Question

If `A = <8 ,9 ,4 >`, `B = <6 ,-9 ,-1 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=33.4^@`

Explanation:

Start by calculating

`vecC=vecA-vecB`

`vecC=〈8,9,4〉-〈6,-9,-1〉=〈2,18,6〉`

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=〈8,9,4〉.〈2,18,6〉=16+162+24=202`

The modulus of `vecA`= `∥〈8,9,4〉∥=sqrt(64+81+16)=sqrt161`

The modulus of `vecC`= `∥〈2,18,6〉∥=sqrt(4+324+36)=sqrt364`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=202/(sqrt161*sqrt364)=0.83`

`theta=arccos(0.83)=33.4^@`