Question

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

Answer 1

The angle is `=56`º

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈5,2,-5〉-〈6,5,9〉=〈-1,-3,-14〉`

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=〈5,2,-5〉.〈-1,-3,-14〉=-5-6+70=59`

The modulus of `vecA`= `∥〈5,2,-5〉∥=sqrt(25+4+25)=sqrt54`

The modulus of `vecC`= `∥〈-1,-3,-14〉∥=sqrt(1+9+196)=sqrt206`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=59/(sqrt54*sqrt206)=0.56`

`theta=56`º