5.1 - Vector geometry and proofs

Use directed routes and scalar multiples to prove relationships between points and lines. You will find vectors to points dividing sides, prove parallelism and collinearity, and use a multiplier to recover a length ratio.

Routes and points on a side (Higher)

Do two medians of a triangle always meet two thirds of the way from each vertex? A neat drawing suggests they might, but one drawing cannot prove a statement about every triangle. We will build a proof from directed movements.

You already know

A backwards arrow changes sign. Vectors add along a joined route. In a ratio $2:3$ , the whole has 5 equal parts, so the first part is $\frac25$ of the whole.

Begin with a concrete route. If $\overrightarrow{OA}=\begin{pmatrix}6\\0\end{pmatrix}$ and $\overrightarrow{OB}=\begin{pmatrix}1\\5\end{pmatrix}$ , going from $A$ back to $O$ , then on to $B$ , gives $\overrightarrow{AB}=\begin{pmatrix}-5\\5\end{pmatrix}$ . We can use exactly the same reasoning with letters, so the result applies to many triangles at once.

Position vector

The position vector of $P$ , relative to origin $O$ , is $\overrightarrow{OP}$ . If $\overrightarrow{OA}=\mathbf a$ and $\overrightarrow{OB}=\mathbf b$ , then $\overrightarrow{AB}=\mathbf b-\mathbf a$ .

Write a route before simplifying. The endpoint letters help you see whether each arrow goes forwards or backwards. If $P$ is on $AB$ , its position vector is the route $O$ to $A$ to $P$ .

Internal division of a side

If $AP:PB=m:n$ , with $m,n>0$ , then

\overrightarrow{OP}=\mathbf a+\frac{m}{m+n}(\mathbf b-\mathbf a)

The fraction multiplies $\overrightarrow{AB}$ , not $\overrightarrow{OB}$ . Not in the Higher formulae sheet: learn the route method.

A midpoint is the simplest internal division, using one half of the side from each end.

Worked example 1: a midpoint position vector

Question. $\overrightarrow{OA}=\mathbf a$ , $\overrightarrow{OB}=\mathbf b$ , and $M$ is the midpoint of $AB$ . Find $\overrightarrow{OM}$ .

Plan. Go to $A$ , then travel half of $AB$ .

\begin{aligned} \overrightarrow{AB}&=\mathbf b-\mathbf a &&\text{route through }O\\ \overrightarrow{OM}&=\mathbf a+\tfrac12(\mathbf b-\mathbf a) &&\text{halfway along }AB\\ &=\tfrac12\mathbf a+\tfrac12\mathbf b &&\text{collect terms} \end{aligned}

Why add $\mathbf a$ ? The half-side movement starts at $A$ , whereas the position vector starts at $O$ .

Answer. $\overrightarrow{OM}=\frac12(\mathbf a+\mathbf b)$ .

Check. Starting from $B$ gives $\mathbf b+\frac12(\mathbf a-\mathbf b)$ , the same expression.

Change only the division ratio from $1:1$ to $1:2$ . The route is unchanged, but its side fraction becomes $\frac13$ .

Worked example 2: unequal division

Question. For the same endpoint vectors, $P$ lies on $AB$ with $AP:PB=1:2$ . Find $\overrightarrow{OP}$ .

Plan. Travel one third of $AB$ after reaching $A$ .

\begin{aligned} \overrightarrow{AP}&=\tfrac13(\mathbf b-\mathbf a) &&\text{one part out of three}\\ \overrightarrow{OP}&=\mathbf a+\tfrac13(\mathbf b-\mathbf a) &&\text{join the route}\\ &=\tfrac23\mathbf a+\tfrac13\mathbf b &&\text{collect terms} \end{aligned}

Why not $\frac12$ ? The ratio compares two pieces; the required fraction compares the first piece with the whole.

Answer. $\overrightarrow{OP}=\frac23\mathbf a+\frac13\mathbf b$ .

Check. $\overrightarrow{PB}=\mathbf b-\overrightarrow{OP}=\frac23(\mathbf b-\mathbf a)=2\overrightarrow{AP}$ .

For an internal point, the coefficients of $\mathbf a$ and $\mathbf b$ are positive and add to 1. This is a useful algebra check, though it does not replace the route.

Watch out

For $AP:PB=1:2$ , the line $\overrightarrow{AP}=\frac12\overrightarrow{AB}$ incorrectly uses the ratio as a fraction of the whole. Use $\frac1{1+2}=\frac13$ .

Change the ratio again, keeping the same two endpoint position vectors.

Proving lines parallel (Higher)

Two nonzero arrows with the same direction, or opposite directions, have parallel supporting lines. The algebraic test is that one vector is a scalar multiple of the other. Equal lengths alone are insufficient: an eastward arrow and a northward arrow may have equal lengths.

Parallel vectors

Nonzero vectors $\mathbf u$ and $\mathbf v$ are parallel if $\mathbf u=k\mathbf v$ for a nonzero scalar $k$ . The sign tells you whether their directions agree or oppose.

A proof must find both directed vectors, show the multiple and state the geometrical conclusion. No vector-proof rule is supplied in the Higher formulae sheet. Keep a useful bracket such as $\mathbf b-\mathbf a$ together so the multiple is visible.

Worked example 3: the midpoint theorem

Question. In a nondegenerate triangle $ABC$ , let $\overrightarrow{AB}=\mathbf a$ and $\overrightarrow{AC}=\mathbf b$ . Points $P$ and $Q$ are the midpoints of $AB$ and $AC$ . Prove $PQ\parallel BC$ and find the length ratio $PQ:BC$ .

Plan. Compare the routes $P$ to $A$ to $Q$ , and $B$ to $A$ to $C$ .

\begin{aligned} \overrightarrow{PQ}&=-\tfrac12\mathbf a+\tfrac12\mathbf b &&\text{back to }A\text{, then to }Q\\ &=\tfrac12(\mathbf b-\mathbf a) &&\text{factorise}\\ \overrightarrow{BC}&=-\mathbf a+\mathbf b &&\text{route through }A\\ \overrightarrow{PQ}&=\tfrac12\overrightarrow{BC} &&\text{compare} \end{aligned}

Why are these vectors nonzero? A nondegenerate triangle has distinct vertices and noncollinear sides, so $\mathbf b-\mathbf a\ne\mathbf0$ .

Answer. The vectors are nonzero scalar multiples, so $PQ\parallel BC$ , with $PQ:BC=1:2$ .

Check. Both points are halfway along their sides, so the common factor $\frac12$ fits both side movements. The proof used arbitrary $\mathbf a,\mathbf b$ , so it applies to every such triangle.

Change only the common fraction along the two sides: each point is now one third of the way from $A$ . The same proof should adapt without assuming a special triangle.

Worked example 4: equal fractions along two sides

Question. In the same general triangle, let $AP=\frac13 AB$ and $AQ=\frac13 AC$ , with $P$ and $Q$ on those sides. Prove $PQ\parallel BC$ and find $PQ:BC$ .

Plan. Use the same two routes with the new common fraction.

\begin{aligned} \overrightarrow{PQ}&=-\tfrac13\mathbf a+\tfrac13\mathbf b &&\text{route }P\text{ to }A\text{ to }Q\\ &=\tfrac13(\mathbf b-\mathbf a) &&\text{factorise}\\ &=\tfrac13\overrightarrow{BC} &&\text{use }\overrightarrow{BC}=\mathbf b-\mathbf a \end{aligned}

Why does equal fraction matter? It produces one common factor multiplying the entire side vector.

Answer. $PQ\parallel BC$ and $PQ:BC=1:3$ , since the compared nonzero vectors have multiplier $\frac13$ .

Check. Reversing both arrows gives $\overrightarrow{QP}=\frac13\overrightarrow{CB}$ , the same relationship.

If the fractions along the two sides differ, you cannot simply claim the join is parallel. You must calculate and compare its vector.

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Number

Algebra

Shape, space and measure

Handling data

5.1 - Vector geometry and proofs

Routes and points on a side (Higher)

Proving lines parallel (Higher)

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