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.

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

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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:

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

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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.

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

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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.

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:

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

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