Question

If `A = <2 ,-8 ,5 >`, `B = <5 ,-7 ,-5 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=59.1º`

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈2,-8,5〉-〈5,-7,-5〉=〈-3,-1,10〉`

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=〈2,-8,5〉.〈-3,-1,10〉=-6+8+50=52`

The modulus of `vecA`= `∥〈2,-8,5〉∥=sqrt(4+64+25)=sqrt93`

The modulus of `vecC`= `∥〈-3,-1,10〉∥=sqrt(9+1+100)=sqrt110`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=52/(sqrt93*sqrt110)=0.51`

`theta=59.1`º