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nobahar
Oct17-10, 04:01 PM
Hello!

I have a quick question regarding the intersection of three planes if the determinant is 0.
If there are solutions, there will be an infinite number of solutions. One of the equations for the plane can be ignored as it is a linear combination of the other two, and can be ignored for the purposes of finding a solution (as far as I am aware).
Okay, this is where I think I may be wrong. The components of a normal vector to the plane are not like a position vector, specifying some point a certain distance from the origin. Because a normal vector can be "moved around": it's dot product with any vector in the plane yields 0. If this is the case, the cross product of two normal vectors (one for each of the two planes), yields some vector with specific values. Is it true that this vector, V, and it's scalar multiples are not necessarily the solutions? Because there is nothing in the normal vectors specifying a location; and the vector that is a product of the cross product of the two normal vectors also does not have any 'information' specifying its location. It will be parallel to the line, but not necessarily on the line. I have seen an example of finding the solutions to the intersection of three planes but no more information was needed (i.e. scalar multiples of the cross product of two normal vectors were the solutions), but the trivial solution (x=y=z=0) was a, erm, solution! Therefore it passes through the origin, and I think that the origin can be used to specify the position of the vector V.
Any help appreciated, sorry if it's not too clear...

tiny-tim
Oct17-10, 04:55 PM
hello nobahar! :smile:

if the determinant is zero, then the three normals are coplanar, so they're all perpendicular to the normal of that plane, so that plane must be parallel to a line in all three planes

either that line is the intersection, or there is no intersection (except "at infinity", at the ends of the line) :wink:

nobahar
Oct17-10, 05:08 PM
hello nobahar! :smile:

if the determinant is zero, then the three normals are coplanar, so they're all perpendicular to the normal of that plane, so that plane must be parallel to a line in all three planes

either that line is the intersection, or there is no intersection (except "at infinity", at the ends of the line) :wink:

Hello TinyTim, you pop up occasionally with words of wisdom!
This is where I am confused. I thought that the normal vector of the coplanar normal vectors (:frown:) will be parallel to the line of intersection.

tiny-tim
Oct17-10, 05:35 PM
For two planes, yes.

But for three planes, you get three lines of intersection, all parallel to each other, so you either get a "toblerone", or the three lines are the same. :wink:

HallsofIvy
Oct18-10, 05:45 AM
Hello TinyTim, you pop up occasionally with words of wisdom!
More often than "occasionally"!

This is where I am confused. I thought that the normal vector of the coplanar normal vectors (:frown:) will be parallel to the line of intersection.

nobahar
Oct18-10, 06:39 AM
More often than "occasionally"!
Hey, how did you managae to take offense to that! That was a compliment. TinyTim shows up whenever I need assitance. Kind of like the Mathematics equivalent of Spiderman (the best superhero, in my opinion).
Okay, so lets say that the three planes share a single line of intersection, and I know this to be true. If I take the cross product of two normal vecotrs. This still will not tell me the equation of the line of intersection, will it?
I need a point on the line, and this, in conjunction with the normal vector, will specify the line. Is this correct?
Thanks for all your help.

tiny-tim
Oct18-10, 06:50 AM
Okay, so lets say that the three planes share a single line of intersection, and I know this to be true. If I take the cross product of two normal vecotrs. This still will not tell me the equation of the line of intersection, will it?
I need a point on the line, and this, in conjunction with the normal vector, will specify the line. Is this correct?

Completely correct. :wink:

(of course, it will also give you a direction along the line, which is irrelevant)

nobahar
Oct18-10, 04:22 PM
Completely correct. :wink:

Phew, I'm slow on the uptake!
Thanks TinyTim, for all your help. I guess your off to help others in mathematical peril (lucky for computers, or you might need an uninterrupted series of bowls to get there!).

P.S. Apologies for the 'not-particularly-good' jokes!

HallsofIvy
Oct19-10, 08:29 AM
Hey, how did you managae to take offense to that!
How did you manage to find any "offense" in that? What I said was also a compliment to TinyTime. Yes, he's my hero also.

That was a compliment. TinyTim shows up whenever I need assitance. Kind of like the Mathematics equivalent of Spiderman (the best superhero, in my opinion).
Okay, so lets say that the three planes share a single line of intersection, and I know this to be true. If I take the cross product of two normal vecotrs. This still will not tell me the equation of the line of intersection, will it?
I need a point on the line, and this, in conjunction with the normal vector, will specify the line. Is this correct?
Thanks for all your help.