Angle θ between two vectors is given by, \(\theta =cos^{-1}( \frac{u
\cdot v}{|u||v|} )\); where \(u \cdot
v=-5(-4)+(-4)(-3)=20+12=32\); \(|u|= \sqrt{ (-5)^{2} + (-4)^{2} } =
\sqrt{25+16} = \sqrt{41}\) and \(|v|=\sqrt{ (-4)^{2} + (-3)^{2} } =
\sqrt{16+9} = \sqrt{25}=5\).
Therefore, \(\theta =cos^{-1}( \frac{32}{ 5\sqrt{41}} )=cos^{-1}0.9995=1.8^o\)
