Question

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

Answer 1

The angle is `=26.6`º

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈6,-7,8〉-〈1,-5,-2〉=〈5,-2,10〉`

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=〈6,-7,8〉.〈5,-2,10〉=30+14+80=124`

The modulus of `vecA`= `∥〈6,-7,8〉∥=sqrt(36+49+64)=sqrt149`

The modulus of `vecC`= `∥〈5,-2,10〉∥=sqrt(25+4+100)=sqrt129`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=124/(sqrt149*sqrt129)=0.894`

`theta=26.6`º