Question

If `A = <6 ,4 ,9 >`, `B = <7 ,-1 ,5 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=42.9^@`

Explanation:

Start by calculating

`vecC=vecA-vecB`

`vecC=〈6,4,9〉-〈7,-1,5〉=〈-1,5,4〉`

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=〈6,4,9〉.〈-1,5,4〉=-6+20+36=50`

The modulus of `vecA`= `∥〈6,4,9〉∥=sqrt(36+16+81)=sqrt133`

The modulus of `vecC`= `∥〈-1,5,4〉∥=sqrt(1+25+9)=sqrt35`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=50/(sqrt133*sqrt35)=0.7328`

`theta=arccos(0.7328)=42.9^@`