Constraints on variables to form orthonormal columns I have a series of equations \begin{align} u_{11} &amp;= \frac{zc_1}{x} \frac{\sin(\delta + \omega)}{\sin \alpha} \\
u_{21} &amp;= \frac{zc_2}{x} \frac{\sin \omega}{\sin \alpha} \\
u_{31} &amp;= \frac{tc_1}{x} \frac{\sin \delta }{\sin \alpha} \\
u_{41} &amp;= 0 \end{align}

and \begin{align} u_{13} &amp;= \frac{zc_1}{y} \Bigg( \cos(\delta+\omega)+\frac{\sin(\delta + \omega)}{\tan \alpha} \Bigg)\\
u_{23} &amp;= \frac{zc_2}{y} \Bigg( \cos(\omega)+\frac{\sin( \omega)}{\tan \alpha} \Bigg) \\
u_{33} &amp;= \frac{tc_1}{y} \Bigg( \cos(\delta)+\frac{\sin(\delta)}{\tan \alpha} \Bigg) \\
u_{43} &amp;= \frac{tc_2}{y} \end{align}

Where I know that $$0 &lt;x,y&lt;1$$ $$0 \leq z,t,c_1,c_2 \leq 1$$ $$0&lt;\alpha&lt;\pi/2$$ $$0 \leq \delta, \omega \leq \pi/2$$ 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 $\delta$.
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 $\delta$.