# 2 Intersecting 3-D planes

• Oct 6th 2011, 04:22 PM
DerekZ10
2 Intersecting 3-D planes
Attachment 22508

I have been at this same problem for a few hours now. The scan of the paper is above. I'm trying to find the parametric and symmetric equations for the line of intersection for the planes:

3x - 3y - 7z = -4 and x - y + 2z = 3.

I thought I had it solved but if i plug the parametric equations into the original equations neither comes out right. It would if one of the signs where different for the 13t in the parametric equations. But I don't know where I dropped it.

Thank you.
• Oct 6th 2011, 04:36 PM
DerekZ10
Re: 2 Intersecting 3-D planes
• Oct 6th 2011, 04:38 PM
Plato
Re: 2 Intersecting 3-D planes
Quote:

Originally Posted by DerekZ10
Attachment 22508
equations for the line of intersection for the planes:
3x - 3y - 7z = -4 and x - y + 2z = 3.
I thought I had it solved but if i plug the parametric equations into the

Can you show that the point $(1,0,1)$ is on both planes?
The direction vector is: $<3,-3,-7>\times<1,-1,2>.$
• Oct 7th 2011, 10:57 AM
HallsofIvy
Re: 2 Intersecting 3-D planes
I would think that a rather obvious thing to do would be to subtract 3 times the second equation from the first:
(3x- 3y- 7z)- 3(x- y+ 2z)= -4- 3(3)
-13z= -13 so that z= 1, a constant.

With z= 1, the two equations become 3x- 3y- 7= -4 and x- y+ 2= 3 which both reduce to x- y= 1.
x= y+ 1. Using y as parameter, x= t+1, y= t, z= 1 are parametric equations for the line of intersection.
• Oct 7th 2011, 05:27 PM
DerekZ10
Re: 2 Intersecting 3-D planes
Is there possibly 2 answers? I went with x = 1 - 13t, y = -13t, and z = 1. Substitution of those worked on both equations. My original mistake was that I dropped neg sign on y = 13t.