I have a series of equations \begin{align} u_{11} &= \frac{zc_1}{x} \frac{\sin(\delta + \omega)}{\sin \alpha} \\ u_{21} &= \frac{zc_2}{x} \frac{\sin \omega}{\sin \alpha} \\ u_{31} &= \frac{tc_1}{x} \frac{\sin \delta }{\sin \alpha} \\ u_{41} &= 0 \end{align}
and \begin{align} u_{13} &= \frac{zc_1}{y} \Bigg( \cos(\delta+\omega)+\frac{\sin(\delta + \omega)}{\tan \alpha} \Bigg)\\ u_{23} &= \frac{zc_2}{y} \Bigg( \cos(\omega)+\frac{\sin( \omega)}{\tan \alpha} \Bigg) \\ u_{33} &= \frac{tc_1}{y} \Bigg( \cos(\delta)+\frac{\sin(\delta)}{\tan \alpha} \Bigg) \\ u_{43} &= \frac{tc_2}{y} \end{align}
Where I know that $$0 <x,y<1$$ $$0 \leq z,t,c_1,c_2 \leq 1$$ $$0<\alpha<\pi/2$$ $$0 \leq \delta, \omega \leq \pi/2$$ and $$x^2+y^2=z^2+t^2=c_1^2+c_2^2=1$$ and all parameters are real. My goal is to find out the constraints needed to make the $(u_{11},u_{21},u_{31},u_{41})$ and $(u_{13},u_{23},u_{33},u_{43})$ orthonormal vectors. That is, given $(x,y,z,t,\alpha,\omega)$, can one find $c_1,c_2,\delta$ that will make these columns orthonormal? We therefore impose that $$u_{11}^2+u_{21}^2+u_{31}^2=1$$ $$u_{13}^2+u_{23}^2+u_{33}^2+u_{43}^2=1$$ and $$u_{11}u_{13}+u_{21}u_{23}+u_{31}u_{33}=0$$ however expanding these expressions do not allow me to solve for $c_1,c_2$ or $\delta$. Is there an efficient way to go about this?
With some additional border conditions, this problem could be directly transformed into the modal solution of a physical problem. Which is the original problem bringing these expressions? That could help to clarify a better solution.
As is, none of the given expression can be simplified in a simple way enabling to give "closed" inequalities on the given variables.
By applying the constraints, it is straightforward to find: $$ \mathscr{C}_{11}: (zc_1)^2(s\delta\omega)^2+(zc_2)^2(s\omega)^2+(tc_1)^2(s\delta)^2 =(xs\alpha)^2\\ \mathscr{C}_{33}: (zc_1)^2(s\delta\omega\alpha)^2+(zc_2)^2(s\omega\alpha)^2+(tc_1)^2(s\delta\alpha)^2+(tc_2)^2(s\alpha)^2 =(ys\alpha)^2\\ \mathscr{C}_{13}: (zc_1)^2s\delta\omega s\delta\omega\alpha+(zc_2)^2s\omega s\omega\alpha+(tc_1)^2s\delta s\delta\alpha =0 $$ where $s\phi_1\cdots\phi_n$ and $c\phi_1\cdots\phi_n$ represent the $\sin$ and $\cos$ functions applied over the sum of the $\phi_i$ angles denoted right in front of it, not to be confused with $c_1$ and $c_2$, variables part of the problem.
Reordering between the $c_1$, $c_2$ terms: $$ \mathscr{C}_{11}: c_1^2 (z^2(s\delta\omega)^2+t^2(s\delta)^2) +c_2^2 z^2(s\omega)^2 = (xs\alpha)^2\\ \mathscr{C}_{33}: c_1^2 (z^2(s\delta\omega\alpha)^2+t^2(s\delta\alpha)^2) +c_2^2 (z^2(s\omega\alpha)^2+t^2(s\alpha)^2) = (ys\alpha)^2\\ \mathscr{C}_{13}: c_1^2(z^2s\delta\omega s\delta\omega\alpha+t^2s\delta s\delta\alpha) +c_2^2(z^2s\omega s\omega\alpha) =0 $$
From Cramer's on $\mathscr{C}_{11}$ and $\mathscr{C}_{33}$: $$ \Delta= (z^2(s\delta\omega)^2+t^2(s\delta)^2) (z^2(s\omega\alpha)^2+t^2(s\alpha)^2)- z^2(s\omega)^2 (z^2(s\delta\omega\alpha)^2+t^2(s\delta\alpha)^2)\\ \Delta_1= (xs\alpha)^2 (z^2(s\omega\alpha)^2+t^2(s\alpha)^2)- (ys\alpha)^2 (z^2(s\delta\omega\alpha)^2+t^2(s\delta\alpha)^2)\\ \Delta_3= (xs\alpha)^2 (z^2(s\delta\omega)^2+t^2(s\delta)^2) - (ys\alpha)^2 z^2(s\omega)^2 $$
The $c_1$ and $c_2$ values are the square roots from: $$ c_1^2={\Delta_1 \over \Delta}= { (xs\alpha)^2 (z^2(s\omega\alpha)^2+t^2(s\alpha)^2)- (ys\alpha)^2 (z^2(s\delta\omega\alpha)^2+t^2(s\delta\alpha)^2) \over (z^2(s\delta\omega)^2+t^2(s\delta)^2)(z^2(s\omega\alpha)^2+t^2(s\alpha)^2)- z^2(s\omega)^2 (z^2(s\delta\omega\alpha)^2+t^2(s\delta\alpha)^2)}\\ c_2^2={\Delta_3 \over \Delta}= { (xs\alpha)^2 ( z^2(s\delta\omega)^2+t^2(s\delta)^2) - (ys\alpha)^2 z^2(s\omega)^2 \over (z^2(s\delta\omega)^2+t^2(s\delta)^2)(z^2(s\omega\alpha)^2+t^2(s\alpha)^2)- z^2(s\omega)^2 (z^2(s\delta\omega\alpha)^2+t^2(s\delta\alpha)^2)} $$
Note that the term $s\delta\omega\alpha$ must be negative in order the signs be preserved (!). This is the only term which can be negative after a $\sin$ operator involving angles in the $(0,pi/2)$ range.
And from $\mathscr{C}_{13}$ we have a expression able to be numerically solved for $\delta$: $$ \mathscr{C}_{13}: \Delta_1^2 (z^2 s\delta\omega s\delta\omega\alpha+t^2 s\delta s\delta\alpha) = -\Delta_3^2 (z^2 s\omega s\omega\alpha) $$ Hence: $$ ( x^2 (z^2 s\omega\alpha^2 +t^2 s\alpha^2)- y^2 (z^2 s\delta\omega\alpha^2 +t^2 s\delta\alpha^2) )^2 (z^2 s\delta\omega s\delta\omega\alpha +t^2 s\delta s\delta\alpha) = -( x^2 (z^2 s\delta\omega^2 +t^2 s\delta^2) - y^2 z^2 s\omega^2 )^2 (z^2 s\omega s\omega\alpha) $$
Which, spite of the closed solution for $c_1(\delta)$ and $c_2(\delta)$, cannot be solved in a closed for the last parameter $\delta$.