Question

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

Answer 1

The angle is `=42.4^@`

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈6,-7,5〉-〈1,-5,-8〉=〈5,-2,13〉`

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,-7,5〉.〈5,-2,13〉=30+14+65=109`

The modulus of `vecA`= `∥〈6,-7,5〉∥=sqrt(36+49+25)=sqrt110`

The modulus of `vecC`= `∥〈5,-2,13〉∥=sqrt(25+4+169)=sqrt198`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=109/(sqrt110*sqrt198)=0.74`

`theta=42.4^@`