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

Jul 19, 2016

$26.58$ degrees

#### Explanation:

We're going to use the dot product here. For two vectors $\vec{u} , \vec{v} \in {\mathbb{R}}^{3}$ where $\vec{u} = \left({u}_{1} , {u}_{2} , {u}_{3}\right) \mathmr{and} \vec{v} = \left({v}_{1} , {v}_{2} , {v}_{3}\right)$ the dot product is given by the following two formulae:

$\vec{u} \cdot \vec{v} = {u}_{1} {v}_{1} + {u}_{2} {v}_{2} + {u}_{3} {v}_{3}$

and

$\vec{u} \cdot \vec{v} = | \vec{u} | | \vec{v} | \cos \theta$

Combining these gives:

$\theta = {\cos}^{- 1} \left(\frac{{u}_{1} {v}_{1} + {u}_{2} {v}_{2} + {u}_{3} {v}_{3}}{| \vec{u} | | \vec{v} |}\right)$

$| \vec{u} | = \sqrt{{5}^{2} + {5}^{2} + {3}^{2}} = \sqrt{59}$

$| \vec{v} | = \sqrt{{4}^{2} + {9}^{2} + {1}^{2}} = \sqrt{98}$

$\theta = {\cos}^{- 1} \left(\frac{5 \cdot 4 + 5 \cdot 9 + 3 \cdot 1}{\sqrt{59} \sqrt{98}}\right) = 26.58$ degrees