Geometric Proofs with Vectors

From routes to geometric facts

In a vector proof, the aim is not just to calculate a movement: it is to use that movement to establish a fact about a diagram. A vector such as $\overrightarrow{AB}$ describes the movement from $A$ to $B$. If $\overrightarrow{OA}=\mathbf a$ and $\overrightarrow{OB}=\mathbf b$, going from $A$ back to $O$ and then on to $B$ gives

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

This ‘finish minus start’ idea lets you find a vector between points even when that vector is not labelled on the diagram. You can choose any route, provided it starts and ends at the right points.

What a scalar multiple proves

Multiplying a non-zero vector by a number changes its length but keeps it on the same line of direction. For example, $2\mathbf a$ points the same way as $\mathbf a$ and is twice as long; $-\mathbf a$ points the opposite way. Therefore, if $\overrightarrow{PQ}=k\overrightarrow{RS}$ for a non-zero , the lines and have the same or opposite direction and are parallel.

For a claim about three points, the common point matters too. If $\overrightarrow{AB}=k\overrightarrow{BC}$, both vectors run through , so , and must lie on one straight line: they are . Scalar multiples alone do not show that points on two separate parallel lines are collinear.

Midpoints and ratios become vectors

If $M$ is the midpoint of $OA$, then $\overrightarrow{OM}=\tfrac12\overrightarrow{OA}$. More generally, if lies on and , the whole segment has equal parts, so

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

You can then follow a route through $P$. For instance, with $\overrightarrow{OA}=\mathbf a$, $\overrightarrow{OB}=\mathbf b$ and ,

$\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}=\mathbf a+\tfrac13(\mathbf b-\mathbf a)=\tfrac23\mathbf a+\tfrac13\mathbf b.$

The coefficients also describe where $P$ sits: it is one third of the way from $A$ towards $B$. This is how a geometric ratio becomes an algebraic expression you can compare with another vector.

A proof that two lines are parallel

In triangle $OAB$, let $M$ and $N$ be the midpoints of $OA$ and $OB$. Write and . Then and . Following the route from through to gives

$\overrightarrow{MN}=-\tfrac12\mathbf a+\tfrac12\mathbf b=\tfrac12(\mathbf b-\mathbf a).$

But $\overrightarrow{AB}=\mathbf b-\mathbf a$, so $\overrightarrow{MN}=\tfrac12\overrightarrow{AB}$. This proves that is parallel to and half its length. Notice how the factorised expression reveals the relationship; a diagram alone would only suggest it.

Using collinearity to find an unknown

A Higher-tier problem may give a straight-line condition and ask you to find an unknown scalar. In the Pearson Edexcel sample diagram, $O,M,A$ and $O,N,B$ are straight lines, as are $A,B,C$. The of is , the midpoint of is , and are . You are given , and .

First, $\overrightarrow{OM}=3\mathbf a$. Since $B$ is the midpoint of $AC$, . Now , so

$\overrightarrow{OC}=\overrightarrow{OB}+\overrightarrow{BC}=6\mathbf b+(6\mathbf b-6\mathbf a)=12\mathbf b-6\mathbf a.$

Find both vectors starting at $M$:

$\overrightarrow{MN}=k\mathbf b-3\mathbf a,\qquad \overrightarrow{MC}=12\mathbf b-9\mathbf a.$

Because $M$, $N$ and $C$ are collinear, $\overrightarrow{MN}=t\overrightarrow{MC}$ for some scalar . Comparing the coefficients of gives , hence . Comparing the coefficients of then gives . The same scalar must work for both components: that is what places on the line .