# What is the angle between <6,9,-2> and < -2,0,4 >?

Dec 12, 2015

$\theta = \arccos \frac{a x + b y + c z}{\sqrt{{x}^{2} + {y}^{2} + {z}^{2}} \sqrt{{a}^{2} + {b}^{2} + {c}^{2}}}$

#### Explanation:

There are many multiplicities of this question, so let me solve all of them.

$v = \left(x , y , z\right)$

$w = \left(a , b , c\right)$

$v \cdot w = | v | \cdot | w | \cdot \cos \theta$

$a x + b y + c z = \sqrt{{x}^{2} + {y}^{2} + {z}^{2}} \sqrt{{a}^{2} + {b}^{2} + {c}^{2}} \cos \theta$

What is left is your arithmetical hability!

$- 12 - 8 = \sqrt{36 + 81 + 4} \sqrt{4 + 16} \cos \theta$

$- \frac{20}{11 \cdot \sqrt{20}} = \cos \theta = - \frac{\sqrt{20}}{11}$