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I want to find parametric and symmetric equations for the line of intersection of the planes $x+y+z=1$ and $x+z$=0. Here's what I have so far:

I need to find a point on this line of intersection. So by setting z=0 in the equation of both planes, I get $x=0, y=1, z=0$. So $(0,1,0)$ is on the line. Now since this line is intersecting both planes, it must be perpendicular to both planes. So to find a line parallel to this line, I take the cross product of the normal vectors of both planes. I get $<1,-1,0>$. And now I'm stuck. I dont know what to do next. Any ideas?

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2 Answers 2

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The line of intersection is in both planes, not perpendicular to them. You have two equations in three unknowns, which should have a solution depending on one parameter, just like a line. So can you solve the two equations to give something like that?

Added: If you subtract the two equations, you get y=1. If you plug that into the first, they become identical, x+z=0. So all points on both planes have to satisfy both of these. A parameterization of the line would then be (t,1,-t)

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Sorry I'm not really understanding your answer.. –  maq Feb 18 '11 at 0:33
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You have said a wrong thing (the line is not perpendicolar to planes, but ot them normal vectors...) but your idea is right. Maybe it was a lapsus. ;-) I think that you calculus of the vector is wrong, because the two normal vectors are:

$$(1,1,1), \ \ (1,0,1)$$

and your vector (1,-1,0) is not perpendicular to them. You can easily find a perpendicular vector by eyes: (1,0,-1), so you don't need to do cross product.

Your line so is parametrically $t(1,0,-1)$ but traslated from the origin (0,0,0) to (0,1,0): that's $(0,1,0)+t(1,0,-1)$. Or, if you prefer this writing:

$$ \left\{ \begin{aligned} &x=t \\ &y=1 \\ &z=-t \end{aligned} \right. $$

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