Question

If `A = <8 ,1 ,-7 >`, `B = <6 ,-2 ,5 >`, and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=39.7`º

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈8,1,-7〉-〈6,-2,5〉=〈2,3,-12〉`

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=〈8,1,-7〉.〈2,3,-12〉=16+3+84=103`

The modulus of `vecA`= `∥〈8,1,-7〉∥=sqrt(64+1+49)=sqrt114`

The modulus of `vecC`= `∥〈2,3,-12〉∥=sqrt(4+9+144)=sqrt157`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=103/(sqrt114*sqrt157)=0.77`

`theta=arccos(0.77)=39.7`º