Intersection of Ellipsoid and Plane

Date: 05/04/2007 at 07:21:58
From: Giorgio
Subject: Ellipsoid and plane intersection

Dear Dr. Math, I need to find out the general equation of the ellipse
resulting from the intersection of an ellipsoid and a plane.  I've
found another answer to a similar question, but it's not the same and
I wasn't able to use that with my problem.  My equations are the
following:

  ellipsoid: x^2/a^2 + y^2/b^2 + z^2/c^2 = 1
  plane: a*x + b*y + c*z + d = 0

The problem is this: I tried to solve the two-equation system, but
this way the result includes a term with x*y, and I can't return to
the canonic equation of an ellipse (x^2/a^2 + y^2/b^2 = 1), which is
essential for me.

I'm a researcher, and the project I'm working on concerns satellite 
orbits and the Earth: for this reason we can assume that both the
ellipsoid and plane pass through the origin O(0,0,0), as will the
ellipse.  I tried to use this information in different ways, but with
no results.  I have no idea on how to solve it in other ways, I hope
you can help me.

Thank you very much.

Date: 05/04/2007 at 08:39:05
From: Doctor Jerry
Subject: Re: Ellipsoid and plane intersection

Hello Giorgio,

Thanks for writing to Dr. Math.

If you have, for example, the ellipsoid

  x^2/2^2 + y^2/1^2 + z^2/3^2 = 1

and the plane

  x + 2y + 3z = 1,

one can eliminate z to find the equation

  85*x^2 + x*(-8 + 16*y) = 320 + 16*y - 340*y^2

This is the equation of the projection of the ellipse into the (x,y)
plane.  One could solve this for y in terms of x, giving, say, y =
f(x).  This leads to parametric equations for the ellipse, something like

  x = x
  y = f(x)
  z = ( 1 - x -2*f(x) )/3

One can only remove the x*y term by some kind of rotation of axes.

If you have further questions about this please write back.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 05/09/2007 at 09:53:44
From: Doctor George
Subject: Re: Ellipsoid and plane intersection

Hi Giorgio,

It sounds like you have may have seen this article from the archive.

     http://mathforum.org/library/drmath/view/63373.html 

I've been thinking about how to build on Dr. Jerry's response to
generate the form of the equation in that article so that you can
determine the locations of the foci, the major axis length, etc.

A closed form solution to the general problem is actually very
difficult to derive.  A closed form solution to your special case 
seems to have the same basic difficulties as the general problem.

But a computational method to find the standard form may be all you
really need.  You can then apply my approach for the general problem 
to your special case.  Here is how I would go about it.

1) Using Dr. Jerry's solution (or rotating the projection in the x,y
plane to eliminate the xy term if you like), compute the (x,y,z)
coordinates of 5 points on the ellipse.

2) Assign a new x'y' coordinate system to the intersecting plane. This
can be chosen arbitrarily using the 5 points.

3) Compute the (x',y') coordinates for each of the points.

4) Substitute each point into the following equation for the ellipse.

      Ax'^2 + Bx'y' + Cy'^2 + Dx' + Ey' = -1

5) We now have 5 linear equations with 5 unknowns. We can compute the
values of A, B, C, D and E.

6) Using standard techniques for the general quadratic we can 
construct a new x",y" system on the intersecting plane and write the
standard form of the ellipse in (x",y") coordinates.

7) From the standard form in the x",y" plane we can find the center
point, the ellipse axes lengths, and their directions in the original
coordinate system.  This allows us to finally write the equation of 
the interesting ellipse in the form from the archive article.

Does that make sense?  Write again if you need more help.

- Doctor George, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 05/17/2007 at 04:43:57
From: Giorgio
Subject: Ellipsoid and plane intersection

First of all, thank you both for your support.  Dr. Jerry's solution
was pretty similar to my situation, so I found it very useful. 
Studying Dr. George's solution took me several days, but finally I 
understood how it works.

I've gone on, but now I'm facing another problem.  When I find the
intersection of ellipsoid and plane I have the equation of an ellipse.
I can't comprehend what ellipse that equation refers to; in other
words, the equation refers to an ellipse some way oriented in
three-dimensional space, but, having only x and y coordinates, have I
to assume that the plane it lies on gets rotated so to become in 2D?

My thought is the following: the ellipse equation has 2 unknowns (x 
and y), but it represents a figure in three dimensions.  Then, it’s 
like that the ellipse plane was rotated so to annihilate z?  If so, I 
should be able to apply the same rotations (in the opposite sense) 
to return to the original plane, correctly oriented in 3D.

If this is correct, my problem comes up: how can I determine the 
inclination angles of the plane containing the ellipse?  My final 
objective is to find the rotation matrix which modifies the 
inclination of the plane containing the ellipse in 3D space.

I tried to solve it this way (remember the plane I work with always
contains the origin of axis).  I found the intersection of the plane
(r*x + s*y + t*z = 0) with xy, xz, and yz, choosing respectively z=0,
y=0, and x=0. Then I had one line lying on each plane: it represents
the intersection of the inclined plane with the Cartesian ones.  I
calculated their inclination angles using arctangent, so to use those
angles with three-dimension rotation matrix.

Although the procedure seems to be correct, I can never arrive at a 
correct result.  Maybe I made some mistakes using the rotation matrix.
I’ll be grateful to you if you could explain to me how they work (and 
if my procedure is correct).

PS: I saw some material explaining how to rotate planes so a point is 
in a specified position in 3D, but my objective is to rotate a plane 
in a particular position: I should need only two (and not three) 
rotations to have the correct inclination of the plane, right?

Date: 05/18/2007 at 07:51:40
From: Doctor George
Subject: Re: Ellipsoid and plane intersection

Hi Giorgio,

You have a global xyz system. Your intent is to find a matrix that
will rotate the x axis to the major axis of the ellipse, the y axis to
the minor axis of the ellipse, and the z axis to the normal vector to
the plane.

From my initial reply, you now have the unit vectors for the major
axis (M), the minor axis (m), and the normal to the plane (N).

If you are storing your points in columns, then your rotation matrix
(R) is composed of M, m and N stored in columns like this.

      | Mx  mx  Nx |
  R = | My  my  Ny |
      | Mz  mz  Nz |

You can see that

    | 1 |   | Mx |
  R | 0 | = | My |
    | 0 |   | Mz |

rotates the global x to your new local x on the ellipse.  Vectors y 
and z are rotated in a similar way.

Does that make sense?  Write again if you need more help.

- Doctor George, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 05/23/2007 at 11:55:59
From: Giorgio
Subject: Thank you (Ellipsoid and plane intersection)

I did it!!!

It took some time, because of several troubles encountered during
implementation, but finally everything works!  I'm so grateful to you
for all your time and support!

All the best,

Giorgio