Question

If `A = <5 ,6 ,-3 >`, `B = <-9 ,6 ,-2 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=51.6^@`

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈5,6,-3〉-〈-9,6,-2〉=〈14,0,-1〉`

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=〈5,6,-3〉.〈14,0,-1〉=70+0+3=73`

The modulus of `vecA`= `∥〈5,6,-3〉∥=sqrt(25+36+9)=sqrt70`

The modulus of `vecC`= `∥〈14,0,-1〉∥=sqrt(196+0+1)=sqrt197`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=73/(sqrt70*sqrt197)=0.62`

`theta=51.6^@`