Lets assume i have a known vector, for example x = [1,0,0]
After 2 rotations, one over the y axis and one over the z axis, i result in a vector which in this example is x' = [0.5774, 0.5774, 0.5774]


Assuming that the rotation angle on the y axis is a, and the rotation angle on z is b, my question is:


i) How can i get the values a & b
ii) a vector perpendicular to x, if submitted to the same rotations will still remain perpendicular, right?



Rotating a known vector over two axis-es to result to another known vector
i) This first question is, in my opinion, simplest with rotation matrices, but may be solved in several different ways.


Let


\begin{equation}
x = \begin{bmatrix}
1 & 0 & 0
\end{bmatrix}
\end{equation}


and


\begin{equation}
x'' = \frac{1}{\sqrt{3}}\begin{bmatrix}
1 & 1 & 1
\end{bmatrix}
\end{equation}


then there exists two rotation matrices, $[A]'$ and $'[B]''$, with rotations $\alpha$ and $\beta$ about the y-axis and z-axis respectively so that


\begin{equation}
x'' = x\;[A]'\;'[B]'' = x\;\begin{bmatrix}
\cos(\alpha) & 0 & -\sin(\alpha) \\
0 & 1 & 0 \\
\sin(\alpha) & 0 & \cos(\alpha)
\end{bmatrix}\begin{bmatrix}
\cos(\beta) & \sin(\beta) & 0 \\
-\sin(\beta) & \cos(\beta) & 0 \\
0 & 0 & 1
\end{bmatrix}
\end{equation}


After evaluation, we find


\begin{equation}
x'' = \begin{bmatrix}
\cos(\alpha)\cos(\beta) & \cos(\alpha)\sin(\beta) & -\sin(\alpha)
\end{bmatrix}
\end{equation}


Note that you actually find two unique solutions to this system of equations, one corresponding to a "forward" rotation sequence and the other a "backwards" rotation sequence. Both end at the same final state and are valid.


ii) Yes. Rotating a system as a whole only changes the coordinates describing it, not the underlying structure.


