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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.

Let an ellipsoid be given with the three positive semiaxes a, b, c

and a plane with the unit normal vector

which contains an interior point of the ellipsoid. A plane spanned by vectors, and containing the point is described in parametric form by

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. Let the scalar 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 and 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 [

Let and be unit vectors orthogonal to the unit normal vector of the plane

and orthogonal to eachother

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

with the center

and the semi-axes

where

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

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 applying to the spar product:

In case Equation (7) is not fulfilled for the initially chosen vectors r and s, i.e., the following transformation may be performed with

The transformed vectors and satisfy the following conditions:, and, which imply conditions (4)-(6). The expression

becomes zero, when choosing such that

holds. This can be reformulated, in case

to

If

holds, can be chosen, leading to

.

Corollary 1: For the unit vectors and orthogonal to each other and the following statement holds:

Statement (17) follows by substituting the definition of and utilising, and. For one obtains for instance:

Theorem 1: Let be the unit normal vector of the plane and let vectors and satisfy, , and condition (7). Putting

and are solutions of the following quadratic equation:

Proof: Utilising (17) one obtains:

Applying diagonality condition (7) and the identity of Lagrange (14) leads to:

For the cross product one obtains:

with the diagonal matrix

According to Grassmann’s expansion theorem for the double cross product (15)

follows, since and. Applying (20), (21), (23) one obtains:

□

A quadratic equation equivalent to (19) is considered in [

Corollary 2: Under the assumptions of Theorem 1 the following three equations are valid:

The first of the three equations was verified in the proof of Theorem 1. The second and the third equation follow analogously.

Theorem 2: Under the assumptions of Theorem 1 the expression for d in (13) is given by:

where is taken from (12).

Proof: The verification of (25) consists of three steps.

Step 1: Applying the identity of Lagrange (14) the following statements hold:

With Corollary 2 and the diagonal matrix

one obtains:

and it follows by substituting (28) into (26)

Introducing expressions

one obtains from (29) using (18) and (30)

Combining both Equations (31) leads to

Step 2: Analogously to the verification of (24) the application of the identity of Lagrange (14) yields:

With the diagonal matrix for the cross product holds:

Therefore one obtains

or

In contrast to the verification of (24), where diagonality condition (7) holds, the analogous expression in (33) need not be zero.

Step 3: Applying the identity of Lagrange (14) again leads to

Substituting the involved cross products according to Corollary 2 and considering diagonality condition (7) one obtains

or

Squaring both sides of (34) and substituting the expressions from (31) leads to:

Substitution of (33) results in equation

or

Substitution of (35) in (32) leads to:

Because of (24)

holds and with (13) one finally obtains relation (25)

Corollary 3: Under the assumptions of Theorem 1 the area of the ellipse obtained by the intersection of the ellipsoid (1) and a plane with unit normal vector and distance 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 becomes zero in case holds; this corresponds to the limiting case, where the cutting plane becomes a tangent plane. This result has been applied in [

Substituting according to (12) in formulars (9) for the coordinates of the center of the ellipse in the plane spanned by and one obtains:

and

The center of the ellipse in is given by:

Theorem 3: Let the assumptions of Theorem 1 be fulfilled. For the center of the ellipse of intersection in holds:

Proof: With diagonal matrices from (27) and

from (22) utilising and (37) one obtains a representation of equivalent to (40):

It is sufficient to show that for the difference

holds. Thus the coefficients in the expansion of in with respect to the orthonormal basis are zero, i.e., is the zero vector.

Applying representation (39) one obtains:

The last expression is zero according to (24). Furthermore one obtains:

and by interchanging the roles of and:

Both previous expressions are zero; this follows by applying diagonality condition (7), the identity of Lagrange (14) and Corollary 2:

Interchanging the roles of and leads to:

□

Corollary 4: The apexes of the ellipse of intersection are given by

where and are denoting the semi-axes according to (10).

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 equalities hold:

This can be shown using in the form (39) and employing condition (7).

Corollary 5: holds if and only if is an interior point of the ellipsoid (1), because of

In the case of, i.e., for the semi-axes (10) of the ellipse of intersection follows. The center (40) of the ellipse of intersection becomes a tangent contact point

of ellipsoid (1) and a tangent plane with normal vector, since holds.

Corollary 6: Describing the ellipse of intersection (8) in parametric form

with, where and are denoting its semi-axes according to (10), leads to a representation as a curve in three dimensional space as indicated in [

This result may be derived substituting the parameters and from the parametric form of the ellipse into Equation (2) of the plane:

or

where is equal to the center of the ellipse as in (39).

As indicated in [

with the unit normal vector:

The distance of the plane from the origin is given by:

According to (25) can be written as:

From (11) it is obvious that holds, as for as an interior point of the ellipsoid is true. Substituting (18) into (10) the semi-axes of the ellipse, the line of intersection of ellipsoid and plane, are given by

where are solutions of Equation (19):

With Theorem 3 one obtains by substituting and from (42) and (43) the formular for the center of the ellipse given by:

Instead of calculating and as solutions of (46) they may be obtained alternatively using the procedure described in 2. Starting with an arbitrary unit vector orthogonal to the unit normal vector given in (42), e.g.

calculating to be orthogonal to both according to and, in case, perform a rotation with angle as described in 2, yielding new vectors and, which are plugged into (18).

A Mathematica program containing both ways of computation of and may be obtained from the author upon request.

In the first special case of a plane containing the origin (see e.g. [

and from (9) holds. Thus Equation (8) of the line of intersection reduces to

A second special case, where holds, was treated in [

and

Because of in (12) holds and (38) reduces to

where and are solutions of the quadratic Equation (46) and vectors and have to be determined as described above according to the procedure shown in 2. Thus Equation (8) of the line of intersection turns into:

The intention of this paper was, to give an elementary closed form solution to the general problem of the intersection of an ellipsoid and a plane.