Question

If `A = <8 ,0 ,4 >`, `B = <6 ,7 ,4 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=75.8º`

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈8,0,4〉-〈6,7,4〉=〈2,-7,0〉`

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=〈8,0,4〉.〈2,-7,0〉=16+0+0=16`

The modulus of `vecA`= `∥〈8,0,4〉∥=sqrt(64+0+16)=sqrt80`

The modulus of `vecC`= `∥〈2,-7,0〉∥=sqrt(4+49+0)=sqrt53`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=16/(sqrt80*sqrt53)=0.25`

`theta=75.8`º