I have a 3D vector from the following shape:
\begin{gather} p = \begin{bmatrix} 0 \\ 0 \\ z \end{bmatrix} \end{gather}
I had like to find a rotation matrix that will make p(1) and p(2) small non zero numbers and p(3) to be close to z value.
basically, by writting,
\begin{gather} p' = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ z\end{bmatrix} = \begin{bmatrix} a_{13} \\ a_{23} \\ a_{33} \end{bmatrix} \cdot z \end{gather}
Since it is rotational matrix I had expect it to have the same magnitude and therefore:
\begin{gather} \sqrt{a_{11}^2 + a_{23}^2 + a_{33}^2} = 1 \end{gather}
The first idea that I had is something from the form of \begin{gather} p' = \begin{bmatrix} 0 & 0 & \sin(\alpha) \\ 0 & 0 & 1 \\ 0 & 0 & \cos(\alpha) \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ z\end{bmatrix} = \begin{bmatrix} \sin(\alpha) \\ 1 \\ \cos(\alpha) \end{bmatrix} \cdot z \qquad, \alpha<1 \end{gather}
However, this matrix is problematic from my understanding because its determinant is zero and also the value of p'(2) is equal to z and I was looking for small value.
Is there any idea how to overcome these problems?
Thank you!