# What is the angle between <-1,9,3 >  and < 5,-8,2 >?

Aug 2, 2018

$\theta = {\cos}^{-} 1 \left(- \frac{71}{\sqrt{8463}}\right)$

#### Explanation:

Let , $\vec{a} = < - 1 , 9 , 3 > \mathmr{and} \vec{b} = < 5 , - 8 , 2 >$ be the two vectors.

$\therefore | \vec{a} | = \sqrt{{\left(- 1\right)}^{2} + {\left(9\right)}^{2} + {\left(3\right)}^{2}} = \sqrt{1 + 81 + 9} = \sqrt{91}$

$\mathmr{and} | \vec{b} | = \sqrt{{\left(5\right)}^{2} + {\left(- 8\right)}^{2} + {\left(2\right)}^{2}} = \sqrt{25 + 64 + 4} = \sqrt{93}$

Dot product : $a \cdot b = \left(- 1\right) \times 5 + 9 \times \left(- 8\right) + 3 \times 2$

$\therefore a \cdot b = - 5 - 72 + 6 = - 71$

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

$\theta = {\cos}^{-} 1 \left(\frac{a \cdot b}{| \vec{a} | | \vec{b} |}\right) = {\cos}^{-} 1 \left(\frac{- 71}{\sqrt{91} \sqrt{93}}\right)$

$\therefore \theta = {\cos}^{-} 1 \left(- \frac{71}{\sqrt{8463}}\right) =$${\left(140.51\right)}^{\circ}$