Question

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

Answer 1

The angle is `=127`º

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈-6,2,1〉-〈-8,8,4〉=〈2,-6,-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=〈-6,2,1〉.〈2,-6,-3〉=-12-12-3=-27`

The modulus of `vecA`= `∥〈-6,2,1〉∥=sqrt(36+4+1)=sqrt41`

The modulus of `vecC`= `∥〈2,-6,-3〉∥=sqrt(4+36+9)=sqrt49=7`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=-27/(sqrt41*7)=-0.6`

`theta=127`º