﻿ Projection of the Semi-Axes of the Ellipse of Intersection

Applied Mathematics
Vol.08 No.09(2017), Article ID:79281,16 pages
10.4236/am.2017.89097

Projection of the Semi-Axes of the Ellipse of Intersection

P. P. Klein

Computing Center, University of Technology Clausthal, Clausthal-Zellerfeld, Germany    Received: August 22, 2017; Accepted: September 22, 2017; Published: September 25, 2017

ABSTRACT

It is well known that the line of intersection of an ellipsoid and a plane is an ellipse (see for instance  ). 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  and also with results obtained by Schrantz  .

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

$n={\left({n}_{1},{n}_{2},{n}_{3}\right)}^{\text{T}},$

which contains an interior point $q={\left({q}_{1},{q}_{2},{q}_{3}\right)}^{\text{T}}$ of the ellipsoid. A plane spanned by vectors $r={\left({r}_{1},{r}_{2},{r}_{3}\right)}^{\text{T}}$ , $s={\left({s}_{1},{s}_{2},{s}_{3}\right)}^{\text{T}}$ and containing the point $q$ is described in parametric form by

$x=q+tr+us\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{with}\text{ }\text{\hspace{0.17em}}x={\left({x}_{1},{x}_{2},{x}_{3}\right)}^{\text{T}}.$ (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 $v={\left({v}_{1},{v}_{2},{v}_{3}\right)}^{\text{T}}$ and $w={\left({w}_{1},{w}_{2},{w}_{3}\right)}^{\text{T}}$ be denoted by

$\left(v,w\right)={v}_{1}{w}_{1}+{v}_{2}{w}_{2}+{v}_{3}{w}_{3}$

and the norm of vector $v$ by

$‖v‖=\sqrt{\left(v,v\right)}.$

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:

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

$\begin{array}{l}\left(r,r\right)={r}_{1}^{2}+{r}_{2}^{2}+{r}_{3}^{2}=1,\\ \left(n,r\right)={n}_{1}{r}_{1}+{n}_{2}{r}_{2}+{n}_{3}{r}_{3}=0,\end{array}$ (4)

$\begin{array}{l}\left(s,s\right)={s}_{1}^{2}+{s}_{2}^{2}+{s}_{3}^{2}=1,\\ \left(n,s\right)={n}_{1}{s}_{1}+{n}_{2}{s}_{2}+{n}_{3}{s}_{3}=0,\end{array}$ (5)

and orthogonal to eachother

$\left(r,s\right)={r}_{1}{s}_{1}+{r}_{2}{s}_{2}+{r}_{3}{s}_{3}=0.$ (6)

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

$\left({D}_{1}r,{D}_{1}s\right)=\frac{{r}_{1}{s}_{1}}{{a}_{1}^{2}}+\frac{{r}_{2}{s}_{2}}{{a}_{2}^{2}}+\frac{{r}_{3}{s}_{3}}{{a}_{3}^{2}}=0$ (7)

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

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

with an angle $\omega$ according to

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

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

$\begin{array}{l}\left({D}_{1}\stackrel{˜}{r},{D}_{1}\stackrel{˜}{r}\right){t}^{2}+\left({D}_{1}\stackrel{˜}{s},{D}_{1}\stackrel{˜}{s}\right){u}^{2}+2\left({D}_{1}q,{D}_{1}\stackrel{˜}{r}\right)t+2\left({D}_{1}q,{D}_{1}\stackrel{˜}{s}\right)u\\ =1-\left({D}_{1}q,{D}_{1}q\right).\end{array}$

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 $\left({t}_{0},{u}_{0}\right)$

${t}_{0}=-\frac{\left({D}_{1}q,{D}_{1}\stackrel{˜}{r}\right)}{\left({D}_{1}\stackrel{˜}{r},{D}_{1}\stackrel{˜}{r}\right)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{and}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}_{0}=-\frac{\left({D}_{1}q,{D}_{1}\stackrel{˜}{s}\right)}{\left({D}_{1}\stackrel{˜}{s},{D}_{1}\stackrel{˜}{s}\right)}$ (11)

and the semi-axes

$A=\sqrt{\frac{1-d}{\left({D}_{1}\stackrel{˜}{r},{D}_{1}\stackrel{˜}{r}\right)}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{and}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}B=\sqrt{\frac{1-d}{\left({D}_{1}\stackrel{˜}{s},{D}_{1}\stackrel{˜}{s}\right)}},$ (12)

where

$d=\left({D}_{1}q,{D}_{1}q\right)-\frac{{\left({D}_{1}q,{D}_{1}\stackrel{˜}{r}\right)}^{2}}{\left({D}_{1}\stackrel{˜}{r},{D}_{1}\stackrel{˜}{r}\right)}-\frac{{\left({D}_{1}q,{D}_{1}\stackrel{˜}{s}\right)}^{2}}{\left({D}_{1}\stackrel{˜}{s},{D}_{1}\stackrel{˜}{s}\right)}.$ (13)

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

Putting

${\beta }_{1}=\left({D}_{1}\stackrel{˜}{r},{D}_{1}\stackrel{˜}{r}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{and}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\beta }_{2}=\left({D}_{1}\stackrel{˜}{s},{D}_{1}\stackrel{˜}{s}\right)$ (14)

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

$A=\sqrt{\frac{1-d}{{\beta }_{1}}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{and}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}B=\sqrt{\frac{1-d}{{\beta }_{2}}}.$ (15)

In  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  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

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

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

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  ) yields

${n}_{3}=\left(n,{e}_{3}\right)=‖n‖‖{e}_{3}‖\mathrm{cos}\Omega =\mathrm{cos}\Omega$ (19)

where $\mathrm{cos}\Omega >0$ 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:

$\begin{array}{l}{x}_{1}={m}_{1}+A\mathrm{cos}\theta {\stackrel{˜}{r}}_{1}+B\mathrm{sin}\theta {\stackrel{˜}{s}}_{1}\\ {x}_{2}={m}_{2}+A\mathrm{cos}\theta {\stackrel{˜}{r}}_{2}+B\mathrm{sin}\theta {\stackrel{˜}{s}}_{2}.\end{array}$ (20)

In general the two dimensional vectors ${\left({\stackrel{˜}{r}}_{1},{\stackrel{˜}{r}}_{2}\right)}^{\text{T}}$ and ${\left({\stackrel{˜}{s}}_{1},{\stackrel{˜}{s}}_{2}\right)}^{\text{T}}$ are not orthogonal because their orthogonality in 3d space implies

${\stackrel{˜}{r}}_{1}{\stackrel{˜}{s}}_{1}+{\stackrel{˜}{r}}_{2}{\stackrel{˜}{s}}_{2}=-{\stackrel{˜}{r}}_{3}{\stackrel{˜}{s}}_{3},$

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:

$\left(\begin{array}{cc}A{\stackrel{˜}{r}}_{1}& B{\stackrel{˜}{s}}_{1}\\ A{\stackrel{˜}{r}}_{2}& B{\stackrel{˜}{s}}_{2}\end{array}\right)\left(\begin{array}{c}\mathrm{cos}\theta \\ \mathrm{sin}\theta \end{array}\right)=\left(\begin{array}{c}{{x}^{\prime }}_{1}\\ {{x}^{\prime }}_{2}\end{array}\right)$ (21)

The determinant of the linear system (21), $AB\left({\stackrel{˜}{r}}_{1}{\stackrel{˜}{s}}_{2}-{\stackrel{˜}{r}}_{2}{\stackrel{˜}{s}}_{1}\right)$ , is different from zero. This can be shown by noting that ${\stackrel{˜}{r}}_{1}{\stackrel{˜}{s}}_{2}-{\stackrel{˜}{r}}_{2}{\stackrel{˜}{s}}_{1}$ is the third component of the vector $\stackrel{˜}{r}×\stackrel{˜}{s}$ . At first this vector is not affected by rotation (8):

$\begin{array}{c}\stackrel{˜}{r}×\stackrel{˜}{s}=\left(\mathrm{cos}\omega r+\mathrm{sin}\omega s\right)×\left(-\mathrm{sin}\omega r+\mathrm{cos}\omega s\right)\\ =\left({\mathrm{cos}}^{2}\omega +{\mathrm{sin}}^{2}\omega \right)\left(r×s\right)=r×s.\end{array}$

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  ):

$r×s=r×\left(n×r\right)=\left(r,r\right)n-\left(r,n\right)r=n$

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

${\stackrel{˜}{r}}_{1}{\stackrel{˜}{s}}_{2}-{\stackrel{˜}{r}}_{2}{\stackrel{˜}{s}}_{1}={r}_{1}{s}_{2}-{r}_{2}{s}_{1}={n}_{3},$ (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

$\mathrm{cos}\theta =\frac{B\left({{x}^{\prime }}_{1}{\stackrel{˜}{s}}_{2}-{{x}^{\prime }}_{2}{\stackrel{˜}{s}}_{1}\right)}{AB\left({\stackrel{˜}{r}}_{1}{\stackrel{˜}{s}}_{2}-{\stackrel{˜}{r}}_{2}{\stackrel{˜}{s}}_{1}\right)},$

$\mathrm{sin}\theta =\frac{A\left({\stackrel{˜}{r}}_{1}{{x}^{\prime }}_{2}-{\stackrel{˜}{r}}_{2}{{x}^{\prime }}_{1}\right)}{AB\left({\stackrel{˜}{r}}_{1}{\stackrel{˜}{s}}_{2}-{\stackrel{˜}{r}}_{2}{\stackrel{˜}{s}}_{1}\right)}.$

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

${B}^{2}{\left({{x}^{\prime }}_{1}{\stackrel{˜}{s}}_{2}-{{x}^{\prime }}_{2}{\stackrel{˜}{s}}_{1}\right)}^{2}+{A}^{2}{\left({\stackrel{˜}{r}}_{1}{{x}^{\prime }}_{2}-{\stackrel{˜}{r}}_{2}{{x}^{\prime }}_{1}\right)}^{2}={A}^{2}{B}^{2}{\left({\stackrel{˜}{r}}_{1}{\stackrel{˜}{s}}_{2}-{\stackrel{˜}{r}}_{2}{\stackrel{˜}{s}}_{1}\right)}^{2}={A}^{2}{B}^{2}{n}_{3}^{2}.$

Expanding the squares on the left side and using the denotations

$\begin{array}{l}{l}_{11}={A}^{2}{\stackrel{˜}{r}}_{2}^{2}+{B}^{2}{\stackrel{˜}{s}}_{2}^{2},\\ {l}_{12}=-\left({A}^{2}{\stackrel{˜}{r}}_{1}{\stackrel{˜}{r}}_{2}+{B}^{2}{\stackrel{˜}{s}}_{1}{\stackrel{˜}{s}}_{2}\right),\\ {l}_{22}={A}^{2}{\stackrel{˜}{r}}_{1}^{2}+{B}^{2}{\stackrel{˜}{s}}_{1}^{2}\end{array}$ (23)

arranged as a $2×2$ matrix $L$

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

$\left({{x}^{\prime }}_{1},{{x}^{\prime }}_{2}\right)L\left(\begin{array}{c}{{x}^{\prime }}_{1}\\ {{x}^{\prime }}_{2}\end{array}\right)={A}^{2}{B}^{2}{n}_{3}^{2}.$ (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

$\left({{x}^{″}}_{1},{{x}^{″}}_{2}\right)=\left({{x}^{\prime }}_{1},{{x}^{\prime }}_{2}\right){S}^{\text{T}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}S\left(\begin{array}{c}{{x}^{\prime }}_{1}\\ {{x}^{\prime }}_{2}\end{array}\right)=\left(\begin{array}{c}{{x}^{″}}_{1}\\ {{x}^{″}}_{2}\end{array}\right)$

the quadratic equation (25) in $\left({{x}^{\prime }}_{1},{{x}^{\prime }}_{2}\right)$ reduces to

$\left({{x}^{″}}_{1},{{x}^{″}}_{2}\right)\text{diag}\left({\lambda }_{1}\left(L\right),{\lambda }_{2}\left(L\right)\right)\left(\begin{array}{c}{{x}^{″}}_{1}\\ {{x}^{″}}_{2}\end{array}\right)={A}^{2}{B}^{2}{n}_{3}^{2}.$ (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):

$\begin{array}{c}\mathrm{det}L={l}_{11}{l}_{22}-{l}_{12}^{2}=\left({A}^{2}{\stackrel{˜}{r}}_{2}^{2}+{B}^{2}{\stackrel{˜}{s}}_{2}^{2}\right)\left({A}^{2}{\stackrel{˜}{r}}_{1}^{2}+{B}^{2}{\stackrel{˜}{s}}_{1}^{2}\right)-{\left({A}^{2}{\stackrel{˜}{r}}_{1}{\stackrel{˜}{r}}_{2}+{B}^{2}{\stackrel{˜}{s}}_{1}{\stackrel{˜}{s}}_{2}\right)}^{2}\\ ={A}^{2}{B}^{2}{\left({\stackrel{˜}{r}}_{1}{\stackrel{˜}{s}}_{2}-{\stackrel{˜}{r}}_{2}{\stackrel{˜}{s}}_{1}\right)}^{2}={A}^{2}{B}^{2}{\left({r}_{1}{s}_{2}-{r}_{2}{s}_{1}\right)}^{2}={A}^{2}{B}^{2}{n}_{3}^{2}.\end{array}$ (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

${A}_{L}=\frac{AB{n}_{3}}{\sqrt{{\lambda }_{1}\left(L\right)}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{B}_{L}=\frac{AB{n}_{3}}{\sqrt{{\lambda }_{2}\left(L\right)}}.$ (28)

With (19) one obtains from (28)

${A}_{L}=\frac{AB\mathrm{cos}\Omega }{\sqrt{{\lambda }_{1}\left(L\right)}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{B}_{L}=\frac{AB\mathrm{cos}\Omega }{\sqrt{{\lambda }_{2}\left(L\right)}}.$ (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:

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

${m}_{11}{x}_{1}^{2}+2{m}_{12}{x}_{1}{x}_{2}+{m}_{22}{x}_{2}^{2}+2{m}_{13}{x}_{1}+2{m}_{23}{x}_{2}+{m}_{33}=0$ (33)

with

$\begin{array}{l}{m}_{11}=\frac{1}{{a}_{1}^{2}}+\frac{{n}_{1}^{2}}{{a}_{3}^{2}{n}_{3}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{m}_{12}=\frac{{n}_{1}{n}_{2}}{{a}_{3}^{2}{n}_{3}^{2}},\\ {m}_{22}=\frac{1}{{a}_{2}^{2}}+\frac{{n}_{2}^{2}}{{a}_{3}^{2}{n}_{3}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{m}_{13}=-\frac{{n}_{1}\kappa }{{a}_{3}^{2}{n}_{3}^{2}},\\ {m}_{23}=-\frac{{n}_{2}\kappa }{{a}_{3}^{2}{n}_{3}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{m}_{33}=\frac{{\kappa }^{2}}{{a}_{3}^{2}{n}_{3}^{2}}-1.\end{array}$ (34)

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

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

$\begin{array}{c}\mathrm{det}M={m}_{11}{m}_{22}-{m}_{12}^{2}=\left(\frac{1}{{a}_{1}^{2}}+\frac{{n}_{1}^{2}}{{a}_{3}^{2}{n}_{3}^{2}}\right)\left(\frac{1}{{a}_{2}^{2}}+\frac{{n}_{2}^{2}}{{a}_{3}^{2}{n}_{3}^{2}}\right)-\frac{{n}_{1}^{2}{n}_{2}^{2}}{{a}_{3}^{4}{n}_{3}^{4}}\\ =\frac{{n}_{3}^{2}}{{a}_{1}^{2}{a}_{2}^{2}{n}_{3}^{2}}+\frac{{n}_{1}^{2}}{{a}_{2}^{2}{a}_{3}^{2}{n}_{3}^{2}}+\frac{{n}_{2}^{2}}{{a}_{1}^{2}{a}_{3}^{2}{n}_{3}^{2}}=\frac{{a}_{1}^{2}{n}_{1}^{2}+{a}_{2}^{2}{n}_{2}^{2}+{a}_{3}^{2}{n}_{3}^{2}}{{a}_{1}^{2}{a}_{2}^{2}{a}_{3}^{2}{n}_{3}^{2}}.\end{array}$ (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:

$\begin{array}{l}{m}_{11}{{x}^{\prime }}_{1}^{2}+2{m}_{12}{{x}^{\prime }}_{1}{{x}^{\prime }}_{2}+{m}_{22}{{x}^{\prime }}_{2}^{2}+2\left({m}_{11}h+{m}_{12}k+{m}_{13}\right){{x}^{\prime }}_{1}\\ +\text{\hspace{0.17em}}2\left({m}_{12}h+{m}_{22}k+{m}_{23}\right){{x}^{\prime }}_{2}+{m}_{11}{h}^{2}+2{m}_{12}hk+{m}_{22}{k}^{2}\\ +\text{\hspace{0.17em}}2{m}_{13}h+2{m}_{23}k+{m}_{33}=0.\end{array}$ (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):

$\begin{array}{l}{m}_{11}{h}^{2}+2{m}_{12}hk+{m}_{22}{k}^{2}+2{m}_{13}h+2{m}_{23}k+{m}_{33}\\ =\frac{{\kappa }^{2}}{{a}_{1}^{2}{n}_{1}^{2}+{a}_{2}^{2}{n}_{2}^{2}+{a}_{3}^{2}{n}_{3}^{2}}-1=-\left(1-d\right).\end{array}$

Thus the quadratic equation (36) reduces to:

$\left({{x}^{\prime }}_{1},{{x}^{\prime }}_{2}\right)M\left(\begin{array}{c}{{x}^{\prime }}_{1}\\ {{x}^{\prime }}_{2}\end{array}\right)=1-d.$ (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

$\left({{x}^{″}}_{1},{{x}^{″}}_{2}\right)=\left({{x}^{\prime }}_{1},{{x}^{\prime }}_{2}\right){T}^{\text{T}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}T\left(\begin{array}{c}{{x}^{\prime }}_{1}\\ {{x}^{\prime }}_{2}\end{array}\right)=\left(\begin{array}{c}{{x}^{″}}_{1}\\ {{x}^{″}}_{2}\end{array}\right)$

the quadratic equation (40) in $\left({{x}^{\prime }}_{1},{{x}^{\prime }}_{2}\right)$ reduces to

$\left({{x}^{″}}_{1},{{x}^{″}}_{2}\right)\text{diag}\left({\lambda }_{1}\left(M\right),{\lambda }_{2}\left(M\right)\right)\left(\begin{array}{c}{{x}^{″}}_{1}\\ {{x}^{″}}_{2}\end{array}\right)=1-d.$ (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

${A}_{M}=\sqrt{\frac{1-d}{{\lambda }_{1}\left(M\right)}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{B}_{M}=\sqrt{\frac{1-d}{{\lambda }_{2}\left(M\right)}}.$ (42)

4. Calculation of Projected Semi-Axes According to Schrantz

In  the ellipse

${x}_{1}=A\mathrm{cos}t,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}_{2}=B\mathrm{sin}t,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}t\in \left[0,2\text{π}\right)$ (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

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

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

${\stackrel{¯}{x}}_{1}=\stackrel{¯}{A}\mathrm{cos}\left(\stackrel{¯}{t}+\psi \right),\text{\hspace{0.17em}}{\stackrel{¯}{x}}_{2}=\stackrel{¯}{B}\mathrm{sin}\stackrel{¯}{t},\text{\hspace{0.17em}}\stackrel{¯}{t}\in \left[0,2\text{π}\right)$ (45)

with

$\stackrel{¯}{A}=\sqrt{{A}^{2}{\mathrm{cos}}^{2}\alpha +{B}^{2}{\mathrm{sin}}^{2}\alpha },$

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

Eliminating parameter $\stackrel{¯}{t}$ from (45) yields a quadratic equation in ${\stackrel{¯}{x}}_{1}$ and ${\stackrel{¯}{x}}_{2}$

${\left(\frac{{\stackrel{¯}{x}}_{1}}{\stackrel{¯}{A}}\right)}^{2}+2\mathrm{sin}\psi \left(\frac{{\stackrel{¯}{x}}_{1}}{\stackrel{¯}{A}}\right)\left(\frac{{\stackrel{¯}{x}}_{2}}{\stackrel{¯}{B}}\right)+{\left(\frac{{\stackrel{¯}{x}}_{2}}{\stackrel{¯}{B}}\right)}^{2}={\mathrm{cos}}^{2}\psi$

or written with the elements

${g}_{11}=\frac{1}{{\stackrel{¯}{A}}^{2}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{g}_{12}=\frac{\mathrm{sin}\psi }{\stackrel{¯}{A}\stackrel{¯}{B}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{g}_{22}=\frac{1}{{\stackrel{¯}{B}}^{2}}$ (47)

forming matrix

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

one obtains

$\left({\stackrel{¯}{x}}_{1},{\stackrel{¯}{x}}_{2}\right)G\left(\begin{array}{c}{\stackrel{¯}{x}}_{1}\\ {\stackrel{¯}{x}}_{2}\end{array}\right)={\mathrm{cos}}^{2}\psi .$ (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

$\left({\stackrel{¯}{\stackrel{¯}{x}}}_{1},{\stackrel{¯}{\stackrel{¯}{x}}}_{2}\right)=\left({\stackrel{¯}{x}}_{1},{\stackrel{¯}{x}}_{2}\right){R}^{\text{T}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}R\left(\begin{array}{c}{\stackrel{¯}{x}}_{1}\\ {\stackrel{¯}{x}}_{2}\end{array}\right)=\left(\begin{array}{c}{\stackrel{¯}{\stackrel{¯}{x}}}_{1}\\ {\stackrel{¯}{\stackrel{¯}{x}}}_{2}\end{array}\right)$

the quadratic equation (48) in $\left({\stackrel{¯}{x}}_{1},{\stackrel{¯}{x}}_{2}\right)$ reduces to

$\left({\stackrel{¯}{\stackrel{¯}{x}}}_{1},{\stackrel{¯}{\stackrel{¯}{x}}}_{2}\right)\text{diag}\left({\lambda }_{1}\left(G\right),{\lambda }_{2}\left(G\right)\right)\left(\begin{array}{c}{\stackrel{¯}{\stackrel{¯}{x}}}_{1}\\ {\stackrel{¯}{\stackrel{¯}{x}}}_{2}\end{array}\right)={\mathrm{cos}}^{2}\psi .$ (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

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

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

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

This is an ellipse in the ${\stackrel{¯}{x}}_{1}-{\stackrel{¯}{x}}_{2}$ plane with the semi-axes

${A}_{G}=\frac{\mathrm{cos}\psi }{\sqrt{{\lambda }_{1}\left(G\right)}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{B}_{G}=\frac{\mathrm{cos}\psi }{\sqrt{{\lambda }_{2}\left(G\right)}}.$ (51)

5. Some Auxiliary Means

Let $H$ stand for the following $2×2$ 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

$\begin{array}{l}{Q}_{L}\left(z\right)={z}^{2}-\left({A}_{L}+{B}_{L}\right)z+{A}_{L}{B}_{L},\\ {Q}_{H}\left(z\right)={z}^{2}-\left({A}_{H}+{B}_{H}\right)z+{A}_{H}{B}_{H},\end{array}$ (53)

have the same coefficients and thus have the same zeros:

$\begin{array}{l}{Q}_{L}\left(z\right)=\left(z-{A}_{L}\right)\left(z-{B}_{L}\right),\\ {Q}_{H}\left(z\right)=\left(z-{A}_{H}\right)\left(z-{B}_{H}\right).\end{array}$ (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 $2{A}_{L}{B}_{L}=2{A}_{H}{B}_{H}$ to both sides of (55) one obtains:

${\left({A}_{L}+{B}_{L}\right)}^{2}={A}_{L}^{2}+2{A}_{L}{B}_{L}+{B}_{L}^{2}={A}_{H}^{2}+2{A}_{H}{B}_{H}+{B}_{H}^{2}={\left({A}_{H}+{B}_{H}\right)}^{2}$

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

$\begin{array}{l}{\lambda }_{1}\left(H\right)+{\lambda }_{2}\left(H\right)={h}_{11}+{h}_{22},\\ {\lambda }_{1}\left(H\right){\lambda }_{2}\left(H\right)={h}_{11}{h}_{22}-{h}_{12}^{2}.\end{array}$ (57)

6. Comparison of the Semi-Axes AL, BL with AM, BM

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 $\stackrel{˜}{r}$ and $\stackrel{˜}{s}$ are fulfilling (4) and (5), the following relations hold:

$\begin{array}{l}{\lambda }_{1}\left(L\right)+{\lambda }_{2}\left(L\right)={l}_{11}+{l}_{22}={A}^{2}\left({\stackrel{˜}{r}}_{1}^{2}+{\stackrel{˜}{r}}_{2}^{2}\right)+{B}^{2}\left({\stackrel{˜}{s}}_{1}^{2}+{\stackrel{˜}{s}}_{2}^{2}\right)\\ ={A}^{2}\left(1-{\stackrel{˜}{r}}_{3}^{2}\right)+{B}^{2}\left(1-{\stackrel{˜}{s}}_{3}^{2}\right)=\frac{1-d}{{\beta }_{1}}\left(1-{\stackrel{˜}{r}}_{3}^{2}\right)+\frac{1-d}{{\beta }_{2}}\left(1-{\stackrel{˜}{s}}_{3}^{2}\right)\\ =\frac{1-d}{{\beta }_{1}{\beta }_{2}}\left({\beta }_{2}\left(1-{\stackrel{˜}{r}}_{3}^{2}\right)+{\beta }_{1}\left(1-{\stackrel{˜}{s}}_{3}^{2}\right)\right)=\frac{1-d}{{\beta }_{1}{\beta }_{2}}\left({\beta }_{1}+{\beta }_{2}-{\beta }_{2}{\stackrel{˜}{r}}_{3}^{2}-{\beta }_{1}{\stackrel{˜}{s}}_{3}^{2}\right)\end{array}$ (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:

${A}_{L}^{2}+{B}_{L}^{2}-\left({A}_{M}^{2}+{B}_{M}^{2}\right)=\frac{1-d}{{\beta }_{1}{\beta }_{2}}\left(\frac{{n}_{1}^{2}}{{a}_{2}^{2}}+\frac{{n}_{2}^{2}}{{a}_{1}^{2}}-{\beta }_{2}{\stackrel{˜}{r}}_{3}^{2}-{\beta }_{1}{\stackrel{˜}{s}}_{3}^{2}\right).$ (70)

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

${n}_{1}={\stackrel{˜}{r}}_{2}{\stackrel{˜}{s}}_{3}-{\stackrel{˜}{r}}_{3}{\stackrel{˜}{s}}_{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{n}_{2}={\stackrel{˜}{r}}_{3}{\stackrel{˜}{s}}_{1}-{\stackrel{˜}{r}}_{1}{\stackrel{˜}{s}}_{3},$

because of $\stackrel{˜}{r}×\stackrel{˜}{s}=r×s=n$ (see Section 2), and

${\beta }_{2}=\left({D}_{1}\stackrel{˜}{s},{D}_{1}\stackrel{˜}{s}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\beta }_{1}=\left({D}_{1}\stackrel{˜}{r},{D}_{1}\stackrel{˜}{r}\right)$

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

$\begin{array}{l}\frac{1}{{a}_{2}^{2}}{\left({\stackrel{˜}{r}}_{2}{\stackrel{˜}{s}}_{3}-{\stackrel{˜}{r}}_{3}{\stackrel{˜}{s}}_{2}\right)}^{2}+\frac{1}{{a}_{1}^{2}}{\left({\stackrel{˜}{r}}_{3}{\stackrel{˜}{s}}_{1}-{\stackrel{˜}{r}}_{1}{\stackrel{˜}{s}}_{3}\right)}^{2}-\left(\frac{{\stackrel{˜}{s}}_{1}^{2}}{{a}_{1}^{2}}+\frac{{\stackrel{˜}{s}}_{2}^{2}}{{a}_{2}^{2}}+\frac{{\stackrel{˜}{s}}_{3}^{2}}{{a}_{3}^{2}}\right){\stackrel{˜}{r}}_{3}^{2}-\left(\frac{{\stackrel{˜}{r}}_{1}^{2}}{{a}_{1}^{2}}+\frac{{\stackrel{˜}{r}}_{2}^{2}}{{a}_{2}^{2}}+\frac{{\stackrel{˜}{r}}_{3}^{2}}{{a}_{3}^{2}}\right){\stackrel{˜}{s}}_{3}^{2}\\ =-2{\stackrel{˜}{r}}_{3}{\stackrel{˜}{s}}_{3}\left(\frac{{\stackrel{˜}{r}}_{1}{\stackrel{˜}{s}}_{1}}{{a}_{1}^{2}}+\frac{{\stackrel{˜}{r}}_{2}{\stackrel{˜}{s}}_{2}}{{a}_{2}^{2}}+\frac{{\stackrel{˜}{r}}_{3}{\stackrel{˜}{s}}_{3}}{{a}_{3}^{2}}\right)=-2{\stackrel{˜}{r}}_{3}{\stackrel{˜}{s}}_{3}\left({D}_{1}\stackrel{˜}{r},{D}_{1}\stackrel{˜}{s}\right)=0,\end{array}$

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

7. Comparison of the Semi-Axes AL, BL with AG, BG

In the first step ${A}_{L}{B}_{L}={A}_{G}{B}_{G}$ will be proven. According to (29) and (51) holds:

${A}_{L}{B}_{L}=\frac{{A}^{2}{B}^{2}{\mathrm{cos}}^{2}\Omega }{\sqrt{{\lambda }_{1}\left(L\right){\lambda }_{2}\left(L\right)}},$ (71)

${A}_{G}{B}_{G}=\frac{{\mathrm{cos}}^{2}\psi }{\sqrt{{\lambda }_{1}\left(G\right){\lambda }_{2}\left(G\right)}}.$ (72)

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

${\lambda }_{1}\left(L\right){\lambda }_{2}\left(L\right)={l}_{11}{l}_{22}-{l}_{12}^{2}={A}^{2}{B}^{2}{\mathrm{cos}}^{2}\Omega .$ (73)

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

${\lambda }_{1}\left(G\right){\lambda }_{2}\left(G\right)={g}_{11}{g}_{22}-{g}_{12}^{2}=\frac{{\mathrm{cos}}^{2}\psi }{{\stackrel{¯}{A}}^{2}{\stackrel{¯}{B}}^{2}}.$ (74)

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

${A}_{L}{B}_{L}-{A}_{G}{B}_{G}=AB\mathrm{cos}\Omega -\stackrel{¯}{A}\stackrel{¯}{B}\mathrm{cos}\psi .$ (75)

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

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

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

$\begin{array}{c}{A}_{L}^{2}+{B}_{L}^{2}={\lambda }_{1}\left(L\right)+{\lambda }_{2}\left(L\right)={l}_{11}+{l}_{22}={A}^{2}\left({\stackrel{˜}{r}}_{1}^{2}+{\stackrel{˜}{r}}_{2}^{2}\right)+{B}^{2}\left({\stackrel{˜}{s}}_{1}^{2}+{\stackrel{˜}{s}}_{2}^{2}\right)\\ ={A}^{2}\left(1-{\stackrel{˜}{r}}_{3}^{2}\right)+{B}^{2}\left(1-{\stackrel{˜}{s}}_{3}^{2}\right)={A}^{2}+{B}^{2}-\left({A}^{2}{\stackrel{˜}{r}}_{3}^{2}+{B}^{2}{\stackrel{˜}{s}}_{3}^{2}\right).\end{array}$ (76)

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

$\begin{array}{c}{A}_{G}^{2}+{B}_{G}^{2}=\frac{{\mathrm{cos}}^{2}\psi }{{\lambda }_{1}\left(G\right)}+\frac{{\mathrm{cos}}^{2}\psi }{{\lambda }_{2}\left(G\right)}=\frac{{\mathrm{cos}}^{2}\psi }{{\lambda }_{1}\left(G\right){\lambda }_{2}\left(G\right)}\left({\lambda }_{2}\left(G\right)+{\lambda }_{1}\left(G\right)\right)\\ ={\stackrel{¯}{A}}^{2}{\stackrel{¯}{B}}^{2}\left({\lambda }_{1}\left(G\right)+{\lambda }_{2}\left(G\right)\right)={\stackrel{¯}{A}}^{2}{\stackrel{¯}{B}}^{2}\left({g}_{11}+{g}_{22}\right)\\ ={\stackrel{¯}{A}}^{2}{\stackrel{¯}{B}}^{2}\left(\frac{1}{{\stackrel{¯}{A}}^{2}}+\frac{1}{{\stackrel{¯}{B}}^{2}}\right)={\stackrel{¯}{B}}^{2}+{\stackrel{¯}{A}}^{2};\end{array}$ (77)

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

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

${A}^{2}{\stackrel{˜}{r}}_{3}^{2}+{B}^{2}{\stackrel{˜}{s}}_{3}^{2}={\mathrm{sin}}^{2}\Omega \left({A}^{2}{\mathrm{sin}}^{2}\alpha +{B}^{2}{\mathrm{cos}}^{2}\alpha \right).$ (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 ${\stackrel{¯}{x}}_{1}$ -axis of the coordinate system in plane ${E}^{\prime }$ . The original plane $E$ thus contains the following three points: $\left(-1,0,0\right)$ , $\left(1,0,0\right)$ , $\left(0,\mathrm{cos}\Omega ,\mathrm{sin}\Omega \right)$ and can therefore be described by the following equation:

$-\mathrm{sin}\Omega \text{ }{\stackrel{¯}{x}}_{2}+\mathrm{cos}\Omega \text{ }{\stackrel{¯}{x}}_{3}=0.$ (80)

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

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

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

$\begin{array}{l}\left(r,r\right)={r}_{1}^{2}+{r}_{2}^{2}+{r}_{3}^{2}=1,\\ \left(n,r\right)=-\mathrm{sin}\Omega \text{ }{r}_{2}+\mathrm{cos}\Omega \text{ }{r}_{3}=0.\end{array}$ (82)

The second equation of (82) yields ${r}_{3}={r}_{2}\mathrm{tan}\Omega$ . Substituting this into the first equation of (82) results in:

${r}_{1}^{2}+{r}_{2}^{2}\left(1+{\mathrm{tan}}^{2}\Omega \right)=1$

or

${r}_{1}^{2}+\frac{{r}_{2}^{2}}{{\mathrm{cos}}^{2}\Omega }=1.$ (83)

If the unit vector $r$ is forming the angle $\alpha$ with the ${\stackrel{¯}{x}}_{1}$ -axis and ${e}_{1}$ is designating a unit vector in ${\stackrel{¯}{x}}_{1}$ -direction according to the definition of the scalar product (see for instance  ) holds

${r}_{1}=\left(r,{e}_{1}\right)=‖r‖‖{e}_{1}‖\mathrm{cos}\alpha =\mathrm{cos}\alpha .$

From (83) one obtains

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

yielding ${r}_{2}=±\mathrm{sin}\alpha \mathrm{cos}\Omega$ and furthermore with the first equation of (82) ${r}_{3}=±\mathrm{sin}\alpha \mathrm{sin}\Omega$ . From

$r=\left(\mathrm{cos}\alpha ,±\mathrm{sin}\alpha \mathrm{cos}\Omega ,±\mathrm{sin}\alpha \mathrm{sin}\Omega \right)$

and $s=n×r$ one obtains

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

By transformation (8) one obtains

${\stackrel{˜}{r}}_{3}=\mathrm{cos}\omega {r}_{3}+\mathrm{sin}\omega {s}_{3}=\mathrm{sin}\left(\omega ±\alpha \right)\mathrm{sin}\Omega ,$

${\stackrel{˜}{s}}_{3}=-\mathrm{sin}\omega {r}_{3}+\mathrm{cos}\omega {s}_{3}=\mathrm{cos}\left(\omega ±\alpha \right)\mathrm{sin}\Omega .$

Thus equation (79) turns into

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

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

major axis of the ellipse (43) and the ${\stackrel{¯}{x}}_{1}$ -axis, is chosen to be $\frac{\omega }{2}$ then (84) is true.

8. Numerical Example

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

${a}_{1}=5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{a}_{2}=4,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{a}_{3}=3$

and let the plane be given by

${x}_{1}+2{x}_{2}+3{x}_{3}+4=0.$

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

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

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

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

According to (17) d can be calculated.

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

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

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

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

$A=4.59157,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}B=3.39705.$

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

${A}_{L}=4.56667,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{B}_{L}=2.73855.$

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.

Cite this paper

Klein, P.P. (2017) Projection of the Semi-Axes of the Ellipse of Intersection. Applied Mathematics, 8, 1320-1335. https://doi.org/10.4236/am.2017.89097

References

1. 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. 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. 3. Schrantz, G.R. (2004) Projections of Ellipses and Circles. Hamline University. (To be found in the internet)

4. 4. Bronshtein, I.N., Semendyayev, K.A., et al. (2007) Handbook of Mathematics. 5th Edition, Springer-Verlag Berlin Heidelberg.