Question

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

Answer 1

The angle is `=131^@`

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈-6,2,1〉-〈-8,7,2〉=〈2,-5,-1〉`

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=〈-6,2,1〉.〈2,-5,-1〉 = (-6)* (2)+(2)* (-5)+(1)* (-1)`

`=-12-10-1=-23`

The modulus of `vecA`= `∥〈-6,2,1〉∥=sqrt( (-6)^2+(2)^2+1^2)=sqrt(36+4+1)=sqrt41`

The modulus of `vecC`= `∥〈2,-5,-1〉∥=sqrt(4+25+1)=sqrt30`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=-23/(sqrt41*sqrt30)=-0.656`

`theta=arccos(-0.656)=131^@`