Intersection of four planes (Gauss-elim?) I need to find the line of intersection going through 4 planes, the equations are of the form:
a is a coefficient.
ax+ay+a*z=a
Now I know how to write the equations in a matrix and use gaussjordan elimination to find x,y,z (and a), but Im not sure now that I have four equations with three variables I dont see how Im gonna get a free parameter which would allow me to find a line of intersection.
I'm sure I must be looking at it backwards somehow.
 A: Further to my comment above.
To find the intersection of two planes defined by the equations $ax+by+cz=d$ and $a'x+b'y+c'z=d'$, eliminate two of the variables, using the third, say $z$, as a parameter. You need to solve a $2\times2$ linear system in $x,y$:
$$\left\{\begin{matrix}ax+by&=&d-cz\\
a'x+b'y&=&d'-c'z\end{matrix}\right.$$
This will yield a parametric equation for the line of interection
$$\left\{\begin{matrix}x=\alpha z+\beta\\y=\alpha' z+\beta'\\z=z\end{matrix}\right.$$
It's not always possible to choose $z$ as the parameter (for instance if the two planes have equations $x=0$ and $z=0$, the parameter will be $y$). So in general, write the parametric equation of the line as
$$\left\{\begin{matrix}x&=&\alpha t+\beta\\y&=&\alpha' t+\beta'\\z&=&\alpha'' t+\beta''\end{matrix}\right.$$
Then, to check that this line lies on any other plane $a''x+b''y+c''z=d''$, simply replace $x,y,z$ by $\alpha t+\beta$, $\alpha' t+\beta'$ and $\alpha'' t+\beta''$ respectively, and verify that the equation holds.
