Question

If `A = <2 ,-5 ,9 >`, `B = <-9 ,1 ,-5 >` and `C=A-B`, what is the angle between A and C?

Answer 1

The angle is `=25.4^@`

Explanation:

Let's start by calculating

`vecC=vecA-vecB`

`vecC=〈2,-5,9〉-〈-9,1,-5> = <11,-6,14>`
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=〈2,-5,9〉.〈11,-6,14〉=22+30+126=178`

The modulus of `vecA`= `∥〈2,-5,9〉∥=sqrt(4+25+81)=sqrt110`

The modulus of `vecC`= #∥〈11,-6,14〉 ∥=sqrt(121+36+196)#

`=sqrt353`

So,

`costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=178/(sqrt110*sqrt353)=0.903`

`theta=25.4^@`