Magnitude and direction of a vector

In a nutshell

You can use Pythagoras' theorem to calculate the magnitude of a vector, $|\textbf{a}|$ . To find a unit vector in the direction of a given vector, divide the vector by its magnitude. A vector can be written in magnitude-direction form, with the direction defined by its angle relative to one of the coordinate axes.

Magnitude of a vector

The diagram shows the vector $\textbf{a}$ inclined at an angle $\theta$ to the positive $x$ -axis.

Maths; Vectors I; KS5 Year 12; Magnitude and direction of a vector

The magnitude of a vector is its length. You can use Pythagoras' theorem to calculate the magnitude of vector $\textbf{a}$ , written as $|\textbf{a}|$ .

If:

Then:

\boxed{\begin{aligned}\textbf{a}&=x\textbf{i}+y\textbf{j}=\begin{pmatrix}x\\y\end{pmatrix}\\\\|\textbf{a}|&=\sqrt{x^2+y^2}\end{aligned}}

Example 1

Find the exact magnitude of $\textbf{r}=\begin{pmatrix}3\\5\end{pmatrix}$ .

$\begin{aligned}|\textbf{r}|&=\sqrt{3^2+5^2}\\&=\sqrt{9+25}\\&=\underline{\sqrt{34}}\end{aligned}$

Note: You are asked for the exact value, so you can leave your answer in surd form.

Unit vectors

A unit vector is any vector with a magnitude of $1$ . You have come across the unit vectors $\textbf{i}$ and $\textbf{j}$ that go along the positive $x$ - and $y$ -axes, respectively. A unit vector of a given vector $\textbf{a}$ can be written as $\^\textbf{a}$ and is found as follows:

$\boxed{\^\textbf{a}=\dfrac{\textbf{a}}{|\textbf{a}|}}$

If the magnitude of $\textbf{a}$ is $5$ , a unit vector of $\textbf{a}$ is:

\^\textbf{a}=\dfrac{\textbf{a}}{5}

Maths; Vectors I; KS5 Year 12; Magnitude and direction of a vector

Example 2

If $\textbf{r}=6\textbf{i}-8\textbf{j}$ , find a unit vector in the direction of $\textbf{r}$ .

First, you need to find the magnitude of $\textbf{r}$ .

$\begin{aligned}|\textbf{r}|&=\sqrt{6^2+(-8)^2}\\&=\sqrt{36+64}\\&=\sqrt{100}\\&=\underline{10}\end{aligned}$ 

For a unit vector in the direction of $\textbf{r}$ :

$\begin{aligned}\^\textbf{r}&=\dfrac{\textbf{r}}{|\textbf{r}|}\\&=\dfrac{{6\textbf{i}-8\textbf{j}}}{10}\\&=\frac{2(3\textbf{i}-4\textbf{j})}{10}\\&=\underline{\frac15(3\textbf{i}-4\textbf{j})}\end{aligned}$

Magnitude-direction form

A vector can be defined by its magnitude and its angle relative to the $x$ - or $y$ -axes. This is its magnitude-direction form.

Maths; Vectors I; KS5 Year 12; Magnitude and direction of a vector

Vector $\textbf{a}$ has magnitude $|\textbf{a}|$ and is at angle $\theta$ to the positive $x$ -axis.

You can use trigonometry for right-angled triangles to find the horizontal and vertical components of $\textbf{a}$ .

$\boxed{\begin{aligned}\textbf{a}=\begin{pmatrix}|\textbf{a}|cos\theta\\|\textbf{a}|sin\theta\end{pmatrix}\end{aligned}}$

Example 3

Vector $\textbf{p}$ has a magnitude of $8$ and makes an angle of $60\degree$ with the positive $x$ -axis. Write $\textbf{p}$ in component form.

It can help to draw the vector.

Maths; Vectors I; KS5 Year 12; Magnitude and direction of a vector

Horizontal component of $\textbf{p}$ :

$\begin{aligned}cos60\degree&=\frac{a}{8}\\&a=8cos60\degree\\&=8\times\frac12\\&=4\end{aligned}$ 

Vertical component of $\textbf{p}$ :

$\begin{aligned}sin60\degree&=\frac{b}{8}\\&b=8sin60\degree\\&=8\times\frac{\sqrt3}2\\&=4\sqrt3\end{aligned}$ 

In component form:

$\textbf{p}=\underline{4\textbf{i}+4\sqrt3\textbf{j}}$