# If A = <2 ,6 ,-8 >, B = <-9 ,4 ,1 > and C=A-B, what is the angle between A and C?

Nov 9, 2016

The angle is theta=43.6º

#### Explanation:

vecC=vecA-vecB=〈2,6,-8〉-〈-9,4,1〉=〈11,2,-9〉
The angle, $\theta$ between $\vec{A}$ and $\vec{C}$ is given by the dot product definition.
vecA.vecC=∥vecA∥*∥vecC∥costheta
The dot product vecA.vecC=〈2,6,-8〉.〈11,2,-9〉=22+12+72=106
The modulus of vecA=∥〈2,6,-8〉∥=sqrt(4+36+64)=sqrt104
The modulus of vecC=∥〈11,2,-9〉∥=sqrt(121+4+81)=sqrt206
$\therefore \cos \theta = \frac{106}{\sqrt{104} \cdot \sqrt{206}} = 0.72$
theta=43.6º