Vector Geometry

A vector is an arrow, not just a number

A vector has a size and a direction, so one number cannot describe it. The vector from A to B is written $\overrightarrow{AB}$: the arrow over the two letters fixes the direction, and the order of the letters is part of the meaning. Going the other way is the same journey reversed, so $\overrightarrow{BA}=-\overrightarrow{AB}$.

On a grid a vector is written as a column vector, $\left(\begin{array}{c}x\\ y\end{array}\right)$, where $x$ is the movement right and $y$ is the movement up. So $\overrightarrow{AB}=\left(\begin{array}{c}3\\ -2\end{array}\right)$ means 3 right and 2 down. When a vector is not tied to particular points, Cambridge prints it in bold as a single letter, $𝐚$ or $𝐛$, and calls it a free vector.

Two vectors are equal when they have the same length and the same direction, wherever they are drawn. In a parallelogram OABC the sides $\overrightarrow{AB}$ and $\overrightarrow{OC}$ are equal vectors, and so are $\overrightarrow{BC}$ and $\overrightarrow{OA}$. That single fact is what lets you replace one side of a parallelogram with the vector named on the opposite side. This is Extended paper work: Core candidates only ever meet a vector as the translation $\left(\begin{array}{c}x\\ y\end{array}\right)$.

• $\overrightarrow{AB}=\left(\begin{array}{c}3\\ -2\end{array}\right)$ means 3 right and 2 down
• $|𝐚|$ and $|\overrightarrow{AB}|$ both mean the length of the vector, its magnitude
• The magnitude of $\left(\begin{array}{c}3\\ 4\end{array}\right)$ is $\sqrt{3^2+4^2}=5$
• $\overrightarrow{AA}=\mathbf{0}$, the zero vector: no length, no direction

Adding and subtracting vectors: the route rule

Vectors add nose to tail. To get from A to C you may travel from A to B and then from B to C, so $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$. The middle letter cancels, exactly as the arrows join up. Any route from the start to the finish gives the same vector, so a longer chain such as $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BD}+\overrightarrow{DC}$ is just as valid.

Subtraction is the rule that turns position vectors into a segment. If O is the origin, $\overrightarrow{OA}=𝐚$ and $\overrightarrow{OB}=𝐛$, then $\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=-𝐚+𝐛=𝐛-𝐚$. In words: the vector from start to finish is the position of the finish minus the position of the start. Say it that way and the sign looks after itself.

Multiplying by a scalar stretches a vector without changing its line. $k𝐚$ is $|k|$ times as long as $𝐚$ and points the same way when $k$ is positive, the opposite way when $k$ is negative. So $\frac{1}{2}𝐚$ is half as long, $3𝐚$ is three times as long, and $-𝐛$ is $𝐛$ reversed.

Expressing a segment in terms of $𝐚$ and $𝐛$

Extended questions almost always hand you two vectors, $𝐚$ and $𝐛$, and ask for a third in terms of them. The method never changes: pick a route from the start of the segment to its end, travel only along vectors you already know, and reverse any arrow you travel against.

Take the parallelogram OABC with $\overrightarrow{OA}=𝐚$ and $\overrightarrow{OC}=𝐛$. Because the opposite sides of a parallelogram are equal vectors, $\overrightarrow{AB}=\overrightarrow{OC}=𝐛$ and $\overrightarrow{BC}=\overrightarrow{OA}=𝐚$. Now walk the routes. $\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}=𝐚+𝐛$, the diagonal from O. And $\overrightarrow{AC}=\overrightarrow{AO}+\overrightarrow{OC}=-𝐚+𝐛=𝐛-𝐚$, the diagonal from A to C.

Collect like terms at the end, but remember that $𝐚$ and $𝐛$ are not like terms. $2𝐚+3𝐛$ cannot be shortened to $5𝐚𝐛$ or to $5(𝐚+𝐛)$: the two vectors point in different directions, so they stay separate for ever.

• $\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}=𝐚+𝐛$, the diagonal from O
• $\overrightarrow{AC}=\overrightarrow{AO}+\overrightarrow{OC}=𝐛-𝐚$, the diagonal from A to C
• $\overrightarrow{AB}=𝐛$, because AB and OC are opposite sides of the parallelogram
• $\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=𝐛-(𝐚+𝐛)=-𝐚$, which is $\overrightarrow{CB}$ reversed

Parallel vectors are scalar multiples

Two vectors are parallel exactly when one is a scalar multiple of the other: $\overrightarrow{PQ}=k\overrightarrow{RS}$ for some number $k$. The scalar does the whole job. If $k=2$ the first is twice as long and points the same way; if $k=-3$ it is three times as long and points the other way; either way the two lie along the same direction.

So a parallel question is really a hunt for the multiple. Work out both vectors in terms of $𝐚$ and $𝐛$, then look for the factor. If $\overrightarrow{PQ}=\frac{1}{2}𝐚+\frac{1}{2}𝐛$ and $\overrightarrow{OB}=𝐚+𝐛$, then $\overrightarrow{PQ}=\frac{1}{2}\overrightarrow{OB}$, so PQ is parallel to OB and half as long. The same scalar gives the length ratio directly: $PQ:OB=1:2$.

Two warnings. Equal lengths on their own prove nothing - $\overrightarrow{AB}$ and $\overrightarrow{DC}$ can be the same length and point opposite ways, and then $\overrightarrow{AB}=-\overrightarrow{DC}$. And the conclusion needs words: the algebra earns the mark for finding the multiple, but the sentence 'so PQ is parallel to OB' is what actually answers the question.

Collinear points

Three points are collinear when they lie on one straight line. With vectors there are two things to show: that two of the segments are parallel, and that those segments share a point. Writing $\overrightarrow{AB}=k\overrightarrow{BC}$ does both at once, because both vectors are written with B, so B is on both segments and the two lines must be the same line.

For example, A, B and C with position vectors $𝐚$, $𝐚+𝐛$ and $𝐚+3𝐛$. Then $\overrightarrow{AB}=(𝐚+𝐛)-𝐚=𝐛$ and $\overrightarrow{BC}=(𝐚+3𝐛)-(𝐚+𝐛)=2𝐛$. Since $\overrightarrow{BC}=2\overrightarrow{AB}$, the segments are parallel and they share the point B, so A, B and C lie on a straight line, with $AB:BC=1:2$.

The shared point is the step students skip. Showing $\overrightarrow{AB}=2\overrightarrow{CD}$ proves that the line AB is parallel to the line CD and nothing more; four points arranged like that are not collinear. The scalar also tells you which point sits in the middle: with $\overrightarrow{AB}=k\overrightarrow{BC}$ and $k$ positive, B is between A and C.

• Parallel: $\overrightarrow{AB}=k\overrightarrow{CD}$, no shared point needed
• Collinear: $\overrightarrow{AB}=k\overrightarrow{BC}$, and the shared point B is essential
• The scalar gives the ratio of the lengths: $k=2$ means $AB:BC=1:2$
• A negative scalar still means parallel and collinear, but the points are then on opposite sides of the shared point

Midpoints and the ratio in which a point divides a segment

The midpoint M of AB has position vector $\overrightarrow{OM}=\frac{1}{2}(𝐚+𝐛)$ when $\overrightarrow{OA}=𝐚$ and $\overrightarrow{OB}=𝐛$. It is the average of the two position vectors, which is why the coordinates of a midpoint are the averages of the coordinates. Halving only one of them is the classic slip: $\frac{1}{2}𝐚+𝐛$ is not the midpoint of AB.

For any other division, turn the ratio into a fraction of the journey. If P divides AB with $AP:PB=m:n$, then $\overrightarrow{AP}=\frac{m}{m+n}\overrightarrow{AB}$, and so $\overrightarrow{OP}=\frac{n}{m+n}𝐚+\frac{m}{m+n}𝐛$. The two fractions swap over: the bigger share of the journey from A belongs with $𝐛$, because P is then nearer to B. With $AP:PB=2:1$, $\overrightarrow{AP}=\frac{2}{3}\overrightarrow{AB}$ and $\overrightarrow{OP}=\frac{1}{3}𝐚+\frac{2}{3}𝐛$.

Working backwards is just as common. A question gives $\overrightarrow{OR}=\frac{2}{5}𝐚+\frac{3}{5}𝐛$ and asks for the ratio $AR:RB$. Put R on the line by writing $\overrightarrow{OR}=(1-t)𝐚+t𝐛$, read off $t=\frac{3}{5}$, and then $AR:RB=t:(1-t)=3:2$. For any point on AB the two coefficients add to 1, which is the quickest check that your answer can be right.

Putting it together: a vector proof, and a length

A vector proof has four steps, and writing them in that order is what earns the marks. Write down the vectors you are given and the parallelogram, midpoint or ratio facts you can use; express the two segments the question is about in terms of $𝐚$ and $𝐛$, one line each; find the scalar that links them; then state the geometric conclusion in words.

If the proof also needs a length, the magnitude formula finishes the job: the vector $\left(\begin{array}{c}x\\ y\end{array}\right)$ has magnitude $\sqrt{x^2+y^2}$. A question that ends 'so OM = 5' is doing two jobs, and both are marked, so never stop at the vector when a length was asked for.

Similarity falls out of the same work. If $\overrightarrow{PQ}=\frac{1}{2}\overrightarrow{OB}$, then triangle APQ is an enlargement of triangle AOB with scale factor $\frac{1}{2}$ about A, so every length in APQ is half the matching length in AOB and the areas are in the ratio $1:4$. Ratios of vectors give ratios of lengths directly; areas go as the square of the scale factor.

• Mark the arrows on your own sketch before you write anything
• Write every new vector on its own line, in terms of $𝐚$ and $𝐛$ only
• Finish with the scalar multiple and then the sentence that names the geometric fact
• Coefficients of $𝐚$ and $𝐛$ never merge: they are different directions
• In a show that question the answer is printed, so the marks are all in the working and the last line must land exactly on the printed expression