Question

If `A = <2 ,-5 ,-4 >`, `B = <-9 ,1 ,-6 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `58.9`º

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈2,-5,-4〉-〈-9,1,-6〉=〈11,-6,2〉`

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=〈2,-5,-4〉.〈11,-6,2〉=22+30-8=44`

The modulus of `vecA`= `∥〈2,-5,-4〉∥=sqrt(4+25+16)=sqrt45`

The modulus of `vecC`= `∥〈11,-6,2〉∥=sqrt(121+36+4)=sqrt161`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=44/(sqrt45*sqrt161)=0.52`

`theta=58.9`º