Question

If A=3i+2j+3k and B=i-j+2k then what is angel between them?

Answer 1

The angle is `=52.5^@`

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=〈3,2,3〉.〈1,-1,2〉=(3)*(1)+(2)*(-1)+(3)*(2)=7`

The modulus of `vecA`= `∥〈3,2,3〉∥=sqrt(9+4+9)=sqrt22`

The modulus of `vecB`= `∥〈1,-1,2〉∥=sqrt(1+1+4)=sqrt6`

So,

`costheta=(vecA.vecB)/(∥vecA∥*∥vecB∥)=7/(sqrt22*sqrt6)=0.609`

`theta=arccos(0.609)=52.5^@`