Vectors in two and three dimensions, magnitude and direction, addition and scalar multiplication, position vectors and unit vectors, and geometric applications including collinearity and the midpoint.

What this dot point is asking

OCR wants you to use vectors in two and three dimensions, write them in column form or in $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ notation, find magnitudes and directions, add and subtract vectors and multiply by a scalar, use position vectors to find displacements, find unit vectors, and apply vectors to geometric problems such as proving points are collinear, finding a midpoint, or working with ratios along a line.

The answer

Notation, magnitude and direction

A vector has both magnitude (size) and direction. In two dimensions it is written $\begin{pmatrix} a \\ b \end{pmatrix}$ or $a\mathbf{i} + b\mathbf{j}$; in three dimensions, $a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$. The magnitude is the length, found by Pythagoras.

Adding, subtracting and scaling

Vectors add and subtract component by component, and scalar multiplication scales each component. Geometrically, addition is "nose to tail" and a scalar multiple keeps the direction (or reverses it if negative) while changing the length.

Position vectors and displacements

The position vector of a point is its vector from the origin. The displacement from $A$ to $B$ is "end minus start": $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$.

Examples in context

Collinearity

Three points are collinear (on one straight line) if two displacements between them are parallel, that is one is a scalar multiple of the other, and they share a common point. This is a standard short proof.

Ratios along a line

A point $P$ dividing $AB$ in the ratio $m : n$ has position vector $\mathbf{p} = \mathbf{a} + \dfrac{m}{m + n}(\mathbf{b} - \mathbf{a})$. Setting $m = n$ recovers the midpoint formula.

Try this

Q1. Find a unit vector in the direction of $6\mathbf{i} + 8\mathbf{j}$. [2 marks]

• Cue. $|6\mathbf{i} + 8\mathbf{j}| = 10$, so the unit vector is $\tfrac{1}{10}(6\mathbf{i} + 8\mathbf{j}) = 0.6\mathbf{i} + 0.8\mathbf{j}$.

Q2. Points $A$ and $B$ have position vectors $\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}$ and $4\mathbf{i} + \mathbf{j} - \mathbf{k}$. Find the distance $AB$. [2 marks]

• Cue. $\overrightarrow{AB} = 3\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}$, so $AB = \sqrt{9 + 9 + 16} = \sqrt{34}$.