Writing and drawing column vectors, adding, subtracting and multiplying vectors by a scalar, and using vectors in geometric proofs at Higher tier.

What this dot point is asking

AQA wants you to write and draw column vectors, add and subtract them, multiply by a scalar, and at Higher tier use vectors to prove geometric facts such as parallel lines and collinear points. Vector arithmetic is straightforward; the Higher proof questions are where the marks (and the difficulty) lie, so building a clear route through a diagram is the key skill.

Column vectors

A column vector $\begin{pmatrix} a \\ b \end{pmatrix}$ has a horizontal component $a$ (right is positive, left negative) and a vertical component $b$ (up is positive, down negative). The vector from point $P(1, 2)$ to point $Q(4, 6)$ is $\overrightarrow{PQ} = \begin{pmatrix} 4 - 1 \\ 6 - 2 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}$. The magnitude (length) of this vector, by Pythagoras, is $\sqrt{3^2 + 4^2} = 5$.

Adding, subtracting and scaling

Add and subtract componentwise. Multiplying by a positive scalar stretches the vector along the same direction; multiplying by a negative scalar reverses it. So $-\begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}$, which points the opposite way and has the same length.

Vectors in geometry proofs at Higher tier

In a labelled diagram with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$, the key relationship is $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a}$ (go back to $O$, then out to $B$). Midpoints, ratios along a line, and parallelograms all build from this. To reach a point $M$ that divides $AB$ in a given ratio, write $\overrightarrow{OM} = \overrightarrow{OA} + \text{(fraction)}\,\overrightarrow{AB}$.

A typical proof asks you to show two vectors are parallel. Express each as a multiple of the base vectors, then show one is a scalar multiple of the other. State explicitly that "since $\overrightarrow{XY} = k\,\overrightarrow{XZ}$, the lines are parallel" to earn the conclusion mark.

Position vectors and finding points

A position vector gives a point's location relative to the origin $O$, so the position vector of $A$ is just $\overrightarrow{OA}$. To reach any point, build a route from $O$ using known vectors: a point $P$ that divides $AB$ in the ratio $1 : 2$ has $\overrightarrow{OP} = \overrightarrow{OA} + \tfrac{1}{3}\overrightarrow{AB}$, because $P$ is one third of the way along $AB$. The fraction of the way is the first ratio part over the total, $\tfrac{1}{1 + 2} = \tfrac{1}{3}$. Setting up the route carefully, origin out to the start then a fraction along, is the dependable method for these ratio-point questions.

Magnitude and direction

While most GCSE vector work is geometric, the magnitude of a column vector is found by Pythagoras: $\begin{pmatrix} a \\ b \end{pmatrix}$ has length $\sqrt{a^2 + b^2}$. So the displacement $\begin{pmatrix} 6 \\ 8 \end{pmatrix}$ has magnitude $\sqrt{36 + 64} = 10$. This links vectors directly to Pythagoras and to coordinate geometry, where the distance between two points is the magnitude of the vector joining them. A vector therefore carries both a size (its magnitude) and a direction (the way it points), which is what distinguishes it from a plain number.