Question
If $|a \times b|\, = 4$ and $|a\,.\,b|\, = 2$, then $|a{|^2}\,\,|b{|^2} = $

Answer

d
(d) Given $|a \times b|\, = 4$ $ \Rightarrow \left| {\,|a||b|\sin \theta \,\,\hat n} \right| = 4$
$ \Rightarrow \left| {\,|a||b|\sin \theta \,} \right| = 4$ …..$(i)$
Also $|a\,.\,b|\, = 2$ $ \Rightarrow \left| {\,|a||b|\cos \theta \,} \right| = 2$
$ \Rightarrow \left| {\,|a||b|\cos \theta \,} \right| = 2$ …..$(ii)$
Now squaring and adding equation $(i)$ and $(ii),$ we get
$|a{|^2}\,.\,|b{|^2}{\sin ^2}\theta \, + |a{|^2}\,.\,|b{|^2}{\cos ^2}\theta = {4^2} + {2^2}$
$\therefore \,|a{|^2}\,.\,|b{|^2}({\sin ^2}\theta + {\cos ^2}\theta ) = 16 + 4$
$ \Rightarrow \,|a{|^2}\,.\,|b{|^2} \times \,1 = 20.$

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