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

Jun 14, 2016

$\hat{\vec{u} , \vec{v}} = 1.46073$ [rad] $\approx {83.70}^{\circ}$

#### Explanation:

Given two vectors $\vec{u} , \vec{v}$ we know that

$\left\langle\vec{u} , \vec{v}\right\rangle = \left\lVert \vec{u} \right\rVert \times \left\lVert \vec{v} \right\rVert \times \cos \left(\hat{\vec{u} , \vec{v}}\right)$

Solving for the angle $\hat{\vec{u} , \vec{v}}$ we have

$\hat{\vec{u} , \vec{v}} = \arccos \left(\frac{\left\langle\vec{u} , \vec{v}\right\rangle}{\left\lVert \vec{u} \right\rVert \times \left\lVert \vec{v} \right\rVert}\right)$

In our example

$\vec{u} = \left\{5 , 5 , - 1\right\}$
$\vec{v} = \left\{4 , - 3 , 1\right\}$

$\left\langle\vec{u} , \vec{v}\right\rangle = 5 \times 4 + 5 \times \left(- 3\right) + \left(- 1\right) \times 1 = 4$
$\left\lVert \vec{u} \right\rVert = \sqrt{{5}^{2} + {5}^{2} + {\left(- 1\right)}^{2}} = \sqrt{51}$
$\left\lVert \vec{v} \right\rVert = \sqrt{{4}^{2} + {\left(- 3\right)}^{2} + {1}^{2}} = \sqrt{26}$

then

$\hat{\vec{u} , \vec{v}} = \arccos \left(\frac{4}{\sqrt{51 \times 26}}\right) = 1.46073$ [rad] $\approx {83.70}^{\circ}$