# If A = <3 ,1 ,-1 >, B = <4 ,-7 ,-1 > and C=A-B, what is the angle between A and C?

Dec 18, 2016

The answer is 79.2º

#### Explanation:

Lets start by calculating

vecC=vecA-vecB=〈3,1,-1〉-〈4,-7,-1〉=〈-1,8,0〉

The angle $\theta$ between $\vec{A}$ an $\vec{C}$ is given by the dot product definition

vecA.vecC=∥vecA∥*vecC∥*cos theta

The dot product =〈3,1,-1〉.〈-1,8,0〉=(3*-1+1*8+0)=5

The modulus of $\vec{A}$ is =∥vecA∥=∥〈3,1,-1〉∥=sqrt(9+1+1)=sqrt11

The modulus of $\vec{C}$ is =∥vecC∥=∥〈-1,8,0〉∥=sqrt(1+64+0)=sqrt65

The angle is costheta=(vecA.vecC)/(∥vecA∥*vecC∥)

$\cos \theta = \frac{5}{\sqrt{11} \cdot \sqrt{65}} = 0.187$

theta=79.2º