What is the angle between <-2,-5,6>  and < -1,-1,2 >?

Mar 6, 2016

$\alpha \cong {15}^{o}$

Explanation:

1)" Find the magnitude of<-2,-5,6>"
$| | A | | = \sqrt{{\left(- 2\right)}^{2} + {\left(- 5\right)}^{2} + {6}^{2}} \text{ } A = \sqrt{65}$
2)" Find the magnitude of <-1,-1,2>"
$| | B | | = \sqrt{{\left(- 1\right)}^{2} + {\left(- 1\right)}^{2} + {2}^{2}} \text{ } B = \sqrt{6}$
3)" Find dot product of <-2,-5,6> and <-1,-1,2>"
$A \cdot B = \left(- 2 \cdot - 1\right) + \left(- 5 \cdot - 1\right) + 6 \cdot 2$
$A \cdot B = + 2 + 5 + 12 = 19$
$A . B = | | A | | \cdot | | B | | \cdot \cos \alpha$
$19 = \sqrt{65} \cdot \sqrt{6} \cdot \cos \alpha$
$19 = \sqrt{65 \cdot 6} \cdot \cos \alpha$
$\cos \alpha = \frac{19}{19 , 749}$
$\cos \alpha = 0 , 962$
$\alpha \cong {15}^{o}$