Represent vectors in column form, add and subtract them and multiply by a scalar, and find the magnitude of a vector

What this dot point is asking

SEAB wants you to represent vectors in column form, add and subtract them, multiply a vector by a scalar, and find a vector's magnitude. Vectors describe quantities with both size and direction, and at O-Level the arithmetic is done component by component.

The answer

What a vector is

A vector has both magnitude (size) and direction, unlike a scalar which has size only. In two dimensions a vector is written as a column:

where $x$ is the movement in the horizontal direction and $y$ the movement in the vertical direction.

Adding and subtracting vectors

Add or subtract vectors component by component, the top with the top and the bottom with the bottom:

Geometrically, adding vectors places them nose to tail, and the result is the single vector from start to finish.

Multiplying by a scalar

Multiplying a vector by a number (scalar) multiplies each component:

This changes the vector's length by the factor $k$ and reverses its direction when $k$ is negative.

Magnitude

The magnitude (length) of a vector is found by Pythagoras from its components:

The magnitude is always a non-negative number, since it is a length.

The vector between two points

A vector linking two points is found by subtracting their position vectors, "destination minus origin". The vector from $A(x_1, y_1)$ to $B(x_2, y_2)$ is $\overrightarrow{AB} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix}$, and its magnitude is the distance $AB$, which is exactly the distance formula in disguise. So from $A(1, 2)$ to $B(4, 6)$, the vector is $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$ with magnitude $5$. Remembering "tip minus tail" both gives the correct direction and ties vectors directly to the coordinate-geometry distance you already know.

Recognising parallel vectors

Two vectors are parallel when one is a scalar multiple of the other, $\mathbf{b} = k\mathbf{a}$ for some number $k$. So $\begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $\begin{pmatrix} 4 \\ 6 \end{pmatrix}$ are parallel because the second is twice the first, and a negative $k$ means they point in opposite directions. This test is the vector equivalent of equal gradients for parallel lines, and it lets you prove points are collinear: if $\overrightarrow{AB}$ is a scalar multiple of $\overrightarrow{AC}$, then $A$, $B$ and $C$ lie on one straight line. Checking for a common scalar factor is the standard way to detect parallel or collinear vectors.

Examples in context

Example 1. Displacement on a grid. Moving $3$ east and $4$ north is the vector $\begin{pmatrix} 3 \\ 4 \end{pmatrix}$, whose magnitude $5$ is the straight-line distance from start to finish. Vectors capture both how far and in which direction.

Example 2. Combining journeys. Two consecutive moves add as vectors, nose to tail, and the resultant is the single equivalent displacement. This is how navigation combines successive legs into a net change of position.

Try this

Q1. Find $\begin{pmatrix} 2 \\ 5 \end{pmatrix} + \begin{pmatrix} 3 \\ -1 \end{pmatrix}$. [1 mark]

• Cue. Add components: $\begin{pmatrix} 5 \\ 4 \end{pmatrix}$.

Q2. Find $3\begin{pmatrix} -2 \\ 4 \end{pmatrix}$. [1 mark]

• Cue. Multiply each component by $3$: $\begin{pmatrix} -6 \\ 12 \end{pmatrix}$.

Q3. Find the magnitude of $\begin{pmatrix} 8 \\ 6 \end{pmatrix}$. [2 marks]

• Cue. $\sqrt{8^2 + 6^2} = \sqrt{100} = 10$.