Question

If `A = <4 ,1 ,5 >`, `B = <6 ,7 ,-2 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=69.9^@`

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈4,1,5〉-〈6,7,-2〉=〈-2,-6,7〉`

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=〈4,1,5〉.〈-2,-6,7〉=-8-6+35=21`

The modulus of `vecA`= `∥〈4,1,5〉∥=sqrt(16+1+25)=sqrt42`

The modulus of `vecC`= `∥〈-2,-6,7〉∥=sqrt(4+36+49)=sqrt89`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=21/(sqrt42*sqrt89)=0.34`

`theta=69.9^@`