Vector geometry

Vectors as directed line segments Extended

A vector can be drawn as a directed line segment: a line with an arrow showing its direction. Its length shows the magnitude. In vector geometry questions the figure is labelled with letters such as $\mathbf{a}$ and $\mathbf{b}$ on some of its sides, and you express other lines in terms of them.

Three facts do most of the work.

• Equal vectors. Any line with the same length and direction is the same vector, wherever it is. In a parallelogram OPQR with $\overrightarrow{OP}=\mathbf{a}$, the opposite side $\overrightarrow{RQ}$ is also $\mathbf{a}$.
• Reversing. Going along a line against its arrow changes the sign: $\overrightarrow{PO}=-\mathbf{a}$.
• Multiples. A line parallel to $\mathbf{a}$ and $k$ times as long, in the same direction, is $k\mathbf{a}$. Half of $\mathbf{a}$ is $\frac{1}{2}\mathbf{a}$.

On a grid, a directed line segment is drawn from its starting point by counting the components: $\left(\begin{array}{c}3\\ -1\end{array}\right)$ is drawn 3 squares right and 1 down, with the arrow at the end.

Routes and position vectors Extended

To find the vector between two points in a figure, choose a route through points whose vectors you know, and add the vectors along it. Any route gives the same answer, because each one describes the same total movement.

For example, in triangle OAB with $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$:
$$\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=-\mathbf{a}+\mathbf{b}=\mathbf{b}-\mathbf{a}$$

The vector from the origin O to a point is called the position vector of the point. If A and B have position vectors $\mathbf{a}$ and $\mathbf{b}$, then
$$\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$$
that is, the position vector of the end minus the position vector of the start. This matches subtracting coordinates: a position vector $\left(\begin{array}{c}x\\ y\end{array}\right)$ points to the point $(x,y)$.

Always simplify your final answer by collecting the $\mathbf{a}$ terms and the $\mathbf{b}$ terms.

In parallelogram OPQR the opposite sides are equal vectors, so $\overrightarrow{OQ}=\mathbf{a}+\mathbf{b}$ by either route, O to P to Q or O to R to Q.

Points that divide a line in a ratio Extended

If P lies on AB with $AP:PB=m:n$, then AB is split into $m+n$ equal parts and AP is $m$ of them:
$$\overrightarrow{AP}=\frac{m}{m+n}\overrightarrow{AB}$$

So the position vector of P is $\overrightarrow{OP}=\overrightarrow{OA}+\frac{m}{m+n}\overrightarrow{AB}$.

The most common case is the midpoint M of AB:
$$\overrightarrow{OM}=\mathbf{a}+\frac{1}{2}(\mathbf{b}-\mathbf{a})=\frac{1}{2}\mathbf{a}+\frac{1}{2}\mathbf{b}=\frac{1}{2}(\mathbf{a}+\mathbf{b})$$

Read ratios carefully. $AP:PB=2:3$ means AP is $\frac{2}{5}$ of AB, not $\frac{2}{3}$. And notice which end you start from: P is nearer A when $m$ is smaller than $n$.

Showing that lines are parallel Extended

Two vectors are parallel when one is a number times the other. To show that two lines are parallel:

1. Find both vectors in terms of $\mathbf{a}$ and $\mathbf{b}$, simplified.
2. Take out a common factor so the same bracket appears in both, for example $\overrightarrow{PQ}=2\mathbf{b}-2\mathbf{a}=2(\mathbf{b}-\mathbf{a})$ and $\overrightarrow{XY}=\mathbf{b}-\mathbf{a}$.
3. Write the conclusion: $\overrightarrow{PQ}=2\overrightarrow{XY}$, so PQ is parallel to XY, and PQ is twice as long.

The number also gives the ratio of the lengths: here $XY:PQ=1:2$. A "show that" answer needs the statement at the end; working alone is not enough.

Showing that three points are collinear Extended

Points are collinear when they lie on one straight line. To show that A, B and C are collinear:

1. Show that $\overrightarrow{AB}$ and $\overrightarrow{BC}$ (or $\overrightarrow{AC}$) are multiples of each other, so the lines are parallel.
2. Point out that the two lines share a point, B.
3. Conclude: parallel lines through a common point are the same line, so A, B and C are collinear.

Both parts of the argument are needed. Parallel on its own is not enough, because two parallel lines can be separate, like the opposite sides of a parallelogram. The multiple also gives a ratio: if $\overrightarrow{BC}=3\overrightarrow{AB}$, then $AB:BC=1:3$.

Ratio and similarity problems Extended

When a line in a triangle is parallel to one side, it cuts off a smaller similar triangle. If C is on OA with $OC=\frac{2}{5}OA$, and CD is parallel to AB with D on OB, then triangle OCD is an enlargement of triangle OAB with scale factor $\frac{2}{5}$ from O. So every vector in the small triangle is $\frac{2}{5}$ of the matching vector in the large one: $\overrightarrow{OD}=\frac{2}{5}\mathbf{b}$ and $\overrightarrow{CD}=\frac{2}{5}(\mathbf{b}-\mathbf{a})$.

Harder questions ask where two lines cross. Call the crossing point X and reach it by two routes, each with an unknown number: for example $\overrightarrow{OX}=\mathbf{a}+\lambda \overrightarrow{AN}$ and $\overrightarrow{OX}=\mathbf{b}+\mu \overrightarrow{BM}$. Because $\mathbf{a}$ and $\mathbf{b}$ are not parallel, the only way two expressions $p\mathbf{a}+q\mathbf{b}$ can be equal is for the coefficients of $\mathbf{a}$ to match and the coefficients of $\mathbf{b}$ to match. That gives two simultaneous equations for $\lambda$ and $\mu$.