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Projection of the Semi-Axes of the Ellipse of Intersection

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DOI: 10.4236/am.2017.89097    379 Downloads   583 Views  
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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].

1. Introduction

Let an ellipsoid be given with the three positive semi-axes a 1 , a 2 , a 3

x 1 2 a 1 2 + x 2 2 a 2 2 + x 3 2 a 3 2 = 1 (1)

and a plane with the unit normal vector

n = ( n 1 , n 2 , n 3 ) T ,

which contains an interior point q = ( q 1 , q 2 , q 3 ) T of the ellipsoid. A plane spanned by vectors r = ( r 1 , r 2 , r 3 ) T , s = ( s 1 , s 2 , s 3 ) T and containing the point q is described in parametric form by

x = q + t r + u s with x = ( x 1 , x 2 , x 3 ) 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 = ( v 1 , v 2 , v 3 ) T and w = ( w 1 , w 2 , w 3 ) T be denoted by

( v , w ) = v 1 w 1 + v 2 w 2 + v 3 w 3

and the norm of vector v by

v = ( v , v ) .

With the diagonal matrix

D 1 = diag ( 1 a 1 , 1 a 2 , 1 a 3 )

the line of intersection has the form:

( t , u ) ( ( D 1 r , D 1 r ) ( D 1 r , D 1 s ) ( D 1 r , D 1 s ) ( D 1 s , D 1 s ) ) ( t u ) + 2 ( ( D 1 q , D 1 r ) , ( D 1 q , D 1 s ) ) ( t u ) = 1 ( D 1 q , D 1 q ) . (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

( r , r ) = r 1 2 + r 2 2 + r 3 2 = 1 , ( n , r ) = n 1 r 1 + n 2 r 2 + n 3 r 3 = 0 , (4)

( s , s ) = s 1 2 + s 2 2 + s 3 2 = 1 , ( n , s ) = n 1 s 1 + n 2 s 2 + n 3 s 3 = 0 , (5)

and orthogonal to eachother

( r , s ) = r 1 s 1 + r 2 s 2 + r 3 s 3 = 0. (6)

If vectors r and s have the additional property

( D 1 r , D 1 s ) = r 1 s 1 a 1 2 + r 2 s 2 a 2 2 + 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 [1] , with vectors r ˜ and s ˜ obtained by a transformation of the form

r ˜ = cos ω r + sin ω s , s ˜ = sin ω r + cos ω s (8)

with an angle ω according to

ω = 1 2 a r c t a n [ 2 ( D 1 r , D 1 s ) ( D 1 r , D 1 r ) ( D 1 s , D 1 s ) ] . (9)

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

( D 1 r ˜ , D 1 r ˜ ) t 2 + ( D 1 s ˜ , D 1 s ˜ ) u 2 + 2 ( D 1 q , D 1 r ˜ ) t + 2 ( D 1 q , D 1 s ˜ ) u = 1 ( D 1 q , D 1 q ) .

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

( t t 0 ) 2 A 2 + ( u u 0 ) 2 B 2 = 1 (10)

with the center ( t 0 , u 0 )

t 0 = ( D 1 q , D 1 r ˜ ) ( D 1 r ˜ , D 1 r ˜ ) and u 0 = ( D 1 q , D 1 s ˜ ) ( D 1 s ˜ , D 1 s ˜ ) (11)

and the semi-axes

A = 1 d ( D 1 r ˜ , D 1 r ˜ ) and B = 1 d ( D 1 s ˜ , D 1 s ˜ ) , (12)

where

d = ( D 1 q , D 1 q ) ( D 1 q , D 1 r ˜ ) 2 ( D 1 r ˜ , D 1 r ˜ ) ( D 1 q , D 1 s ˜ ) 2 ( D 1 s ˜ , D 1 s ˜ ) . (13)

Because of 1 d 1 ( D 1 q , D 1 q ) > 0 the numerator 1 d in (12) is positive.

Putting

β 1 = ( D 1 r ˜ , D 1 r ˜ ) and β 2 = ( D 1 s ˜ , D 1 s ˜ ) (14)

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

A = 1 d β 1 and B = 1 d β 2 . (15)

In [1] it is shown that β 1 and β 2 according to (14) are solutions of the following quadratic equation

β 2 [ n 1 2 ( 1 a 2 2 + 1 a 3 2 ) + n 2 2 ( 1 a 1 2 + 1 a 3 2 ) + n 3 2 ( 1 a 1 2 + 1 a 2 2 ) ] β + n 1 2 a 2 2 a 3 2 + n 2 2 a 1 2 a 3 2 + n 3 2 a 1 2 a 2 2 = 0. (16)

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

d = κ 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 + ( A cos θ ) r ˜ + ( B sin θ ) s ˜ (18)

with center m = q + t 0 r ˜ + u 0 s ˜ , where t 0 and u 0 are from (11), semi-axes A and B from (12), θ [ 0,2 π ) and vectors r ˜ and 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 Ω . 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 = ( n , e 3 ) = n e 3 cos Ω = cos Ω (19)

where cos Ω > 0 holds for 0 Ω < 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 Ω < 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:

x 1 = m 1 + A cos θ r ˜ 1 + B sin θ s ˜ 1 x 2 = m 2 + A cos θ r ˜ 2 + B sin θ s ˜ 2 . (20)

In general the two dimensional vectors ( r ˜ 1 , r ˜ 2 ) T and ( s ˜ 1 , s ˜ 2 ) T are not orthogonal because their orthogonality in 3d space implies

r ˜ 1 s ˜ 1 + r ˜ 2 s ˜ 2 = r ˜ 3 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 1 = x 1 m 1 and x 2 = x 2 m 2 is treated:

( A r ˜ 1 B s ˜ 1 A r ˜ 2 B s ˜ 2 ) ( cos θ sin θ ) = ( x 1 x 2 ) (21)

The determinant of the linear system (21), A B ( r ˜ 1 s ˜ 2 r ˜ 2 s ˜ 1 ) , is different from zero. This can be shown by noting that r ˜ 1 s ˜ 2 r ˜ 2 s ˜ 1 is the third component of the vector r ˜ × s ˜ . At first this vector is not affected by rotation (8):

r ˜ × s ˜ = ( cos ω r + sin ω s ) × ( sin ω r + cos ω s ) = ( cos 2 ω + sin 2 ω ) ( r × s ) = r × s .

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 × s = r × ( n × r ) = ( r , r ) n ( r , n ) r = n

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

r ˜ 1 s ˜ 2 r ˜ 2 s ˜ 1 = r 1 s 2 r 2 s 1 = n 3 , (22)

which is positive because of (19) for angles Ω with 0 Ω < π 2 .

Solving the linear system (21) leads to

cos θ = B ( x 1 s ˜ 2 x 2 s ˜ 1 ) A B ( r ˜ 1 s ˜ 2 r ˜ 2 s ˜ 1 ) ,

sin θ = A ( r ˜ 1 x 2 r ˜ 2 x 1 ) A B ( r ˜ 1 s ˜ 2 r ˜ 2 s ˜ 1 ) .

Since cos 2 θ + sin 2 θ = 1 together with (22) the following quadratic equation in x 1 and x 2 is obtained:

B 2 ( x 1 s ˜ 2 x 2 s ˜ 1 ) 2 + A 2 ( r ˜ 1 x 2 r ˜ 2 x 1 ) 2 = A 2 B 2 ( r ˜ 1 s ˜ 2 r ˜ 2 s ˜ 1 ) 2 = A 2 B 2 n 3 2 .

Expanding the squares on the left side and using the denotations

l 11 = A 2 r ˜ 2 2 + B 2 s ˜ 2 2 , l 12 = ( A 2 r ˜ 1 r ˜ 2 + B 2 s ˜ 1 s ˜ 2 ) , l 22 = A 2 r ˜ 1 2 + B 2 s ˜ 1 2 (23)

arranged as a 2 × 2 matrix L

L = ( l 11 l 12 l 12 l 22 ) (24)

leads to

( x 1 , x 2 ) L ( x 1 x 2 ) = 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 λ 1 ( L ) , λ 2 ( L ) :

L = S 1 diag ( λ 1 ( L ) , λ 2 ( L ) ) S

with a nonsingular transformation matrix S , being orthogonal, i.e. S 1 = S T , the inverse of S is equal to the transpose of S . Putting

( x 1 , x 2 ) = ( x 1 , x 2 ) S T , S ( x 1 x 2 ) = ( x 1 x 2 )

the quadratic equation (25) in ( x 1 , x 2 ) reduces to

( x 1 , x 2 ) diag ( λ 1 ( L ) , λ 2 ( L ) ) ( x 1 x 2 ) = A 2 B 2 n 3 2 . (26)

The eigenvalues λ 1 ( L ) , λ 2 ( L ) 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):

det L = l 11 l 22 l 12 2 = ( A 2 r ˜ 2 2 + B 2 s ˜ 2 2 ) ( A 2 r ˜ 1 2 + B 2 s ˜ 1 2 ) ( A 2 r ˜ 1 r ˜ 2 + B 2 s ˜ 1 s ˜ 2 ) 2 = A 2 B 2 ( r ˜ 1 s ˜ 2 r ˜ 2 s ˜ 1 ) 2 = A 2 B 2 ( r 1 s 2 r 2 s 1 ) 2 = A 2 B 2 n 3 2 . (27)

Dividing (26) by A 2 B 2 n 3 2 yields

λ 1 ( L ) A 2 B 2 n 3 2 ( x 1 ) 2 + λ 2 ( L ) A 2 B 2 n 3 2 ( x 2 ) 2 = 1.

This is an ellipse projected from 3d space (18) onto the x 1 x 2 plane with the semi-axes

A L = A B n 3 λ 1 ( L ) , B L = A B n 3 λ 2 ( L ) . (28)

With (19) one obtains from (28)

A L = A B cos Ω λ 1 ( L ) , B L = A B cos Ω λ 2 ( L ) . (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 = 1 A 1 2 + A 2 2 + A 3 2 ( A 1 , A 2 , A 3 ) . (31)

The distance between the plane and the origin is given by

κ = A 4 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 κ = 0.

Without loss of generality A 3 0 shall be assumed. Then n 3 0 holds:

x 3 = 1 n 3 ( κ n 1 x 1 n 2 x 2 ) .

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

m 11 = 1 a 1 2 + n 1 2 a 3 2 n 3 2 , m 12 = n 1 n 2 a 3 2 n 3 2 , m 22 = 1 a 2 2 + n 2 2 a 3 2 n 3 2 , m 13 = n 1 κ a 3 2 n 3 2 , m 23 = n 2 κ a 3 2 n 3 2 , m 33 = κ 2 a 3 2 n 3 2 1. (34)

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

M = ( m 11 m 12 m 12 m 22 )

det M = m 11 m 22 m 12 2 = ( 1 a 1 2 + n 1 2 a 3 2 n 3 2 ) ( 1 a 2 2 + n 2 2 a 3 2 n 3 2 ) n 1 2 n 2 2 a 3 4 n 3 4 = n 3 2 a 1 2 a 2 2 n 3 2 + n 1 2 a 2 2 a 3 2 n 3 2 + n 2 2 a 1 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 a 1 2 a 2 2 a 3 2 n 3 2 . (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 1 + h , x 2 = x 2 + k with parameters h and k to be determined later. After substitution into (33) one obtains:

m 11 x 1 2 + 2 m 12 x 1 x 2 + m 22 x 2 2 + 2 ( m 11 h + m 12 k + m 13 ) x 1 + 2 ( m 12 h + m 22 k + m 23 ) x 2 + m 11 h 2 + 2 m 12 h k + m 22 k 2 + 2 m 13 h + 2 m 23 k + m 33 = 0. (36)

The terms x 1 and x 2 in (36) vanish if h and k are determined by the linear system:

m 11 h + m 12 k = m 13 , m 12 h + m 22 k = m 23 . (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 0 . Solving the linear system (37) yields:

h = m 13 m 22 + m 23 m 12 m 11 m 22 m 12 2 , k = m 11 m 23 + m 12 m 13 m 11 m 22 m 12 2 . (38)

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

h = a 1 2 n 1 κ a 1 2 n 1 2 + a 2 2 n 2 2 + a 3 2 n 3 2 , k = a 2 2 n 2 κ a 1 2 n 1 2 + a 2 2 n 2 2 + a 3 2 n 3 2 . (39)

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

m 11 h 2 + 2 m 12 h k + m 22 k 2 + 2 m 13 h + 2 m 23 k + m 33 = κ 2 a 1 2 n 1 2 + a 2 2 n 2 2 + a 3 2 n 3 2 1 = ( 1 d ) .

Thus the quadratic equation (36) reduces to:

( x 1 , x 2 ) M ( x 1 x 2 ) = 1 d . (40)

M as a real symmetric matrix can be diagonalized and thus is similar to the diagonal matrix of its eigenvalues λ 1 ( M ) , λ 2 ( M ) :

M = T 1 diag ( λ 1 ( M ) , λ 2 ( M ) ) T

with a nonsingular transformation matrix T , being orthogonal, i.e. T 1 = T T , the inverse of T is equal to the transpose of T . Putting

( x 1 , x 2 ) = ( x 1 , x 2 ) T T , T ( x 1 x 2 ) = ( x 1 x 2 )

the quadratic equation (40) in ( x 1 , x 2 ) reduces to

( x 1 , x 2 ) diag ( λ 1 ( M ) , λ 2 ( M ) ) ( x 1 x 2 ) = 1 d . (41)

The eigenvalues λ 1 ( M ) , λ 2 ( M ) 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

λ 1 ( M ) 1 d ( x 1 ) 2 + λ 2 ( M ) 1 d ( x 2 ) 2 = 1.

This is an ellipse in the x 1 x 2 plane with the semi-axes

A M = 1 d λ 1 ( M ) , B M = 1 d λ 2 ( M ) . (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 [ 0 , 2 π ) (43)

with the semi-axes A and B is projected from plane E onto plane E . As in

Section 2 the angle between the two planes is denoted by Ω , with 0 Ω π 2 . Let α , with 0 α 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

( E E ) and let ψ be a phase-shift with 0 ψ π 2 and ψ = τ σ where

the angles τ and σ are determined by

cos σ = A cos α A 2 cos 2 α + B 2 sin 2 α , sin σ = B sin α A 2 cos 2 α + B 2 sin 2 α , cos τ = B cos α A 2 sin 2 α + B 2 cos 2 α , sin τ = A sin α A 2 sin 2 α + B 2 cos 2 α . (44)

The projected ellipse in the plane E is given by

x ¯ 1 = A ¯ cos ( t ¯ + ψ ) , x ¯ 2 = B ¯ sin t ¯ , t ¯ [ 0 , 2 π ) (45)

with

A ¯ = A 2 cos 2 α + B 2 sin 2 α ,

B ¯ = cos Ω A 2 sin 2 α + B 2 cos 2 α . (46)

Eliminating parameter t ¯ from (45) yields a quadratic equation in x ¯ 1 and x ¯ 2

( x ¯ 1 A ¯ ) 2 + 2 sin ψ ( x ¯ 1 A ¯ ) ( x ¯ 2 B ¯ ) + ( x ¯ 2 B ¯ ) 2 = cos 2 ψ

or written with the elements

g 11 = 1 A ¯ 2 , g 12 = sin ψ A ¯ B ¯ , g 22 = 1 B ¯ 2 (47)

forming matrix

G = ( g 11 g 12 g 12 g 22 )

one obtains

( x ¯ 1 , x ¯ 2 ) G ( x ¯ 1 x ¯ 2 ) = cos 2 ψ . (48)

G as a real symmetric matrix can be diagonalized and thus is similar to the diagonal matrix of its eigenvalues λ 1 ( G ) , λ 2 ( G ) :

G = R 1 diag ( λ 1 ( G ) , λ 2 ( G ) ) R

with a nonsingular transformation matrix R , being orthogonal, i.e. R 1 = R T , the inverse of R is equal to the transpose of R . Putting

( x ¯ ¯ 1 , x ¯ ¯ 2 ) = ( x ¯ 1 , x ¯ 2 ) R T , R ( x ¯ 1 x ¯ 2 ) = ( x ¯ ¯ 1 x ¯ ¯ 2 )

the quadratic equation (48) in ( x ¯ 1 , x ¯ 2 ) reduces to

( x ¯ ¯ 1 , x ¯ ¯ 2 ) diag ( λ 1 ( G ) , λ 2 ( G ) ) ( x ¯ ¯ 1 x ¯ ¯ 2 ) = cos 2 ψ . (49)

The eigenvalues λ 1 ( G ) , λ 2 ( G ) 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 = 1 A ¯ 2 B ¯ 2 sin 2 ψ A ¯ 2 B ¯ 2 = cos 2 ψ A ¯ 2 B ¯ 2 (50)

is positive for 0 ψ < π 2 . Dividing (49) by cos 2 ψ for 0 ψ < π 2 yields

λ 1 ( G ) cos 2 ψ ( x ¯ ¯ 1 ) 2 + λ 2 ( G ) cos 2 ψ ( x ¯ ¯ 2 ) 2 = 1.

This is an ellipse in the x ¯ 1 x ¯ 2 plane with the semi-axes

A G = cos ψ λ 1 ( G ) , B G = cos ψ λ 2 ( G ) . (51)

5. Some Auxiliary Means

Let H stand for the following 2 × 2 matrix:

H = ( h 11 h 12 h 12 h 22 ) (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

Q L ( z ) = z 2 ( A L + B L ) z + A L B L , Q H ( z ) = z 2 ( A H + B H ) z + A H B H , (53)

have the same coefficients and thus have the same zeros:

Q L ( z ) = ( z A L ) ( z B L ) , Q H ( z ) = ( z A H ) ( z B H ) . (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:

( A L + B L ) 2 = A L 2 + 2 A L B L + B L 2 = A H 2 + 2 A H B H + B H 2 = ( A H + B H ) 2

which yields A L + B L = A H + B H since the semi-axes are positive.

λ 1 ( L ) , λ 2 ( L ) are the zeros of the characteristic polynomial of L . This can be expressed in two ways:

P L ( λ ) = ( l 11 λ ) ( l 22 λ ) l 12 2 = λ 2 ( l 11 + l 22 ) λ + l 11 l 22 l 12 2 ,

P L ( λ ) = ( λ λ 1 ( L ) ) ( λ λ 2 ( L ) ) = λ 2 ( λ 1 ( L ) + λ 2 ( L ) ) λ + λ 1 ( L ) λ 2 ( L ) .

Comparing the coefficients one obtains

λ 1 ( L ) + λ 2 ( L ) = l 11 + l 22 , λ 1 ( L ) λ 2 ( L ) = l 11 l 22 l 12 2 . (56)

Similarly the results for matrix H instead of L are

λ 1 ( H ) + λ 2 ( H ) = h 11 + h 22 , λ 1 ( H ) λ 2 ( H ) = h 11 h 22 h 12 2 . (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 = A 2 B 2 n 3 2 λ 1 ( L ) λ 2 ( L ) , (58)

A M B M = 1 d λ 1 ( M ) λ 2 ( M ) . (59)

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

λ 1 ( L ) λ 2 ( L ) = 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:

λ 1 ( M ) λ 2 ( M ) = m 11 m 22 m 12 2 = 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 β 1 and β 2 are solutions of (16)

β 1 β 2 = n 1 2 a 2 2 a 3 2 + n 2 2 a 1 2 a 3 2 + n 3 2 a 1 2 a 2 2 = 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)

λ 1 ( L ) λ 2 ( L ) = 1 d β 1 1 d β 2 n 3 2 = ( 1 d ) 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 = ( 1 d ) 2 λ 1 ( M ) λ 2 ( M ) . (63)

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

A L B L = A 2 B 2 n 3 2 λ 1 ( L ) λ 2 ( L ) = λ 1 ( L ) λ 2 ( L ) λ 1 ( L ) λ 2 ( L ) = λ 1 ( L ) λ 2 ( L ) = 1 d λ 1 ( M ) λ 2 ( M ) = A M B M .

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

A L 2 + B L 2 = A 2 B 2 n 3 2 ( 1 λ 1 ( L ) + 1 λ 2 ( L ) ) = A 2 B 2 n 3 2 λ 1 ( L ) λ 2 ( L ) ( λ 2 ( L ) + λ 1 ( L ) ) = λ 1 ( L ) + λ 2 ( L ) . (64)

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

A M 2 + B M 2 = 1 d λ 1 ( M ) + 1 d λ 2 ( M ) = 1 d λ 1 ( M ) λ 2 ( M ) ( λ 2 ( M ) + λ 1 ( M ) ) = ( 1 d ) 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 ( λ 1 ( M ) + λ 2 ( M ) ) = ( 1 d ) n 3 2 β 1 β 2 ( λ 1 ( M ) + λ 2 ( M ) ) . (65)

Together with

λ 1 ( M ) + λ 2 ( M ) = m 11 + m 22 = 1 n 3 2 ( n 3 2 a 1 2 + n 3 2 a 2 2 + n 1 2 + n 2 2 a 3 2 ) (66)

(65) yields

A M 2 + B M 2 = ( 1 d ) β 1 β 2 ( n 3 2 a 1 2 + n 3 2 a 2 2 + n 1 2 + n 2 2 a 3 2 ) . (67)

In continuation of (64), because r ˜ and s ˜ are fulfilling (4) and (5), the following relations hold:

λ 1 ( L ) + λ 2 ( L ) = l 11 + l 22 = A 2 ( r ˜ 1 2 + r ˜ 2 2 ) + B 2 ( s ˜ 1 2 + s ˜ 2 2 ) = A 2 ( 1 r ˜ 3 2 ) + B 2 ( 1 s ˜ 3 2 ) = 1 d β 1 ( 1 r ˜ 3 2 ) + 1 d β 2 ( 1 s ˜ 3 2 ) = 1 d β 1 β 2 ( β 2 ( 1 r ˜ 3 2 ) + β 1 ( 1 s ˜ 3 2 ) ) = 1 d β 1 β 2 ( β 1 + β 2 β 2 r ˜ 3 2 β 1 s ˜ 3 2 ) (68)

with

β 1 + β 2 = n 1 2 ( 1 a 2 2 + 1 a 3 2 ) + n 2 2 ( 1 a 1 2 + 1 a 3 2 ) + n 3 2 ( 1 a 1 2 + 1 a 2 2 ) (69)

because β 1 and β 2 are solutions of (16). Combining (64), (68), (69) and (67) one obtains:

A L 2 + B L 2 ( A M 2 + B M 2 ) = 1 d β 1 β 2 ( n 1 2 a 2 2 + n 2 2 a 1 2 β 2 r ˜ 3 2 β 1 s ˜ 3 2 ) . (70)

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

n 1 = r ˜ 2 s ˜ 3 r ˜ 3 s ˜ 2 , n 2 = r ˜ 3 s ˜ 1 r ˜ 1 s ˜ 3 ,

because of r ˜ × s ˜ = r × s = n (see Section 2), and

β 2 = ( D 1 s ˜ , D 1 s ˜ ) , β 1 = ( D 1 r ˜ , D 1 r ˜ )

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

1 a 2 2 ( r ˜ 2 s ˜ 3 r ˜ 3 s ˜ 2 ) 2 + 1 a 1 2 ( r ˜ 3 s ˜ 1 r ˜ 1 s ˜ 3 ) 2 ( s ˜ 1 2 a 1 2 + s ˜ 2 2 a 2 2 + s ˜ 3 2 a 3 2 ) r ˜ 3 2 ( r ˜ 1 2 a 1 2 + r ˜ 2 2 a 2 2 + r ˜ 3 2 a 3 2 ) s ˜ 3 2 = 2 r ˜ 3 s ˜ 3 ( r ˜ 1 s ˜ 1 a 1 2 + r ˜ 2 s ˜ 2 a 2 2 + r ˜ 3 s ˜ 3 a 3 2 ) = 2 r ˜ 3 s ˜ 3 ( D 1 r ˜ , D 1 s ˜ ) = 0,

because r ˜ and 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 = A 2 B 2 cos 2 Ω λ 1 ( L ) λ 2 ( L ) , (71)

A G B G = cos 2 ψ λ 1 ( G ) λ 2 ( G ) . (72)

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

λ 1 ( L ) λ 2 ( L ) = l 11 l 22 l 12 2 = A 2 B 2 cos 2 Ω . (73)

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

λ 1 ( G ) λ 2 ( G ) = g 11 g 22 g 12 2 = cos 2 ψ A ¯ 2 B ¯ 2 . (74)

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

A L B L A G B G = A B cos Ω A ¯ B ¯ cos ψ . (75)

According to the definition of ψ = τ σ given in the beginning of Section 4 together with (44) and (46) one obtains:

cos ψ = cos ( τ σ ) = A B cos Ω A ¯ 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

A L 2 + B L 2 = λ 1 ( L ) + λ 2 ( L ) = l 11 + l 22 = A 2 ( r ˜ 1 2 + r ˜ 2 2 ) + B 2 ( s ˜ 1 2 + s ˜ 2 2 ) = A 2 ( 1 r ˜ 3 2 ) + B 2 ( 1 s ˜ 3 2 ) = A 2 + B 2 ( A 2 r ˜ 3 2 + B 2 s ˜ 3 2 ) . (76)

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

A G 2 + B G 2 = cos 2 ψ λ 1 ( G ) + cos 2 ψ λ 2 ( G ) = cos 2 ψ λ 1 ( G ) λ 2 ( G ) ( λ 2 ( G ) + λ 1 ( G ) ) = A ¯ 2 B ¯ 2 ( λ 1 ( G ) + λ 2 ( G ) ) = A ¯ 2 B ¯ 2 ( g 11 + g 22 ) = A ¯ 2 B ¯ 2 ( 1 A ¯ 2 + 1 B ¯ 2 ) = B ¯ 2 + A ¯ 2 ; (77)

(77) is continued by substituting B ¯ and A ¯ from (46)

cos 2 Ω ( A 2 sin 2 α + B 2 cos 2 α ) + A 2 cos 2 α + B 2 sin 2 α = A 2 ( cos 2 α + cos 2 Ω sin 2 α ) + B 2 ( sin 2 α + cos 2 Ω cos 2 α ) = A 2 ( cos 2 α + ( 1 sin 2 Ω ) sin 2 α ) + B 2 ( sin 2 α + ( 1 sin 2 Ω ) cos 2 α ) = A 2 ( cos 2 α + sin 2 α sin 2 Ω sin 2 α ) + B 2 ( sin 2 α + cos 2 α sin 2 Ω cos 2 α ) = A 2 ( 1 sin 2 Ω sin 2 α ) + B 2 ( 1 sin 2 Ω cos 2 α ) = A 2 + B 2 sin 2 Ω ( A 2 sin 2 α + B 2 cos 2 α ) (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 r ˜ 3 2 + B 2 s ˜ 3 2 = sin 2 Ω ( A 2 sin 2 α + B 2 cos 2 α ) . (79)

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

angle Ω with 0 Ω π 2 . Without loss of generality the intersection of E

and E , E E , shall be the x ¯ 1 -axis of the coordinate system in plane E . The original plane E thus contains the following three points: ( 1,0,0 ) , ( 1,0,0 ) , ( 0 , cos Ω , sin Ω ) and can therefore be described by the following equation:

sin Ω x ¯ 2 + cos Ω x ¯ 3 = 0. (80)

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

n = ( 0 , sin Ω , cos Ω ) . (81)

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

( r , r ) = r 1 2 + r 2 2 + r 3 2 = 1 , ( n , r ) = sin Ω r 2 + cos Ω r 3 = 0. (82)

The second equation of (82) yields r 3 = r 2 tan Ω . Substituting this into the first equation of (82) results in:

r 1 2 + r 2 2 ( 1 + tan 2 Ω ) = 1

or

r 1 2 + r 2 2 cos 2 Ω = 1. (83)

If the unit vector r is forming the angle α with the x ¯ 1 -axis and e 1 is designating a unit vector in x ¯ 1 -direction according to the definition of the scalar product (see for instance [4] ) holds

r 1 = ( r , e 1 ) = r e 1 cos α = cos α .

From (83) one obtains

r 2 2 = ( 1 cos 2 α ) cos 2 Ω = sin 2 α cos 2 Ω ,

yielding r 2 = ± sin α cos Ω and furthermore with the first equation of (82) r 3 = ± sin α sin Ω . From

r = ( cos α , ± sin α cos Ω , ± sin α sin Ω )

and s = n × r one obtains

s = ( sin α , cos α cos Ω , cos α sin Ω ) .

By transformation (8) one obtains

r ˜ 3 = cos ω r 3 + sin ω s 3 = sin ( ω ± α ) sin Ω ,

s ˜ 3 = sin ω r 3 + cos ω s 3 = cos ( ω ± α ) sin Ω .

Thus equation (79) turns into

( A 2 sin 2 ( ω ± α ) + B 2 cos 2 ( ω ± α ) ) sin 2 Ω = sin 2 Ω ( A 2 sin 2 α + B 2 cos 2 α ) . (84)

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

major axis of the ellipse (43) and the x ¯ 1 -axis, is chosen to be ω 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 + 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 = 1 1 2 + 2 2 + 3 2 ( 1,2,3 ) .

Furthermore in (32) the distance κ of the plane to the origin is given

κ = 4 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 = 1 1 2 + 2 2 ( 2, 1,0 ) T ,

calculating s to be orthogonal to both according to s = n × r and, as ( D 1 r , D 1 s ) 0 , perform a rotation with angle ω given in (9), yielding new vectors r ˜ and s ˜ according to (8), which are plugged into ( D 1 r ˜ , D 1 r ˜ ) and ( D 1 s ˜ , D 1 s ˜ ) .

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

A = 4.59157 , B = 3.39705.

Furthermore having calculated the eigenvalues λ 1 ( L ) and λ 2 ( L ) 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 = 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.

Cite this paper

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

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.

  
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