Vector Geometry and Proof

1 The Two Results Everything Depends On

Vector proof questions almost always ask you to show that lines are parallel or that points are collinear (lie on a straight line). Both come from the same single idea.

Result 1 — Parallel

If $\overrightarrow{AB} = k\,\overrightarrow{CD}$ for some number $k$,
then $AB$ is parallel to $CD$

Result 2 — Collinear

If $\overrightarrow{AB} = k\,\overrightarrow{BC}$ (sharing the point $B$),
then $A$, $B$ and $C$ lie on a straight line

Parallel vs collinear. Two vectors that are multiples of each other point along the same direction. If they are separate line segments, the lines are parallel. If they share a point, the three points must all lie on one straight line — they are collinear.

2 The Method

Write down the vectors you are given, usually $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$.
Find a route for each vector you need, and express it in terms of $\mathbf{a}$ and $\mathbf{b}$.
Simplify each expression fully.
Compare the two: factorise to show one is a multiple of the other.
State the conclusion clearly, quoting the reason.

Factorising is the whole trick. An answer like $3\mathbf{a} + 6\mathbf{b}$ does not obviously relate to $\mathbf{a} + 2\mathbf{b}$ — until you write it as $3(\mathbf{a} + 2\mathbf{b})$. Always take out the common factor at the end.

Write a proper concluding sentence. Finding that $\overrightarrow{PQ} = 2\overrightarrow{RS}$ scores the method marks, but you must add: "so $PQ$ is parallel to $RS$" (and often "and twice as long") to score the final mark.

3 Proving Lines Are Parallel

Worked Example 1 — The midpoint theorem

In triangle $OAB$, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $M$ is the midpoint of $OA$ and $N$ is the midpoint of $OB$. Prove that $MN$ is parallel to $AB$ and half its length.

①$\overrightarrow{OM} = \tfrac{1}{2}\mathbf{a}$ and $\overrightarrow{ON} = \tfrac{1}{2}\mathbf{b}$.

②$\overrightarrow{MN} = \overrightarrow{MO} + \overrightarrow{ON} = -\tfrac{1}{2}\mathbf{a} + \tfrac{1}{2}\mathbf{b}$

③Factorise: $\overrightarrow{MN} = \tfrac{1}{2}(\mathbf{b} - \mathbf{a})$

④$\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}$

⑤So $\overrightarrow{MN} = \tfrac{1}{2}\overrightarrow{AB}$.

Since $\overrightarrow{MN}$ is a scalar multiple of $\overrightarrow{AB}$, the lines $MN$ and $AB$ are parallel; and the multiple is $\tfrac{1}{2}$, so $MN$ is half the length of $AB$. ∎

This is the midpoint theorem: the line joining the midpoints of two sides of a triangle is parallel to the third side and half its length. Vectors prove it in four lines.

Worked Example 2 — A quadrilateral

In quadrilateral $OABC$, $\overrightarrow{OA} = 2\mathbf{a}$, $\overrightarrow{OB} = 2\mathbf{a} + 4\mathbf{b}$ and $\overrightarrow{OC} = 6\mathbf{b}$. Show that $AB$ is parallel to $OC$.

①$\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -2\mathbf{a} + (2\mathbf{a} + 4\mathbf{b})$

②$= 4\mathbf{b}$

③$\overrightarrow{OC} = 6\mathbf{b}$

④$\overrightarrow{AB} = \tfrac{4}{6}\overrightarrow{OC} = \tfrac{2}{3}\overrightarrow{OC}$

$\overrightarrow{AB}$ is a scalar multiple of $\overrightarrow{OC}$, so $AB \parallel OC$. ∎

Since only one pair of sides is parallel, $OABC$ is a trapezium.

4 Proving Points Are Collinear

The two things you must show

(1) One vector is a multiple of the other, and
(2) they share a common point

Both parts are needed. Two parallel vectors that do not share a point give parallel lines, not collinear points. Always say which point is shared.

Worked Example 3 — Collinear points

$\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$. $P$ is the point with $\overrightarrow{OP} = \mathbf{a} + 2\mathbf{b}$ and $Q$ is the point with $\overrightarrow{OQ} = 3\mathbf{a} + 6\mathbf{b}$. Show that $O$, $P$ and $Q$ are collinear.

①$\overrightarrow{OP} = \mathbf{a} + 2\mathbf{b}$

②$\overrightarrow{OQ} = 3\mathbf{a} + 6\mathbf{b} = 3(\mathbf{a} + 2\mathbf{b})$

③So $\overrightarrow{OQ} = 3\overrightarrow{OP}$.

$\overrightarrow{OQ}$ is a scalar multiple of $\overrightarrow{OP}$, and both start at the common point $O$. Therefore $O$, $P$ and $Q$ lie on the same straight line. ∎

Furthermore $OQ$ is three times as long as $OP$, so $P$ lies one third of the way from $O$ to $Q$.

Worked Example 4 — Collinear with a shared middle point

$\overrightarrow{XY} = 2\mathbf{p} - \mathbf{q}$ and $\overrightarrow{YZ} = 6\mathbf{p} - 3\mathbf{q}$. Show that $X$, $Y$ and $Z$ are collinear, and find the ratio $XY : YZ$.

①Factorise the second: $\overrightarrow{YZ} = 3(2\mathbf{p} - \mathbf{q})$

②So $\overrightarrow{YZ} = 3\overrightarrow{XY}$.

③The two vectors are parallel and share the point $Y$.

Therefore $X$, $Y$ and $Z$ are collinear. ∎

Since $YZ$ is three times $XY$, the ratio $XY : YZ = 1 : 3$.

5 Harder Ratio Problems

Worked Example 5 — Finding a ratio

$OACB$ is a parallelogram with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $D$ is the point on $AB$ with $AD : DB = 1 : 2$. Show that $O$, $D$ and $C$ are not collinear, where $\overrightarrow{OC} = \mathbf{a} + \mathbf{b}$.

①$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$, so $\overrightarrow{AD} = \tfrac{1}{3}(\mathbf{b} - \mathbf{a})$.

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

③Factorise: $\overrightarrow{OD} = \tfrac{1}{3}(2\mathbf{a} + \mathbf{b})$

④$\overrightarrow{OC} = \mathbf{a} + \mathbf{b}$. Is $2\mathbf{a} + \mathbf{b}$ a multiple of $\mathbf{a} + \mathbf{b}$?

⑤For a multiple $k$ we would need $2 = k$ and $1 = k$ at the same time — impossible.

So $\overrightarrow{OD}$ is not a scalar multiple of $\overrightarrow{OC}$, and the three points are not collinear. ∎

Worked Example 6 — Two routes to the same point

In triangle $OAB$, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $M$ is the midpoint of $AB$ and $G$ lies on $OM$ with $OG : GM = 2 : 1$. Show that $\overrightarrow{OG} = \tfrac{1}{3}(\mathbf{a} + \mathbf{b})$.

①$\overrightarrow{OM} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b})$ (the midpoint result).

②$OG : GM = 2 : 1$, so $G$ is $\tfrac{2}{3}$ of the way along $OM$.

③$\overrightarrow{OG} = \tfrac{2}{3}\overrightarrow{OM} = \tfrac{2}{3} \times \tfrac{1}{2}(\mathbf{a} + \mathbf{b})$

④$= \tfrac{1}{3}(\mathbf{a} + \mathbf{b})$ ∎

$G$ is the centroid of the triangle — the point where all three medians meet. The symmetry of $\tfrac{1}{3}(\mathbf{a} + \mathbf{b})$ hints at why all three medians pass through it.

6 Common Mistakes

Mistake 1 — Getting the direction wrong. $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$, not $\mathbf{a} - \mathbf{b}$. Remember: finish minus start.

Mistake 2 — Not factorising. Leaving the answer as $\tfrac{2}{3}\mathbf{a} + \tfrac{1}{3}\mathbf{b}$ hides the structure. Write $\tfrac{1}{3}(2\mathbf{a} + \mathbf{b})$ so the comparison is obvious.

Mistake 3 — Forgetting the conclusion. "$\overrightarrow{PQ} = 2\overrightarrow{RS}$" is not a proof by itself. Finish with "therefore $PQ$ is parallel to $RS$".

Mistake 4 — Confusing parallel with collinear. Say explicitly which point is shared when claiming collinearity.

A reliable habit: for every vector you need, write the route first (e.g. "$\overrightarrow{MN} = \overrightarrow{MO} + \overrightarrow{ON}$") before substituting. It keeps the signs straight.

7 Quick Reference

Parallel

$\overrightarrow{AB} = k\overrightarrow{CD}$ for some scalar $k$.

Collinear

A scalar multiple and a shared point.

Route first

Write $\overrightarrow{MN} = \overrightarrow{MO} + \overrightarrow{ON}$ before substituting.

Finish minus start

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

Midpoint

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

Ratio $m:n$

$\overrightarrow{AP} = \dfrac{m}{m+n}\overrightarrow{AB}$.

Factorise

Always take out the common factor to reveal the multiple.

Conclude

Write the sentence: "therefore … is parallel to …".

Not a multiple?

If two coefficients demand different values of $k$, no multiple exists.

8 Practice Questions

Question 1

$\overrightarrow{AB} = 3\mathbf{a} + 6\mathbf{b}$ and $\overrightarrow{CD} = \mathbf{a} + 2\mathbf{b}$. Show that $AB$ is parallel to $CD$ and state the ratio of their lengths.

▶ Show solution

$\overrightarrow{AB} = 3\mathbf{a} + 6\mathbf{b} = 3(\mathbf{a} + 2\mathbf{b}) = 3\overrightarrow{CD}$

$\overrightarrow{AB}$ is a scalar multiple of $\overrightarrow{CD}$, so $AB \parallel CD$.

The multiple is $3$, so $AB : CD = 3 : 1$.

Question 2

$\overrightarrow{OP} = 2\mathbf{a} - \mathbf{b}$ and $\overrightarrow{OQ} = 8\mathbf{a} - 4\mathbf{b}$. Show that $O$, $P$ and $Q$ are collinear.

▶ Show solution

$\overrightarrow{OQ} = 8\mathbf{a} - 4\mathbf{b} = 4(2\mathbf{a} - \mathbf{b}) = 4\overrightarrow{OP}$

$\overrightarrow{OQ}$ is a scalar multiple of $\overrightarrow{OP}$, and both vectors start at the shared point $O$.

Therefore $O$, $P$ and $Q$ are collinear. ∎

Question 3

Are $\overrightarrow{XY} = 4\mathbf{a} + 6\mathbf{b}$ and $\overrightarrow{ZW} = 6\mathbf{a} + 8\mathbf{b}$ parallel? Justify your answer.

▶ Show solution

For them to be parallel we would need $\overrightarrow{ZW} = k\overrightarrow{XY}$.

Comparing $\mathbf{a}$: $6 = 4k$, so $k = 1.5$.

Comparing $\mathbf{b}$: $8 = 6k$, so $k = \tfrac{4}{3}$.

The two values of $k$ disagree, so no, they are not parallel.

Question 4

In triangle $OAB$, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $C$ is the midpoint of $OA$ and $D$ is the midpoint of $OB$. Find $\overrightarrow{CD}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

▶ Show solution

$\overrightarrow{OC} = \tfrac{1}{2}\mathbf{a}$ and $\overrightarrow{OD} = \tfrac{1}{2}\mathbf{b}$.

$\overrightarrow{CD} = \overrightarrow{CO} + \overrightarrow{OD} = -\tfrac{1}{2}\mathbf{a} + \tfrac{1}{2}\mathbf{b}$

$= \tfrac{1}{2}(\mathbf{b} - \mathbf{a})$

Question 5

$\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$, and $P$ lies on $AB$ with $AP : PB = 3 : 1$. Find $\overrightarrow{OP}$.

▶ Show solution

Total parts $= 4$, so $\overrightarrow{AP} = \tfrac{3}{4}(\mathbf{b} - \mathbf{a})$.

$\overrightarrow{OP} = \mathbf{a} + \tfrac{3}{4}(\mathbf{b} - \mathbf{a}) = \tfrac{1}{4}\mathbf{a} + \tfrac{3}{4}\mathbf{b}$

$= \tfrac{1}{4}(\mathbf{a} + 3\mathbf{b})$

Question 6

$OABC$ is a parallelogram with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. $M$ is the midpoint of $AB$. Show that $\overrightarrow{OM} = \mathbf{a} + \tfrac{1}{2}\mathbf{c}$.

▶ Show solution

In a parallelogram, $\overrightarrow{AB} = \overrightarrow{OC} = \mathbf{c}$.

$M$ is the midpoint of $AB$, so $\overrightarrow{AM} = \tfrac{1}{2}\mathbf{c}$.

$\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \tfrac{1}{2}\mathbf{c}$ ∎

Question 7

$\overrightarrow{PQ} = \mathbf{u} + 3\mathbf{v}$ and $\overrightarrow{QR} = 2\mathbf{u} + 6\mathbf{v}$. Show that $P$, $Q$ and $R$ are collinear, and find $PQ : QR$.

▶ Show solution

$\overrightarrow{QR} = 2\mathbf{u} + 6\mathbf{v} = 2(\mathbf{u} + 3\mathbf{v}) = 2\overrightarrow{PQ}$

The vectors are scalar multiples and share the point $Q$.

Therefore $P$, $Q$ and $R$ are collinear, and $PQ : QR = 1 : 2$. ∎

Question 8

In triangle $OAB$, $\overrightarrow{OA} = 3\mathbf{a}$ and $\overrightarrow{OB} = 3\mathbf{b}$. $P$ lies on $OA$ with $OP : PA = 1 : 2$, and $Q$ lies on $OB$ with $OQ : QB = 1 : 2$. Prove that $PQ$ is parallel to $AB$.

▶ Show solution

$OP : PA = 1 : 2$ means $P$ is $\tfrac{1}{3}$ of the way along $OA$, so $\overrightarrow{OP} = \tfrac{1}{3}(3\mathbf{a}) = \mathbf{a}$.

Similarly $\overrightarrow{OQ} = \mathbf{b}$.

$\overrightarrow{PQ} = \overrightarrow{PO} + \overrightarrow{OQ} = -\mathbf{a} + \mathbf{b} = \mathbf{b} - \mathbf{a}$

$\overrightarrow{AB} = -3\mathbf{a} + 3\mathbf{b} = 3(\mathbf{b} - \mathbf{a})$

So $\overrightarrow{AB} = 3\overrightarrow{PQ}$: a scalar multiple, therefore $PQ \parallel AB$ (and $AB$ is three times as long). ∎

Question 9

$ABCD$ is a quadrilateral. $P$, $Q$, $R$ and $S$ are the midpoints of $AB$, $BC$, $CD$ and $DA$ respectively. Let $\overrightarrow{AB} = 2\mathbf{p}$, $\overrightarrow{BC} = 2\mathbf{q}$ and $\overrightarrow{CD} = 2\mathbf{r}$.

(a) Find $\overrightarrow{PQ}$. (b) Find $\overrightarrow{SR}$. (c) What does this prove about $PQRS$?

▶ Show solution

(a) $\overrightarrow{PB} = \tfrac{1}{2}\overrightarrow{AB} = \mathbf{p}$ and $\overrightarrow{BQ} = \tfrac{1}{2}\overrightarrow{BC} = \mathbf{q}$.

$\overrightarrow{PQ} = \mathbf{p} + \mathbf{q}$

(b) $\overrightarrow{DA} = -(\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD}) = -(2\mathbf{p} + 2\mathbf{q} + 2\mathbf{r})$, since the four sides return to the start.

Take the route $\overrightarrow{SR} = \overrightarrow{SD} + \overrightarrow{DR}$.

$\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} = 2\mathbf{p} + 2\mathbf{q} + 2\mathbf{r}$, and $S$ is the midpoint of $AD$, so

$\overrightarrow{SD} = \tfrac{1}{2}\overrightarrow{AD} = \mathbf{p} + \mathbf{q} + \mathbf{r}$

$\overrightarrow{DR} = -\tfrac{1}{2}\overrightarrow{CD} = -\mathbf{r}$

$\overrightarrow{SR} = (\mathbf{p} + \mathbf{q} + \mathbf{r}) - \mathbf{r} = \mathbf{p} + \mathbf{q}$

(c) $\overrightarrow{PQ} = \overrightarrow{SR}$, so $PQ$ and $SR$ are equal in length and parallel.

A quadrilateral with one pair of sides equal and parallel is a parallelogram, so $PQRS$ is a parallelogram — whatever shape $ABCD$ was. ∎

Question 10

In triangle $OAB$, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $M$ is the midpoint of $OB$. The point $N$ lies on $AB$ with $AN : NB = 2 : 1$.

(a) Find $\overrightarrow{AM}$. (b) Find $\overrightarrow{AN}$. (c) Find $\overrightarrow{MN}$, fully simplified. (d) The point $X$ has $\overrightarrow{OX} = \tfrac{2}{3}\mathbf{a} + \mathbf{b}$. Determine whether $M$, $N$ and $X$ are collinear.

▶ Show solution

(a) $\overrightarrow{OM} = \tfrac{1}{2}\mathbf{b}$, so $\overrightarrow{AM} = \overrightarrow{AO} + \overrightarrow{OM} = -\mathbf{a} + \tfrac{1}{2}\mathbf{b}$.

(b) $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$, and $N$ is $\tfrac{2}{3}$ of the way along.

$\overrightarrow{AN} = \tfrac{2}{3}(\mathbf{b} - \mathbf{a})$

(c) $\overrightarrow{MN} = \overrightarrow{MA} + \overrightarrow{AN} = \left(\mathbf{a} - \tfrac{1}{2}\mathbf{b}\right) + \tfrac{2}{3}\mathbf{b} - \tfrac{2}{3}\mathbf{a}$

$= \tfrac{1}{3}\mathbf{a} + \tfrac{1}{6}\mathbf{b} = \tfrac{1}{6}(2\mathbf{a} + \mathbf{b})$

(d) $\overrightarrow{ON} = \overrightarrow{OA} + \overrightarrow{AN} = \mathbf{a} + \tfrac{2}{3}\mathbf{b} - \tfrac{2}{3}\mathbf{a} = \tfrac{1}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{b}$

$\overrightarrow{NX} = \overrightarrow{OX} - \overrightarrow{ON} = \left(\tfrac{2}{3}\mathbf{a} + \mathbf{b}\right) - \left(\tfrac{1}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{b}\right) = \tfrac{1}{3}\mathbf{a} + \tfrac{1}{3}\mathbf{b}$

$= \tfrac{1}{3}(\mathbf{a} + \mathbf{b})$

Compare with $\overrightarrow{MN} = \tfrac{1}{6}(2\mathbf{a} + \mathbf{b})$. For these to be parallel we would need $\mathbf{a} + \mathbf{b} = k(2\mathbf{a} + \mathbf{b})$, giving $1 = 2k$ and $1 = k$ — a contradiction.

So $\overrightarrow{NX}$ is not a multiple of $\overrightarrow{MN}$, and $M$, $N$ and $X$ are not collinear.

Vector Geometry & Proof (G25) · GCSE Maths Revision · Created with MathJax