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

Nov 11, 2016

The angle is =77.6º

#### Explanation:

Let'scalculate vecC=vecA-vecB=〈1,6,-8〉-〈-9,4,-8〉=〈10,2,0〉

The angle is given by the dot product definition
vecA.vecC=∥vecA∥*∥vecC∥costheta
where $\theta$ is the angle between the two vector.

vecA.vecC=〈1,6,-8〉.〈10,2,0〉=10+12-0=22

The modulus of vecA=∥〈1,6,-8〉∥=sqrt(1+36+64)=sqrt101

The modulus of vecC=∥〈10,2,0〉∥=sqrt(100+4)=sqrt104

So, $\cos \theta = \frac{22}{\sqrt{101} \cdot \sqrt{104}} = 0.214$

theta=77.6º