# If A = <8 ,3 ,2 >, B = <6 ,-4 ,5 >, and C=A-B, what is the angle between A and C?

Nov 7, 2016

The angle is 63.3º

#### Explanation:

The angle between two vectors is given by the dot product.
veca.vecc=∥veca∥*∥vecc∥costheta
where $\theta$ is the angle between the two vectors
vecc=veca-vecb=〈8,3,2〉-〈6,-4,5〉=〈2,7,-3〉
The dot product is veca.vecc=〈8,3,2〉.〈2,7,-3〉=16+21-6=31
The modulus of $\vec{a}$ is =∥veca∥=∥〈8,3,2〉∥=sqrt(64+9+4)=sqrt77
The modulus of $\vec{c}$ is =∥vecc∥=∥〈2,7,-3〉∥=sqrt(4+49+9)=sqrt62

So $\cos \theta = \frac{31}{\sqrt{77} \cdot \sqrt{62}} = 0.449$
theta=63.3º