Question

What is the angle between `<0,9,-6 >` and `<1,3,5>`?

Answer 1

The angle is `=92.7^@`

Explanation:

The vectors are `vecA= < 0,9,-6>` and `vecB= <1,3,5>`

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=〈0,9,-6〉.〈1,3,5〉=(0xx1)+(9xx3)+(-6xx5)=0+27-30=-3`

The modulus of `vecA`= `∥〈0,9,-6〉∥=sqrt(0+9^2+(-6)^2)=sqrt(81+36)=sqrt(117)`

The modulus of `vecB`= `∥〈1,3,5〉∥=sqrt(1^2+3^2+5^2)=sqrt(1+9+25)=sqrt35`

So,

`costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=-3/(sqrt117*sqrt35)=-0.0469`

`theta=arccos(-0.0469)=92.7^@`