Question

Find the angle between u=<7,6> and v=<4,2>

Answer 1

\(u\cdot v=7*4+6*2=28+12=40\)

also

\(u\cdot v=||u||||v||\cos(\theta)\)

so

\(40=\sqrt{7^2+6^2}\sqrt{4^2+2^2}\cos(\theta)\\40=\sqrt{75}\sqrt{20}\cos(\theta)\\\frac{40}{\sqrt{75}\sqrt{20}}=\cos(\theta)\\\theta=\cos^{-1}(\frac{40}{\sqrt{75}\sqrt{20}})\).

Answer 2

This is the nice thing about having two definitions for the dot product.

\(v=<v_1,v_2>,u=<u_1,u_2>\)

then \(u\circ v=v_1u_1+v_1v_2\)

and

\(u\circ v=||u||||v||\cos(\theta)\)