# If A = <5 ,6 ,-3 >, B = <-9 ,6 ,-5 > and C=A-B, what is the angle between A and C?

Nov 14, 2016

The angle is $= 57.3$º

#### Explanation:

Let's start calculating $\vec{C}$

vecC=vecA-VecB=〈5,6,-3〉-〈-9,6,-5〉
=〈14,0,2〉

To calculate the angle $\theta$ between $\vec{A}$ and $\vec{C}$,
we use the dot product definition:

vecA.vecC=∥vecA∥*∥vecC∥costheta

The dot product =〈5,6,-3〉.〈14,0,2〉=70+0-6=64

The modulus of vecA=∥〈5,6,-3〉∥=sqrt(25+36+9)=sqrt70

The modulus of vecC=∥〈14,0,2〉∥=sqrt(196+0+4)=sqrt200

:. cos theta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=64/(sqrt70sqrt200)

$\cos \theta = 0.54$

$\theta = 57.3$º