# Magnitude and Direction of Vectors

### Magnitude of a Vector

The magnitude of a vector $\stackrel{\to }{PQ}$ is the distance between the initial point $P$ and the end point $Q$. In symbols the magnitude of $\stackrel{\to }{PQ}$ is written as $|\text{\hspace{0.17em}}\stackrel{\to }{PQ}\text{\hspace{0.17em}}|$.

If the coordinates of the initial point and the end point of a vector is given, the Distance Formula can be used to find its magnitude.

$|\text{\hspace{0.17em}}\stackrel{\to }{PQ}\text{\hspace{0.17em}}|=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$

Example :

Find the magnitude of the vector $\stackrel{\to }{PQ}$ whose initial point $P$ is at $\left(1,1\right)$ and end point is at $Q$ is at $\left(5,3\right)$.

Solution:

Use the Distance Formula.

Substitute the values of ${x}_{1}$, ${y}_{1}$, ${x}_{2}$, and ${y}_{2}$.

$\begin{array}{l}|\text{\hspace{0.17em}}\stackrel{\to }{PQ}\text{\hspace{0.17em}}|=\sqrt{{\left(5-1\right)}^{2}+{\left(3-1\right)}^{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{{4}^{2}+{2}^{2}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{16+4}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\sqrt{20}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\approx 4.5\end{array}$

The magnitude of $\stackrel{\to }{PQ}$ is about $4.5$.

### Direction of a Vector

The direction of a vector is the measure of the angle it makes with a horizontal line.

One of the following formulas can be used to find the direction of a vector:

$\mathrm{tan}\theta =\frac{y}{x}$, where $x$ is the horizontal change and $y$ is the vertical change

or

$\mathrm{tan}\theta =\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$, where $\left({x}_{1},{y}_{1}\right)$ is the initial point and $\left({x}_{2},{y}_{2}\right)$ is the terminal point.

Example :

Find the direction of the vector $\stackrel{\to }{PQ}$ whose initial point $P$ is at $\left(2,3\right)$ and end point is at $Q$ is at $\left(5,8\right)$.

The coordinates of the initial point and the terminal point are given. Substitute them in the formula $\mathrm{tan}\theta =\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$.

$\begin{array}{l}\mathrm{tan}\theta =\frac{8-3}{5-2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{5}{3}\end{array}$

Find the inverse tan, then use a calculator.

$\begin{array}{l}\theta ={\mathrm{tan}}^{-1}\left(\frac{5}{3}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\approx 59°\end{array}$

The vector $\stackrel{\to }{PQ}$ has a direction of about $59°$.