Question

If `A = <1 ,6 ,9 >`, `B = <-9 ,-6 ,7 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=54.2`º

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈1,6,9〉-〈-9,-6,7〉=〈10,12,2〉`

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=〈1,6,9〉.〈10,12,2〉=10+72+18=100`

The modulus of `vecA`= `∥〈1,6,9〉∥=sqrt(1+36+81)=sqrt118`

The modulus of `vecC`= `∥〈10,12,2〉∥=sqrt(100+144+4)=sqrt248`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=100/(sqrt118*sqrt248)=0.58`

`theta=54.2`º