How do I find the plane at which two hyperplanes intersect?

Question

In the case of finding the line at which two planes intersect, you need to take the cross product of the normal of the two planes. This cross product is simply taking the determinant of matrix:

i  j  k
x1 y1 z1
x2 y2 z2

Where (x, y, z) is the normal vector of each plane. The result is a vector parallel to the intersection line. From there you need to find a point which lies on both planes. The two parts combined give you a fully defined line.

How can this be extended to hyperplanes intersecting at a plane? I would assume I would need to take the determinant of a similar matrix, but the matrix I think of:

h  i  j  k
w1 x1 y1 z1
w2 x2 y2 z2

Is not a square matrix. Also, I don't know how to find a point which lies on both hyperplanes.

Can anyone explain to me how to find the intersection plane of the hyperplanes?

Thanks for your time!

Answer 1

You don't have to calculate a determinant for that, just perform a simple variable replacement and you will get the intersection plane. For instance, if you have two hyperplanes:

3x + 4y + 2z - 7w = 10
2x - 3y + 2z + 1w = 2

You can then isolate w (or any other variable):

w = 2 - 2x + 3y - 2z

And replace it in the first equation:

3x + 4y + 2z - 7(2 - 2x + 3y - 2z) = 10

Which results in:

17x - 17y + 16z - 14 = 10

And now you have your intersection plane. Just simple math.

The complete 4D plane representation is based on both equations, first you find (x, y, z) values that solve 17x - 17y + 16z - 14 = 10 and then you calculate w using w = 2 - 2x + 3y - 2z.

Answer 2

The answer with a simple variable replacement is incorrect. 3x + 4y + 2z - 7(2 - 2x + 3y - 2z) = 10 is itself a three-dimensional hyperplane in four-dimensional space and it does not represent the intersection of the two given three-dimensional hyperplanes in four-dimensional space. The fact that the equation has one less variable does not decrease the dimensionality of the object.

For analogy: y=7 is still a one-dimensional line in 2d, just like y=x+7 is. And z+y=5 is still a 2d plane in 3d, just like x+y+z=5 is.

Variable replacement does not work in 3D, we do the cross-product as outlined, and it does not work in 4D. It takes 2 equations to represent a 2D object in 4D (the intersection of two 3D hyperplanes is a 2D object.) For analogy, tell me the single "equation" that maps as a point in 2D. y=5x+2 is a line, y=x is a line, x=6 is a line, y=0 is a line. Even the simple equation y=1 is a 3D hyperplane if we are in 4D. Removing a variable is not the way to get the equation for a 0D point in 2D, or for a 1D line in 3D, or for our 2D intersection-of-two-3D-hyperplanes in 4D. All of these require exactly two simultaneously true equations to define them. Cannot just replace variables.

Answer 3

You need to setup a matrix system (Ax=b) corresponding to the hyper-planes and then look at the rank of the solution. That will tell whether it even has a solution and if so whether its a point/line/plane/etc.

I have a question: Is it true that " there is n 3-dim hyperlanes in R^4 such that the intersection between them is a plane, for all positive integer n"