Question

If `A = <4 ,2 ,1 >`, `B = <-8 ,4 ,9 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The answer is `=57.3`º

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈4,2,1〉-〈-8,4,9〉=〈12,-2,-8〉`

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=〈4,2,1〉.〈12,-2,-8〉=48-4-8=36`

The modulus of `vecA`= `∥〈4,2,1〉∥=sqrt(16+4+1)=sqrt21`

The modulus of `vecC`= `∥〈12,-2,-8〉∥=sqrt(144+4+64)=sqrt212`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=36/(sqrt21*sqrt212)=0.54`

`theta=57.3`º