A) \[a=\pm \frac{1}{\sqrt{2}};b=\pm \frac{1}{\sqrt{6}};c=\pm \frac{1}{\sqrt{3}}\]
B) \[a=\pm \frac{1}{\sqrt{2}};b=\pm \frac{1}{\sqrt{3}};c=\pm \frac{1}{\sqrt{6}}\]
C) \[a=\pm \frac{1}{\sqrt{6}};b=\pm \frac{1}{\sqrt{2}};c=\pm \frac{1}{\sqrt{3}}\]
D) \[a=\pm \frac{1}{\sqrt{3}};b=\pm \frac{1}{\sqrt{2}};c=\pm \frac{1}{\sqrt{6}}\]
Correct Answer: A
Solution :
[a] Let \[A=\left[ \begin{matrix} 0 & 2b & c \\ a & b & -c \\ a & -b & c \\ \end{matrix} \right]\], Now, \[{{A}^{T}}=\left[ \begin{matrix} 0 & a & a \\ 2b & b & -b \\ c & -c & c \\ \end{matrix} \right]\] \[\because \] A is orthogonal \[\therefore \,A{{A}^{T}}=I\] \[\Rightarrow \left[ \begin{matrix} 0 & 2b & c \\ a & b & -c \\ a & -b & c \\ \end{matrix} \right]\left[ \begin{matrix} 0 & a & a \\ 2b & b & -b \\ c & -c & c \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\] Equating the corresponding elements, we get \[4{{b}^{2}}+{{c}^{2}}=1...(1)\] \[2{{b}^{2}}-{{c}^{2}}=0...(2)\] \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1...(3)\] On solving (1), (2) and (3), we get \[a=\pm \frac{1}{\sqrt{2}};b=\pm \frac{1}{\sqrt{6}};c=\pm \frac{1}{\sqrt{3}}\]
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