Question

If `A = <3 ,-1 ,-8 >`, `B = <5 ,6 ,-9 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=96.3^@`

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈3,-1,-8〉-〈5,6,-9〉=〈-2,-7,1〉`

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=〈3,-1,-8〉.〈-2-7,1〉=-6+7-8=-7`

The modulus of `vecA`= `∥〈3,-1,-8〉∥=sqrt(9+1+64)=sqrt74`

The modulus of `vecC`= `∥〈-2,-7,1〉∥=sqrt(4+49+1)=sqrt54`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=-7/(sqrt74*sqrt54)=-0.11`

`theta=arccos (-0.11)=96.3^@`