# What is the angle between <-1,8,2 > and < 0,-3,1>?

May 5, 2018

Approximately
$2.563 \textcolor{w h i t e}{l} \text{rad}$

or equivalently,
${147}^{\textsf{\text{o}}}$.

#### Explanation:

Let $\theta$ be the angle between the two vectors . By the geometric definition of scalar products,

$\text{Angle between vector " bb(a)" and } \boldsymbol{b} = \arccos \left(\frac{\boldsymbol{a} \cdot \boldsymbol{b}}{\left\mid \boldsymbol{a} \right\mid \cdot \left\mid \boldsymbol{b} \right\mid}\right)$

Where $\boldsymbol{a} \cdot \boldsymbol{b}$ the dot product of $\boldsymbol{a}$ and $\boldsymbol{b}$, and $\left\mid \boldsymbol{a} \right\mid \cdot \left\mid \boldsymbol{b} \right\mid$ the product of their magnitude.

Let $\boldsymbol{a} = < - 1 , 8 , 2 >$ and $\boldsymbol{b} = < 0 , - 3 , 1 >$,

$\theta = \arccos \left(\frac{< - 1 , 8 , 2 > \cdot < 0 , - 3 , 1 >}{\sqrt{1 + {8}^{2} + {2}^{2}} \cdot \sqrt{{3}^{2} + {1}^{2}}}\right)$
$\textcolor{w h i t e}{\theta} = \arccos \left(\frac{- 24 + 2}{\sqrt{69} \cdot \sqrt{10}}\right)$
$\textcolor{w h i t e}{\theta} = \arccos \left(- \frac{22}{\sqrt{690}}\right)$
color(white)(theta)~~2.563color(white)(l)"rad"=147^sf("o")