# How Can You Find the Angle Between Two Vectors?

The angle between two vectors is always an angle $\alpha \in \phantom{\rule{-0.17em}{0ex}}\left[0\text{°},180\text{°}\right]$.

Formula

### TheAngleBetweenTwoVectors

 $\mathrm{cos}\alpha =\frac{\stackrel{\to }{u}\cdot \stackrel{\to }{v}}{\phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{u}|\cdot \phantom{\rule{-0.17em}{0ex}}|\stackrel{\to }{v}|},\phantom{\rule{2em}{0ex}}\alpha \in \phantom{\rule{-0.17em}{0ex}}\left[0\text{°},180\text{°}\right]$

Example 1

Find the angle between the vectors $\phantom{\rule{-0.17em}{0ex}}\left(4,1\right)$ and $\phantom{\rule{-0.17em}{0ex}}\left(2,-3\right)$.

You insert the numbers into the formula, which gives you

$\begin{array}{llll}\hfill \mathrm{cos}\alpha & =\frac{\phantom{\rule{-0.17em}{0ex}}\left(4,1\right)\cdot \phantom{\rule{-0.17em}{0ex}}\left(2,-3\right)}{\phantom{\rule{-0.17em}{0ex}}|\phantom{\rule{-0.17em}{0ex}}\left(4,1\right)|\cdot \phantom{\rule{-0.17em}{0ex}}|\phantom{\rule{-0.17em}{0ex}}\left(2,-3\right)|}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{8-3}{\sqrt{{4}^{2}+{1}^{2}}\cdot \sqrt{{2}^{2}+\phantom{\rule{-0.17em}{0ex}}{\left(-3\right)}^{2}}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{5}{\sqrt{17}\cdot \sqrt{13}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx \frac{5}{14.86}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 0.336.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

That means the angle is

 $\alpha ={\mathrm{cos}}^{-1}\phantom{\rule{-0.17em}{0ex}}\left(0.336\right)\approx 70.3\text{°}.$