# What is the angle between <-3,9,-7 >  and < 4,-2,8 >?

Feb 20, 2016

$\theta \cong 2.49$ radians

#### Explanation:

Note: The angel between two nonzero vector u and v, where $0 \le \theta \le \pi$ is define as

$\vec{u} = < {u}_{1} , {u}_{2} , {u}_{3} >$

$\vec{v} = < {v}_{1} , {v}_{2} , {v}_{3} >$

cos theta= (u *v )/(||u|| " ||v||

Where as: $\text{ } u \cdot v = \left({u}_{1} {v}_{1}\right) + \left({u}_{2} {v}_{2}\right) + \left({u}_{3} {v}_{3}\right)$

$| | u | | = \sqrt{{\left({u}_{1}\right)}^{2} + {\left({u}_{2}\right)}^{2} + {\left({u}_{3}\right)}^{2}}$

$| | v | | = \sqrt{{\left({v}_{1}\right)}^{2} + {\left({v}_{2}\right)}^{2} + {\left({v}_{3}\right)}^{2}}$

Step 1: Let

$\vec{u} = < - 3 , 9 , - 7 >$ and
$\vec{v} = < 4 , - 2 , 8 >$

Step 2: Let's find $\textcolor{red}{u \cdot v}$

$\textcolor{red}{u \cdot v} = \left(- 3\right) \left(4\right) + \left(9\right) \left(- 2\right) + \left(- 7\right) \left(8\right)$

$= - 12 - 18 - 56$
$= \textcolor{red}{- 86}$

Step 3: Let find $\textcolor{b l u e}{| | u | |}$

$\vec{u} = < - 3 , 9 - 7 >$

$\textcolor{b l u e}{| | u | |} = \sqrt{{\left(- 3\right)}^{2} + {\left(9\right)}^{2} + {\left(- 7\right)}^{2}}$

$= \sqrt{9 + 81 + 49}$

$= \textcolor{b l u e}{\sqrt{139}}$

Step 4 Let find $\textcolor{p u r p \le}{| | v | |}$

$\vec{v} = < 4 , - 2 , 8 >$

$\textcolor{p u r p \le}{| | v | |} = \sqrt{{\left(4\right)}^{2} + {\left(- 2\right)}^{2} + {\left(8\right)}^{2}}$

= sqrt(16 + 4 + 64) =color(purple)(sqrt84)

Step 5; Let substitute it back to the formula given above, and find $\theta$

cos theta= (u *v)/(||u|| " ||v||)

cos theta = color(red)(-86)/((color(blue)sqrt(139))color(purple)((sqrt84))

$\cos \theta = \frac{\textcolor{red}{- 86}}{\sqrt{11676}}$

$\theta = {\cos}^{- 1} \left(- \frac{86}{\sqrt{11676}}\right)$

$\theta \cong 2.49$ radians

**note: this is because $u \cdot v < 0$