Vectors Proof Questions

What you'll learn

How to turn a geometry diagram into vector algebra.
How to use ratios on a line segment to find position vectors.
How to prove points lie on the same straight line.
How to handle harder questions with extensions, midpoints, parallelograms and regular hexagons.

1. Vector routes: the starting point

A vector describes a movement: it has a size and a direction. In proof questions, you usually express every movement using the given vectors, such as $a$ and $b$ .

Definition

Vector notation

$AB⃗\vec{AB}$ means the vector from point A to point B.
A position vector starts from the origin point O, for example $OA⃗\vec{OA}$ .
A scalar is a number multiplying a vector, such as 3 in $3 a$ .

The most useful route fact is:

AB⃗=OB⃗−OA⃗\vec{AB}=\vec{OB}-\vec{OA}

That means: to go from A to B, go from A back to O, then from O to B.

Example

Finding a side vector

In triangle OAB, suppose $OA⃗=4a\vec{OA}=4a$ and $OB⃗=3b\vec{OB}=3b$ . Find $AB⃗\vec{AB}$ and $BA⃗\vec{BA}$ .

Triangle OAB showing the position vectors from O to A and B, and the side vectors between A and B.

To travel from A to B, subtract the position vector of A from the position vector of B.
$AB⃗=3b−4a\vec{AB}=3b-4a$
The reverse vector has the opposite sign.
$BA⃗=4a−3b\vec{BA}=4a-3b$

Key Idea

Routes must join up

When you write a vector route, the end of one vector must be the start of the next. For example, $AO⃗+OB⃗=AB⃗\vec{AO}+\vec{OB}=\vec{AB}$ .

2. Points dividing a line in a ratio

If P lies on AB and $AP:PB=m:n\text{AP}:\text{PB}=m:n$ , then P is $mm+n\frac{m}{m+n}$ of the way from A to B.

Point P divides AB in the ratio AP:PB = m:n, so P is m out of m+n parts from A.

So:

OP⃗=OA⃗+mm+nAB⃗\vec{OP}=\vec{OA}+\frac{m}{m+n}\vec{AB}

You can also use the weighted form:

OP⃗=nm+nOA⃗+mm+nOB⃗\vec{OP}=\frac{n}{m+n}\vec{OA}+\frac{m}{m+n}\vec{OB}

Tip

Opposite weights

In the weighted formula, the coefficient of $OA⃗\vec{OA}$ uses the part near B, and the coefficient of $OB⃗\vec{OB}$ uses the part near A.

Example

Finding $k$ from a ratio

In triangle OAB, $OA⃗=3a\vec{OA}=3a$ and $OB⃗=2b\vec{OB}=2b$ . Point P lies on AB with $AP:PB=2:1\text{AP}:\text{PB}=2:1$ . Given that $OP⃗=k(3a+4b)\vec{OP}=k(3a+4b)$ , find $k$ .

Point P is two thirds of the way from A to B because AP:PB = 2:1.

Since AP is 2 out of 3 equal parts, P is $23\frac{2}{3}$ of the way from A to B.
$OP⃗=OA⃗+23AB⃗\vec{OP}=\vec{OA}+\frac{2}{3}\vec{AB}$
Substitute $OA⃗=3a\vec{OA}=3a$ and $AB⃗=2b−3a\vec{AB}=2b-3a$ .
$OP⃗=3a+23(2b−3a)=3a+43b−2a=a+43b\begin{aligned} \vec{OP}&=3a+\frac{2}{3}(2b-3a)\\ &=3a+\frac{4}{3}b-2a\\ &=a+\frac{4}{3}b \end{aligned}$
Factor the expression so it matches the given form.
$OP⃗=13(3a+4b)\vec{OP}=\frac{1}{3}(3a+4b)$
Therefore $k=13k=\frac{1}{3}$ .

Common Mistake

Swapping the ratio

If $AP:PB=2:1\text{AP}:\text{PB}=2:1$ , P is $23\frac{2}{3}$ of the way from A, not $13\frac{1}{3}$ of the way from A.

3. Proving points are on the same straight line

Three points are collinear if they lie on one straight line.

Definition

Collinear points

To prove A, B and C are collinear, show that two vectors on the same line are scalar multiples of each other, such as $AB⃗=λAC⃗\vec{AB}=\lambda\vec{AC}$ .

Collinear points can be proved by showing vectors from the same starting point are scalar multiples.

The cleanest method is:

Find position vectors for the points.
Choose one point as a common starting point.
Form two vectors from that point.
Show one is a scalar multiple of the other.

Example

Showing three points are collinear

In triangle OAB, $OA⃗=6a\vec{OA}=6a$ and $OB⃗=4b\vec{OB}=4b$ . Point C lies on OA with $OC:CA=2:1\text{OC}:\text{CA}=2:1$ . M is the midpoint of AB. Point D lies on OB extended with $OB:OD=1:2\text{OB}:\text{OD}=1:2$ . Show that C, M and D are collinear.

The setup shows C on OA, M at the midpoint of AB, and D on OB extended for the collinearity proof.

Find the position vectors of C, M and D.
$OC⃗=23(6a)=4aOM⃗=12(6a+4b)=3a+2bOD⃗=2(4b)=8b\begin{aligned} \vec{OC}&=\frac{2}{3}(6a)=4a\\ \vec{OM}&=\frac{1}{2}(6a+4b)=3a+2b\\ \vec{OD}&=2(4b)=8b \end{aligned}$
Use C as the common starting point.
$CM⃗=OM⃗−OC⃗=(3a+2b)−4a=−a+2b\begin{aligned} \vec{CM}&=\vec{OM}-\vec{OC}\\ &=(3a+2b)-4a\\ &=-a+2b \end{aligned}$
Find the vector from C to D.
$CD⃗=OD⃗−OC⃗=8b−4a=−4a+8b\begin{aligned} \vec{CD}&=\vec{OD}-\vec{OC}\\ &=8b-4a\\ &=-4a+8b \end{aligned}$
Compare the two vectors.
$CD⃗=4CM⃗\vec{CD}=4\vec{CM}$
Since $CD⃗\vec{CD}$ is a scalar multiple of $CM⃗\vec{CM}$ , the points C, M and D are collinear.

Common Mistake

Parallel is not always enough

To prove points are on the same straight line, compare vectors that share a point, or clearly state that the parallel vectors pass through a common point.

4. Using shape facts

For a parallelogram, opposite sides are parallel and equal. If O is one corner and A and B are adjacent corners, the opposite corner C has:

In a parallelogram, the diagonal position vector to C is the sum of the two adjacent position vectors.

OC⃗=OA⃗+OB⃗\vec{OC}=\vec{OA}+\vec{OB}

For a regular hexagon, all sides are equal and all angles are equal. With centre O, opposite vertices have opposite position vectors.

Example

Regular hexagon proof

PQRSTU is a regular hexagon with centre O. Suppose $OP⃗=p\vec{OP}=p$ and $OQ⃗=q\vec{OQ}=q$ . N is the midpoint of QR. Y lies on PQ extended so that $PQ:QY=3:2\text{PQ}:\text{QY}=3:2$ . Prove that T, N and Y are collinear.

Regular hexagon PQRSTU with N as the midpoint of QR and Y on the extension of PQ for the collinearity proof.

Use regular hexagon facts. Since P and Q are adjacent vertices, the next vertex has position vector $OR⃗=q−p\vec{OR}=q-p$ , and the opposite vertex has $OT⃗=−q\vec{OT}=-q$ .
Find the midpoint N of QR.
$ON⃗=12(OQ⃗+OR⃗)=12(q+q−p)=−12p+q\begin{aligned} \vec{ON}&=\frac{1}{2}(\vec{OQ}+\vec{OR})\\ &=\frac{1}{2}(q+q-p)\\ &=-\frac{1}{2}p+q \end{aligned}$
Since $PQ:QY=3:2\text{PQ}:\text{QY}=3:2$ , the whole vector from P to Y is $53PQ⃗\frac{5}{3}\vec{PQ}$ .
$OY⃗=OP⃗+53PQ⃗=p+53(q−p)=−23p+53q\begin{aligned} \vec{OY}&=\vec{OP}+\frac{5}{3}\vec{PQ}\\ &=p+\frac{5}{3}(q-p)\\ &=-\frac{2}{3}p+\frac{5}{3}q \end{aligned}$
Compare vectors starting from T.
$TN⃗=ON⃗−OT⃗=−12p+2qTY⃗=OY⃗−OT⃗=−23p+83q\begin{aligned} \vec{TN}&=\vec{ON}-\vec{OT}\\ &=-\frac{1}{2}p+2q\\ \vec{TY}&=\vec{OY}-\vec{OT}\\ &=-\frac{2}{3}p+\frac{8}{3}q \end{aligned}$
Show one is a scalar multiple of the other.
$TY⃗=43TN⃗\vec{TY}=\frac{4}{3}\vec{TN}$
Therefore T, N and Y are collinear.

5. Finding an unknown extension vector

Sometimes the question tells you that three points are collinear and asks you to find an unknown vector.

The trick is to introduce a variable, then compare coefficients of $a$ and $b$ .

Definition

Independent vectors

Vectors $a$ and $b$ are independent if they are not parallel. That means if two expressions in $a$ and $b$ are equal, the coefficient of $a$ must match and the coefficient of $b$ must match.

Example

Finding an unknown vector

In triangle OAB, $OA⃗=6a\vec{OA}=6a$ and $OB⃗=3b\vec{OB}=3b$ . Point C lies on OA with $OC:CA=5:1\text{OC}:\text{CA}=5:1$ . Point D lies on AB with $AD:DB=1:2\text{AD}:\text{DB}=1:2$ . The line OB is extended to E. Given that C, D and E are collinear, find $BE⃗\vec{BE}$ .

The unknown extension BE is found using C, D and E lying on the same straight line.

Find the position vector of C.
$OC⃗=56(6a)=5a\vec{OC}=\frac{5}{6}(6a)=5a$
Find the position vector of D using the ratio on AB.
$OD⃗=23(6a)+13(3b)=4a+b\begin{aligned} \vec{OD}&=\frac{2}{3}(6a)+\frac{1}{3}(3b)\\ &=4a+b \end{aligned}$
Let $BE⃗=tb\vec{BE}=tb$ , so $OE⃗=(3+t)b\vec{OE}=(3+t)b$ .
Find $CD⃗\vec{CD}$ and $CE⃗\vec{CE}$ .
$CD⃗=(4a+b)−5a=−a+bCE⃗=(3+t)b−5a=−5a+(3+t)b\begin{aligned} \vec{CD}&=(4a+b)-5a=-a+b\\ \vec{CE}&=(3+t)b-5a=-5a+(3+t)b \end{aligned}$
Since C, D and E are collinear, write $CE⃗=λCD⃗\vec{CE}=\lambda\vec{CD}$ and compare the $a$ coefficients.
$−5=−λ⇒λ=5-5=-\lambda \Rightarrow \lambda=5$
Compare the $b$ coefficients: $3 + t = 5$ , so $t = 2$ .
Therefore $BE⃗=2b\vec{BE}=2b$ .

Exam technique

In the exam

Start by writing position vectors for every important point.
For a ratio, decide whether you are moving from the first point or using the weighted formula.
To prove collinearity, form two vectors with the same starting point and show one is a scalar multiple of the other.
If a point is on an extension, introduce a variable and solve by comparing coefficients.

Self review

Check yourself

Can you find $AB⃗\vec{AB}$ quickly from $OA⃗\vec{OA}$ and $OB⃗\vec{OB}$ ?
If $AP:PB=3:2\text{AP}:\text{PB}=3:2$ , what fraction of the way from A to B is P?
When proving three points are collinear, which two vectors would you compare?

Vectors Proof Questions

What you'll learn

How to turn a geometry diagram into vector algebra.
How to use ratios on a line segment to find position vectors.
How to prove points lie on the same straight line.
How to handle harder questions with extensions, midpoints, parallelograms and regular hexagons.

1. Vector routes: the starting point

A vector describes a movement: it has a size and a direction. In proof questions, you usually express every movement using the given vectors, such as $a$ and $b$ .

Definition

Vector notation

$AB⃗\vec{AB}$ means the vector from point A to point B.
A position vector starts from the origin point O, for example $OA⃗\vec{OA}$ .
A scalar is a number multiplying a vector, such as 3 in $3 a$ .

The most useful route fact is:

AB⃗=OB⃗−OA⃗\vec{AB}=\vec{OB}-\vec{OA}

That means: to go from A to B, go from A back to O, then from O to B.

Example

Finding a side vector

In triangle OAB, suppose $OA⃗=4a\vec{OA}=4a$ and $OB⃗=3b\vec{OB}=3b$ . Find $AB⃗\vec{AB}$ and $BA⃗\vec{BA}$ .

Triangle OAB showing the position vectors from O to A and B, and the side vectors between A and B.

To travel from A to B, subtract the position vector of A from the position vector of B.
$AB⃗=3b−4a\vec{AB}=3b-4a$
The reverse vector has the opposite sign.
$BA⃗=4a−3b\vec{BA}=4a-3b$

Key Idea

Routes must join up

When you write a vector route, the end of one vector must be the start of the next. For example, $AO⃗+OB⃗=AB⃗\vec{AO}+\vec{OB}=\vec{AB}$ .

2. Points dividing a line in a ratio

If P lies on AB and $AP:PB=m:n\text{AP}:\text{PB}=m:n$ , then P is $mm+n\frac{m}{m+n}$ of the way from A to B.

Point P divides AB in the ratio AP:PB = m:n, so P is m out of m+n parts from A.

So:

OP⃗=OA⃗+mm+nAB⃗\vec{OP}=\vec{OA}+\frac{m}{m+n}\vec{AB}

You can also use the weighted form:

OP⃗=nm+nOA⃗+mm+nOB⃗\vec{OP}=\frac{n}{m+n}\vec{OA}+\frac{m}{m+n}\vec{OB}

Tip

Opposite weights

In the weighted formula, the coefficient of $OA⃗\vec{OA}$ uses the part near B, and the coefficient of $OB⃗\vec{OB}$ uses the part near A.

Example

Finding $k$ from a ratio

In triangle OAB, $OA⃗=3a\vec{OA}=3a$ and $OB⃗=2b\vec{OB}=2b$ . Point P lies on AB with $AP:PB=2:1\text{AP}:\text{PB}=2:1$ . Given that $OP⃗=k(3a+4b)\vec{OP}=k(3a+4b)$ , find $k$ .

Point P is two thirds of the way from A to B because AP:PB = 2:1.

Since AP is 2 out of 3 equal parts, P is $23\frac{2}{3}$ of the way from A to B.
$OP⃗=OA⃗+23AB⃗\vec{OP}=\vec{OA}+\frac{2}{3}\vec{AB}$
Substitute $OA⃗=3a\vec{OA}=3a$ and $AB⃗=2b−3a\vec{AB}=2b-3a$ .
$OP⃗=3a+23(2b−3a)=3a+43b−2a=a+43b\begin{aligned} \vec{OP}&=3a+\frac{2}{3}(2b-3a)\\ &=3a+\frac{4}{3}b-2a\\ &=a+\frac{4}{3}b \end{aligned}$
Factor the expression so it matches the given form.
$OP⃗=13(3a+4b)\vec{OP}=\frac{1}{3}(3a+4b)$
Therefore $k=13k=\frac{1}{3}$ .

Common Mistake

Swapping the ratio

If $AP:PB=2:1\text{AP}:\text{PB}=2:1$ , P is $23\frac{2}{3}$ of the way from A, not $13\frac{1}{3}$ of the way from A.

3. Proving points are on the same straight line

Three points are collinear if they lie on one straight line.

Definition

Collinear points

To prove A, B and C are collinear, show that two vectors on the same line are scalar multiples of each other, such as $AB⃗=λAC⃗\vec{AB}=\lambda\vec{AC}$ .

Collinear points can be proved by showing vectors from the same starting point are scalar multiples.

The cleanest method is:

Find position vectors for the points.
Choose one point as a common starting point.
Form two vectors from that point.
Show one is a scalar multiple of the other.

Example

Showing three points are collinear

The setup shows C on OA, M at the midpoint of AB, and D on OB extended for the collinearity proof.

Find the position vectors of C, M and D.
$OC⃗=23(6a)=4aOM⃗=12(6a+4b)=3a+2bOD⃗=2(4b)=8b\begin{aligned} \vec{OC}&=\frac{2}{3}(6a)=4a\\ \vec{OM}&=\frac{1}{2}(6a+4b)=3a+2b\\ \vec{OD}&=2(4b)=8b \end{aligned}$
Use C as the common starting point.
$CM⃗=OM⃗−OC⃗=(3a+2b)−4a=−a+2b\begin{aligned} \vec{CM}&=\vec{OM}-\vec{OC}\\ &=(3a+2b)-4a\\ &=-a+2b \end{aligned}$
Find the vector from C to D.
$CD⃗=OD⃗−OC⃗=8b−4a=−4a+8b\begin{aligned} \vec{CD}&=\vec{OD}-\vec{OC}\\ &=8b-4a\\ &=-4a+8b \end{aligned}$
Compare the two vectors.
$CD⃗=4CM⃗\vec{CD}=4\vec{CM}$
Since $CD⃗\vec{CD}$ is a scalar multiple of $CM⃗\vec{CM}$ , the points C, M and D are collinear.

Common Mistake

Parallel is not always enough

To prove points are on the same straight line, compare vectors that share a point, or clearly state that the parallel vectors pass through a common point.

4. Using shape facts

For a parallelogram, opposite sides are parallel and equal. If O is one corner and A and B are adjacent corners, the opposite corner C has:

In a parallelogram, the diagonal position vector to C is the sum of the two adjacent position vectors.

OC⃗=OA⃗+OB⃗\vec{OC}=\vec{OA}+\vec{OB}

For a regular hexagon, all sides are equal and all angles are equal. With centre O, opposite vertices have opposite position vectors.

Example

Regular hexagon proof

Regular hexagon PQRSTU with N as the midpoint of QR and Y on the extension of PQ for the collinearity proof.

Use regular hexagon facts. Since P and Q are adjacent vertices, the next vertex has position vector $OR⃗=q−p\vec{OR}=q-p$ , and the opposite vertex has $OT⃗=−q\vec{OT}=-q$ .
Find the midpoint N of QR.
$ON⃗=12(OQ⃗+OR⃗)=12(q+q−p)=−12p+q\begin{aligned} \vec{ON}&=\frac{1}{2}(\vec{OQ}+\vec{OR})\\ &=\frac{1}{2}(q+q-p)\\ &=-\frac{1}{2}p+q \end{aligned}$
Since $PQ:QY=3:2\text{PQ}:\text{QY}=3:2$ , the whole vector from P to Y is $53PQ⃗\frac{5}{3}\vec{PQ}$ .
$OY⃗=OP⃗+53PQ⃗=p+53(q−p)=−23p+53q\begin{aligned} \vec{OY}&=\vec{OP}+\frac{5}{3}\vec{PQ}\\ &=p+\frac{5}{3}(q-p)\\ &=-\frac{2}{3}p+\frac{5}{3}q \end{aligned}$
Compare vectors starting from T.
$TN⃗=ON⃗−OT⃗=−12p+2qTY⃗=OY⃗−OT⃗=−23p+83q\begin{aligned} \vec{TN}&=\vec{ON}-\vec{OT}\\ &=-\frac{1}{2}p+2q\\ \vec{TY}&=\vec{OY}-\vec{OT}\\ &=-\frac{2}{3}p+\frac{8}{3}q \end{aligned}$
Show one is a scalar multiple of the other.
$TY⃗=43TN⃗\vec{TY}=\frac{4}{3}\vec{TN}$
Therefore T, N and Y are collinear.

5. Finding an unknown extension vector

Sometimes the question tells you that three points are collinear and asks you to find an unknown vector.

The trick is to introduce a variable, then compare coefficients of $a$ and $b$ .

Definition

Independent vectors

Vectors $a$ and $b$ are independent if they are not parallel. That means if two expressions in $a$ and $b$ are equal, the coefficient of $a$ must match and the coefficient of $b$ must match.

Example

Finding an unknown vector

The unknown extension BE is found using C, D and E lying on the same straight line.

Find the position vector of C.
$OC⃗=56(6a)=5a\vec{OC}=\frac{5}{6}(6a)=5a$
Find the position vector of D using the ratio on AB.
$OD⃗=23(6a)+13(3b)=4a+b\begin{aligned} \vec{OD}&=\frac{2}{3}(6a)+\frac{1}{3}(3b)\\ &=4a+b \end{aligned}$
Let $BE⃗=tb\vec{BE}=tb$ , so $OE⃗=(3+t)b\vec{OE}=(3+t)b$ .
Find $CD⃗\vec{CD}$ and $CE⃗\vec{CE}$ .
$CD⃗=(4a+b)−5a=−a+bCE⃗=(3+t)b−5a=−5a+(3+t)b\begin{aligned} \vec{CD}&=(4a+b)-5a=-a+b\\ \vec{CE}&=(3+t)b-5a=-5a+(3+t)b \end{aligned}$
Since C, D and E are collinear, write $CE⃗=λCD⃗\vec{CE}=\lambda\vec{CD}$ and compare the $a$ coefficients.
$−5=−λ⇒λ=5-5=-\lambda \Rightarrow \lambda=5$
Compare the $b$ coefficients: $3 + t = 5$ , so $t = 2$ .
Therefore $BE⃗=2b\vec{BE}=2b$ .

Exam technique

In the exam

Start by writing position vectors for every important point.
For a ratio, decide whether you are moving from the first point or using the weighted formula.
To prove collinearity, form two vectors with the same starting point and show one is a scalar multiple of the other.
If a point is on an extension, introduce a variable and solve by comparing coefficients.

Self review

Check yourself

Can you find $AB⃗\vec{AB}$ quickly from $OA⃗\vec{OA}$ and $OB⃗\vec{OB}$ ?
If $AP:PB=3:2\text{AP}:\text{PB}=3:2$ , what fraction of the way from A to B is P?
When proving three points are collinear, which two vectors would you compare?

Vectors Proof Questions

What you'll learn

1. Vector routes: the starting point

2. Points dividing a line in a ratio

3. Proving points are on the same straight line

4. Using shape facts

5. Finding an unknown extension vector

Recap questions

You've reached the end

Vectors Proof Questions

What you'll learn

1. Vector routes: the starting point

2. Points dividing a line in a ratio

3. Proving points are on the same straight line

4. Using shape facts

5. Finding an unknown extension vector

Recap questions

You've reached the end

Vectors Proof Questions

What you'll learn

1. Vector routes: the starting point

2. Points dividing a line in a ratio

3. Proving points are on the same straight line

4. Using shape facts

5. Finding an unknown extension vector

Recap questions

You've reached the end

Vectors Proof Questions Revision Guide

Vectors Proof Questions

What you'll learn

1. Vector routes: the starting point

2. Points dividing a line in a ratio

3. Proving points are on the same straight line

4. Using shape facts

5. Finding an unknown extension vector

Recap questions

You've reached the end

Vectors Proof Questions Revision Guide