# If A = <-5 ,2 ,1 >, B = <-8 ,4 ,9 > and C=A-B, what is the angle between A and C?

Nov 17, 2016

The angle is $= 124.2$º

#### Explanation:

Let's calculate $\vec{C}$

$\vec{C} = \vec{A} - \vec{B}$

Therefore, vecC=〈-5,2,1〉-〈-8,4,9〉=〈3,-2,-8〉

The angle $\theta$ between $\vec{A}$ and $\vec{C}$ is calculated from the dot product definition.

vecA.vecC=∥vecA∥.∥vecC∥costheta

The dot product is
vecA.vecC=〈-5,2,1〉.〈3,-2,-8〉=-15-4-8=-27

The modulus of vecA=∥〈-5,2,1〉∥=sqrt(25+4+1)=sqrt30

The modulus of vecC=∥〈3,-2,-8〉∥=sqrt(9+4+64)=sqrt77

Therefore, costheta=vecA.vecC/(∥vecA∥.∥vecC∥)

$= - \frac{27}{\sqrt{30} \cdot \sqrt{77}} = - 0.56$

$\theta = 124.2$º