Question
The values of $$a, b, c$$ if \[\left[ {\begin{array}{*{20}{c}}
0&{2b}&c\\
a&b&{ - c}\\
a&{ - b}&c
\end{array}} \right]\] is orthogonal are
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.
$$a = \pm \frac{1}{{\sqrt 3 }};b = \pm \frac{1}{{\sqrt 2 }};c = \pm \frac{1}{{\sqrt 6 }}$$
Answer :
$$a = \pm \frac{1}{{\sqrt 2 }};b = \pm \frac{1}{{\sqrt 6 }};c = \pm \frac{1}{{\sqrt 3 }}$$
Solution :
Let, \[A = \left[ {\begin{array}{*{20}{c}}
0&{2b}&c\\
a&b&{ - c}\\
a&{ - b}&c
\end{array}} \right]\]
Now, \[{A^T} = \left[ {\begin{array}{*{20}{c}}
0&a&a\\
{2b}&b&{ - b}\\
c&{ - c}&c
\end{array}} \right]\]
∵ $$A$$ is orthogonal
∴ $$A{A^T} = I$$
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}
0&{2b}&c\\
a&b&{ - c}\\
a&{ - b}&c
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0&a&a\\
{2b}&b&{ - b}\\
c&{ - c}&c
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&0&0\\
0&1&0\\
0&0&1
\end{array}} \right]\]
Equating the corresponding elements, we get,
$$4{b^2} + {c^2} = 1\,\,\,\,.....\left( 1 \right)$$
$$\eqalign{
& 2{b^2} - {c^2} = 0\,\,\,\,.....\left( 2 \right) \cr
& {a^2} + {b^2} + {c^2} = 1\,\,\,\,.....\left( 3 \right) \cr} $$
On solving (1), (2) and (3), we get,
$$a = \pm \frac{1}{{\sqrt 2 }};b = \pm \frac{1}{{\sqrt 6 }};c = \pm \frac{1}{{\sqrt 3 }}$$