Question

If `A = <1 ,6 ,-8 >`, `B = <-9 ,4 ,0 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=48.7`º

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈1,6,-8〉-〈-9,4,0〉=〈10,2,-8〉`

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=〈1,6,-8〉.〈10,2,-8〉=10+12+64=86`

The modulus of `vecA`= `∥〈1,6,-8〉∥=sqrt(1+36+64)=sqrt101`

The modulus of `vecC`= `∥〈10,2,-8〉∥=sqrt(100+4+64)=sqrt168`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=86/(sqrt101*sqrt168)=0.66`

`theta=48.7`º