Question

Find the angle between the two vectors A=2.00i+3.00j+1.00k and B=4.00i+2.00j-1.00k

Answer 1

$\theta=\arccos\frac{\vec A\vec B}{|\vec A||\vec B|}=\arccos\frac{2\cdot 4+3\cdot 2-1\cdot 1}{\sqrt{2^2+3^2+1^2}\sqrt{4^2+2^2+(-1)^2}}=40.7°.$