1. Introduction
Let an ellipsoid be given with the three positive semiaxes a, b, c
(1)
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
(2)
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:
(3)
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 [3] a generalization from three to p-dimensional space is discussed.
Let 
and 
be unit vectors orthogonal to the unit normal vector 
of the plane
(4)
(5)
and orthogonal to eachother
(6)
Furthermore vectors 
and 
may be chosen such that
(7)
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
(8)
with the center 
(9)
and the semi-axes
(10)
where
(11)
In order to show that the semi-axes (10) are independent of the choice of 
this vector may be decomposed orthogonally with respect to
:
(12)
where 
is the distance of plane (2) from the origin. Substituting 
into (11) one obtains employing (4), (5), (6) and (7)
(13)
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
(14)
and the Grassmann expansion theorem for the double cross product
(15)
2. 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
(16)
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:
(17)
Statement (17) follows by substituting the definition of 
and utilising
, 
and
. For 
one obtains for instance:
3. A Quadratic Equation
Theorem 1: Let 
be the unit normal vector of the plane and let vectors 
and 
satisfy
, 
, 
and condition (7). Putting
(18)
and 
are solutions of the following quadratic equation:
(19)
Proof: Utilising (17) one obtains:
Applying diagonality condition (7) and the identity of Lagrange (14) leads to:
(20)
For the cross product 
one obtains:
(21)
with the diagonal matrix
(22)
According to Grassmann’s expansion theorem for the double cross product (15)
(23)
follows, since 
and
. Applying (20), (21), (23) one obtains:
(24)
□
A quadratic equation equivalent to (19) is considered in [5].
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.
4. A Formular for d
Theorem 2: Under the assumptions of Theorem 1 the expression for d in (13) is given by:
(25)
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:
(26)
With Corollary 2 and the diagonal matrix
(27)
one obtains:
(28)
and it follows by substituting (28) into (26)
(29)
Introducing expressions
(30)
one obtains from (29) using (18) and (30)
(31)
Combining both Equations (31) leads to
(32)
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
(33)
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
(34)
Squaring both sides of (34) and substituting the expressions from (31) leads to:
Substitution of (33) results in equation
or
(35)
Substitution of (35) in (32) leads to:
(36)
Because of (24)
(37)
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 [6].
5. The Center of the Ellipse
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
(38)
The center 
of the ellipse in 
is given by:
(39)
Theorem 3: Let the assumptions of Theorem 1 be fulfilled. For the center 
of the ellipse of intersection in 
holds:
(40)
Proof: With diagonal matrices 
from (27) and 
from (22) utilising 
and (37) one obtains a representation of 
equivalent to (40):
(41)
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 [7]
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).
6. Applications
As indicated in [2], viewing 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 isodose lines, radio-therapy treatment of the eye can be planned. For this purpose the line of intersection of ellipsoid (1) and the plane, having the normal vector 
and containing the point
, situated in the interior of (1), is determined. The plane has the form:
with the unit normal vector:
(42)
The distance of the plane from the origin is given by:
(43)
According to (25) 
can be written as:
(44)
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
(45)
where 
are solutions of Equation (19):
(46)
With Theorem 3 one obtains by substituting 
and 
from (42) and (43) the formular for the center 
of the ellipse given by:
(47)
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. [1]), i.e. 
is the zero vector, it follows by (43), (44) and (47) that
, 
and 
is the zero vector also. Furthermore the semi-axes of the ellipse in (45) reduce to
and from (9) 
holds. Thus Equation (8) of the line of intersection reduces to
A second special case, where 
holds, was treated in [2]. Then the above formulas (43), (44) and (47) reduce to:
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:
7. Conclusion
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.