1}{2}\mathbf{b}=3\left(\tfrac{2}{3}\mathbf{a}-\tfrac{1}{6}\mathbf{b}\right)=3\,\overrightarrow{QP}

Since $\overrightarrow{QR}=3\,\overrightarrow{QP}$ and both vectors start from the common point $Q$ , the points $P$ , $Q$ and $R$ lie on one straight line, they are collinear.

(c) From $\overrightarrow{QR}=3\,\overrightarrow{QP}$ , the length $QR$ is $3$ times $QP$ , so $QP:QR=1:3$ . As $P$ lies between $Q$ and $R$ , $PR=QR-QP$ corresponds to $3-1=2$ parts, giving $QP:PR=1:2$ .

Check. The single scalar $3$ reproduces both components: $3\times\tfrac{2}{3}=2$ matches the $\mathbf{a}$ term of $\overrightarrow{QR}$ , and $3\times(-\tfrac{1}{6})=-\tfrac{1}{2}$ matches the $\mathbf{b}$ term.

One scalar fitting both coefficients is exactly what collinearity requires, so the argument is sound.

Finding a sensible first step

When a vectors proof looks unfamiliar, do not try to 'see' the answer on the diagram, set up the algebra that always works. The reliable first step is to write every point named in the question as a position vector from the single origin $O$ , in terms of the two base vectors $\mathbf{a}$ and $\mathbf{b}$ .

A point dividing a segment in a given ratio comes straight from $\overrightarrow{OP}=\overrightarrow{OA}+\tfrac{m}{m+n}\overrightarrow{AB}$ ; a midpoint is the average of its two ends. Once each point is a position vector, decide what you must prove and pick the two vectors that carry it.

For collinearity, form two vectors from one shared point and test whether one is a scalar multiple of the other. Because $\mathbf{a}$ and $\mathbf{b}$ are not parallel, matching coefficients is a valid, mark-earning argument.

What markers reward

Marking is analytic, so method marks are awarded line by line. On a vector-proof question a marker looks for:

Each point written as a position vector from a single origin, in terms of $\mathbf{a}$ and $\mathbf{b}$ .
The division ratio applied correctly, for example $\overrightarrow{OP}=\mathbf{a}+\tfrac{1}{3}(\mathbf{b}-\mathbf{a})$ for $AP:PB=1:2$ .
Two vectors formed from a common point, such as $\overrightarrow{QP}$ and $\overrightarrow{QR}$ , before any comparison.
One vector shown explicitly as a scalar multiple of the other, $\overrightarrow{QR}=3\,\overrightarrow{QP}$ , with the scalar stated.
A clear statement that a shared point plus a scalar multiple means the three points are collinear.
The ratio read from that scalar and given in simplest form, $QP:PR=1:2$ .

How a teacher helps

Vector proofs reward a habit more than a trick: put everything on one origin and let the coefficients do the work. In a one-to-one lesson our teachers build that habit with you, we label the figure, write each point as a position vector, and choose the two vectors from a shared point before comparing them, so 'prove collinear' stops feeling like guesswork.

We also make you state why matching coefficients is allowed, because markers reward that reasoning. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

Message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

How do I prove three points are collinear with vectors?

Form two vectors that start from the same point, say $\overrightarrow{QP}$ and $\overrightarrow{QR}$ . If you can write one as a scalar multiple of the other, $\overrightarrow{QR}=k\,\overrightarrow{QP}$ , then the three points lie on one straight line.

Here $k=3$ , so $P$ , $Q$ and $R$ are collinear.

How do I find the position vector of a point that divides a segment?

Use $\overrightarrow{OP}=\overrightarrow{OA}+\tfrac{m}{m+n}\overrightarrow{AB}$ for $AP:PB=m:n$ . For $AP:PB=1:2$ this gives $\overrightarrow{OP}=\mathbf{a}+\tfrac{1}{3}(\mathbf{b}-\mathbf{a})=\tfrac{2}{3}\mathbf{a}+\tfrac{1}{3}\mathbf{b}$ .

A midpoint is the special case of ratio $1:1$ .

Why can I compare the coefficients of a and b?

Because $\mathbf{a}$ and $\mathbf{b}$ are non-zero and not parallel, every vector has exactly one expression $p\,\mathbf{a}+q\,\mathbf{b}$ . So if two expressions are equal, their $\mathbf{a}$ -coefficients match and their $\mathbf{b}$ -coefficients match.

That uniqueness is what lets you solve for an unknown scalar or confirm a multiple.

How is Add Math Paper 2 marked on vector proofs?

Paper 2 is 2 hours 30 minutes and 100 marks; marking is analytic, so method marks are awarded line by line. Correct position vectors, a clean scalar-multiple step and a stated collinearity conclusion earn marks even if the final ratio is simplified only at the end.

Source:SRC-DSKP-EN SRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026