Question

If `A = <7 ,-5 ,-2 >`, `B = <4 ,-8 ,9 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=74.4^@`

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈7,-5,-2〉-〈4,-8,9〉=〈3,3,-11〉`

The angle between the `2` vectors, `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=〈7,-5,-2〉.〈3,3,-11〉=21-15+22=28`

The modulus of `vecA`= `∥〈7,-5,-2〉∥=sqrt(49+25+4)=sqrt78`

The modulus of `vecC`= `∥〈3,3,-11〉∥=sqrt(9+9+121)=sqrt139`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=28/(sqrt78*sqrt139)=0.27`

`theta=arccos(0.27)=74.4^@`