Vectors Proof Questions

What you'll learn

How to write position vectors using two base vectors such as $a$ and $b$ .
How to handle points that divide a line in a given ratio.
How to prove that three points are on the same straight line.
How to use collinearity to find an unknown vector on an extended line.

1. Vector basics you need first

In these questions, you usually start from a fixed point called $O$ . You then describe every other point by its position from $O$ .

Definition

Vector and position vector

A vector has a size and a direction.
$AB⃗\vec{AB}$ means the movement from point A to point B.
A position vector gives the position of a point from the origin, for example $OA⃗\vec{OA}$ .

The most important rule is:

AB⃗=OB⃗−OA⃗\vec{AB}=\vec{OB}-\vec{OA}

Think: end minus start.

Example

Finding a midpoint position vector

Suppose $OA⃗=4a\vec{OA}=4a$ and $OB⃗=7b\vec{OB}=7b$ . Point N is the midpoint of AB. Find $AB⃗\vec{AB}$ and $ON⃗\vec{ON}$ .

Use end minus start to find $AB⃗\vec{AB}$ .
$AB⃗=7b−4a\vec{AB}=7b-4a$
Since N is halfway from A to B, add half of $AB⃗\vec{AB}$ to $OA⃗\vec{OA}$ .
$ON⃗=4a+12(7b−4a)=4a+72b−2a=2a+72b\begin{aligned} \vec{ON}&=4a+\frac{1}{2}(7b-4a)\\ &=4a+\frac{7}{2}b-2a\\ &=2a+\frac{7}{2}b \end{aligned}$
So $AB⃗=7b−4a\vec{AB}=7b-4a$ and $ON⃗=2a+72b\vec{ON}=2a+\frac{7}{2}b$ .

2. Points dividing a line in a ratio

If P lies on AB, the ratio tells you how far along the line P is.

For example, if $A P : P B = 2 : 3$ , then the whole line AB has 5 equal parts, and P is 2 of those parts from A.

This is the standard triangle setup for ratio questions.

Vector triangle showing P dividing AB in a ratio

If $A P : P B = m : n$ , then:

AP⃗=mm+nAB⃗\vec{AP}=\frac{m}{m+n}\vec{AB}

So:

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

Key Idea

Use the fraction from the starting end

If P is measured from A, use the AP part of the ratio. In $A P : P B = m : n$ , P is $mm+n\frac{m}{m+n}$ of the way from A to B.

Example

Finding the value of k

In triangle OAB, $OA⃗=2a\vec{OA}=2a$ and $OB⃗=5b\vec{OB}=5b$ . Point P lies on AB so that $A P : P B = 2 : 3$ . Given that $OP⃗=k(3a+5b)\vec{OP}=k(3a+5b)$ , find $k$ .

First find the vector from A to B.
$AB⃗=5b−2a\vec{AB}=5b-2a$
Since $A P : P B = 2 : 3$ , P is $25\frac{2}{5}$ of the way from A to B.
$OP⃗=OA⃗+25AB⃗=2a+25(5b−2a)=2a+2b−45a=65a+2b\begin{aligned} \vec{OP}&=\vec{OA}+\frac{2}{5}\vec{AB}\\ &=2a+\frac{2}{5}(5b-2a)\\ &=2a+2b-\frac{4}{5}a\\ &=\frac{6}{5}a+2b \end{aligned}$
Factor the answer to match the form in the question.
$65a+2b=25(3a+5b)\frac{6}{5}a+2b=\frac{2}{5}(3a+5b)$
Therefore $k=25k=\frac{2}{5}$ .

Common Mistake

Using the wrong part of the ratio

If $A P : P B = 2 : 3$ , do not use $35\frac{3}{5}$ from A. The 3 parts are from P to B, not from A to P.

3. Proving three points are on the same straight line

Three points on one straight line are called collinear.

To prove collinearity, you usually show that two vectors are scalar multiples of each other.

Definition

Scalar multiple

A scalar is an ordinary number. If one vector is a scalar multiple of another, such as $AD⃗=λAE⃗\vec{AD}=\lambda\vec{AE}$ , then the two vectors are parallel and lie along the same line.

The target is to compare vectors that start from the same point.

Collinearity vector proof diagram

Key Idea

Same start point, same line

To prove A, D and E are collinear, try to show $AD⃗=λAE⃗\vec{AD}=\lambda\vec{AE}$ , or $AE⃗=λAD⃗\vec{AE}=\lambda\vec{AD}$ , for some number $λ\lambda$ .

Example

A parallelogram collinearity proof

OABC is a parallelogram with adjacent sides $OA⃗=3a\vec{OA}=3a$ and $OB⃗=3b\vec{OB}=3b$ . Point D lies on OC so that $O D : D C = 2 : 1$ . Point E is the midpoint of BC. Show that A, D and E lie on one straight line.

In a parallelogram, the diagonal position vector is found by adding the adjacent sides.
$OC⃗=3a+3b\vec{OC}=3a+3b$
Since $O D : D C = 2 : 1$ , D is $23\frac{2}{3}$ of the way from O to C.
$OD⃗=23(3a+3b)=2a+2b\vec{OD}=\frac{2}{3}(3a+3b)=2a+2b$
E is the midpoint of B and C.
$OE⃗=12(OB⃗+OC⃗)=12(3b+3a+3b)=32a+3b\begin{aligned} \vec{OE}&=\frac{1}{2}(\vec{OB}+\vec{OC})\\ &=\frac{1}{2}(3b+3a+3b)\\ &=\frac{3}{2}a+3b \end{aligned}$
Now compare vectors starting from A.
$AD⃗=OD⃗−OA⃗=(2a+2b)−3a=−a+2b\begin{aligned} \vec{AD}&=\vec{OD}-\vec{OA}\\ &=(2a+2b)-3a\\ &=-a+2b \end{aligned}$
Also find $AE⃗\vec{AE}$ .
$AE⃗=OE⃗−OA⃗=(32a+3b)−3a=−32a+3b=32(−a+2b)\begin{aligned} \vec{AE}&=\vec{OE}-\vec{OA}\\ &=\left(\frac{3}{2}a+3b\right)-3a\\ &=-\frac{3}{2}a+3b\\ &=\frac{3}{2}(-a+2b) \end{aligned}$
Since $AE⃗=32AD⃗\vec{AE}=\frac{3}{2}\vec{AD}$ , the vectors are scalar multiples. Therefore A, D and E are collinear.

Common Mistake

Comparing position vectors directly

To prove A, D and E are collinear, do not compare $OD⃗\vec{OD}$ and $OE⃗\vec{OE}$ . That would only tell you about a line through O. Use vectors such as $AD⃗\vec{AD}$ and $AE⃗\vec{AE}$ .

4. Regular hexagon vector facts

In a regular hexagon, the centre-to-vertex vectors are very useful. If two adjacent position vectors are $p$ and $q$ , the next vertex can often be written using $q - p$ .

The diagram shows the common pattern.

Regular hexagon vector positions

For a regular hexagon PQRSTU with centre O:

If $OP⃗=p\vec{OP}=p$ and $OQ⃗=q\vec{OQ}=q$ , then $OR⃗=q−p\vec{OR}=q-p$ .
Opposite vertices have opposite position vectors, so $OT⃗=−q\vec{OT}=-q$ and $OS⃗=−p\vec{OS}=-p$ .

Example

Proving collinearity in a regular hexagon

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

Use the regular hexagon fact for the next vertex.
$OR⃗=q−p\vec{OR}=q-p$
N is the midpoint of Q and R.
$ON⃗=12(OQ⃗+OR⃗)=12(q+q−p)=q−12p\begin{aligned} \vec{ON}&=\frac{1}{2}(\vec{OQ}+\vec{OR})\\ &=\frac{1}{2}(q+q-p)\\ &=q-\frac{1}{2}p \end{aligned}$
Since Y is beyond Q and $P Q : Q Y = 3 : 2$ , use $QY⃗=23PQ⃗\vec{QY}=\frac{2}{3}\vec{PQ}$ .
$OY⃗=OQ⃗+QY⃗=q+23(q−p)=53q−23p\begin{aligned} \vec{OY}&=\vec{OQ}+\vec{QY}\\ &=q+\frac{2}{3}(q-p)\\ &=\frac{5}{3}q-\frac{2}{3}p \end{aligned}$
The vertex T is opposite Q, so $OT⃗=−q\vec{OT}=-q$ .
Compare vectors from T.
$TN⃗=ON⃗−OT⃗=(q−12p)−(−q)=2q−12p\begin{aligned} \vec{TN}&=\vec{ON}-\vec{OT}\\ &=\left(q-\frac{1}{2}p\right)-(-q)\\ &=2q-\frac{1}{2}p \end{aligned}$
Now find $TY⃗\vec{TY}$ .
$TY⃗=OY⃗−OT⃗=(53q−23p)−(−q)=83q−23p=43(2q−12p)\begin{aligned} \vec{TY}&=\vec{OY}-\vec{OT}\\ &=\left(\frac{5}{3}q-\frac{2}{3}p\right)-(-q)\\ &=\frac{8}{3}q-\frac{2}{3}p\\ &=\frac{4}{3}\left(2q-\frac{1}{2}p\right) \end{aligned}$
Since $TY⃗=43TN⃗\vec{TY}=\frac{4}{3}\vec{TN}$ , T, N and Y are on the same straight line.

5. Using collinearity to find an unknown vector

Sometimes the question tells you that three points are collinear, and you must find an unknown vector. The method is the same, but you introduce an unknown scalar.

Tip

Let the extension be unknown

If E is on OB extended, write something like $BE⃗=xb\vec{BE}=xb$ . Then $OE⃗=OB⃗+BE⃗\vec{OE}=\vec{OB}+\vec{BE}$ .

Example

Finding a vector on an extension

In triangle OAB, $OA⃗=8a\vec{OA}=8a$ and $OB⃗=3b\vec{OB}=3b$ . Point C lies on OA so that $O C : C A = 3 : 1$ . Point D lies on AB so that $A D : D B = 1 : 2$ . The line OB is extended to E. Given that C, D and E are collinear, find $BE⃗\vec{BE}$ .

Find the position vector of C.
$OC⃗=34(8a)=6a\vec{OC}=\frac{3}{4}(8a)=6a$
Find the position vector of D. Since $A D : D B = 1 : 2$ , D is $13\frac{1}{3}$ of the way from A to B.
$OD⃗=OA⃗+13AB⃗=8a+13(3b−8a)=163a+b\begin{aligned} \vec{OD}&=\vec{OA}+\frac{1}{3}\vec{AB}\\ &=8a+\frac{1}{3}(3b-8a)\\ &=\frac{16}{3}a+b \end{aligned}$
Let $BE⃗=xb\vec{BE}=xb$ , so $OE⃗=(3+x)b\vec{OE}=(3+x)b$ .
Since C, D and E are collinear, write $CD⃗=λCE⃗\vec{CD}=\lambda\vec{CE}$ .
$CD⃗=(163a+b)−6a=−23a+b\vec{CD}=\left(\frac{16}{3}a+b\right)-6a=-\frac{2}{3}a+b$
Also write $CE⃗\vec{CE}$ in terms of $x$ .
$CE⃗=(3+x)b−6a=−6a+(3+x)b\vec{CE}=(3+x)b-6a=-6a+(3+x)b$
Compare coefficients in $CD⃗=λCE⃗\vec{CD}=\lambda\vec{CE}$ .
$−23=−6λ-\frac{2}{3}=-6\lambda$
So $λ=19\lambda=\frac{1}{9}$ . Now compare the coefficients of $b$ .
$1=19(3+x)1=\frac{1}{9}(3+x)$
Solve for $x$ .
$3 + x = 9$
Therefore $x = 6$ , so $BE⃗=6b\vec{BE}=6b$ .

Exam technique

In the exam

Write down the position vector of every important point from O.
For a ratio, turn it into a fraction of the whole line segment.
To prove a straight line, compare two vectors with the same starting point.
If a point is on an extension, introduce an unknown such as $BE⃗=xb\vec{BE}=xb$ and compare coefficients.

Self review

Check yourself

If $A P : P B = 2 : 5$ , what fraction of $AB⃗\vec{AB}$ is $AP⃗\vec{AP}$ ?
Why is comparing $OD⃗\vec{OD}$ and $OE⃗\vec{OE}$ not enough to prove A, D and E are collinear?
In a regular hexagon, if adjacent position vectors are $p$ and $q$ , what is the position vector of the next vertex?

Vectors Proof Questions

What you'll learn

How to write position vectors using two base vectors such as $a$ and $b$ .
How to handle points that divide a line in a given ratio.
How to prove that three points are on the same straight line.
How to use collinearity to find an unknown vector on an extended line.

1. Vector basics you need first

In these questions, you usually start from a fixed point called $O$ . You then describe every other point by its position from $O$ .

Definition

Vector and position vector

A vector has a size and a direction.
$AB⃗\vec{AB}$ means the movement from point A to point B.
A position vector gives the position of a point from the origin, for example $OA⃗\vec{OA}$ .

The most important rule is:

AB⃗=OB⃗−OA⃗\vec{AB}=\vec{OB}-\vec{OA}

Think: end minus start.

Example

Finding a midpoint position vector

Suppose $OA⃗=4a\vec{OA}=4a$ and $OB⃗=7b\vec{OB}=7b$ . Point N is the midpoint of AB. Find $AB⃗\vec{AB}$ and $ON⃗\vec{ON}$ .

Use end minus start to find $AB⃗\vec{AB}$ .
$AB⃗=7b−4a\vec{AB}=7b-4a$
Since N is halfway from A to B, add half of $AB⃗\vec{AB}$ to $OA⃗\vec{OA}$ .
$ON⃗=4a+12(7b−4a)=4a+72b−2a=2a+72b\begin{aligned} \vec{ON}&=4a+\frac{1}{2}(7b-4a)\\ &=4a+\frac{7}{2}b-2a\\ &=2a+\frac{7}{2}b \end{aligned}$
So $AB⃗=7b−4a\vec{AB}=7b-4a$ and $ON⃗=2a+72b\vec{ON}=2a+\frac{7}{2}b$ .

2. Points dividing a line in a ratio

If P lies on AB, the ratio tells you how far along the line P is.

For example, if $A P : P B = 2 : 3$ , then the whole line AB has 5 equal parts, and P is 2 of those parts from A.

This is the standard triangle setup for ratio questions.

Vector triangle showing P dividing AB in a ratio

If $A P : P B = m : n$ , then:

AP⃗=mm+nAB⃗\vec{AP}=\frac{m}{m+n}\vec{AB}

So:

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

Key Idea

Use the fraction from the starting end

If P is measured from A, use the AP part of the ratio. In $A P : P B = m : n$ , P is $mm+n\frac{m}{m+n}$ of the way from A to B.

Example

Finding the value of k

In triangle OAB, $OA⃗=2a\vec{OA}=2a$ and $OB⃗=5b\vec{OB}=5b$ . Point P lies on AB so that $A P : P B = 2 : 3$ . Given that $OP⃗=k(3a+5b)\vec{OP}=k(3a+5b)$ , find $k$ .

First find the vector from A to B.
$AB⃗=5b−2a\vec{AB}=5b-2a$
Since $A P : P B = 2 : 3$ , P is $25\frac{2}{5}$ of the way from A to B.
$OP⃗=OA⃗+25AB⃗=2a+25(5b−2a)=2a+2b−45a=65a+2b\begin{aligned} \vec{OP}&=\vec{OA}+\frac{2}{5}\vec{AB}\\ &=2a+\frac{2}{5}(5b-2a)\\ &=2a+2b-\frac{4}{5}a\\ &=\frac{6}{5}a+2b \end{aligned}$
Factor the answer to match the form in the question.
$65a+2b=25(3a+5b)\frac{6}{5}a+2b=\frac{2}{5}(3a+5b)$
Therefore $k=25k=\frac{2}{5}$ .

Common Mistake

Using the wrong part of the ratio

If $A P : P B = 2 : 3$ , do not use $35\frac{3}{5}$ from A. The 3 parts are from P to B, not from A to P.

3. Proving three points are on the same straight line

Three points on one straight line are called collinear.

To prove collinearity, you usually show that two vectors are scalar multiples of each other.

Definition

Scalar multiple

A scalar is an ordinary number. If one vector is a scalar multiple of another, such as $AD⃗=λAE⃗\vec{AD}=\lambda\vec{AE}$ , then the two vectors are parallel and lie along the same line.

The target is to compare vectors that start from the same point.

Collinearity vector proof diagram

Key Idea

Same start point, same line

To prove A, D and E are collinear, try to show $AD⃗=λAE⃗\vec{AD}=\lambda\vec{AE}$ , or $AE⃗=λAD⃗\vec{AE}=\lambda\vec{AD}$ , for some number $λ\lambda$ .

Example

A parallelogram collinearity proof

In a parallelogram, the diagonal position vector is found by adding the adjacent sides.
$OC⃗=3a+3b\vec{OC}=3a+3b$
Since $O D : D C = 2 : 1$ , D is $23\frac{2}{3}$ of the way from O to C.
$OD⃗=23(3a+3b)=2a+2b\vec{OD}=\frac{2}{3}(3a+3b)=2a+2b$
E is the midpoint of B and C.
$OE⃗=12(OB⃗+OC⃗)=12(3b+3a+3b)=32a+3b\begin{aligned} \vec{OE}&=\frac{1}{2}(\vec{OB}+\vec{OC})\\ &=\frac{1}{2}(3b+3a+3b)\\ &=\frac{3}{2}a+3b \end{aligned}$
Now compare vectors starting from A.
$AD⃗=OD⃗−OA⃗=(2a+2b)−3a=−a+2b\begin{aligned} \vec{AD}&=\vec{OD}-\vec{OA}\\ &=(2a+2b)-3a\\ &=-a+2b \end{aligned}$
Also find $AE⃗\vec{AE}$ .
$AE⃗=OE⃗−OA⃗=(32a+3b)−3a=−32a+3b=32(−a+2b)\begin{aligned} \vec{AE}&=\vec{OE}-\vec{OA}\\ &=\left(\frac{3}{2}a+3b\right)-3a\\ &=-\frac{3}{2}a+3b\\ &=\frac{3}{2}(-a+2b) \end{aligned}$
Since $AE⃗=32AD⃗\vec{AE}=\frac{3}{2}\vec{AD}$ , the vectors are scalar multiples. Therefore A, D and E are collinear.

Common Mistake

Comparing position vectors directly

To prove A, D and E are collinear, do not compare $OD⃗\vec{OD}$ and $OE⃗\vec{OE}$ . That would only tell you about a line through O. Use vectors such as $AD⃗\vec{AD}$ and $AE⃗\vec{AE}$ .

4. Regular hexagon vector facts

In a regular hexagon, the centre-to-vertex vectors are very useful. If two adjacent position vectors are $p$ and $q$ , the next vertex can often be written using $q - p$ .

The diagram shows the common pattern.

Regular hexagon vector positions

For a regular hexagon PQRSTU with centre O:

If $OP⃗=p\vec{OP}=p$ and $OQ⃗=q\vec{OQ}=q$ , then $OR⃗=q−p\vec{OR}=q-p$ .
Opposite vertices have opposite position vectors, so $OT⃗=−q\vec{OT}=-q$ and $OS⃗=−p\vec{OS}=-p$ .

Example

Proving collinearity in a regular hexagon

Use the regular hexagon fact for the next vertex.
$OR⃗=q−p\vec{OR}=q-p$
N is the midpoint of Q and R.
$ON⃗=12(OQ⃗+OR⃗)=12(q+q−p)=q−12p\begin{aligned} \vec{ON}&=\frac{1}{2}(\vec{OQ}+\vec{OR})\\ &=\frac{1}{2}(q+q-p)\\ &=q-\frac{1}{2}p \end{aligned}$
Since Y is beyond Q and $P Q : Q Y = 3 : 2$ , use $QY⃗=23PQ⃗\vec{QY}=\frac{2}{3}\vec{PQ}$ .
$OY⃗=OQ⃗+QY⃗=q+23(q−p)=53q−23p\begin{aligned} \vec{OY}&=\vec{OQ}+\vec{QY}\\ &=q+\frac{2}{3}(q-p)\\ &=\frac{5}{3}q-\frac{2}{3}p \end{aligned}$
The vertex T is opposite Q, so $OT⃗=−q\vec{OT}=-q$ .
Compare vectors from T.
$TN⃗=ON⃗−OT⃗=(q−12p)−(−q)=2q−12p\begin{aligned} \vec{TN}&=\vec{ON}-\vec{OT}\\ &=\left(q-\frac{1}{2}p\right)-(-q)\\ &=2q-\frac{1}{2}p \end{aligned}$
Now find $TY⃗\vec{TY}$ .
$TY⃗=OY⃗−OT⃗=(53q−23p)−(−q)=83q−23p=43(2q−12p)\begin{aligned} \vec{TY}&=\vec{OY}-\vec{OT}\\ &=\left(\frac{5}{3}q-\frac{2}{3}p\right)-(-q)\\ &=\frac{8}{3}q-\frac{2}{3}p\\ &=\frac{4}{3}\left(2q-\frac{1}{2}p\right) \end{aligned}$
Since $TY⃗=43TN⃗\vec{TY}=\frac{4}{3}\vec{TN}$ , T, N and Y are on the same straight line.

5. Using collinearity to find an unknown vector

Sometimes the question tells you that three points are collinear, and you must find an unknown vector. The method is the same, but you introduce an unknown scalar.

Tip

Let the extension be unknown

If E is on OB extended, write something like $BE⃗=xb\vec{BE}=xb$ . Then $OE⃗=OB⃗+BE⃗\vec{OE}=\vec{OB}+\vec{BE}$ .

Example

Finding a vector on an extension

Find the position vector of C.
$OC⃗=34(8a)=6a\vec{OC}=\frac{3}{4}(8a)=6a$
Find the position vector of D. Since $A D : D B = 1 : 2$ , D is $13\frac{1}{3}$ of the way from A to B.
$OD⃗=OA⃗+13AB⃗=8a+13(3b−8a)=163a+b\begin{aligned} \vec{OD}&=\vec{OA}+\frac{1}{3}\vec{AB}\\ &=8a+\frac{1}{3}(3b-8a)\\ &=\frac{16}{3}a+b \end{aligned}$
Let $BE⃗=xb\vec{BE}=xb$ , so $OE⃗=(3+x)b\vec{OE}=(3+x)b$ .
Since C, D and E are collinear, write $CD⃗=λCE⃗\vec{CD}=\lambda\vec{CE}$ .
$CD⃗=(163a+b)−6a=−23a+b\vec{CD}=\left(\frac{16}{3}a+b\right)-6a=-\frac{2}{3}a+b$
Also write $CE⃗\vec{CE}$ in terms of $x$ .
$CE⃗=(3+x)b−6a=−6a+(3+x)b\vec{CE}=(3+x)b-6a=-6a+(3+x)b$
Compare coefficients in $CD⃗=λCE⃗\vec{CD}=\lambda\vec{CE}$ .
$−23=−6λ-\frac{2}{3}=-6\lambda$
So $λ=19\lambda=\frac{1}{9}$ . Now compare the coefficients of $b$ .
$1=19(3+x)1=\frac{1}{9}(3+x)$
Solve for $x$ .
$3 + x = 9$
Therefore $x = 6$ , so $BE⃗=6b\vec{BE}=6b$ .

Exam technique

In the exam

Write down the position vector of every important point from O.
For a ratio, turn it into a fraction of the whole line segment.
To prove a straight line, compare two vectors with the same starting point.
If a point is on an extension, introduce an unknown such as $BE⃗=xb\vec{BE}=xb$ and compare coefficients.

Self review

Check yourself

If $A P : P B = 2 : 5$ , what fraction of $AB⃗\vec{AB}$ is $AP⃗\vec{AP}$ ?
Why is comparing $OD⃗\vec{OD}$ and $OE⃗\vec{OE}$ not enough to prove A, D and E are collinear?
In a regular hexagon, if adjacent position vectors are $p$ and $q$ , what is the position vector of the next vertex?

Vectors Proof Questions

What you'll learn

1. Vector basics you need first

2. Points dividing a line in a ratio

3. Proving three points are on the same straight line

4. Regular hexagon vector facts

5. Using collinearity to find an unknown vector

Recap questions

You've reached the end

Vectors Proof Questions

What you'll learn

1. Vector basics you need first

2. Points dividing a line in a ratio

3. Proving three points are on the same straight line

4. Regular hexagon vector facts

5. Using collinearity to find an unknown vector

Recap questions

You've reached the end

Vectors Proof Questions

What you'll learn

1. Vector basics you need first

2. Points dividing a line in a ratio

3. Proving three points are on the same straight line

4. Regular hexagon vector facts

5. Using collinearity to find an unknown vector

Recap questions

You've reached the end

Vectors Proof Questions Revision Guide

Vectors Proof Questions

What you'll learn

1. Vector basics you need first

2. Points dividing a line in a ratio

3. Proving three points are on the same straight line

4. Regular hexagon vector facts

5. Using collinearity to find an unknown vector

Recap questions

You've reached the end

Vectors Proof Questions Revision Guide