Question

find the angle between the given vectors to the nearest tenth of a degree. u = <2, -4>, v = <3, -8>

Answer 1

Angle θ between two vectors is given by, $\theta =cos^{-1}( \frac{u \cdot v}{|u||v|} )$; where $u \cdot v=2(3)+(-4)(-8)=6+32=38$; $|u|= \sqrt{ 2^{2} + (-4)^{2} } = \sqrt{4+16} = \sqrt{20}$ and $|v|=\sqrt{ 3^{2} + (-8)^{2} } = \sqrt{9+64} = \sqrt{73}$.
Therefore, $\theta =cos^{-1}( \frac{38}{ \sqrt{20} \times \sqrt{73} } )=cos^{-1}0.9945=6.0^o$