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

Aug 2, 2018

$\theta = {\cos}^{-} 1 \left(- \frac{78}{\sqrt{6205}}\right)$

#### Explanation:

Let , $\vec{a} = < 0 , - 7 , 6 > \mathmr{and} \vec{b} = < - 1 , 6 , - 6 >$ be the two vectors.

$\therefore | \vec{a} | = \sqrt{{\left(0\right)}^{2} + {\left(- 7\right)}^{2} + {\left(6\right)}^{2}} = \sqrt{0 + 49 + 36} = \sqrt{85}$

$\mathmr{and} | \vec{b} | = \sqrt{{\left(- 1\right)}^{2} + {\left(6\right)}^{2} + {\left(- 6\right)}^{2}} = \sqrt{1 + 36 + 36}$=$\sqrt{73}$

Dot product : $a \cdot b = < 0 , - 7 , 6 > \cdot < - 1 , 6 , - 6 >$

$\therefore a \cdot b = \left(0\right) \times \left(- 1\right) + \left(- 7\right) \times \left(6\right) + \left(6\right) \times \left(- 6\right)$

$\therefore a \cdot b = 0 - 42 - 36 = - 78$

Now the angle between $\vec{a} \mathmr{and} \vec{b}$ is:

$\theta = {\cos}^{-} 1 \left(\frac{a \cdot b}{| \vec{a} | | \vec{b} |}\right) = {\cos}^{-} 1 \left(\frac{- 78}{\sqrt{85} \sqrt{73}}\right)$

$\therefore \theta = {\cos}^{-} 1 \left(- \frac{78}{\sqrt{6205}}\right) =$${\left(171.97\right)}^{\circ}$