Intersections of Planes Example

A question in class today pointed out some minor detail that I left out in finding the equation of a line which is the intersection of two planes. Lets look at an example where the method in class does not work, see how to fix it, and then look at what went wrong.

Lets consider the planes $2x+y+z=2$ and $4x+2y-3z=4$. If we set $z=0$, we are left with the two equations $2x+y=2$ and $4x+2y=4$, which have infinitely many solutions. We can't continue.

A good way around this is simply to set $x=0$ or $y=0$ and solve the way we did in class. If we set $x=0$, we are given the equations $y+z=2$ and $2y-3z=4$. These have a solution of $y=2$ and $z=0$, giving us the point $(0,2,0)$. Now we are all set. We can cross product the normal vectors from each plane to get a vector pointing in the direction of the line and finish by writing the equation.

What went wrong? Geometrically, setting z=0 is equivalent to looking for the intersection of the three planes $2x+y+z=2$, $4x+2y-3z=4$, and $z=0$. In this case the line of intersection of $2x+y+z=2$ and $4x+2y-3z=4$ is contained in the plane $z=0$, so we gain no information by setting $z=0$. In the picture below $2x+y+z=2$ is the red plane, $4x+2y-3z=4$ is the blue plane, and $z=0$ is the yellow plane. Notice how the line of intersection of the red and blue planes is contained in the yellow plane.

We fixed this by setting $x=0$. Which is illustrated in the following graph(the red and blue are as above and $x=0$ is yellow.

While the intersection of the planes is not in the frame it is clear that they will intersect in a point, giving us a point which is on our line.