Question

If `A = <4 ,2 ,5 >`, `B = <6 ,2 ,-7 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `50.4`º

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈4,2,5〉-〈6,2,-7〉=〈-2,0,12〉`

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=〈4,2,5〉.〈-2,0,12〉=-8+0+60=52`

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

The modulus of `vecC`= `∥〈-2,0,12〉∥=sqrt(4+0+144)=sqrt148`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=52/(sqrt45*sqrt148)=0.637`

`theta=50.4`º