On the Ellipsoid and Plane Intersection Equation

It is well known that the line of intersection of an ellipsoid and a plane is an ellipse. In this note simple formulas for the semi-axes and the center of the ellipse are given, involving only the semi-axes of the ellipsoid, the componentes of the unit normal vector of the plane and the distance of the plane from the center of coordinates. This topic is relatively common to study, but, as indicated in [1], a closed form solution to the general problem is actually very difficult to derive. This is attemped here. As applications problems are treated, which were posed in the internet [1,2], pertaining to satellite orbits in space and to planning radio-therapy treatment of eyes.


Introduction
Let an ellipsoid be given with the three positive semiaxes a, b, c and a plane with the unit normal vector   , , q q q  q r r  r   ellipsoid.A plane spanned by vectors , and containing the point is described q in parametric form by  T 1 2 3 with , , .
Inserting the components of into the equation of the ellipsoid (1) leads to the line of intersection as a quadratic form in the variables and u .Let the scalar x t product in for two vectors and be denoted by With the diagonal matrix the line of intersection has the form: .
As is an interior point of the ellipsoid the righthand side of Equation ( 2) is positive.The  matrix in Equation (3) is a Gram matrix.If the vectors 1 and 1 are linearly independent, this is equivalent with the linear independence of the vectors and , the matrix in (3) is positive definite and the line of intersection is an ellipse.In [3] a generalization from three to p-dimensional space is discussed.

D s r D s r
Let and be unit vectors orthogonal to the unit normal vector of the plane n s (5) and orthogonal to eachother (6) Furthermore vectors and may be chosen such that holds.This will be shown in the next section.Condition (7) ensures that the matrix in (3) has diagonal form.Then the line of intersection reduces to an ellipse in translational form and  , , and the semi-axes where , q s q q r r s s


In order to show that the semi-axes (10) are independent of the choice of this vector may be decomposed orthogonally with respect to : where is the distance of plane (2) from the origin.Substituting into (11) one obtains employing (4), ( 5), ( 6) and ( 7) The following rules of computation for the cross product in ([4, p.147]) will be applied later on repeatedly.For vectors of the identity of Lagrange holds and the Grassmann expansion theorem for the double cross product

Construction of Vectors r and s
Let be a unit vector orthogonal to the unit normal vector of the plane, so that Equations (4) hold.A suitable vector is obtained as a cross product Then Equations ( 5) and ( 6) are fulfilled: is a unit vector, as can be shown by the identity of Lagrange (14), utilising , and   : Furthermore one obtains according to the rules apply-in holds.This can be reformulated, in case  can be chosen π 4 , leading to holds, Corollary 1: For the un vectors and orthogo  1 for 1, 2,3.
s n   ains for in ance: Copyright © 2012 SciRes.AM

A Quadratic Equation
Theorem 1: Let be the unit normal vector of the plane and let vectors and satisfy  and 2  are solutions of the following quadratic equation: Proof: Utilising (17) one obtains: ) and the identity of Lagrange (14) leads to: For the cross product one obtains: Corollary 2: Under the assumptions following three equations are valid:  r he three equations was ve h nd and th

A Formular for d
Theorem 2: Under the assumptions of Theorem 1 t ex A quadratic equation eq of Theorem 1 the The first of t rified in the proof of Theorem 1. T e seco e third equation follow analogously.he pression for d in (13) is given by: where  is taken from (12).
Proof: The verification of (25) consists of three steps.
Step 2: Analogously to the verification of application of the identity of Lagrange ( 14 In contrast to the verification of (24), where diago   , D D r s in (33) need not be zero.
Step 3: Applying the identity of Lagrange (14) again leads to

r n s n n r r s s
Corollary 3: Under the assumptions of Theorem 1 the area F soid of the ellipse obtained by the intersection of the ellip (1) and a plane with unit no distance rmal vector n and  from the origin is given by: This is proven by the formula for the area of an ellipse: and by applying (25) and (37).The area of intersection F becomes zero in case t    holds; this corresponds nt Substituting according to (12) in formulars (9) for the to the limiting case, where the cutting plane becomes a tange plane.This result has been applied in [6].

The Center of the Ellipse q
Copyright © 2012 SciRes.AM coordinates of the center of the ellipse in the plane span and one obtains: The center m of the ellipse in is given by: Theorem 3: Let the assumptions of Th The last expre is according to (24).Furtherm and by interchangi e roles of r and : Both previous expressions are zero; this fo ws by applying diagonality condition (7), the identity of Lagrange (14) and Corollary 2: Interchanging the roles of and leads to: where A and B are denoting the semi-axes according to (10).A  m r Clearly and are points of the plane cutting the ellipsoid.In order to show that they are belonging to the ellipse of intersection, it has to be verified that they are situated on the ellipsoid, i.e. the following eq B  m s ualities hold: This can be shown using in the form (39) and employing condition (7).

   x m r s
This result may be derived substituting the parameters t and u from the parametric form of the ellipse into Equation (2) of the plan where 0 0 t u   q r s is equal to the center m of the ellipse as in (39).

Applications
dicated in [2] win isodose lines, radio-therapy treatment of th e can be plann el determ has the form: with the unit normal vector: As in , vie g a section through an ellipsoidal eye from a viewpoint normal to the intersection plane and displaying the intersection on that plane along with a projection of the eye structures and e ey ed.For this purpose the line of intersection of lipsoid (1) and the plane, having the normal vector   T , , i j k and containing the point , , q q q  q , situated in the interior of (1), is ined.The plane e plane from the origin is given by: According to (25) d can be written as: From (11) it is obvious that 1 d  holds, as for q as an interior point of the ellipsoid true.Substituting (18) into (10) th the line of intersection of ellipsoid and plane, are given ) ting and With Theorem 3 one obtains by substitu n  from (42) and (43) the formular for the center of the ellipse given by: The intention of this paper was, to give an elem closed form solution to the general problem of the interon of an ellipsoid and a plane.
zero, when choosing  such that other and  s = n r the following state- ment holds: ) and substituting the expressions from (31) leads to:


is the zero Applying representation (39) on btains:

4 :
The apexes of the ellipse of intersection are

m 5 : 1 dm
Corollary  holds if and only if m is an interior point of the ellipsoid (1), because of of ellipsoid (1) and a tangent plane with normal vector , since n