A) \[{{\sin }^{-1}}\left\{ \cos \left( {{\sin }^{-1}}\sqrt{\frac{2-\sqrt{3}}{4}} \right. \right.\]
B) \[\left. \left. +{{\cos }^{-1}}\frac{\sqrt{12}}{4}+{{\sec }^{-1}}\sqrt{2} \right) \right\}\]
C) \[\frac{\pi }{4}\]
D) None of these
Correct Answer: C
Solution :
The lines are parallel to the vectors \[{{b}_{1}}=\hat{i}+2\hat{j}-\hat{k}\]and\[{{b}_{2}}=-\hat{i}+\hat{j}-2\hat{k}\] \[\therefore \]The plane normal to the vector n is \[{{b}_{1}}\times {{b}_{2}}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & 2 & -1 \\ -1 & 1 & -2 \\ \end{matrix} \right|=-3\hat{i}+3\hat{j}+3\hat{k}\] The required plane passes through \[(\hat{i}+\hat{j})\] and is normal to the vector n. \[\therefore \] Its equation is r ? n = a - n \[\Rightarrow \]\[r.(-3\hat{i}+3\hat{j}+3\hat{k})=(\hat{i}+\hat{j}).(-3\hat{i}+3\hat{j}+3\hat{k})\] \[\Rightarrow \]\[r.(-3\hat{i}+3\hat{j}+3\hat{k})=-3+3\] \[\Rightarrow \]\[r.(-\hat{i}+\hat{j}+\hat{k})=0\]
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