Question

Find the angle between two vectors a and b with magnitudes $\sqrt{3}$ and 2 respectively, having $a.b=\sqrt{6}$

Answer 1

It is given that $|a|=\sqrt{3},|b|=2$ and $a.b=\sqrt{6}$
Let $\theta$ be the required angle, then
$cos\theta =\frac{a.b}{|a||b|}=\frac{\sqrt{6}}{\sqrt{3}\times 2}=\frac{1}{\sqrt{2}}\Rightarrow \theta =co{s}^{-1}\left(\frac{1}{\sqrt{2}}\right)=co{s}^{-1}\left(cos\frac{\pi }{4}\right)=\frac{\pi }{4}$
Hence, the angle between the given vectors a and b is $\frac{\pi }{4}$