Question

If `A = <4 ,1 ,8 >`, `B = <6 ,7 ,-2 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=51.7^@`

Explanation:

Start by calculating

`vecC=vecA-vecB`

`vecC=〈4,1,8〉-〈6,7,-2〉=〈-2,-6,10〉`

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=〈4,1,8〉.〈-2,-6,10〉=-8-6+80=66`

The modulus of `vecA`= `∥〈4,1,8〉∥=sqrt(16+1+64)=sqrt81=9`

The modulus of `vecC`= `∥〈-2,-6,10〉∥=sqrt(4+36+100)=sqrt140`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=66/(9*sqrt140)=0.6198`

`theta=arccos(0.6198)=51.7^@`