A size and a direction, written four ways.

Four ways to write a vector.

A vector has a size and a direction. The vector \(\mathbf{a}\) in Fig. 1 means 'move 3 right and 1 up'. Cambridge accepts any of the forms below, but you must write it correctly: in print a vector is bold, \(\mathbf{a}\); by hand you underline it, \(\underline{a}\), or draw an arrow over the letters of its end points, \(\overrightarrow{OA}\). The unit vectors \(\mathbf{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}\) and \(\mathbf{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}\) point one unit along the \(x\)- and \(y\)-axes, so any vector is a number of \(\mathbf{i}\) steps plus a number of \(\mathbf{j}\) steps.

To add or subtract vectors, add or subtract the components; to multiply by a scalar, multiply both components:

\[ \begin{pmatrix} p \\ q \end{pmatrix} + \begin{pmatrix} r \\ s \end{pmatrix} = \begin{pmatrix} p + r \\ q + s \end{pmatrix}, \qquad k\begin{pmatrix} p \\ q \end{pmatrix} = \begin{pmatrix} kp \\ kq \end{pmatrix} \]

Geometrically, \(\mathbf{a} + \mathbf{b}\) is the journey 'do \(\mathbf{a}\), then do \(\mathbf{b}\)', drawn tip to tail as in Fig. 1. The sum of several vectors is their resultant, and a chain of moves joins up: \(\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} = \overrightarrow{AD}\).

Magnitude and unit vectors.

The magnitude (length) of a vector is written \(|\mathbf{a}|\) or \(|\overrightarrow{AB}|\). The components are the two shorter sides of a right-angled triangle, so Pythagoras gives the length. Dividing a vector by its own magnitude keeps its direction but makes its length 1: this is the unit vector in that direction, sometimes written \(\hat{\mathbf{a}}\). Multiply the unit vector by \(k\) and you have a vector of magnitude \(k\) in the same direction.

\[ |x\mathbf{i} + y\mathbf{j}| = \sqrt{x^{2} + y^{2}}, \qquad \hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|}, \qquad \text{vector of magnitude } k \text{ along } \mathbf{a} = \frac{k}{|\mathbf{a}|}\,\mathbf{a} \]

In Fig. 1, \(|\mathbf{a}| = \sqrt{3^{2} + 1^{2}} = \sqrt{10}\), \(|\mathbf{b}| = \sqrt{5}\) and \(|\mathbf{a} + \mathbf{b}| = \sqrt{4^{2} + 3^{2}} = 5\). Since \(\sqrt{10} + \sqrt{5} \approx 5.40\), the magnitude of a sum is NOT the sum of the magnitudes unless the vectors point the same way.

Position vectors, parallel vectors and equating components.

The position vector of a point \(A\) is \(\overrightarrow{OA}\), the vector from the origin \(O\) to \(A\), usually called \(\mathbf{a}\). The point \(A(2, 5)\) has position vector \(2\mathbf{i} + 5\mathbf{j}\). To go from \(A\) to \(B\), go back from \(A\) to \(O\) and then out to \(B\): \(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\mathbf{a} + \mathbf{b}\).

• Between two points: \(\overrightarrow{AB} = \mathbf{b} - \mathbf{a}\), 'end minus start'.
• Parallel: \(\mathbf{u}\) and \(\mathbf{v}\) are parallel when \(\mathbf{u} = k\mathbf{v}\) for some scalar \(k\). A negative \(k\) means opposite directions. For \(\begin{pmatrix} p \\ q \end{pmatrix}\) and \(\begin{pmatrix} r \\ s \end{pmatrix}\) this means the SAME multiplier on both components.
• Equating like vectors: if \(x_{1}\mathbf{i} + y_{1}\mathbf{j} = x_{2}\mathbf{i} + y_{2}\mathbf{j}\), then \(x_{1} = x_{2}\) and \(y_{1} = y_{2}\). One vector equation gives two ordinary equations, enough to find two unknowns.