# What is the angle between <1 , 3 , -4 >  and  < 5 , -2 , 2 > ?

Oct 2, 2016

$\theta \approx 1.833 r a \mathrm{di} a n s$

#### Explanation:

Let $\overline{A} = < 1 , 3 , - 4 >$
Let $\overline{B} = < 5 , - 2 , 2 >$

bar A•bar B = (1)(5) + (3)(-2) + (-4)(2)

bar A•bar B = -9

|barA| = sqrt(1² + 3² + (-4)²)

$| \overline{A} | = \sqrt{26}$

|barB| = sqrt(5²+ (-2)² + 2²)

$| \overline{B} | = \sqrt{33}$

bar A•bar B = |barA||barB|cos(theta) where $\theta$ is the angle between the vectors

Substitute the known values and then solve for $\theta$:

$- 9 = \sqrt{26} \sqrt{33} \cos \left(\theta\right)$

$\theta = {\cos}^{-} 1 \left(- \frac{9}{\sqrt{26} \sqrt{33}}\right)$

$\theta \approx 1.833 r a \mathrm{di} a n s$