Question

What is the angle between `<-1,0,6 >` and `< 2,-1,5 >`?

Answer 1

The angle is `=32.8^@`

Explanation:

The angle between `vecA` and `vecB` is given by the dot product definition.

`vecA.vecB=∥vecA∥*∥vecB∥costheta`

Where `theta` is the angle between `vecA` and `vecB`

The dot product is

`vecA.vecB=〈-1,0,6〉.〈2,-1,5〉=-2+0+30=28`

The modulus of `vecA`= `∥〈-1,0,6〉∥=sqrt(1+0+36)=sqrt37`

The modulus of `vecB`= `∥〈2,-1,5〉∥=sqrt(4+1+25)=sqrt30`

So,

`costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=28/(sqrt37*sqrt30)=0.84`

`theta=arccos(0.84)=32.8^@`