Find orthogonal operator to satisfy the transformation everyone,
here I have a question as shown in the figure.
firstly ,I assume the standard  matrix for the operator to be $A=[a_1, a_2, a_3]$
,and I know the property that 
transpose of A=inverse of A
However,I don't know the next step to solve A
can anyone help me?thanks!

 A: Let
$$ \vec{i}=\left[ \begin{array}{c} 1\\0\\0\end{array}\right],\quad 
\vec{j}=\left[ \begin{array}{c} 0\\1\\0\end{array}\right],\quad \text{and}\quad
\vec{k}=\left[ \begin{array}{c} 0\\0\\1\end{array}\right].$$
Also, let
$$ \vec{u}=\left[ \begin{array}{c} \frac{1}{\sqrt{2}}\\0\\-\frac{1}{\sqrt{2}}\end{array}\right]. $$
Note that these vectors have norm $1.$ We are looking for an orthogonal matrix $T$ such that
$T\vec{u}=\vec{j}$. What we have to do is to find two more vectors with norm $1$, say $\vec{v}$ and $\vec{w}$, such that $\vec{u}, \vec{v}, \vec{w}$ are pairwise orthogonal. A matrix which maps three orthonormal vectors into three orthonormal vectors must be orthogonal.
It is obvious that $\vec{j}$ is orthogonal to $\vec{u}$. So let $\vec{v}=\vec{j}$. It is
easy now to determine $\vec{w}$. It has to be orthogonal to $\vec{u}$ and $\vec{v}$. But vector with this property is $\vec{u}\times \vec{v}$. Hence, let
$$ \vec{u}=\left[ \begin{array}{c} \frac{1}{\sqrt{2}}\\0\\-\frac{1}{\sqrt{2}}\end{array}\right], \quad \vec{v}=\left[ \begin{array}{c} 0\\1\\0\end{array}\right],\quad \text{and}\quad
\vec{w}=\left[ \begin{array}{c} \frac{1}{\sqrt{2}}\\0\\\frac{1}{\sqrt{2}}\end{array}\right]. $$
These are three orthonormal vectors and we want that $T$ maps them to $\vec{i}, \vec{j}, \vec{k}$ in such a way that $T\vec{u}=\vec{j}$. Hence we some freedom to choose where $\vec{v}$ and $\vec{w}$ are mapped. Let $T\vec{v}=\vec{i}$ and $T\vec{w}=\vec{k}$. This $T$ is an orthogonal matrix as it maps one orthonormal basis into another orthonormal basis. Hence the transpose maps as follows:
$$ T^T\vec{i}=\vec{v}, \quad T^T\vec{j}=\vec{u}, \quad \text{and}\quad T\vec{k}=\vec{w}. $$
From this we see that 
$$ T^T=\left[ \begin{array}{ccc} 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\
1 & 0 & 0 \\
0 &-\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}
\end{array} \right] $$
and consequently
$$ T=\left[ \begin{array}{ccc} 0 & 1 & 0\\
\frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\
\end{array} \right]. $$
Note that this is just one of the possible matrices $T$. 
