A) \[\tan {{\,}^{-1}}\left( \frac{1}{\sqrt{3}} \right);\,\tan {{\,}^{-1}}\left( \frac{1}{2} \right);\,\tan {{\,}^{-1}}\left( \frac{\sqrt{3}+2}{1-2\sqrt{3}} \right)\]
B) \[\tan {{\,}^{-1}}\left( \sqrt{3} \right);\,\tan {{\,}^{-1}}\left( \frac{1}{\sqrt{3}} \right);\,\cot {{\,}^{-1}}\left( 0 \right)\]
C) \[\tan {{\,}^{-1}}\left( \sqrt{3} \right);\,\tan {{\,}^{-1}}\left( 2 \right);\,\tan {{\,}^{-1}}\left( \frac{\sqrt{3}+2}{2\sqrt{3}-1} \right)\]
D) \[\tan {{\,}^{-1}}\left( \sqrt{3} \right);\,\tan {{\,}^{-1}}\left( \sqrt{2} \right);\,\tan {{\,}^{-1}}\left( \frac{\sqrt{2}+3}{3\sqrt{2}-1} \right)\]
Correct Answer: B
Solution :
[b] Let \[\overset{\to }{\mathop{x}}\,=\sqrt{3}(\overset{\to }{\mathop{a}}\,\times \overset{\to }{\mathop{b}}\,)\] and \[\overset{\to }{\mathop{y}}\,=\overset{\to }{\mathop{b}}\,-(\overset{\to }{\mathop{a}}\,\cdot \overset{\to }{\mathop{b}}\,)\overset{\to }{\mathop{a}}\,\] Clearly \[\overset{\to }{\mathop{x}}\,\cdot \overset{\to }{\mathop{y}}\,=0\Rightarrow \overset{\to }{\mathop{x}}\,\] and \[\overset{\to }{\mathop{y}}\,\] are perpendicular So, one angle is \[\frac{\pi }{2}\]. Also \[|\overset{\to }{\mathop{x}}\,|=\sqrt{3}|b\,\,sin\,\,\theta |\], where \[\theta \] is angle between vectors \[\overset{\to }{\mathop{a}}\,\] and \[\overset{\to }{\mathop{b}}\,\] \[(|\vec{a}|=1)\] \[|\overset{\to }{\mathop{y}}\,|=\sqrt{{{\left\{ \vec{b}-(\vec{a}.\vec{b})\vec{a} \right\}}^{2}}}=\sqrt{{{b}^{2}}-{{(\vec{a}.\vec{b})}^{2}}}\] \[=\sqrt{{{b}^{2}}-{{b}^{2}}\cos \theta }=\left| b\,\,\sin \,\,\theta \right|\]
\[\therefore \frac{|\overset{\to }{\mathop{x}}\,|}{|\overset{\to }{\mathop{y}}\,|}=\sqrt{3}=\tan \alpha \Rightarrow \alpha =\frac{\pi }{3}\]. So, \[\beta =\frac{\pi }{6}\]
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