What is the angle between <5,0,-2 >  and <-8,3,-6> ?

Jun 22, 2016

Reqd. Angle $= \arccos \left(- \frac{28}{\sqrt{3161}}\right) ,$ or $\pi - \arccos \left(\frac{28}{\sqrt{3161}}\right) .$

Explanation:

If $\theta$ is the reqd. angle between the given vectors, say $\vec{x}$=$< 5 , 0 , - 2 >$ and $\vec{y}$=$< - 8 , 3 , - 6 > ,$ then, we have by defn. of Dot Product

$\vec{x} . \vec{y} = | | \vec{x} | | | | \vec{y} | | \cos \theta \ldots \ldots \ldots . \left(1\right)$

Now $\vec{x} . \vec{y}$=$< 5 , 0 , - 2 >$.$< - 8 , 3 , - 6 >$=$5 \cdot \left(- 8\right) + 0 \cdot 3 + \left(- 2\right) \cdot \left(- 6\right) = - 40 + 0 + 12 = - 28.$
$| | \vec{x} | | = \sqrt{{5}^{2} + {0}^{2} + {\left(- 2\right)}^{2}} = \sqrt{29.}$
$| | \vec{y} | \setminus = \sqrt{{\left(- 8\right)}^{2} + {3}^{2} + {\left(- 6\right)}^{2}} = \sqrt{109.}$

Substituting these values in (1), we get, $\cos \theta = - \frac{28}{\sqrt{29} \cdot \sqrt{109}} = - \frac{28}{\sqrt{3161.}}$

Hence, Reqd. Angle $= \arccos \left(- \frac{28}{\sqrt{3161}}\right) ,$ or $\pi - \arccos \left(\frac{28}{\sqrt{3161}}\right) .$