# What is the angle between <-1,8,6 > and < -6,3,0>?

Dec 31, 2016

$\theta \approx {63.58}^{o}$

#### Explanation:

The equation for the angle between two vectors is given by:

$\cos \left(\theta\right) = \frac{\vec{a} \cdot \vec{b}}{| \vec{a} | \cdot | \vec{b} |}$

Where $\vec{a} = < - 1 , 8 , 6 >$ and $\vec{b} = < - 6 , 3 , 0 >$

First we find the dot product of the two vectotrs:

$\vec{a} \cdot \vec{b} = < - 1 , 8 , 6 > \cdot < - 6 , 3 , 0 >$

$\implies \left(- 1 \cdot - 6\right) + \left(8 \cdot 3\right) + \left(6 \cdot 0\right)$

$\implies 6 + 24 = 30$

Next, we find the product of the magnitude of each vector:

$| \vec{a} | = \sqrt{{\left({a}_{x}\right)}^{2} + {\left({a}_{y}\right)}^{2} + {\left({a}_{z}\right)}^{2}}$

$\implies \sqrt{{\left(- 1\right)}^{2} + {\left(8\right)}^{2} + {\left(6\right)}^{2}}$

$\implies \sqrt{1 + 64 + 36}$

$\implies \sqrt{101}$

$| \vec{b} | = \sqrt{{\left({b}_{x}\right)}^{2} + {\left({b}_{y}\right)}^{2} + {\left({b}_{z}\right)}^{2}}$

$\implies \sqrt{{\left(- 6\right)}^{2} + {\left(3\right)}^{2} + {\left(0\right)}^{2}}$

$\implies \sqrt{36 + 9 + 0}$

$\implies \sqrt{45}$

And $\sqrt{45} \cdot \sqrt{101} = \sqrt{4545}$

Revisiting our original equation, we can solve for $\theta$ by taking the inverse cosine:

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

Substituting in our values found above:

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

$\implies \theta = {\cos}^{-} 1 \left(\frac{30}{\sqrt{4545}}\right)$

This comes out to $\approx {63.58}^{o}$ or $\approx 1.11 r a d$

Explanation of equation:

We can derive the given equation using a geometric interpretation of the vectors.

Given vectors $\vec{a}$ and $\vec{b}$:

By the law of cosines, given:

${c}^{2} = {a}^{2} + {b}^{2} - 2 a b \cos \left(\theta\right)$

Applying this to our vector-made triangle:

$| \vec{a} - \vec{b} {|}^{2} = | \vec{a} {|}^{2} + | \vec{b} {|}^{2} - 2 | \vec{a} | \cdot | \vec{b} | \cos \left(\theta\right)$

Simplifying:

$\implies \left(\vec{a} - \vec{b}\right) \cdot \left(\vec{a} - \vec{b}\right) = \left(\vec{a} \cdot \vec{a}\right) + \left(\vec{b} \cdot \vec{b}\right) - 2 | \vec{a} | \cdot \vec{|} b | \cos \left(\theta\right)$

$\implies \cancel{{\vec{a}}^{2}} - 2 \left(\vec{a} \cdot \vec{b}\right) \cancel{+ {\vec{b}}^{2}} = \cancel{{\vec{a}}^{2}} + \cancel{{\vec{b}}^{2}} - 2 | \vec{a} | \cdot \vec{|} b | \cos \left(\theta\right)$

$\implies - 2 \left(\vec{a} \cdot \vec{b}\right) = - 2 | \vec{a} | \cdot \vec{|} b | \cos \left(\theta\right)$

$\implies \left(\vec{a} \cdot \vec{b}\right) = | \vec{a} | \cdot \vec{|} b | \cos \left(\theta\right)$

Solve for $\cos \left(\theta\right)$:

$\implies \cos \left(\theta\right) = \frac{\vec{a} \cdot \vec{b}}{| \vec{a} | \cdot \vec{|} b |}$