Find the angle between two vectors a and b with magnitudes `sqrt(3)` and 2 respectively, having `a.b=sqrt(6)`.

Explanation

It is given that `a|=sqrt(3),|b|=2 and ab=sqrt(6)`.
Let `theta` be the required angle , then
`cos theta =(a.b)/(|a||b|)=(sqrt(6))/(sqrt(3)xx2)=(1)/(sqrt(2))rArr theta=cos^(-1)((1)/(sqrt(2)))=cos^(-1)("cos"(pi)/(4))=(pi)/(4)`.
Hence , the angle between the given vectors a and b is `pi//4`.