# What is the angle between <0,-7,6>  and <-1,9,-3>?

Oct 19, 2016

$\theta = \arccos \left(- \frac{81}{\sqrt{7735}}\right) \approx 2.7414$

#### Explanation:

We'll use the fact that the dot product $\vec{u} \cdot \vec{v}$ can be calculated in two ways:

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

Dividing by $| | \vec{u} | | \cdot | | \vec{v} | |$, we get

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

or

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

In our case, we have $\vec{u} = < 0 , - 7 , 6 >$ and $\vec{v} = < - 1 , 9 , - 3 >$.

If we calculate their magnitudes, we get

$| | \vec{u} | | = \sqrt{{0}^{2} + {\left(- 7\right)}^{2} + {6}^{2}} = \sqrt{85}$
and
$| | \vec{v} | | = \sqrt{{\left(- 1\right)}^{2} + {9}^{2} + {\left(- 3\right)}^{2}} = \sqrt{91}$

So, the formula we derived gives us

$\theta = \arccos \left(\frac{0 \left(- 1\right) + \left(- 7\right) \left(9\right) + 6 \left(- 3\right)}{\sqrt{85} \cdot \sqrt{91}}\right)$

$= \arccos \left(- \frac{81}{\sqrt{7735}}\right)$

$\approx 2.7414$