Need Help on Solving Mixed Linear/Nonlinear System of Equations I am trying to solve the following mixed linear/nonlinear system of equations. Could someone kindly give me some tips on how I can solve it numerically?
$a_i$ and $c_i$ are known and $x_i$, $y_i$, $z_i$, $r_i$, $s_i$, $t_i$ are unknown. There are 18 equations and 18 unknowns.
$$
a_1x_1 + a_2x_2 + a_3x_3 + r_{1} = 0 \tag{1-3}\\
a_1y_1 + a_2y_2 + a_3y_3 + r_{2} = 0 \\
a_1z_1 + a_2z_2 + a_3z_3 + r_{3} = 0
$$
$$
a_4x_1 + a_5x_2 + a_6x_3 + s_{1} = 0\tag{4-6}\\
a_4y_1 + a_5y_2 + a_6y_3 + s_{2} = 0\\
a_4z_1 + a_5z_2 + a_6z_3 + s_{3} = 0
$$
$$
a_7x_1 + a_8x_2 + a_9x_3 + t_{1} = 0\tag{7-9}\\
a_7y_1 + a_8y_2 + a_9y_3 + t_{2} = 0\\
a_7z_1 + a_8z_2 + a_9z_3 + t_{3} = 0
$$
$$
r_{1}^2 + r_{2}^2 + r_{3}^2 = c_{1}^2\tag{10-12}\\
s_{1}^2 + s_{2}^2 + s_{3}^2 = c_{2}^2\\
t_{1}^2 + t_{2}^2 + t_{3}^2 = c_{3}^2\\
$$
$$
r_1s_1 + r_2s_2 + r_3s_3 = c_1c_2\tag{13-15}\\
r_1t_1 + r_2t_2 + r_3t_3 = c_1c_3\\
s_1t_1 + s_2t_2 + s_3t_3 = c_2c_3\\
$$
$$
t_1 = r_1 + s_1\tag{16-18}\\
t_2 = r_2 + s_2\\
t_3 = r_3 + s_3\\
$$
UPDATE:
Relationship among some of the constants:
$$c_1 + c_2 = c_3\tag{19}$$
 A: Hint:   assuming $\,c_j \ne 0\,$, introduce the new variables $\,u_j=r_j/c_1, v_j = s_j / c_2, w_j=t_j / c_3\,$, then dividing equations $\,(10)$ - $(12)\,$ and $\,(13)$ - $(15)\,$ by the respective RHSs:
$$
\begin{align}
u_{1}^2 + u_{2}^2 + u_{3}^2 = 1 \tag{10.a} \\
v_{1}^2 + v_{2}^2 + v_{3}^2 = 1 \tag{11.a} \\
w_{1}^2 + w_{2}^2 + w_{3}^2 = 1 \tag{12.a} \\
\end{align}
$$
$$
\begin{align}
u_1v_1 + u_2v_2 + u_3v_3 = 1 \tag{13.a} \\
u_1w_1 + u_2w_2 + u_3w_3 = 1 \tag{14.a} \\
v_1w_1 + v_2w_2 + v_3w_3 = 1 \tag{15.a} \\
\end{align}
$$
Combining $\,(10.a)+(11.a)-2 \cdot (13.a)\,$ gives:
$$
(u_1 - v_1)^2 + (u_2 - v_2)^2 + (u_3 - v_3)^2=0
$$
Therefore $\,u_j=v_j\,$, and by symmetry $\,u_j=v_j=w_j=\lambda_j\,$.
Since $\,c_3=c_1+c_2\,$, it follows that the last three equations $\,(16)$-$(18)\,$ are satisfied automatically:
$$
t_j = \lambda_jc_3 = \lambda_j c_1+\lambda_jc_2=r_j+s_j
$$
Therefore the non-linear part $\,(10)$-$(18)\,$ of the system has the general solution $\,r_j=\lambda_jc_1\,$, $\,s_j=\lambda_jc_2\,$, $\,t_j=\lambda_jc_3\,$, where $\,\lambda_1, \lambda_2, \lambda_3\,$ are some constants such that $\,\lambda_1^2+\lambda_2^2+\lambda_3^2=1\,$.
Substituting these in equations $\,(1)$-$(9)\,$ leaves a linear system of $9$ equations in $9$ unknowns which can be solved by the usual methods.
