How do I find the angle θ between two vectors?

My givens are this:
u = <3, 2>
v = <4, 0>

1 Answer

theta=cos^-1(3/sqrt13)

Explanation:

Given:
vecu=3i+2j
vecv=4i+0j

Consider the dot product
vecu.vecv=|vecu|.|vecv|.costheta
where theta represents the angle between the vectors vecu and vecv

vecu.vecv=(3)(4)+(2)(0)=12+0=12

|vecu|=sqrt(3^2+2^2)=sqrt(9+4)=sqrt13

|vecv|=sqrt(4^2+0^2)=sqrt(16+0)=sqrt16=4

Substituting

12=(sqrt13)(4)costheta

Interchanging lhs and rhs

(sqrt13)(4)costheta=12

costheta=12/(4sqrt13)=3/sqrt13

costheta=3/sqrt13

Hence,
theta=cos^-1(3/sqrt13)