2D Vectors - Extended Maths

What a vector is

A vector has both size and direction, unlike a plain number (a scalar) which has size only. In two dimensions we usually write a vector as a column vector: the top number is how far it moves in the \(x\) direction and the bottom number is how far it moves in the \(y\) direction.

Column vector: A vector written as \(\begin{pmatrix} x \\ y \end{pmatrix}\), where \(x\) is the horizontal step and \(y\) is the vertical step.

So \(\mathbf{v} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}\) means "3 to the right and 2 up". A negative number simply reverses that direction: \(\begin{pmatrix} -3 \\ 2 \end{pmatrix}\) means "3 to the left and 2 up". Vectors are often written in bold (\(\mathbf{v}\)) or with an arrow above the letter.

Position is not the point

A vector describes a movement, not a fixed place. The same \(\begin{pmatrix} 3 \\ 2 \end{pmatrix}\) starting anywhere on the plane is the same vector, because the step is identical.

Adding and subtracting

To add or subtract vectors, work on each row on its own: add the top numbers together, then add the bottom numbers together. Geometrically, adding \(\mathbf{a} + \mathbf{b}\) means "do movement \(\mathbf{a}\), then movement \(\mathbf{b}\)", landing you at the combined step.

\[ \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} + \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \end{pmatrix} \]

Worked example

Given \(\mathbf{a} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} 3 \\ -1 \end{pmatrix}\), find \(\mathbf{a} + \mathbf{b}\) and \(\mathbf{a} - \mathbf{b}\).

Add row by row: \(\begin{pmatrix} 2 + 3 \\ 5 + (-1) \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \end{pmatrix}\).

Subtract row by row: \(\begin{pmatrix} 2 - 3 \\ 5 - (-1) \end{pmatrix} = \begin{pmatrix} -1 \\ 6 \end{pmatrix}\).

\(\mathbf{a} + \mathbf{b} = \begin{pmatrix} 5 \\ 4 \end{pmatrix}, \quad \mathbf{a} - \mathbf{b} = \begin{pmatrix} -1 \\ 6 \end{pmatrix}\)

Watch the signs when subtracting a negative: \(5 - (-1)\) becomes \(5 + 1 = 6\). That double negative is the most common slip here.

Scalar multiplication

Multiplying a vector by an ordinary number (a scalar) stretches or shrinks it. Every component is multiplied by that number. A scalar of \(2\) doubles the length; a scalar of \(-1\) keeps the length but flips the direction to the exact opposite.

\[ k\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} kx \\ ky \end{pmatrix} \]

Worked example

If \(\mathbf{v} = \begin{pmatrix} 4 \\ -3 \end{pmatrix}\), find \(3\mathbf{v}\) and \(-2\mathbf{v}\).

Multiply each component by 3: \(3\mathbf{v} = \begin{pmatrix} 3 \times 4 \\ 3 \times (-3) \end{pmatrix} = \begin{pmatrix} 12 \\ -9 \end{pmatrix}\).

Multiply each component by \(-2\): \(-2\mathbf{v} = \begin{pmatrix} -2 \times 4 \\ -2 \times (-3) \end{pmatrix} = \begin{pmatrix} -8 \\ 6 \end{pmatrix}\).

\(3\mathbf{v} = \begin{pmatrix} 12 \\ -9 \end{pmatrix}, \quad -2\mathbf{v} = \begin{pmatrix} -8 \\ 6 \end{pmatrix}\)

Because \(3\mathbf{v}\) is just \(\mathbf{v}\) scaled up, it points the same way but is three times as long. Two vectors that are scalar multiples of each other are parallel.

Magnitude (the length)

The magnitude of a vector is its length, written \(|\mathbf{v}|\). Because the horizontal and vertical steps meet at a right angle, the vector is the hypotenuse of a right-angled triangle, so we use Pythagoras.

\[ |\mathbf{v}| = \sqrt{x^2 + y^2} \]

Worked example

Find the magnitude of \(\mathbf{v} = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\), then of \(\mathbf{w} = \begin{pmatrix} -5 \\ 12 \end{pmatrix}\).

Square each component and add: \(3^2 + 4^2 = 9 + 16 = 25\).

Take the square root: \(|\mathbf{v}| = \sqrt{25} = 5\).

For \(\mathbf{w}\): \((-5)^2 + 12^2 = 25 + 144 = 169\), so \(|\mathbf{w}| = \sqrt{169} = 13\).

\(|\mathbf{v}| = 5\) and \(|\mathbf{w}| = 13\)

Where this is assessed

Column-vector arithmetic and magnitude show up under Criterion A (knowing and applying the method) and Criterion C (setting out your working clearly with correct notation). Squaring a negative always gives a positive, so a minus sign inside the vector never makes the length negative.

Check yourself

1. Find the magnitude of \(\mathbf{a} = \begin{pmatrix} 4 \\ -3 \end{pmatrix}\). +

Square and add: \(4^2 + (-3)^2 = 16 + 9 = 25\). Take the root: \(|\mathbf{a}| = \sqrt{25} = \mathbf{5}\).

2. Work out \(\begin{pmatrix} 1 \\ 2 \end{pmatrix} + 2\begin{pmatrix} 3 \\ -1 \end{pmatrix}\). +

First scale: \(2\begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ -2 \end{pmatrix}\). Then add row by row: \(\begin{pmatrix} 1+6 \\ 2+(-2) \end{pmatrix} = \mathbf{\begin{pmatrix} 7 \\ 0 \end{pmatrix}}\).

3. If \(\mathbf{p} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}\) and \(\mathbf{q} = \begin{pmatrix} 2 \\ 7 \end{pmatrix}\), find \(\mathbf{p} - \mathbf{q}\) and its magnitude. +

Subtract row by row: \(\mathbf{p} - \mathbf{q} = \begin{pmatrix} 5-2 \\ 3-7 \end{pmatrix} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}\). Its magnitude is \(\sqrt{3^2 + (-4)^2} = \sqrt{9+16} = \sqrt{25} = \mathbf{5}\).

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