Question

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

Answer 1

The angle is `=66.6`º

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈4,2,5〉-〈-8,2,8〉=〈12,0,-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=〈4,2,5〉.〈12,0,-3〉=48+0-15=33`

The modulus of `vecA`= `∥〈4,2,5〉∥=sqrt(16+4+25)=sqrt45`

The modulus of `vecC`= `∥〈12,0,-3〉∥=sqrt(144+0+9)=sqrt153`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=33/(sqrt45*sqrt153)=0.4`

`theta=66.6`º