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

Jan 3, 2017

The angle is $= 64.7$º

#### Explanation:

We use the dot product definition to calculate the angle $\theta$ between 2 vectors

veca.vecb=∥veca∥*∥vecb∥*cos theta

cos theta=(veca.vecb)/(∥veca∥*∥vecb∥)

We start by calculating the dot product,

veca.vecb=〈-2,1,1〉*〈1,3,8〉=(-2+3+8)=9

The modulus of veca=∥〈-2,1,1〉∥=sqrt(4+1+1)=sqrt6

The modulus of vecb=∥〈1,3,8〉∥=sqrt(1+9+64)=sqrt74

$\cos \theta = \frac{9}{\sqrt{6} \cdot \sqrt{74}} = 0.427$

$\theta = 64.7$º