Projection of the Semi-Axes of the Ellipse of Intersection

P. P. Klein

Computing Center, University of Technology Clausthal, Clausthal-Zellerfeld, Germany.

DOI: 10.4236/am.2017.89097                

Abstract

It is well known that the line of intersection of an ellipsoid and a plane is an ellipse (see for instance [1]). In this note the semi-axes of the ellipse of intersection will be projected from 3d space onto a 2d plane. It is shown that the projected semi-axes agree with results of a method used by Bektas [2] and also with results obtained by Schrantz [3].

Keywords

Ellipsoid and Plane Intersection, Projection of the Semi-Axes of the Ellipse of Intersection

1. Introduction

Let an ellipsoid be given with the three positive semi-axes $a_1$ , $a_2$ , $a_3$

$\frac{x_1^2}{a_1^2}+\frac{x_2^2}{a_2^2}+\frac{x_3^2}{a_3^2}=1$ (1)

and a plane with the unit normal vector

$n={\left(n_1\mathrm{,}{n}_2\mathrm{,}{n}_3\right)}^{\text{T}}\mathrm{,}$

which contains an interior point $q={\left(q_1\mathrm{,}{q}_2\mathrm{,}{q}_3\right)}^{\text{T}}$ of the ellipsoid. A plane spanned by vectors $r={\left(r_1\mathrm{,}{r}_2\mathrm{,}{r}_3\right)}^{\text{T}}$ , $s={\left(s_1\mathrm{,}{s}_2\mathrm{,}{s}_3\right)}^{\text{T}}$ and containing the point $q$ is described in parametric form by

$x=q+tr+us\text{with}x={\left(x_1\mathrm{,}{x}_2\mathrm{,}{x}_3\right)}^{\text{T}}\mathrm{.}$ (2)

Inserting the components of $x$ into the equation of the ellipsoid (1) leads to the line of intersection as a quadratic form in the variables t and u. Let the scalar product in $R^3$ for two vectors $v={\left(v_1\mathrm{,}{v}_2\mathrm{,}{v}_3\right)}^{\text{T}}$ and $w={\left(w_1\mathrm{,}{w}_2\mathrm{,}{w}_3\right)}^{\text{T}}$ be denoted by

$\left(v\mathrm{,}w\right)=v_1{w}_1+v_2{w}_2+v_3{w}_3$

and the norm of vector $v$ by

$\Vert v\Vert =\sqrt{\left(v\mathrm{,}v\right)}\mathrm{.}$

With the diagonal matrix

$D_1=\text{diag}\left(\frac{1}{a_1},\frac{1}{a_2},\frac{1}{a_3}\right)$

the line of intersection has the form:

$\begin{array}{l}\left(t\mathrm{,}u\right)\left(\begin{array}{cc}\left(D_{1}r\mathrm{,}{D}_{1}r\right)& \left(D_{1}r\mathrm{,}{D}_{1}s\right)\\ \left(D_{1}r\mathrm{,}{D}_{1}s\right)& \left(D_{1}s\mathrm{,}{D}_{1}s\right)\end{array}\right)\left(\begin{array}{c}t\\ u\end{array}\right)\\ +2\left(\left(D_{1}q\mathrm{,}{D}_{1}r\right)\mathrm{,}\left(D_{1}q\mathrm{,}{D}_{1}s\right)\right)\left(\begin{array}{c}t\\ u\end{array}\right)\\ =1-\left(D_{1}q\mathrm{,}{D}_{1}q\right)\mathrm{.}\end{array}$ (3)

As $q$ is an interior point of the ellipsoid the right-hand side of Equation (3) is positive.

Let $r$ and $s$ be unit vectors orthogonal to the unit normal vector $n$ of the plane

$\begin{array}{l}\left(r\mathrm{,}r\right)=r_1^2+r_2^2+r_3^2=1,\\ \left(n\mathrm{,}r\right)=n_1{r}_1+n_2{r}_2+n_3{r}_3=0,\end{array}$ (4)

$\begin{array}{l}\left(s\mathrm{,}s\right)=s_1^2+s_2^2+s_3^2=1,\\ \left(n\mathrm{,}s\right)=n_1{s}_1+n_2{s}_2+n_3{s}_3=0,\end{array}$ (5)

and orthogonal to eachother

$\left(r\mathrm{,}s\right)=r_1{s}_1+r_2{s}_2+r_3{s}_3=0.$ (6)

If vectors $r$ and $s$ have the additional property

$\left(D_{1}r\mathrm{,}{D}_{1}s\right)=\frac{r_1{s}_1}{a_1^2}+\frac{r_2{s}_2}{a_2^2}+\frac{r_3{s}_3}{a_3^2}=0$ (7)

the $2\times 2$ matrix in (3) has diagonal form. If condition (7) does not hold for vectors $r$ and $s$ , it can be fulfilled, as shown in [1] , with vectors $\tilde{r}$ and $\tilde{s}$ obtained by a transformation of the form

$\begin{array}{l}\tilde{r}=\cos \omega r+\sin \omega s\mathrm{,}\\ \tilde{s}=-\sin \omega r+\cos \omega s\end{array}$ (8)

with an angle $\omega$ according to

$\omega =\frac{1}{2}arctan\left[\frac{2\left(D_{1}r\mathrm{,}{D}_{1}s\right)}{\left(D_{1}r\mathrm{,}{D}_{1}r\right)-\left(D_{1}s\mathrm{,}{D}_{1}s\right)}\right]\mathrm{.}$ (9)

Relations (4), (5) and (6) hold for the transformed vectors $\tilde{r}$ and $\tilde{s}$ instead of $r$ and $s$ . If plane (2) is written instead of vectors $r$ and $s$ with the transformed vectors $\tilde{r}$ and $\tilde{s}$ the $2\times 2$ matrix in (3) has diagonal form because of condition (7):

$\begin{array}{l}\left(D_1\tilde{r}\mathrm{,}{D}_1\tilde{r}\right)t^2+\left(D_1\tilde{s}\mathrm{,}{D}_1\tilde{s}\right)u^2+2\left(D_{1}q\mathrm{,}{D}_1\tilde{r}\right)t+2\left(D_{1}q\mathrm{,}{D}_1\tilde{s}\right)u\\ =1-\left(D_{1}q\mathrm{,}{D}_{1}q\right)\mathrm{.}\end{array}$

Then the line of intersection reduces to an ellipse in translational form

$\frac{{\left(t-t_0\right)}^2}{A^2}+\frac{{\left(u-u_0\right)}^2}{B^2}=1$ (10)

with the center $\left(t_0\mathrm{,}{u}_0\right)$

$t_0=-\frac{\left(D_{1}q\mathrm{,}{D}_1\tilde{r}\right)}{\left(D_1\tilde{r}\mathrm{,}{D}_1\tilde{r}\right)}\text{and}{u}_0=-\frac{\left(D_{1}q\mathrm{,}{D}_1\tilde{s}\right)}{\left(D_1\tilde{s}\mathrm{,}{D}_1\tilde{s}\right)}$ (11)

and the semi-axes

$A=\sqrt{\frac{1-d}{\left(D_1\tilde{r}\mathrm{,}{D}_1\tilde{r}\right)}}\text{and}B=\sqrt{\frac{1-d}{\left(D_1\tilde{s}\mathrm{,}{D}_1\tilde{s}\right)}}\mathrm{,}$ (12)

where

$d=\left(D_{1}q\mathrm{,}{D}_{1}q\right)-\frac{{\left(D_{1}q\mathrm{,}{D}_1\tilde{r}\right)}^2}{\left(D_1\tilde{r}\mathrm{,}{D}_1\tilde{r}\right)}-\frac{{\left(D_{1}q\mathrm{,}{D}_1\tilde{s}\right)}^2}{\left(D_1\tilde{s}\mathrm{,}{D}_1\tilde{s}\right)}\mathrm{.}$ (13)

Because of $1-d\ge 1-\left(D_{1}q\mathrm{,}{D}_{1}q\right)>0$ the numerator $1-d$ in (12) is positive.

Putting

${\beta }_1=\left(D_1\tilde{r}\mathrm{,}{D}_1\tilde{r}\right)\text{and}{\beta }_2=\left(D_1\tilde{s}\mathrm{,}{D}_1\tilde{s}\right)$ (14)

the semi-axes A, B given in (12) can be rewritten as

$A=\sqrt{\frac{1-d}{{\beta }_1}}\text{and}B=\sqrt{\frac{1-d}{{\beta }_2}}.$ (15)

In [1] it is shown that ${\beta }_1$ and ${\beta }_2$ according to (14) are solutions of the following quadratic equation

$\begin{array}{l}{\beta }^2-\left[n_1^2\left(\frac{1}{a_2^2}+\frac{1}{a_3^2}\right)+n_2^2\left(\frac{1}{a_1^2}+\frac{1}{a_3^2}\right)+n_3^2\left(\frac{1}{a_1^2}+\frac{1}{a_2^2}\right)\right]\beta \\ +\frac{n_1^2}{a_2^2{a}_3^2}+\frac{n_2^2}{a_1^2{a}_3^2}+\frac{n_3^2}{a_1^2{a}_2^2}=0.\end{array}$ (16)

Furthermore it is proven in [1] that d according to (13) satisfies

$d=\frac{{\kappa }^2}{a_1^2{n}_1^2+a_2^2{n}_2^2+a_3^2{n}_3^2}.$ (17)

2. Projection of the Ellipse of Intersection onto a 2-d Plane

The curve of intersection in 3d space can be described by

$x=m+\left(A\cos \theta \right)\tilde{r}+\left(B\sin \theta \right)\tilde{s}$ (18)

with center $m=q+t_0\tilde{r}+u_0\tilde{s}$ , where $t_0$ and $u_0$ are from (11), semi-axes A and B from (12), $\theta \in \left[\mathrm{0,2}\text{π}\right)$ and vectors $\tilde{r}$ and $\tilde{s}$ obtained after a suitable rotation (8) starting from initial vectors $r$ and $s$ (see for instance [1] ).

Without loss of generality the plane of projection of the ellipse (18) shall be the $x_1-x_2$ plane. The angle between the plane of intersection (2) containing the ellipse (18) and the plane of projection is denoted by $\Omega$ . The same angle is to be found between the unit normal $n$ of the plane of intersection (2) and the $x_3$ -direction, normal to the plane of projection. Denoting the unit vector in $x_3$ -direction by $e_3$ the definition of the scalar product (see for instance [4] ) yields

$n_3=\left(n\mathrm{,}{e}_3\right)=\Vert n\Vert \Vert e_3\Vert \cos \Omega =\cos \Omega$ (19)

where $\cos \Omega >0$ holds for $0\le \Omega <\frac{\text{π}}{2}$ .

Let us assume that the plane of intersection (2) is not perpendicular to the

plane of projection, the $x_1-x_2$ plane. This means that $0\le \Omega <\frac{\text{π}}{2}$ is valid and

according to (19) $n_3>0$ holds.

The ellipse of intersection (18) projected from 3d space onto the $x_1-x_2$ plane has the following form:

$\begin{array}{l}{x}_1=m_1+A\cos \theta {\tilde{r}}_1+B\sin \theta {\tilde{s}}_1\\ x_2=m_2+A\cos \theta {\tilde{r}}_2+B\sin \theta {\tilde{s}}_2.\end{array}$ (20)

In general the two dimensional vectors ${\left({\tilde{r}}_1\mathrm{,}{\tilde{r}}_2\right)}^{\text{T}}$ and ${\left({\tilde{s}}_1\mathrm{,}{\tilde{s}}_2\right)}^{\text{T}}$ are not orthogonal because their orthogonality in 3d space implies

${\tilde{r}}_1{\tilde{s}}_1+{\tilde{r}}_2{\tilde{s}}_2=-{\tilde{r}}_3{\tilde{s}}_3,$

which need not be zero. In order to calculate the lenghts of the semi-axes A and B projected from 3d space onto the $x_1-x_2$ plane the following linear system deduced from (20) with the abbreviations ${x^{\prime }}_1=x_1-m_1$ and ${x^{\prime }}_2=x_2-m_2$ is treated:

$\left(\begin{array}{cc}A{\tilde{r}}_1& B{\tilde{s}}_1\\ A{\tilde{r}}_2& B{\tilde{s}}_2\end{array}\right)\left(\begin{array}{c}\cos \theta \\ \sin \theta \end{array}\right)=\left(\begin{array}{c}{x^{\prime }}_1\\ {x^{\prime }}_2\end{array}\right)$ (21)

The determinant of the linear system (21), $AB\left({\tilde{r}}_1{\tilde{s}}_2-{\tilde{r}}_2{\tilde{s}}_1\right)$ , is different from zero. This can be shown by noting that ${\tilde{r}}_1{\tilde{s}}_2-{\tilde{r}}_2{\tilde{s}}_1$ is the third component of the vector $\tilde{r}\times \tilde{s}$ . At first this vector is not affected by rotation (8):

$\begin{array}{c}\tilde{r}\times \tilde{s}=\left(\cos \omega r+\sin \omega s\right)\times \left(-\sin \omega r+\cos \omega s\right)\\ =\left({\cos }^2\omega +{\sin }^2\omega \right)\left(r\times s\right)=r\times s.\end{array}$

This result was obtained by applying the rules for the cross product in $R^3$ . Furthermore one obtains employing the Grassman expansion theorem (see for instance [4] ):

$r\times s=r\times \left(n\times r\right)=\left(r\mathrm{,}r\right)n-\left(r\mathrm{,}n\right)r=n$

because of $\left(r\mathrm{,}r\right)=1$ and $\left(r\mathrm{,}n\right)=0$ . Thus one ends up with

${\tilde{r}}_1{\tilde{s}}_2-{\tilde{r}}_2{\tilde{s}}_1=r_1{s}_2-r_2{s}_1=n_3,$ (22)

which is positive because of (19) for angles $\Omega$ with $0\le \Omega <\frac{\text{π}}{2}$ .

Solving the linear system (21) leads to

$\cos \theta =\frac{B\left({x^{\prime }}_1{\tilde{s}}_2-{x^{\prime }}_2{\tilde{s}}_1\right)}{AB\left({\tilde{r}}_1{\tilde{s}}_2-{\tilde{r}}_2{\tilde{s}}_1\right)},$

$\sin \theta =\frac{A\left({\tilde{r}}_1{x^{\prime }}_2-{\tilde{r}}_2{x^{\prime }}_1\right)}{AB\left({\tilde{r}}_1{\tilde{s}}_2-{\tilde{r}}_2{\tilde{s}}_1\right)}.$

Since ${\cos }^2\theta +{\sin }^2\theta =1$ together with (22) the following quadratic equation in ${x^{\prime }}_1$ and ${x^{\prime }}_2$ is obtained:

$B^2{\left({x^{\prime }}_1{\tilde{s}}_2-{x^{\prime }}_2{\tilde{s}}_1\right)}^2+A^2{\left({\tilde{r}}_1{x^{\prime }}_2-{\tilde{r}}_2{x^{\prime }}_1\right)}^2=A^2{B}^2{\left({\tilde{r}}_1{\tilde{s}}_2-{\tilde{r}}_2{\tilde{s}}_1\right)}^2=A^2{B}^2{n}_3^2\mathrm{.}$

Expanding the squares on the left side and using the denotations

$\begin{array}{l}{l}_{11}=A^2{\tilde{r}}_2^2+B^2{\tilde{s}}_2^2,\\ l_{12}=-\left(A^2{\tilde{r}}_1{\tilde{r}}_2+B^2{\tilde{s}}_1{\tilde{s}}_2\right),\\ l_{22}=A^2{\tilde{r}}_1^2+B^2{\tilde{s}}_1^2\end{array}$ (23)

arranged as a $2\times 2$ matrix $L$

$L=\left(\begin{array}{cc}{l}_{11}& l_{12}\\ l_{12}& l_{22}\end{array}\right)$ (24)

leads to

$\left({x^{\prime }}_1\mathrm{,}{x^{\prime }}_2\right)L\left(\begin{array}{c}{x^{\prime }}_1\\ {x^{\prime }}_2\end{array}\right)=A^2{B}^2{n}_3^2\mathrm{.}$ (25)

$L$ as a real symmetric matrix can be diagonalized and thus is similar to the diagonal matrix of its eigenvalues ${\lambda }_1\left(L\right)$ , ${\lambda }_2\left(L\right)$ :

$L=S^{-1}\text{diag}\left({\lambda }_1\left(L\right),{\lambda }_2\left(L\right)\right)S$

with a nonsingular transformation matrix $S$ , being orthogonal, i.e. $S^{-1}=S^{\text{T}}$ , the inverse of $S$ is equal to the transpose of $S$ . Putting

$\left({x^{″}}_1\mathrm{,}{x^{″}}_2\right)=\left({x^{\prime }}_1\mathrm{,}{x^{\prime }}_2\right)S^{\text{T}}\mathrm{,}S\left(\begin{array}{c}{x^{\prime }}_1\\ {x^{\prime }}_2\end{array}\right)=\left(\begin{array}{c}{x^{″}}_1\\ {x^{″}}_2\end{array}\right)$

the quadratic equation (25) in $\left({x^{\prime }}_1\mathrm{,}{x^{\prime }}_2\right)$ reduces to

$\left({x^{″}}_1\mathrm{,}{x^{″}}_2\right)\text{diag}\left({\lambda }_1\left(L\right),{\lambda }_2\left(L\right)\right)\left(\begin{array}{c}{x^{″}}_1\\ {x^{″}}_2\end{array}\right)=A^2{B}^2{n}_3^2.$ (26)

The eigenvalues ${\lambda }_1\left(L\right)$ , ${\lambda }_2\left(L\right)$ are positive because $L$ is positive definite; this is true since the terms $l_{11}$ and $l_{11}{l}_{22}-l_{12}^2$ are positive. For $l_{11}$ this is clear; for the second term, the determinant of $L$ , holds because of (22):

$\begin{array}{c}\det L=l_{11}{l}_{22}-l_{12}^2=\left(A^2{\tilde{r}}_2^2+B^2{\tilde{s}}_2^2\right)\left(A^2{\tilde{r}}_1^2+B^2{\tilde{s}}_1^2\right)-{\left(A^2{\tilde{r}}_1{\tilde{r}}_2+B^2{\tilde{s}}_1{\tilde{s}}_2\right)}^2\\ =A^2{B}^2{\left({\tilde{r}}_1{\tilde{s}}_2-{\tilde{r}}_2{\tilde{s}}_1\right)}^2=A^2{B}^2{\left(r_1{s}_2-r_2{s}_1\right)}^2=A^2{B}^2{n}_3^2.\end{array}$ (27)

Dividing (26) by $A^2{B}^2{n}_3^2$ yields

$\frac{{\lambda }_1\left(L\right)}{A^2{B}^2{n}_3^2}{\left({x^{″}}_1\right)}^2+\frac{{\lambda }_2\left(L\right)}{A^2{B}^2{n}_3^2}{\left({x^{″}}_2\right)}^2=1.$

This is an ellipse projected from 3d space (18) onto the $x_1-x_2$ plane with the semi-axes

$A_L=\frac{AB{n}_3}{\sqrt{{\lambda }_1\left(L\right)}},B_L=\frac{AB{n}_3}{\sqrt{{\lambda }_2\left(L\right)}}.$ (28)

With (19) one obtains from (28)

$A_L=\frac{AB\cos \Omega }{\sqrt{{\lambda }_1\left(L\right)}},B_L=\frac{AB\cos \Omega }{\sqrt{{\lambda }_2\left(L\right)}}.$ (29)

3. Calculation of Semi-Axes According to a Method Used by Bektas

Let the ellipsoid (1) be given and a plane in the form

$A_1{x}_1+A_2{x}_2+A_3{x}_3+A_4=0.$ (30)

The unit normal vector of the plane is:

$n=\frac{1}{\sqrt{A_1^2+A_2^2+A_3^2}}\left(A_1\mathrm{,}{A}_2\mathrm{,}{A}_3\right)\mathrm{.}$ (31)

The distance between the plane and the origin is given by

$\kappa =-\frac{A_4}{\sqrt{A_1^2+A_2^2+A_3^2}}.$ (32)

The plane written in Hessian normal form then reads:

$n_1{x}_1+n_2{x}_2+n_3{x}_3-\kappa =0.$

Without loss of generality $A_3\ne 0$ shall be assumed. Then $n_3\ne 0$ holds:

$x_3=\frac{1}{n_3}\left(\kappa -n_1{x}_1-n_2{x}_2\right).$

Forming $x_3^2$ and substituting into equation (1) gives:

$m_{11}{x}_1^2+2m_{12}{x}_1{x}_2+m_{22}{x}_2^2+2m_{13}{x}_1+2m_{23}{x}_2+m_{33}=0$ (33)

with

$\begin{array}{l}{m}_{11}=\frac{1}{a_1^2}+\frac{n_1^2}{a_3^2{n}_3^2},m_{12}=\frac{n_1{n}_2}{a_3^2{n}_3^2},\\ m_{22}=\frac{1}{a_2^2}+\frac{n_2^2}{a_3^2{n}_3^2},m_{13}=-\frac{n_1\kappa }{a_3^2{n}_3^2},\\ m_{23}=-\frac{n_2\kappa }{a_3^2{n}_3^2},m_{33}=\frac{{\kappa }^2}{a_3^2{n}_3^2}-1.\end{array}$ (34)

In the sequel the determinant of the following matrix will be needed:

$M=\left(\begin{array}{cc}{m}_{11}& m_{12}\\ m_{12}& m_{22}\end{array}\right)$

$\begin{array}{c}\det M=m_{11}{m}_{22}-m_{12}^2=\left(\frac{1}{a_1^2}+\frac{n_1^2}{a_3^2{n}_3^2}\right)\left(\frac{1}{a_2^2}+\frac{n_2^2}{a_3^2{n}_3^2}\right)-\frac{n_1^2{n}_2^2}{a_3^4{n}_3^4}\\ =\frac{n_3^2}{a_1^2{a}_2^2{n}_3^2}+\frac{n_1^2}{a_2^2{a}_3^2{n}_3^2}+\frac{n_2^2}{a_1^2{a}_3^2{n}_3^2}=\frac{a_1^2{n}_1^2+a_2^2{n}_2^2+a_3^2{n}_3^2}{a_1^2{a}_2^2{a}_3^2{n}_3^2}.\end{array}$ (35)

In order to get rid of the linear terms $x_1$ and $x_2$ in (33) the following translation can be performed: $x_1={x^{\prime }}_1+h$ , $x_2={x^{\prime }}_2+k$ with parameters h and k to be determined later. After substitution into (33) one obtains:

$\begin{array}{l}{m}_{11}{x^{\prime }}_1^2+2m_{12}{x^{\prime }}_1{x^{\prime }}_2+m_{22}{x^{\prime }}_2^2+2\left(m_{11}h+m_{12}k+m_{13}\right){x^{\prime }}_1\\ +2\left(m_{12}h+m_{22}k+m_{23}\right){x^{\prime }}_2+m_{11}{h}^2+2m_{12}hk+m_{22}{k}^2\\ +2m_{13}h+2m_{23}k+m_{33}=0.\end{array}$ (36)

The terms ${x^{\prime }}_1$ and ${x^{\prime }}_2$ in (36) vanish if h and k are determined by the linear system:

$\begin{array}{l}{m}_{11}h+m_{12}k=-m_{13},\\ m_{12}h+m_{22}k=-m_{23}.\end{array}$ (37)

The linear system (37) has $M$ as matrix of coefficients, the determinant of which is given in (35). It is nonzero because of the assumption $n_3\ne 0$ . Solving the linear system (37) yields:

$\begin{array}{l}h=\frac{-m_{13}{m}_{22}+m_{23}{m}_{12}}{m_{11}{m}_{22}-m_{12}^2},\\ k=\frac{-m_{11}{m}_{23}+m_{12}{m}_{13}}{m_{11}{m}_{22}-m_{12}^2}.\end{array}$ (38)

Substituting the terms (34) into (38) gives the result:

$\begin{array}{l}h=\frac{a_1^2{n}_1\kappa }{a_1^2{n}_1^2+a_2^2{n}_2^2+a_3^2{n}_3^2},\\ k=\frac{a_2^2{n}_2\kappa }{a_1^2{n}_1^2+a_2^2{n}_2^2+a_3^2{n}_3^2}.\end{array}$ (39)

With the terms h and k from (39) the constant term in (36) turns out to be, together with (17):

$\begin{array}{l}{m}_{11}{h}^2+2m_{12}hk+m_{22}{k}^2+2m_{13}h+2m_{23}k+m_{33}\\ =\frac{{\kappa }^2}{a_1^2{n}_1^2+a_2^2{n}_2^2+a_3^2{n}_3^2}-1=-\left(1-d\right).\end{array}$

Thus the quadratic equation (36) reduces to:

$\left({x^{\prime }}_1\mathrm{,}{x^{\prime }}_2\right)M\left(\begin{array}{c}{x^{\prime }}_1\\ {x^{\prime }}_2\end{array}\right)=1-d\mathrm{.}$ (40)

$M$ as a real symmetric matrix can be diagonalized and thus is similar to the diagonal matrix of its eigenvalues ${\lambda }_1\left(M\right)$ , ${\lambda }_2\left(M\right)$ :

$M=T^{-1}\text{diag}\left({\lambda }_1\left(M\right),{\lambda }_2\left(M\right)\right)T$

with a nonsingular transformation matrix $T$ , being orthogonal, i.e. $T^{-1}=T^{\text{T}}$ , the inverse of $T$ is equal to the transpose of $T$ . Putting

$\left({x^{″}}_1\mathrm{,}{x^{″}}_2\right)=\left({x^{\prime }}_1,{x^{\prime }}_2\right)T^{\text{T}},T\left(\begin{array}{c}{x^{\prime }}_1\\ {x^{\prime }}_2\end{array}\right)=\left(\begin{array}{c}{x^{″}}_1\\ {x^{″}}_2\end{array}\right)$

the quadratic equation (40) in $\left({x^{\prime }}_1,{x^{\prime }}_2\right)$ reduces to

$\left({x^{″}}_1\mathrm{,}{x^{″}}_2\right)\text{diag}\left({\lambda }_1\left(M\right),{\lambda }_2\left(M\right)\right)\left(\begin{array}{c}{x^{″}}_1\\ {x^{″}}_2\end{array}\right)=1-d.$ (41)

The eigenvalues ${\lambda }_1\left(M\right)$ , ${\lambda }_2\left(M\right)$ are positive because $M$ is positive definite; this is true since the terms $m_{11}$ and $m_{11}{m}_{22}-m_{12}^2$ are positive. For $m_{11}$ this is clear; the second term, the determinant of $M$ , is given in (35). If a point of the plane (30) exists which is an interior point of the ellipsoid (1), then $1-d$ is positive (see Section 1). Dividing (41) by $1-d$ yields

$\frac{{\lambda }_1\left(M\right)}{1-d}{\left({x^{″}}_1\right)}^2+\frac{{\lambda }_2\left(M\right)}{1-d}{\left({x^{″}}_2\right)}^2=1.$

This is an ellipse in the $x_1-x_2$ plane with the semi-axes

$A_M=\sqrt{\frac{1-d}{{\lambda }_1\left(M\right)}},B_M=\sqrt{\frac{1-d}{{\lambda }_2\left(M\right)}}.$ (42)

4. Calculation of Projected Semi-Axes According to Schrantz

In [3] the ellipse

$x_1=A\cos t,x_2=B\sin t,t\in \left[0,2\text{π}\right)$ (43)

with the semi-axes A and B is projected from plane E onto plane $E^{\prime }$ . As in

Section 2 the angle between the two planes is denoted by $\Omega$ , with $0\le \Omega \le \frac{\text{π}}{2}$ . Let $\alpha$ , with $0\le \alpha \le \frac{\text{π}}{2}$ , be the angle between the major axis of the original

ellipse (43) and the straight line of intersection of the two planes E and $E^{\prime }$

$\left(E\cap E^{\prime }\right)$ and let $\psi$ be a phase-shift with $0\le \psi \le \frac{\text{π}}{2}$ and $\psi =\tau -\sigma$ where

the angles $\tau$ and $\sigma$ are determined by

$\begin{array}{l}\cos \sigma =\frac{A\cos \alpha }{\sqrt{A^2{\cos }^2\alpha +B^2{\sin }^2\alpha }},\\ \sin \sigma =\frac{B\sin \alpha }{\sqrt{A^2{\cos }^2\alpha +B^2{\sin }^2\alpha }},\\ \cos \tau =\frac{B\cos \alpha }{\sqrt{A^2{\sin }^2\alpha +B^2{\cos }^2\alpha }},\\ \sin \tau =\frac{A\sin \alpha }{\sqrt{A^2{\sin }^2\alpha +B^2{\cos }^2\alpha }}.\end{array}$ (44)

The projected ellipse in the plane $E^{\prime }$ is given by

${\bar{x}}_1=\bar{A}\cos \left(\bar{t}+\psi \right),{\bar{x}}_2=\bar{B}\sin \bar{t},\bar{t}\in \left[0,2\text{π}\right)$ (45)

with

$\bar{A}=\sqrt{A^2{\cos }^2\alpha +B^2{\sin }^2\alpha },$

$\bar{B}=\cos \Omega \sqrt{A^2{\sin }^2\alpha +B^2{\cos }^2\alpha }.$ (46)

Eliminating parameter $\bar{t}$ from (45) yields a quadratic equation in ${\bar{x}}_1$ and ${\bar{x}}_2$

${\left(\frac{{\bar{x}}_1}{\bar{A}}\right)}^2+2\sin \psi \left(\frac{{\bar{x}}_1}{\bar{A}}\right)\left(\frac{{\bar{x}}_2}{\bar{B}}\right)+{\left(\frac{{\bar{x}}_2}{\bar{B}}\right)}^2={\cos }^2\psi$

or written with the elements

$g_{11}=\frac{1}{{\bar{A}}^2},g_{12}=\frac{\sin \psi }{\bar{A}\bar{B}},g_{22}=\frac{1}{{\bar{B}}^2}$ (47)

forming matrix

$G=\left(\begin{array}{cc}{g}_{11}& g_{12}\\ g_{12}& g_{22}\end{array}\right)$

one obtains

$\left({\bar{x}}_1,{\bar{x}}_2\right)G\left(\begin{array}{c}{\bar{x}}_1\\ {\bar{x}}_2\end{array}\right)={\cos }^2\psi .$ (48)

$G$ as a real symmetric matrix can be diagonalized and thus is similar to the diagonal matrix of its eigenvalues ${\lambda }_1\left(G\right)$ , ${\lambda }_2\left(G\right)$ :

$G=R^{-1}\text{diag}\left({\lambda }_1\left(G\right),{\lambda }_2\left(G\right)\right)R$

with a nonsingular transformation matrix $R$ , being orthogonal, i.e. $R^{-1}=R^{\text{T}}$ , the inverse of $R$ is equal to the transpose of $R$ . Putting

$\left({\overline{\stackrel{¯}{x}}}_1\mathrm{,}{\overline{\stackrel{¯}{x}}}_2\right)=\left({\bar{x}}_1\mathrm{,}{\bar{x}}_2\right)R^{\text{T}}\mathrm{,}R\left(\begin{array}{c}{\bar{x}}_1\\ {\bar{x}}_2\end{array}\right)=\left(\begin{array}{c}{\overline{\stackrel{¯}{x}}}_1\\ {\overline{\stackrel{¯}{x}}}_2\end{array}\right)$

the quadratic equation (48) in $\left({\bar{x}}_1\mathrm{,}{\bar{x}}_2\right)$ reduces to

$\left({\overline{\stackrel{¯}{x}}}_1,{\overline{\stackrel{¯}{x}}}_2\right)\text{diag}\left({\lambda }_1\left(G\right),{\lambda }_2\left(G\right)\right)\left(\begin{array}{c}{\overline{\stackrel{¯}{x}}}_1\\ {\overline{\stackrel{¯}{x}}}_2\end{array}\right)={\cos }^2\psi .$ (49)

The eigenvalues ${\lambda }_1\left(G\right)$ , ${\lambda }_2\left(G\right)$ are positive, if G is positive definite; this is the case if the terms $g_{11}$ and $g_{11}{g}_{22}-g_{12}^2$ are positive. For $g_{11}$ this is true; the second term, the determinant of G, given by

$\det G=g_{11}{g}_{22}-g_{12}^2=\frac{1}{{\bar{A}}^2{\bar{B}}^2}-\frac{{\sin }^2\psi }{{\bar{A}}^2{\bar{B}}^2}=\frac{{\cos }^2\psi }{{\bar{A}}^2{\bar{B}}^2}$ (50)

is positive for $0\le \psi <\frac{\text{π}}{2}$ . Dividing (49) by ${\cos }^2\psi$ for $0\le \psi <\frac{\text{π}}{2}$ yields

$\frac{{\lambda }_1\left(G\right)}{{\cos }^2\psi }{\left({\overline{\stackrel{¯}{x}}}_1\right)}^2+\frac{{\lambda }_2\left(G\right)}{{\cos }^2\psi }{\left({\overline{\stackrel{¯}{x}}}_2\right)}^2=1.$

This is an ellipse in the ${\bar{x}}_1-{\bar{x}}_2$ plane with the semi-axes

$A_G=\frac{\cos \psi }{\sqrt{{\lambda }_1\left(G\right)}},B_G=\frac{\cos \psi }{\sqrt{{\lambda }_2\left(G\right)}}.$ (51)

5. Some Auxiliary Means

Let $H$ stand for the following $2\times 2$ matrix:

$H=\left(\begin{array}{cc}{h}_{11}& h_{12}\\ h_{12}& h_{22}\end{array}\right)$ (52)

and be a place holder for the matrices $M$ and $G$ used above. The semi-axes $A_L$ , $B_L$ projected onto the $x_1-x_2$ plane, given in (28), are compared with the semi-axes $A_H$ , $B_H$ . It will be shown that the two polynomials

$\begin{array}{l}{Q}_L\left(z\right)=z^2-\left(A_L+B_L\right)z+A_L{B}_L\mathrm{,}\\ Q_H\left(z\right)=z^2-\left(A_H+B_H\right)z+A_H{B}_H\mathrm{,}\end{array}$ (53)

have the same coefficients and thus have the same zeros:

$\begin{array}{l}{Q}_L\left(z\right)=\left(z-A_L\right)\left(z-B_L\right)\mathrm{,}\\ Q_H\left(z\right)=\left(z-A_H\right)\left(z-B_H\right)\mathrm{.}\end{array}$ (54)

In the first step $A_L{B}_L=A_H{B}_H$ will be proven. In the second step

$A_L^2+B_L^2=A_H^2+B_H^2$ (55)

will be shown. This is sufficient, since by adding $2A_L{B}_L=2A_H{B}_H$ to both sides of (55) one obtains:

${\left(A_L+B_L\right)}^2=A_L^2+2A_L{B}_L+B_L^2=A_H^2+2A_H{B}_H+B_H^2={\left(A_H+B_H\right)}^2$

which yields $A_L+B_L=A_H+B_H$ since the semi-axes are positive.

${\lambda }_1\left(L\right)$ , ${\lambda }_2\left(L\right)$ are the zeros of the characteristic polynomial of $L$ . This can be expressed in two ways:

$P_L\left(\lambda \right)=\left(l_{11}-\lambda \right)\left(l_{22}-\lambda \right)-l_{12}^2={\lambda }^2-\left(l_{11}+l_{22}\right)\lambda +l_{11}{l}_{22}-l_{12}^2,$

$P_L\left(\lambda \right)=\left(\lambda -{\lambda }_1\left(L\right)\right)\left(\lambda -{\lambda }_2\left(L\right)\right)={\lambda }^2-\left({\lambda }_1\left(L\right)+{\lambda }_2\left(L\right)\right)\lambda +{\lambda }_1\left(L\right){\lambda }_2\left(L\right).$

Comparing the coefficients one obtains

$\begin{array}{l}{\lambda }_1\left(L\right)+{\lambda }_2\left(L\right)=l_{11}+l_{22},\\ {\lambda }_1\left(L\right){\lambda }_2\left(L\right)=l_{11}{l}_{22}-l_{12}^2.\end{array}$ (56)

Similarly the results for matrix $H$ instead of $L$ are

$\begin{array}{l}{\lambda }_1\left(H\right)+{\lambda }_2\left(H\right)=h_{11}+h_{22}\mathrm{,}\\ {\lambda }_1\left(H\right){\lambda }_2\left(H\right)=h_{11}{h}_{22}-h_{12}^2\mathrm{.}\end{array}$ (57)

6. Comparison of the Semi-Axes A_L, B_L with A_M, B_M

In the first step $A_L{B}_L=A_M{B}_M$ will be proven. According to (28) and (42) holds:

$A_L{B}_L=\frac{A^2{B}^2{n}_3^2}{\sqrt{{\lambda }_1\left(L\right){\lambda }_2\left(L\right)}},$ (58)

$A_M{B}_M=\frac{1-d}{\sqrt{{\lambda }_1\left(M\right){\lambda }_2\left(M\right)}}.$ (59)

In the case of matrix $L$ combining (56) and (27) yields:

${\lambda }_1\left(L\right){\lambda }_2\left(L\right)=l_{11}{l}_{22}-l_{12}^2=A^2{B}^2{n}_3^2.$ (60)

In the case of matrix $M$ combining (57), where $M$ is substituted for $H$ ,and (35) leads to:

${\lambda }_1\left(M\right){\lambda }_2\left(M\right)=m_{11}{m}_{22}-m_{12}^2=\frac{a_1^2{n}_1^2+a_2^2{n}_2^2+a_3^2{n}_3^2}{a_1^2{a}_2^2{a}_3^2{n}_3^2}.$ (61)

Because ${\beta }_1$ and ${\beta }_2$ are solutions of (16)

${\beta }_1{\beta }_2=\frac{n_1^2}{a_2^2{a}_3^2}+\frac{n_2^2}{a_1^2{a}_3^2}+\frac{n_3^2}{a_1^2{a}_2^2}=\frac{a_1^2{n}_1^2+a_2^2{n}_2^2+a_3^2{n}_3^2}{a_1^2{a}_2^2{a}_3^2}$ (62)

holds and because of (60), (15), (62) and (61)

${\lambda }_1\left(L\right){\lambda }_2\left(L\right)=\frac{1-d}{{\beta }_1}\frac{1-d}{{\beta }_2}{n}_3^2=\frac{{\left(1-d\right)}^2{a}_1^2{a}_2^2{a}_3^2{n}_3^2}{a_1^2{n}_1^2+a_2^2{n}_2^2+a_3^2{n}_3^2}=\frac{{\left(1-d\right)}^2}{{\lambda }_1\left(M\right){\lambda }_2\left(M\right)}.$ (63)

Thus with (58), (60), (63) and (59) one concludes

$\begin{array}{c}{A}_L{B}_L=\frac{A^2{B}^2{n}_3^2}{\sqrt{{\lambda }_1\left(L\right){\lambda }_2\left(L\right)}}=\frac{{\lambda }_1\left(L\right){\lambda }_2\left(L\right)}{\sqrt{{\lambda }_1\left(L\right){\lambda }_2\left(L\right)}}=\sqrt{{\lambda }_1\left(L\right){\lambda }_2\left(L\right)}\\ =\frac{1-d}{\sqrt{{\lambda }_1\left(M\right){\lambda }_2\left(M\right)}}=A_M{B}_M.\end{array}$

In the second step because of (28) and (60) holds

$\begin{array}{c}{A}_L^2+B_L^2=A^2{B}^2{n}_3^2\left(\frac{1}{{\lambda }_1\left(L\right)}+\frac{1}{{\lambda }_2\left(L\right)}\right)\\ =\frac{A^2{B}^2{n}_3^2}{{\lambda }_1\left(L\right){\lambda }_2\left(L\right)}\left({\lambda }_2\left(L\right)+{\lambda }_1\left(L\right)\right)={\lambda }_1\left(L\right)+{\lambda }_2\left(L\right).\end{array}$ (64)

Because of (42), (61) and (62) holds

$\begin{array}{c}{A}_M^2+B_M^2=\frac{1-d}{{\lambda }_1\left(M\right)}+\frac{1-d}{{\lambda }_2\left(M\right)}=\frac{1-d}{{\lambda }_1\left(M\right){\lambda }_2\left(M\right)}\left({\lambda }_2\left(M\right)+{\lambda }_1\left(M\right)\right)\\ =\frac{\left(1-d\right)a_1^2{a}_2^2{a}_3^2{n}_3^2}{a_1^2{n}_1^2+a_2^2{n}_2^2+a_3^2{n}_3^2}\left({\lambda }_1\left(M\right)+{\lambda }_2\left(M\right)\right)\\ =\frac{\left(1-d\right)n_3^2}{{\beta }_1{\beta }_2}\left({\lambda }_1\left(M\right)+{\lambda }_2\left(M\right)\right).\end{array}$ (65)

Together with

${\lambda }_1\left(M\right)+{\lambda }_2\left(M\right)=m_{11}+m_{22}=\frac{1}{n_3^2}\left(\frac{n_3^2}{a_1^2}+\frac{n_3^2}{a_2^2}+\frac{n_1^2+n_2^2}{a_3^2}\right)$ (66)

(65) yields

$A_M^2+B_M^2=\frac{\left(1-d\right)}{{\beta }_1{\beta }_2}\left(\frac{n_3^2}{a_1^2}+\frac{n_3^2}{a_2^2}+\frac{n_1^2+n_2^2}{a_3^2}\right).$ (67)

In continuation of (64), because $\tilde{r}$ and $\tilde{s}$ are fulfilling (4) and (5), the following relations hold:

$\begin{array}{l}{\lambda }_1\left(L\right)+{\lambda }_2\left(L\right)=l_{11}+l_{22}=A^2\left({\tilde{r}}_1^2+{\tilde{r}}_2^2\right)+B^2\left({\tilde{s}}_1^2+{\tilde{s}}_2^2\right)\\ =A^2\left(1-{\tilde{r}}_3^2\right)+B^2\left(1-{\tilde{s}}_3^2\right)=\frac{1-d}{{\beta }_1}\left(1-{\tilde{r}}_3^2\right)+\frac{1-d}{{\beta }_2}\left(1-{\tilde{s}}_3^2\right)\\ =\frac{1-d}{{\beta }_1{\beta }_2}\left({\beta }_2\left(1-{\tilde{r}}_3^2\right)+{\beta }_1\left(1-{\tilde{s}}_3^2\right)\right)=\frac{1-d}{{\beta }_1{\beta }_2}\left({\beta }_1+{\beta }_2-{\beta }_2{\tilde{r}}_3^2-{\beta }_1{\tilde{s}}_3^2\right)\end{array}$ (68)

with

${\beta }_1+{\beta }_2=n_1^2\left(\frac{1}{a_2^2}+\frac{1}{a_3^2}\right)+n_2^2\left(\frac{1}{a_1^2}+\frac{1}{a_3^2}\right)+n_3^2\left(\frac{1}{a_1^2}+\frac{1}{a_2^2}\right)$ (69)

because ${\beta }_1$ and ${\beta }_2$ are solutions of (16). Combining (64), (68), (69) and (67) one obtains:

$A_L^2+B_L^2-\left(A_M^2+B_M^2\right)=\frac{1-d}{{\beta }_1{\beta }_2}\left(\frac{n_1^2}{a_2^2}+\frac{n_2^2}{a_1^2}-{\beta }_2{\tilde{r}}_3^2-{\beta }_1{\tilde{s}}_3^2\right).$ (70)

To simplify the term in round brackets of (70) the following relations are used:

$n_1={\tilde{r}}_2{\tilde{s}}_3-{\tilde{r}}_3{\tilde{s}}_2,n_2={\tilde{r}}_3{\tilde{s}}_1-{\tilde{r}}_1{\tilde{s}}_3,$

because of $\tilde{r}\times \tilde{s}=r\times s=n$ (see Section 2), and

${\beta }_2=\left(D_1\tilde{s}\mathrm{,}{D}_1\tilde{s}\right)\mathrm{,}{\beta }_1=\left(D_1\tilde{r}\mathrm{,}{D}_1\tilde{r}\right)$

according to (14). The term in round brackets of (70) thus becomes:

$\begin{array}{l}\frac{1}{a_2^2}{\left({\tilde{r}}_2{\tilde{s}}_3-{\tilde{r}}_3{\tilde{s}}_2\right)}^2+\frac{1}{a_1^2}{\left({\tilde{r}}_3{\tilde{s}}_1-{\tilde{r}}_1{\tilde{s}}_3\right)}^2-\left(\frac{{\tilde{s}}_1^2}{a_1^2}+\frac{{\tilde{s}}_2^2}{a_2^2}+\frac{{\tilde{s}}_3^2}{a_3^2}\right){\tilde{r}}_3^2-\left(\frac{{\tilde{r}}_1^2}{a_1^2}+\frac{{\tilde{r}}_2^2}{a_2^2}+\frac{{\tilde{r}}_3^2}{a_3^2}\right){\tilde{s}}_3^2\\ =-2{\tilde{r}}_3{\tilde{s}}_3\left(\frac{{\tilde{r}}_1{\tilde{s}}_1}{a_1^2}+\frac{{\tilde{r}}_2{\tilde{s}}_2}{a_2^2}+\frac{{\tilde{r}}_3{\tilde{s}}_3}{a_3^2}\right)=-2{\tilde{r}}_3{\tilde{s}}_3\left(D_1\tilde{r}\mathrm{,}{D}_1\tilde{s}\right)=\mathrm{0,}\end{array}$

because $\tilde{r}$ and $\tilde{s}$ have been chosen in such a way that condition (7) is fulfilled.

7. Comparison of the Semi-Axes A_L, B_L with A_G, B_G

In the first step $A_L{B}_L=A_G{B}_G$ will be proven. According to (29) and (51) holds:

$A_L{B}_L=\frac{A^2{B}^2{\cos }^2\Omega }{\sqrt{{\lambda }_1\left(L\right){\lambda }_2\left(L\right)}},$ (71)

$A_G{B}_G=\frac{{\cos }^2\psi }{\sqrt{{\lambda }_1\left(G\right){\lambda }_2\left(G\right)}}.$ (72)

In the case of matrix $L$ combining (56), (27) and (19) yields:

${\lambda }_1\left(L\right){\lambda }_2\left(L\right)=l_{11}{l}_{22}-l_{12}^2=A^2{B}^2{\cos }^2\Omega .$ (73)

In the case of matrix $G$ combining (57), where $G$ is substituted for $H$ ,and (50) leads to:

${\lambda }_1\left(G\right){\lambda }_2\left(G\right)=g_{11}{g}_{22}-g_{12}^2=\frac{{\cos }^2\psi }{{\bar{A}}^2{\bar{B}}^2}.$ (74)

Substitution of (73) into (71) and (74) into (72) yield

$A_L{B}_L-A_G{B}_G=AB\cos \Omega -\bar{A}\bar{B}\cos \psi .$ (75)

According to the definition of $\psi =\tau -\sigma$ given in the beginning of Section 4 together with (44) and (46) one obtains:

$\cos \psi =\cos \left(\tau -\sigma \right)=\frac{AB\cos \Omega }{\bar{A}\bar{B}}.$

Substituting this into (75) one ends up with $A_L{B}_L-A_G{B}_G=0$ .

In the second step because of (64), (56) and (23) holds

$\begin{array}{c}{A}_L^2+B_L^2={\lambda }_1\left(L\right)+{\lambda }_2\left(L\right)=l_{11}+l_{22}=A^2\left({\tilde{r}}_1^2+{\tilde{r}}_2^2\right)+B^2\left({\tilde{s}}_1^2+{\tilde{s}}_2^2\right)\\ =A^2\left(1-{\tilde{r}}_3^2\right)+B^2\left(1-{\tilde{s}}_3^2\right)=A^2+B^2-\left(A^2{\tilde{r}}_3^2+B^2{\tilde{s}}_3^2\right).\end{array}$ (76)

Because of (51), (74), (57), where matrix $G$ is substituted for matrix $H$ ,and (47) holds

$\begin{array}{c}{A}_G^2+B_G^2=\frac{{\cos }^2\psi }{{\lambda }_1\left(G\right)}+\frac{{\cos }^2\psi }{{\lambda }_2\left(G\right)}=\frac{{\cos }^2\psi }{{\lambda }_1\left(G\right){\lambda }_2\left(G\right)}\left({\lambda }_2\left(G\right)+{\lambda }_1\left(G\right)\right)\\ ={\bar{A}}^2{\bar{B}}^2\left({\lambda }_1\left(G\right)+{\lambda }_2\left(G\right)\right)={\bar{A}}^2{\bar{B}}^2\left(g_{11}+g_{22}\right)\\ ={\bar{A}}^2{\bar{B}}^2\left(\frac{1}{{\bar{A}}^2}+\frac{1}{{\bar{B}}^2}\right)={\bar{B}}^2+{\bar{A}}^2;\end{array}$ (77)

(77) is continued by substituting $\bar{B}$ and $\bar{A}$ from (46)

$\begin{array}{l}{\cos }^2\Omega \left(A^2{\sin }^2\alpha +B^2{\cos }^2\alpha \right)+A^2{\cos }^2\alpha +B^2{\sin }^2\alpha \\ =A^2\left({\cos }^2\alpha +{\cos }^2\Omega {\sin }^2\alpha \right)+B^2\left({\sin }^2\alpha +{\cos }^2\Omega {\cos }^2\alpha \right)\\ =A^2\left({\cos }^2\alpha +\left(1-{\sin }^2\Omega \right){\sin }^2\alpha \right)+B^2\left({\sin }^2\alpha +\left(1-{\sin }^2\Omega \right){\cos }^2\alpha \right)\\ =A^2\left({\cos }^2\alpha +{\sin }^2\alpha -{\sin }^2\Omega {\sin }^2\alpha \right)+B^2\left({\sin }^2\alpha +{\cos }^2\alpha -{\sin }^2\Omega {\cos }^2\alpha \right)\\ =A^2\left(1-{\sin }^2\Omega {\sin }^2\alpha \right)+B^2\left(1-{\sin }^2\Omega {\cos }^2\alpha \right)\\ =A^2+B^2-{\sin }^2\Omega \left(A^2{\sin }^2\alpha +B^2{\cos }^2\alpha \right)\end{array}$ (78)

Comparing (76) and (78), in order to show equality $A_L^2+B_L^2=A_G^2+B_G^2$ , it has to be proven:

$A^2{\tilde{r}}_3^2+B^2{\tilde{s}}_3^2={\sin }^2\Omega \left(A^2{\sin }^2\alpha +B^2{\cos }^2\alpha \right).$ (79)

As already described in the beginning of Section 4 the ellipse (43) is projected from the original plane $E$ onto the plane $E^{\prime }$ . Both planes are forming an

angle $\Omega$ with $0\le \Omega \le \frac{\text{π}}{2}$ . Without loss of generality the intersection of $E$

and $E^{\prime }$ , $E\cap E^{\prime }$ , shall be the ${\bar{x}}_1$ -axis of the coordinate system in plane $E^{\prime }$ . The original plane $E$ thus contains the following three points: $\left(-\mathrm{1,0,0}\right)$ , $\left(\mathrm{1,0,0}\right)$ , $\left(0,\cos \Omega ,\sin \Omega \right)$ and can therefore be described by the following equation:

$-\sin \Omega {\bar{x}}_2+\cos \Omega {\bar{x}}_3=0.$ (80)

The unit normal vector $n$ of plane (80) given by (31) is

$n=\left(0,-\sin \Omega ,\cos \Omega \right)\mathrm{.}$ (81)

In order to describe a unit vector $r$ in the plane $E$ the equations (4) must hold:

$\begin{array}{l}\left(r\mathrm{,}r\right)=r_1^2+r_2^2+r_3^2=1,\\ \left(n\mathrm{,}r\right)=-\sin \Omega r_2+\cos \Omega r_3=0.\end{array}$ (82)

The second equation of (82) yields $r_3=r_2\tan \Omega$ . Substituting this into the first equation of (82) results in:

$r_1^2+r_2^2\left(1+{\tan }^2\Omega \right)=1$

or

$r_1^2+\frac{r_2^2}{{\cos }^2\Omega }=1.$ (83)

If the unit vector $r$ is forming the angle $\alpha$ with the ${\bar{x}}_1$ -axis and $e_1$ is designating a unit vector in ${\bar{x}}_1$ -direction according to the definition of the scalar product (see for instance [4] ) holds

$r_1=\left(r\mathrm{,}{e}_1\right)=\Vert r\Vert \Vert e_1\Vert \cos \alpha =\cos \alpha .$

From (83) one obtains

$r_2^2=\left(1-{\cos }^2\alpha \right){\cos }^2\Omega ={\sin }^2\alpha {\cos }^2\Omega ,$

yielding $r_2=\pm \sin \alpha \cos \Omega$ and furthermore with the first equation of (82) $r_3=\pm \sin \alpha \sin \Omega$ . From

$r=\left(\cos \alpha ,\pm \sin \alpha \cos \Omega ,\pm \sin \alpha \sin \Omega \right)$

and $s=n\times r$ one obtains

$s=\left(\mp \sin \alpha ,\cos \alpha \cos \Omega ,\cos \alpha \sin \Omega \right)\mathrm{.}$

By transformation (8) one obtains

${\tilde{r}}_3=\cos \omega r_3+\sin \omega s_3=\sin \left(\omega \pm \alpha \right)\sin \Omega ,$

${\tilde{s}}_3=-\sin \omega r_3+\cos \omega s_3=\cos \left(\omega \pm \alpha \right)\sin \Omega .$

Thus equation (79) turns into

$\begin{array}{l}\left(A^2{\sin }^2\left(\omega \pm \alpha \right)+B^2{\cos }^2\left(\omega \pm \alpha \right)\right){\sin }^2\Omega \\ ={\sin }^2\Omega \left(A^2{\sin }^2\alpha +B^2{\cos }^2\alpha \right).\end{array}$ (84)

Equation (84) is fulfilled if $\omega \pm \alpha =\alpha$ holds. The $+$ -case leads to $\omega =0$ , which means that (84) is fulfilled if transformation (8) is the identity, i.e. $\tilde{r}=r$ , $\tilde{s}=s$ ; the $-$ -case leads to $\omega =2\alpha$ , meaning that if $\alpha$ , the angle between the

major axis of the ellipse (43) and the ${\bar{x}}_1$ -axis, is chosen to be $\frac{\omega }{2}$ then (84) is true.

8. Numerical Example

The following numerical example is taken from [2] . Let the semi-axes of the ellipsoid (1) be

$a_1=5,a_2=4,a_3=3$

and let the plane be given by

$x_1+2x_2+3x_3+4=0.$

The following calculations have been performed with Mathematica. According to (31) the unit normal vector $n$ of the plane is

$n=\frac{1}{\sqrt{1^2+2^2+3^2}}\left(\mathrm{1,2,3}\right)\mathrm{.}$

Furthermore in (32) the distance $\kappa$ of the plane to the origin is given

$\kappa =-\frac{4}{\sqrt{1^2+2^2+3^2}}\mathrm{.}$

According to (17) d can be calculated.

Starting with an arbitrary unit vector $r$ orthogonal to the unit normal vector $n$ , for instance

$r=\frac{1}{\sqrt{1^2+2^2}}{\left(\mathrm{2,}-\mathrm{1,0}\right)}^{\text{T}}\mathrm{,}$

calculating $s$ to be orthogonal to both according to $s=n\times r$ and, as $\left(D_{1}r\mathrm{,}{D}_{1}s\right)\ne 0$ , perform a rotation with angle $\omega$ given in (9), yielding new vectors $\tilde{r}$ and $\tilde{s}$ according to (8), which are plugged into $\left(D_1\tilde{r}\mathrm{,}{D}_1\tilde{r}\right)$ and $\left(D_1\tilde{s}\mathrm{,}{D}_1\tilde{s}\right)$ .

The semi-axes A and B in 3d space according to (12) can be calculated to be

$A=4.59157,B=\mathrm{3.39705.}$

Furthermore having calculated the eigenvalues ${\lambda }_1\left(L\right)$ and ${\lambda }_2\left(L\right)$ the semi-axes $A_L$ and $B_L$ projected onto the $x_1-x_2$ plane according to (28) are

$A_L=4.56667,B_L=\mathrm{2.73855.}$

The same results are obtained calculating $A_M$ and $B_M$ according to (42) by the method used by Bektas.

9. Conclusion

The intention of this paper was, to show that the semi-axes of the ellipse of intersection projected from 3d space onto a 2d plane are the same as those calculated by a method used by Bektas. Furthermore they are also equal to the semi-axes of the projected ellipse obtained by Schrantz.

Conflicts of Interest

The authors declare no conflicts of interest.

References

[1]	Klein, P.P. (2012) On the Ellipsoid and Plane Intersection Equation. Applied Mathematics, 3, 1634-1640. https://doi.org/10.4236/am.2012.311226
[2]	Bektas, S. (2016) On the Intersection of an Ellipsoid and a Plane. International Journal of Research in Engineering and Applied Sciences, 6, 273-283.
[3]	Schrantz, G.R. (2004) Projections of Ellipses and Circles. Hamline University. (To be found in the internet)
[4]	Bronshtein, I.N., Semendyayev, K.A., et al. (2007) Handbook of Mathematics. 5th Edition, Springer-Verlag Berlin Heidelberg.