Question

If `A = <2 ,6 ,-8 >`, `B = <-9 ,4 ,1 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `theta=43.6º`

Explanation:

`vecC=vecA-vecB=〈2,6,-8〉-〈-9,4,1〉=〈11,2,-9〉`
The angle, `theta` between `vecA` and `vecC` is given by the dot product definition.
`vecA.vecC=∥vecA∥*∥vecC∥costheta`
The dot product `vecA.vecC=〈2,6,-8〉.〈11,2,-9〉=22+12+72=106`
The modulus of `vecA=∥〈2,6,-8〉∥=sqrt(4+36+64)=sqrt104`
The modulus of `vecC=∥〈11,2,-9〉∥=sqrt(121+4+81)=sqrt206`
`:. costheta=106/(sqrt104*sqrt206)=0.72`
`theta=43.6º`