# What is the angle between <3 , 2 , -1 >  and  < 1, -3 , 6 > ?

##### 1 Answer
Mar 29, 2016

$\alpha \cong 69 , 23$

#### Explanation:

$A = < 3 , 2 , - 1 > \text{ } B = < 1 , - 3 , 6 >$
$\text{steps...}$
$1 - \text{ find magnitude of A}$
$2 - \text{ find magnitude of B}$
$3 - \text{ find dot product } A \cdot B$
$4 - \text{ use : } A \cdot B = | | A | | \cdot | | B | | \cdot \cos \alpha$

$| | A | | = \sqrt{{3}^{2} + {2}^{2} + {\left(- 1\right)}^{2}} = \sqrt{9 + 4 + 1} = \sqrt{14}$
$| | B | | = \sqrt{{1}^{2} + {\left(- 3\right)}^{2} + {6}^{2}} = \sqrt{1 + 9 + 36} = \sqrt{46}$

$A \cdot B = {A}_{x} \cdot {B}_{x} + {A}_{y} \cdot {B}_{y} + {A}_{z} \cdot {B}_{z}$
$A \cdot B = 3 \cdot 1 + \left(2 \cdot \left(- 3\right)\right) + \left(- 1 \cdot 6\right)$
$A \cdot B = 3 - 6 - 6$
$A \cdot B = - 9$

$A \cdot B = | | A | | \cdot | | B | | \cdot \cos \alpha$
$- 9 = \sqrt{14} \cdot \sqrt{46} \cdot \cos \alpha$
$\cos \alpha = - \frac{9}{\sqrt{14 \cdot 46}}$

$\cos \alpha = - \frac{9}{25 , 38}$
$\cos \alpha = 0 , 3546099291$
$\alpha \cong 69 , 23$