Question

If `A = <6 ,8 ,-3 >`, `B = <7 ,1 ,-4 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `50.9º`

Explanation:

Let's start by calculating `vecC=vecA-vecB`
`vecC=〈6,8,-3〉-〈7,1,-4〉=〈-1,7,1〉`
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` an `vecC`

`:. cos theta=vecA.vecC/(∥vecA∥*∥vecC∥)`
The dot product is `vecA.vecC=〈6,8,-3〉.〈-1,7,1〉=-6+56-3=47`
The modulus of `vecA=∥vecA∥=sqrt(36+64+9)=sqrt109`
The modulus of `vecC=∥vecC∥=sqrt(1+49+1)=sqrt51`
So `costheta=47/(sqrt109*sqrt51)=0.63`
`theta=50.9º`