# Line of intersection of two planes

• Apr 24th 2013, 12:07 PM
flametag3
Line of intersection of two planes
I am a little confused as to how to find the line of intersection of two planes

x + y + z -1 = 0
2x + y -z + 3 = 0

should i be using gaussian elimination for this ?but from what i am reading about gaussian elimination the top row wil multiple answers
• Apr 24th 2013, 01:29 PM
Plato
Re: Line of intersection of two planes
Quote:

Originally Posted by flametag3
I am a little confused as to how to find the line of intersection of two planes

x + y + z -1 = 0
2x + y -z + 3 = 0

should i be using gaussian elimination for this ?but from what i am reading about gaussian elimination the top row wil multiple answers

Find any point which is on both planes.

Then find $\displaystyle <1,1,1>\times<2,1,-1>$, the cross product of the two normals.

Now the line contains that point and has direction the cross product.
• Apr 24th 2013, 01:50 PM
HallsofIvy
Re: Line of intersection of two planes
Quote:

Originally Posted by flametag3
I am a little confused as to how to find the line of intersection of two planes

x + y + z -1 = 0
2x + y -z + 3 = 0

should i be using gaussian elimination for this ?but from what i am reading about gaussian elimination the top row wil multiple answers

Yes, it will, and it should, because the intersection of two planes is not a single point, it is a line- there are an infinite number of points lying on that line that will satisfy both equations.

The simplest thing to do, I think, is to add the two equations, eliminating z: 3x+2y+ 2= 0 so that 2y= -2- 3x and then y= -1- (3/2)x. Putting that into the first equation, x+ (-1 - (3/2)x)+ z- 1= -(1/2)x+ z= 0 so that z= (1/2)x.

The line or intersection is given by y= -1- (3/2)x or z= (1/2)x. Or if you prefer, take t= (1/2)x so that x= 2t, y= -1- 3t, z= t.