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

Nov 12, 2016

The angle is $56.1$º

#### Explanation:

Let's calculate vecC=vecA-vecB=〈2,-5,-8〉-〈-9,4,-2〉@
=〈11,-9,-6〉

To calculate the angle $\theta$ between the 2 vectors, we use the dot product definition.

vecA.vecC=∥vecA∥*∥vecC∥*costheta

The dot product is vecA.vecC=〈2,-5,-8〉.〈11,-9,-2〉
$= 22 + 45 + 16 = 83$

The modulus of $\vec{A}$ = ∥vecA∥=∥〈2,-5,-8〉∥=sqrt(4+25+64)=sqrt93

The modulus of $\vec{C}$ = ∥vecC∥=∥〈11,-9,-6〉∥=sqrt(121+81+36)=sqrt238

$\cos \theta = \frac{83}{\sqrt{93} \cdot \sqrt{238}} = 0.56$

$\theta = 56.1$º