Question

What is the angle between `<6,-2,9>` and `< -2,1,-8 >`?

Answer 1

The angle is `=160.3^@`

Explanation:

The angle between `2` vectors, `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=〈6,-2,9〉.〈-2,1,-8〉=-12-2-72=-86`

The modulus of `vecA`= `∥〈6,-2,9〉∥=sqrt(36+4+81)=sqrt121=11`

The modulus of `vecB`= `∥〈-2,1,-8〉∥=sqrt(4+1+64)=sqrt69`

So,

`costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=-86/(11*sqrt69)=-0.94`

`theta=arccos(-0.94)=160.3^@`