Question

If `A = <6 ,-7 ,8 >`, `B = <9 ,-5 ,-2 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=54.1`º

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈6,-7,8〉-〈9,-5,-2〉=〈-3,-2,10〉`

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,-7,8〉.〈-3,-2,10〉=-18+14+80=76`

The modulus of `vecA`= `∥〈6,-7,8〉∥=sqrt(36+49+64)=sqrt149`

The modulus of `vecC`= `∥〈-3,-2,10〉∥=sqrt(9+4+100)=sqrt113`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=76/(sqrt149*sqrt113)=0.59`

`theta=54.1`º