Question

If `A = <4 ,-9 ,2 >`, `B = <1 ,-3 ,-8 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=44.7^@`

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈4,-9,2〉-〈1,-3,-8〉=〈3,-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,-9,2〉.〈3,-6,10〉=12+54+20=86`

The modulus of `vecA`= `∥〈4,-9,2〉∥=sqrt(16+81+4)=sqrt101`

The modulus of `vecC`= `∥〈3,-6,10〉∥=sqrt(9+36+100)=sqrt145`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=86/(sqrt101*sqrt145)=0.74`

`theta=44.7^@`