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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 aaa and bbb.

Definition

Vector notation

  • AB⃗\vec{AB}AB means the vector from point A to point B.
  • A position vector starts from the origin point O, for example OA⃗\vec{OA}OA.
  • A scalar is a number multiplying a vector, such as 3 in 3a3a3a.

The most useful route fact is:

AB⃗=OB⃗−OA⃗\vec{AB}=\vec{OB}-\vec{OA}AB=OB−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}=4aOA=4a and OB⃗=3b\vec{OB}=3bOB=3b. Find AB⃗\vec{AB}AB and BA⃗\vec{BA}BA.

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

  1. To travel from A to B, subtract the position vector of A from the position vector of B.

    AB⃗=3b−4a\vec{AB}=3b-4aAB=3b−4a

  2. The reverse vector has the opposite sign.

    BA⃗=4a−3b\vec{BA}=4a-3bBA=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}AO+OB=AB.

2. Points dividing a line in a ratio

If P lies on AB and AP:PB=m:n\text{AP}:\text{PB}=m:nAP:PB=m:n, then P is mm+n\frac{m}{m+n}m+nm​ 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}OP=OA+m+nm​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}OP=m+nn​OA+m+nm​OB
Tip

Opposite weights

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

Example

Finding k from a ratio

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

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

  1. Since AP is 2 out of 3 equal parts, P is 23\frac{2}{3}32​ of the way from A to B.

    OP⃗=OA⃗+23AB⃗\vec{OP}=\vec{OA}+\frac{2}{3}\vec{AB}OP=OA+32​AB

  2. Substitute OA⃗=3a\vec{OA}=3aOA=3a and AB⃗=2b−3a\vec{AB}=2b-3aAB=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}OP​=3a+32​(2b−3a)=3a+34​b−2a=a+34​b​

  3. Factor the expression so it matches the given form.

    OP⃗=13(3a+4b)\vec{OP}=\frac{1}{3}(3a+4b)OP=31​(3a+4b)

  4. Therefore k=13k=\frac{1}{3}k=31​.

Common Mistake

Swapping the ratio

If AP:PB=2:1\text{AP}:\text{PB}=2:1AP:PB=2:1, P is 23\frac{2}{3}32​ of the way from A, not 13\frac{1}{3}31​ 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}AB=λAC.

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

The cleanest method is:

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

Showing three points are collinear

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

  1. 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}OCOMOD​=32​(6a)=4a=21​(6a+4b)=3a+2b=2(4b)=8b​

  2. 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}CM​=OM−OC=(3a+2b)−4a=−a+2b​

  3. 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}CD​=OD−OC=8b−4a=−4a+8b​

  4. Compare the two vectors.

    CD⃗=4CM⃗\vec{CD}=4\vec{CM}CD=4CM

  5. Since CD⃗\vec{CD}CD is a scalar multiple of CM⃗\vec{CM}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}OC=OA+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}=pOP=p and OQ⃗=q\vec{OQ}=qOQ​=q. N is the midpoint of QR. Y lies on PQ extended so that PQ:QY=3:2\text{PQ}:\text{QY}=3:2PQ: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.

  1. Use regular hexagon facts. Since P and Q are adjacent vertices, the next vertex has position vector OR⃗=q−p\vec{OR}=q-pOR=q−p, and the opposite vertex has OT⃗=−q\vec{OT}=-qOT=−q.

  2. 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}ON​=21​(OQ​+OR)=21​(q+q−p)=−21​p+q​

  3. Since PQ:QY=3:2\text{PQ}:\text{QY}=3:2PQ:QY=3:2, the whole vector from P to Y is 53PQ⃗\frac{5}{3}\vec{PQ}35​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}OY​=OP+35​PQ​=p+35​(q−p)=−32​p+35​q​

  4. 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}TNTY​=ON−OT=−21​p+2q=OY−OT=−32​p+38​q​

  5. Show one is a scalar multiple of the other.

    TY⃗=43TN⃗\vec{TY}=\frac{4}{3}\vec{TN}TY=34​TN

  6. 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 aaa and bbb.

Definition

Independent vectors

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

Example

Finding an unknown vector

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

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

  1. Find the position vector of C.

    OC⃗=56(6a)=5a\vec{OC}=\frac{5}{6}(6a)=5aOC=65​(6a)=5a

  2. 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}OD​=32​(6a)+31​(3b)=4a+b​

  3. Let BE⃗=tb\vec{BE}=tbBE=tb, so OE⃗=(3+t)b\vec{OE}=(3+t)bOE=(3+t)b.

  4. Find CD⃗\vec{CD}CD and CE⃗\vec{CE}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}CDCE​=(4a+b)−5a=−a+b=(3+t)b−5a=−5a+(3+t)b​

  5. Since C, D and E are collinear, write CE⃗=λCD⃗\vec{CE}=\lambda\vec{CD}CE=λCD and compare the aaa coefficients.

    −5=−λ⇒λ=5-5=-\lambda \Rightarrow \lambda=5−5=−λ⇒λ=5

  6. Compare the bbb coefficients: 3+t=53+t=53+t=5, so t=2t=2t=2.

  7. Therefore BE⃗=2b\vec{BE}=2bBE=2b.

Exam technique

In the exam

  1. Start by writing position vectors for every important point.
  2. For a ratio, decide whether you are moving from the first point or using the weighted formula.
  3. To prove collinearity, form two vectors with the same starting point and show one is a scalar multiple of the other.
  4. 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}AB quickly from OA⃗\vec{OA}OA and OB⃗\vec{OB}OB?
  • If AP:PB=3:2\text{AP}:\text{PB}=3:2AP: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?

Recap questions

Test yourself with 5 quick questions on this guide. Answer them all correctly to complete it.

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