# What is the angle between <5 , 2 , -1 >  and  < 1, -3 , 1 > ?

$\cos \theta = \frac{< 5 , 2 , - 1 > \cdot < 1 , - 3 , 1 >}{\sqrt{{5}^{2} + {2}^{2} + {\left(- 1\right)}^{2}} \sqrt{{1}^{2} + {\left(- 3\right)}^{2} + {1}^{2}}} \to \cos \theta = \frac{5 \left(1\right) + 2 \left(- 3\right) + \left(- 1\right) 1}{\sqrt{30} \sqrt{11}} \to \theta = {\cos}^{-} 1 \left(- \frac{2}{\sqrt{330}}\right)$
$\theta = {96.32}^{\circ}$
To find the angle between the vectors use the formula cos theta=(a*b)/(|a||b|. Note that a and b are vectors and $a \cdot b$is the dot product of the vectors