System of two equations with 3 unknowns and parameters Is there a way to solve this for $c_1$, $c_2$, $c_3$ in terms of $a$'s and $b$'s?
$$
\begin{cases}
a_1c_1+a_2c_2+a_3c_3=0 \\ 
b_1c_1+b_2c_2+b_3c_3=0
\end{cases}
$$
 A: In general, this system (with only two equations but three unknown) will have an infinite number of solutions. You can choose one of the $c$'s as a free variable and solve the system in terms of the $a$'s, $b$'s and the free variable as a parameter.
For example, solving for $c_1$ and $c_2$ (see Wolfram|Alpha) yields:
$$c_1 = \frac{c_3 (a_3 b_2 - a_2 b_3)}{a_2 b_1 - a_1 b_2} \quad , \quad c_2 = \frac{c_3 (a_3 b_1 - a_1 b_3)}{a_1 b_2 - a_2 b_1}$$
provided that $a_2 b_1 \ne a_1 b_2$ and $b_2 \ne 0$.

Based on your comment; I'll add this. We now have an infinite number of solutions of the form:
$$(c_1,c_2,c_3) = \left( \frac{c_3 (a_3 b_2 - a_2 b_3)}{a_2 b_1 - a_1 b_2} \;,\;  \frac{c_3 (a_3 b_1 - a_1 b_3)}{a_1 b_2 - a_2 b_1}\;,\; c_3 \right)$$
where you can take $c_3 \in \mathbb{R}$ arbitrarily. For a 'nice' solution, choose $c_3 = a_1 b_2 - a_2 b_1$ to get:
$$\left(a_2 b_3 - a_3 b_2 \;,\; a_3 b_1 - a_1 b_3 \;,\;  a_1 b_2 - a_2 b_1 \right)$$
A: The system tells you that the vector $\mathbf c$ shall be normal to the vectors $\mathbf a$ and $\mathbf b$. Therefore put $\mathbf c= \lambda \mathbf a \times \mathbf b$.
