# What is the angle between <1 , 5 , 3 >  and  < 4, 9 , 4 > ?

Mar 20, 2016

$\alpha = 14 , {07}^{o}$

#### Explanation:

$\text{find dot product of <1,5,3> and <4,9,4> like animation above}$
$A . B = 1 \cdot 4 + 5 \cdot 9 + 3 \cdot 4$
$A . B = 4 + 45 + 12$
$A . B = 61$

$\text{ now find the magnitude of A and B}$

$| | A | | = \sqrt{{1}^{2} + {5}^{2} + {3}^{2}} = \sqrt{1 + 25 + 9} = \sqrt{35}$
$| | B | | = \sqrt{{4}^{2} + {9}^{2} + {4}^{2}} = \sqrt{16 + 81 + 16} = \sqrt{113}$

$\text{Dot product is determined by:}$
$A . B = | | A | | \cdot | | B | | \cdot \cos \alpha$
$A \cdot B = 61$
$| | A | | = \sqrt{35}$
$| | B | | = \sqrt{113}$
$61 = \sqrt{35} \cdot \sqrt{113} \cdot \cos \alpha$
$61 = \sqrt{35 \cdot 113} \cdot \cos \alpha$
$\cos \alpha = \frac{61}{\sqrt{35 \cdot 113}}$
$\cos \alpha = 0 , 9699661782$
$\alpha = 14 , {07}^{o}$