# What is the angle between <-3,5,0 >  and < 4,0,-2>?

Mar 8, 2016

$\alpha \cong {21}^{o}$

#### Explanation:

$A = < - 3 , 5 , 0 > \text{ } B = < 4 , 0 , - 2 >$
$\text{step-1: find the magnitudes:} | | C | | = \sqrt{{a}^{2} + {b}^{2} + {c}^{2}}$
$| | A | | = \sqrt{{\left(- 3\right)}^{2} + {5}^{2} + {0}^{2}} \text{ } | | A | | = \sqrt{36}$
$| | B | | = \sqrt{{4}^{2} + {0}^{2} + {\left(- 2\right)}^{2}} \text{ } | | B | | = \sqrt{20}$
$\text{step-2:find dot product of A and B }$
$A . B = \left(- 3 \cdot 4\right) + \left(5 \cdot 0\right) + \left(- 2 \cdot 0\right) \text{ } A . B = - 12$
$\text{use dot product formula } A . B = | | A | | \cdot | | B | | \cdot \cos \alpha$
$- 12 = \sqrt{36} \cdot \sqrt{20} \cdot \cos \alpha$
$- 12 = \sqrt{36 \cdot 20} \cdot \cos \alpha$
$\cos \alpha = - \frac{12}{\sqrt{36 \cdot 20}}$
$\cos \alpha = - 0 , 93510265$
$\alpha \cong {21}^{o}$