# What is the angle between <-1,8,8 >  and <-1,7,3> ?

Angle
$\theta = {21.8036678}^{\circ}$
$\theta = {21}^{\circ} 48 ' 13.2 ' '$

#### Explanation:

Let the vector $v = - i + 8 j + 8 k$
Let the vector $w = - i + 7 j + 3 k$
Let $\theta$ be the angle between the vectors

Use the formula

$\cos \theta = \frac{v \cdot w}{| \setminus | v | \setminus | \text{ } | \setminus | w | \setminus |}$

$v \cdot w = \left(- 1\right) \left(- 1\right) + 8 \left(7\right) + 8 \left(3\right) = 1 + 56 + 24 = 81$

$| \setminus | v | \setminus | = \sqrt{{\left(- 1\right)}^{2} + {8}^{2} + {8}^{2}} = \sqrt{1 + 64 + 64} = \sqrt{129}$

$| \setminus | w | \setminus | = \sqrt{{\left(- 1\right)}^{2} + {7}^{2} + {3}^{2}} = \sqrt{1 + 49 + 9} = \sqrt{59}$

Let us continue

$\cos \theta = \frac{v \cdot w}{| \setminus | v | \setminus | \text{ } | \setminus | w | \setminus |}$

$\cos \theta = \frac{81}{\sqrt{129} \cdot \sqrt{59}}$

$\theta = {\cos}^{-} 1 \left(\frac{81}{\sqrt{129} \cdot \sqrt{59}}\right)$

$\theta = {21.8036678}^{\circ}$
$\theta = {21}^{\circ} 48 ' 13.2 ' '$

God bless....I hope the explanation is useful.