Line of Intersection of two Planes

[mattmns]

How do I find the line of intersection of two planes? I have an idea, but both of the planes have a -2z

ie. Plane 1: 10x-4y-2z=4 Plane 2: 14x+7y-2z

If I set them both equal to each other, I lose the z part. So, is there some other way to solve this, or am I missing something? Thanks!

[dextercioby]

The equation for the second plane is incomplete

Daniel.

P.S.If "z" goes,then that line is in the Oxy plane,right...?

[mattmns]

The equation for the second plane is incomplete

Daniel.

P.S.If "z" goes,then that line is in the Oxy plane,right...?

woops.

Just add =#. I am just curious about the general case of how to solve such a problem.


ie. Plane 1: 10x-4y-2z=4 Plane 2: 14x+7y-2z=6

Also, if z is removed what would the symmetric form of that line be? Would it just not have a z part?

[dextercioby]

What do you mean symmetric form...?Just set them equal and find the equation giving the line.It should read as linear combo of "x" & "y" yielding 0.

Daniel.

[mattmns]

Ok here is the whole problem.

I am trying to find an equation for a line that passes through a point P(x,y,z) and is parallel to the line of interestion of the planes p1, and p2.

For example: Find an equation for the line that passes through the point (0,1,-1) and is parallel to the line of intersection of the planes 2x + y - 2z = 5 and 3x - 6y - 2z = 7.


edit.. Symmetric form of a line in R3.

If L is a line that contains the point (x_0, y_0, z_0) and is parallel to the vector v = Ai + Bj + Ck then the point (x, y, z,) is on L if and only if its cooridnates satisfy

\frac{x-x_0}{A} = \frac{y-y_0}{B} = \frac{z-z_0}{C}


edit2... fixed the latex

[dextercioby]

Here's the parametric equation of the intersection of the 2 planes

\left\{\begin{array}{c}z=\frac{15y-1}{2}\\x=7y+2 \end{array}\right

I hope it's easy from now.

Daniel.

[mattmns]

Killing a fly with a sledgehammer as my hs math teacher would say. How about this, forget the whole finding the equation of the line. Find a point A that is on the interesecting line, and then a point B on the intersecting line, then using the vector AB? I think that will work. Now back to bed. :cool:


edit... And to find the actual equation of the intersecting line doing the same thing would probably work, just replace the point with a point that it is on the intersecting line.

[snoble]

I don't see how the parametric form is a sledgehammer in this case. You still have to find these two points and the easiest way to find them is to write x and z in terms of y then choose two y's. This is exactly what daniel has given you.

Steven

[Benny]

I'm probably missing something but do you really need the equation to the line of intersection to find the line you are seeking? The cross product of the normals to the two planes gives you a vector which is parallel to the line of intersection. The vector equation of the line you want is then r = (vector representing given point) + t(cross product) where t is a scalar.

The symmetric equation for that line can then be easily found.

[mattmns]

I don't see how the parametric form is a sledgehammer in this case. You still have to find these two points and the easiest way to find them is to write x and z in terms of y then choose two y's. This is exactly what daniel has given you.

Steven
Yes, you are right the points will still be needed. I meant sledgehammer in the way I was trying to solve the problem, considering that there seem to be much easier routes, not about dex helping me out.

I'm probably missing something but do you really need the equation to the line of intersection to find the line you are seeking? The cross product of the normals to the two planes gives you a vector which is parallel to the line of intersection. The vector equation of the line you want is then r = (vector representing given point) + t(cross product) where t is a scalar.

The symmetric equation for that line can then be easily found.
That seems like it will work too, I never thought about that, thanks.