# What is the angle between <1 , 5 , 9 >  and  < -2 , 3 , 2 > ?

Oct 23, 2016

$\theta \approx 0.76 r a \mathrm{di} a n s$

#### Explanation:

Let $\overline{A} = < 1 , 5 , 9 >$
Let $\overline{B} = < - 2 , 3 , 2 >$

The $\overline{A} \cdot \overline{B}$ is:

$\overline{A} \cdot \overline{B} = \left(1\right) \left(- 2\right) + \left(5\right) \left(3\right) + \left(9\right) \left(2\right) = 31$

$| \overline{A} | = \sqrt{{1}^{2} + {5}^{2} + {9}^{2}} = \sqrt{107}$

$| \overline{B} | = \sqrt{{\left(- 2\right)}^{2} + {3}^{2} + {2}^{2}} = \sqrt{17}$

Use $\overline{A} \cdot \overline{B} = | \overline{A} | | \overline{B} | \cos \left(\theta\right)$

Solve for $\theta$

$\theta = {\cos}^{-} 1 \left(\frac{\overline{A} \cdot \overline{B}}{| \overline{A} | | \overline{B} |}\right)$

$\theta = {\cos}^{-} 1 \left(\frac{31}{\sqrt{107} \sqrt{17}}\right)$

$\theta \approx 0.76 r a \mathrm{di} a n s$