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

Nov 15, 2016

The angle is $49.5$º

#### Explanation:

Let's calculate $\vec{C}$

vecC=vecA-vecB=〈2,-5,-8〉-〈-9,4,-5〉

=〈11,-9,-3〉

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

vecA.vecC=∥vecA∥*∥vecC∥costheta

The dot product vecA.vecC=〈2,-5,-8〉.〈11,-9,-3〉

$= 2 \cdot 11 + 5 \cdot 9 + 8 \cdot 3 = 22 + 45 + 24 = 91$

The modulus of ∥vecA∥=∥〈2,-5,-8〉∥=sqrt(4+25+64)=sqrt93

The modulus of ∥vecC∥=∥〈11,-9,-3〉∥=sqrt(121+81+9)=sqrt211

Therefore, $\cos \theta = \frac{91}{\sqrt{93} \cdot \sqrt{211}} = 0.65$

$\theta = 49.5$º