Question

If `A = <8 ,3 ,2 >`, `B = <6 ,-1 ,5 >`, and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=62.3`º

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈8,3,2〉-〈6,-1,5〉=〈2,4,-3〉`

The angle between `vecA` and `vecC` is given by the dot product definition.

`vecA.vecC=∥vecA∥*∥vecC∥costheta`

Where `theta` is the angle between `vecA` and `vecC`

The dot product is

`vecA.vecC=〈8,3,2〉.〈2,4,-3〉=16+12-6=22`

The modulus of `vecA`= `∥〈8,3,2〉∥=sqrt(64+9+4)=sqrt77`

The modulus of `vecC`= `∥〈2,4,-3〉∥=sqrt(4+16+9)=sqrt29`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=22/(sqrt77*sqrt29)=0.47`

`theta=62.3`º