Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

2 Answers 2

up vote 1 down vote accepted

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)

share|improve this answer
    
Sorry I'm not really understanding your answer.. –  maq Feb 18 '11 at 0:33

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:

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.