# What is the angle between <7 , 2 , -3 >  and  < 8 , 6 , 0 > ?

Apr 19, 2016

approximately $0.528 \text{ radians}$ or ${30.28}^{\circ}$

#### Explanation:

The angle, $\theta$, between two vectors $u$ and $v$
is such that
$\textcolor{w h i t e}{\text{XXX}} \cos \left(\theta\right) = \frac{u \cdot v}{\left\mid \left\mid u \right\mid \right\mid \cdot \left\mid \left\mid v \right\mid \right\mid}$

For
$\textcolor{w h i t e}{\text{XXX}} u = < {u}_{1} , {u}_{2} , {u}_{3} > = < 7 , 2 , - 3 >$
and
$\textcolor{w h i t e}{\text{XXX}} v = < {v}_{1} , {v}_{2} , {v}_{3} > = < 8 , 6 , 0 >$

$u \cdot v = {u}_{1} \cdot {v}_{1} + {u}_{2} \cdot {v}_{2} + {u}_{3} \cdot {v}_{3}$
$\textcolor{w h i t e}{\text{XXx}} = 7 \times 8 + 2 \times 6 + \left(- 3\right) \times 0 = 68$

$\left\mid \left\mid u \right\mid \right\mid = \sqrt{{u}_{1}^{2} + {u}_{2}^{2} + {u}_{3}^{2}}$
$\textcolor{w h i t e}{\text{XXx}} = \sqrt{{7}^{2} + {2}^{2} + {\left(- 3\right)}^{2}} = \sqrt{62}$

$\left\mid \left\mid v \right\mid \right\mid = \sqrt{{v}_{1}^{2} + {v}_{2}^{2} + {v}_{3}^{2}}$
$\textcolor{w h i t e}{\text{XXx}} = \sqrt{{8}^{2} + {6}^{2} + {0}^{2}} = \sqrt{100} = 10$

Therefore
$\textcolor{w h i t e}{\text{XXX}} \cos \left(\theta\right) = \frac{68}{10 \sqrt{62}}$
$\textcolor{w h i t e}{\text{XXXXXXX}} \approx 0.863601$ (using a calculator)
and
$\textcolor{w h i t e}{\text{XXX}} \theta = \arccos \left(0.863601\right) \approx 0.528426$ (radians)