# Find the angle between two vectors a and b with magnitudes sqrt3? and 2, respectively having a?b=sqrt6

$It\;is\;given\;that\\*|\vec{a}|=\sqrt3,\;|\vec{b}|=2\;and\;\vec{a}.\vec{b}=\sqrt6\\* \Rightarrow \vec{a}.\vec{b}=|a||b|\cos \theta\\*\therefore \sqrt6=\sqrt3\times 2 \cos \theta\Rightarrow \cos \theta=\frac{\sqrt6}{\sqrt3\times 2}=\frac{1}{\sqrt2}\\*\Rightarrow \cos \theta=\frac{1}{\sqrt2}\\*\Rightarrow \theta=\frac{\pi}{4}\\*Hence,\;the\;angle\;between\;the\;given\;vectors\; \vec{a}\;and\;\vec{b}\;is\;\frac{\pi}{4}$

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