# What is the angle between <7,2,-4 >  and < -3,6,-2>?

Jul 21, 2017

The angle is $= {91}^{\circ}$

#### Explanation:

Let $\vec{a} = < 7 , 2 , - 4 >$

$\vec{b} = < - 3 , 6 , - 2 >$

The angle is calculated with the dot product.

$\vec{a} . \vec{b} = | | \vec{a} | | \cdot | | \vec{b} | | \cdot \cos \theta$

Where $\theta$ is the angle between the $2$ vectors

The dot product is

$\vec{a} . \vec{b} = < 7 , 2 , - 4 > . < - 3 , 6 , - 2 > = - 21 + 12 + 8 = - 1$

The modulus of $\vec{a}$ is

$= | | \vec{a} | | = | | < 7 , 2 , - 4 > | | = \sqrt{49 + 4 + 16} = \sqrt{69}$

The modulus of $\vec{b}$ is

$= | | \vec{b} | | = | | < - 3 , 6 , - 2 > | | = \sqrt{9 + 36 + 4} = \sqrt{49}$

The angle is

$\theta = {\cos}^{-} 1 \left(\frac{\vec{a} . \vec{b}}{| | \vec{a} | | \cdot | | \vec{b} | |}\right) = {\cos}^{-} 1 \left(- \frac{1}{\sqrt{69} \sqrt{49}}\right)$

$= {\cos}^{-} 1 \left(- 0.017\right) = {91}^{\circ}$