Question

If `A = <-9 ,2 ,5 >`, `B = <-8 ,3 ,4 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=48.7^@`

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

#vecC=〈-9,2,5〉-〈-8,3,4〉=〈-1,-1,1>

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=〈-9,2,5〉.〈-1,-1,1〉=9-2+5=12`

The modulus of `vecA`= `∥〈-9,2,5〉∥=sqrt(81+4+25)=sqrt110`

The modulus of `vecC`= `∥〈-1,-1,1〉∥=sqrt(1+1+1)=sqrt3`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=12/(sqrt110*sqrt3)=0.66`

`theta=48.7^@`