$\,|\vec a - \,\,\vec b {|^2}\,\, = \,\,|\,\vec a {|^2}\,\, + \;\,|\,\,\vec b {|^2}\,\, - \,\,2\,\,\left( {\vec a .\,\,\vec b } \right)$
$ \Rightarrow \,\,|\vec a - \,\,\vec b {|^2}\,\, = \,\,|\,\vec a {|^2}\,\, + \;\,|\,\,\vec b {|^2}\,\, - \,\,2\,\,\,|\,\vec a |\,\,\,\,|\,\,\vec b |\,\,\,\cos \,\,2\theta $
$ \Rightarrow \,|\vec a - \,\,\vec b {|^2}\,\, = \,\,2\,\, - \,\,2\,\cos \,\,2\theta \,\,\left[ {\because \,\,|\vec a |\,\, = \,|\,\,\vec b |\,\, = \,\,1\,} \right]$
$ \Rightarrow \,|\vec a - \,\,\vec b {|^2}\,\, = \,\,4{\sin ^2}\,\theta $
હવે , $|\vec a - \,\,\vec b {|^2}\,\, = \,\,2\,\,|\sin \,\,\theta |\,\,$
$\,\,\,|\vec a - \,\,\vec b |\,\,\, < \,\,1\,$
$ \Rightarrow \,\,2\,\,|\sin \,\,\theta |\,\, < \,\,1\,\,\,\, \Rightarrow \,\,|\sin \,\,\theta |\,\, < \,\,\frac{1}{2}$
$ \Rightarrow \,\,\theta \,\, \in \,\,[0,\,\,\pi /6)\,\, \cup \,\,(5\pi /6,\,\,\pi ]$
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