A) \[\sin {{\,}^{-1}}\frac{1}{3}\]
B) \[\cos {{\,}^{-1}}\frac{1}{14}\]
C) \[sin{{\,}^{-1}}\frac{9}{14}\]
D) \[sin{{\,}^{-1}}\frac{2}{3}\]
Correct Answer: C
Solution :
[c] \[\vec{b}\times \vec{c}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & 1 & 2 \\ 1 & 3 & 3 \\ \end{matrix} \right|=-3\hat{i}-\hat{j}+2\hat{k}\] If \[\theta \] is the angle between \[\overset{\to }{\mathop{a}}\,\] and the plane containing \[\overset{\to }{\mathop{b}}\,\] and \[\overset{\to }{\mathop{c}}\,\], then \[\cos (90{}^\circ -\theta )=\left| \frac{\overset{\to }{\mathop{a}}\,.(\overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{c}}\,)}{|\overset{\to }{\mathop{a}}\,||\overset{\to }{\mathop{b}}\,\times \overset{\to }{\mathop{c}}\,|} \right|\] \[=\frac{1}{\sqrt{14}}.\frac{1}{\sqrt{14}}|(-9-2+2)|=\frac{9}{14}\] \[\Rightarrow \sin \theta =\frac{9}{14}\Rightarrow \theta ={{\sin }^{-1}}\left( \frac{9}{14} \right).\]
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