Geometry for the HP48S/SX/G/GX

Overview

1°) Angle between 2 Vectors
2°) Cross Product ( 7D-Vectors )
3°) UxV ( n-dimensional Vectors )
4°) Distance Point-StraightLine
5°) Distance Point-HyperPlane
6°) Line-Line Distance
7°) Distance Point-HyperEllipsoid
8°) Bisecting Hyperplanes
9°) Hypersurface Area of a Hyperellipsoid
10°) Equation of a Hyperplane
11°) Equation of a Hypersphere
12°) Heron Formula
13°) Brahmagupta Formula
14°) Tetrahedron Volume
15°) Generating Integer Tetrahedra
16°) Quadratic Hypersurface
17°) Jacobi's Method

1°) Angle between 2 Vectors

'ANGL' takes the coordinates of 2 vectors X & Y and returns the angle A

STACK	INPUTS	OUTPUTS
Level 2	[ x₁ , ........ , x_n ]	/
Level 1	[ y₁ , ........ , y_n ]	A

Example:

[ 3 4 1 7 ] ENTER
[ 4 6 8 9 ] ANGL -> 28°324038159

2°) Cross Product ( 7D-Vectors )

U(x₁,x₂,....,x₇) and V(y₁,y₂,....,y₇) are two 7D-vectors.

CROS7 calculates the cross product W(z₁,z₂,....,z₇) by the formula given here

STACK	INPUTS	OUTPUTS
Level 2	[ x₁ , ........ , x_n ]	/
Level 1	[ y₁ , ........ , y_n ]	[ z₁ , ........ , z_n ]

Example: U( 6 2 4 7 10 13 12 ) V( 4 6 3 2 5 7 8 )

[ 6 2 4 7 10 13 12 ] ENTER
[ 4 6 3 2 5 7 8 ] CROS7 -> [ -37 26 73 59 -40 33 -47 ]

3°) UxV ( N-Dimensional Vectors )

U(x₁,x₂,....,x_n) and V(y₁,y₂,....,y_n) are 2 vectors.

In n-dimensional spaces, the "cross" product may also be defined as an antisymmetric Tensor P = [P_i,j] where P_i,j = x_i.y_j - x_j.y_i i , j = 1 , 2 , ...... , n

STACK	INPUTS	OUTPUTS
Level 2	[ x₁ , ........ , x_n ]	/
Level 1	[ y₁ , ........ , y_n ]	[P_i,j]

Example: U(2,3,7,1) V(3,1,4,6)

[ 2 3 7 1 ] ENTER
[ 3 1 4 6 ] UxV gives the antisymmetrix matrix:

[[ 0 -7 -13 9 ]
[ 7 0 5 17 ]
[ 13 -5 0 38 ]
[ -9 -17 -38 0 ]]

4°) Distance Point-StraightLine

-The point is determined by its coordinates: P( x₁ , ...... , x_n )
-The line (L) is defined by 2 points A( a₁ , ...... , a_n ) B( b₁ , ...... , b_n )

STACK	INPUTS	OUTPUTS
Level 3	[ x₁ , ........ , x_n ]	/
Level 2	[ a₁ , ........ , a_n ]	/
Level 1	[ b₁ , ........ , b_n ]	D

Example: (L) is defined by A(2,3,4,5) & B(3,7,11,14) and P(1,3,7,9)

[ 1 3 7 9 ] ENTER
[ 2 3 4 5 ] ENTER
[ 3 7 11 14 ] PLN -> 2.16024689947

5°) Distance Point-Hyperplane

-The point is determined by its coordinates: P( x₁ , ...... , x_n )
-The hyperplane (H) by one of its equations: a₁.x₁ + ...... + a_n.x_n + b = 0

STACK	INPUTS	OUTPUTS
Level 2	[ x₁ , ........ , x_n ]	/
Level 1	[ a₁ , ........ , a_n , b ]	D

Example: (H): 2x + 3y + 4z + 7t - 9 = 0 P(1,3,9,6)

[ 1 3 9 6 ] ENTER
[ 2 3 4 7 -9 ] PPLN -> 9.05821627315

6°) Line-Line Distance

(L) is defined by 2 points A( a₁ , ...... , a_n ) and B( b₁ , ...... , b_n )
(L') is defined by 2 points C( c₁ , ...... , c_n ) and D( d₁ , ...... , d_n )

STACK	INPUTS	OUTPUTS
Level 4	[ a₁ , ........ , a_n ]	/
Level 3	[ b₁ , ........ , b_n ]	/
Level 2	[ c₁ , ........ , c_n ]	/
Level 1	[ d₁ , ........ , d_n ]	Dist

Example: A(2,3,4,5) & B(3,7,11,7) C(1,2,3,7) & D(3,7,4,10)

[ 2 3 4 5 ] ENTER
[ 3 7 11 7 ] ENTER
[ 1 2 3 7 ] ENTER
[ 3 7 4 10] LLN -> 2.42892317172

7°) Distance Point-HyperEllipsoid

DPHE calculates the distance between a point P( x₁ , ...... , x_n ) and a hyperellipsoid defined by (x₁/a₁)² + ..... + (x_n/a_n)² = 1

STACK	INPUTS	OUTPUTS
Level 2	[ x₁ , ........ , x_n ]	[ y₁ , ........ , y_n ]
Level 1	[ a₁ , ........ , a_n ]	D

where Q( y₁ , ........ , y_n ) is the point on the hyperellipsoid such that PQ = D

Example: (E):   x² / 49 + y² / 36 + z² / 25 + t² / 16 = 1      P(10,11,12,13)

[ 10 11 12 13 ] ENTER
[ 7 6 5 4 ] DPHE -> 17.81606768 and [ y₁ , ........ , y_n ] = [ 3.46444028983 , 3.08322396764 , 2.55456142811 , 1.91816467489 ]

8°) Bisecting Hyperplanes

They are defined by the equations (H) : a₁ x₁ + ........ + a_n x_n + b = 0 ,   (H') :   a'₁ x₁ + ........ + a'_n x_n + b' = 0

BISHP returns the equations of the bisecting hyperplanes    (H") : c₁ x₁ + ........ + c_n x_n + d = 0 ,   (H'") :   c'₁ x₁ + ........ + c'_n x_n + d' = 0

STACK	INPUTS	OUTPUTS
Level 2	[ a₁ , ........ , a_n , b ]	[ c₁ , ........ , c_n , d ]
Level 1	[ a'₁ , ........ , a'_n , b' ]	[ c'₁ , ........ , c'_n , d' ]

Example: (H): x + y + 7 z + 7 t + 4 = 0 (H'): 2 x + 4 y + 4 z + 8 t + 3 = 0

[ 1 1 7 7 4 ] ENTER
[ 2 4 4 8 3 ] BISHP -> [ -0.1 -0.3 0.3 -0.1 0.1 ] & [ 0.3 0.5 1.1 1.5 0.7 ] which may be rewritten:

(H"): - x - 3 y + 3 z - t + 1 = 0 & (H'"): 3 x + 5 y + 11 z + 15 t + 7 = 0

9°) Hypersurface Area of a HyperEllipsoid

-The area A of an hypersurface defined by an equation of the form x_N = f( x₁ , ............. , x_N-1 ) may be calculated by the (N-1) multiple integral:

A = § ...... § [ 1 + ( ¶f / ¶x₁ )² + ................. + ( ¶f / ¶x_N-1 )² ]^1/2 dx₁ ............. dx_N-1 ( § = Integral )

-Applying this formula to an hyperellipsoid x₁² / a₁² + ............... + x_N² / a_N² = 1 leads to ( after several changes of variables )

A = 2^N a₁ ........ a_N-1 §₀^PI/2 ..... §₀^PI/2 Cos^N-2 µ₁ Cos^N-3 µ₂ .......... Cos µ_N-1 [ g( µ₁ , ....... , µ_N-1 ) ]^1/2 dµ₁ .......... dµ_N-1

where g( µ₁ , ....... , µ_N-1 ) = 1 + (a_N²/a₁² - 1) Sin² µ₁ + (a_N²/a₂² - 1) Cos² µ₁ Sin² µ₂ + ...... + (a_N²/a_N-1² - 1) Cos² µ₁ Cos² µ₂ .... Sin² µ_N-1

-For N = 4 , we have to compute a triple integral which remains possible in a "reasonnable" time
-For larger N-values, the execution time rapidly becomes prohibitive.

-Fortunately, in reference [1], a method is described to transform the (N-1) multiple integral into a univariate integral.
-With minor modifications and a few simplifications, it yields:

A = (PI)^(N-1)/2 [ a₂ ........ a_N / Gamma((N+1)/2) ] §₀¹ f(u) du

with f(u) = (1-u)^(N-1)/2 u^-1/2 [ 1 + SUM_i=2,...,N (a₁²/a_i²/b_i) ] ( b₂ ..... b_N )^-1/2 where b_i = 1 - u + u (a₁²/a_i²)

SAHE uses a Gauss-Chebyshev formula, namely:

§₀¹ [ (1-x) / x ]^1/2 g(x) dx = SUM_i=1,2,...,n w_i g(x_i)

-Unlike many Gaussian formulae, the abscissas & weights may be easily calculated:

x_i = Sin² ([ (2i-1) / (2n+1) ] PI/2) & w_i = [ 2.PI / (2n+1) ] Cos² ([ (2i-1) / (2n+1) ] PI/2)

-So, we can choose n large enough to get an almost perfect accuracy.

STACK	INPUTS	OUTPUTS
Level 2	n	/
Level 1	[ a₁ , ........ , a_N ]	Area

where n = number of points employed by Gauss-Chebyshev formula

Example: x² / 16 + y² / 25 + z² / 36 + t² / 49 = 1

-With n = 4

4 ENTER
[ 4 5 6 7 ] SAHE -> 3194.72464771

-With n = 8

8 ENTER
[ 4 5 6 7 ] SAHE -> 3194.58486195

-With n = 16

16 ENTER
[ 4 5 6 7 ] SAHE -> 3194.58485708

-With n = 32

32 ENTER
[ 4 5 6 7 ] SAHE -> 3194.58485708

10°) Equation of a HyperPlane

The hyperplane in an n-dimensional space is defined by n points

STACK	INPUTS	OUTPUTS
Level n	[ xⁿ₁ , ........ , xⁿ_n ]	/
..........	...................	/
Level 1	[ x¹₁ , ........ , x¹_n ]	[ a₁ , ........ , a_n , b ]

Example: A( 1 2 7 4 ) B( 3 1 5 2 ) C( 2 4 7 1 ) D( 4 1 8 6 )

[ 1 2 7 4 ] ENTER
[ 3 1 5 2 ] ENTER
[ 2 4 7 1 ] ENTER
[ 4 1 8 6 ] PLAN -> [ 5 26 -27 19 56 ] So (H): 5 x + 26 y - 27 z + 19 t + 56 = 0

11°) Equation of a HyperSphere

The equation of a hypersphere is: x₁² + x₂² + .............. + x_n² + a₁ x₁ + a₂ x₂ + ............. + a_n x_n + b = 0

SPHR takes the coordinates of (n+1) points and returns the coordinates of the center, the radius and [ a₁ , ........ , a_n , b ]

STACK	INPUTS	OUTPUTS
Level n	[ xⁿ⁺¹₁ , ........ , xⁿ⁺¹_n ]	/
..........	...................	/
Level 3	....................	[ a₁ , ........ , a_n , b ]
Level 2	...................	R
Level 1	[ x¹₁ , ........ , x¹_n ]	[ a₁ , ........ , a_n , b ]

Example:

-Find an equation of the sphere passing through the 4 points: A(-1,4,7) B(2,4,6) C(5,1,0) D(3,-3,-4)

[ -1 4 7 ] ENTER
[ 2 4 6 ] ENTER
[ 5 1 0 ] ENTER
[ 3 -3 -4 ] SPHR -> [ 0.66666666664 -8.3333333333 6.9999999995 ] and R = 12.4454364683 and [ -1.33333333328 16.6666666666 -13.9999999999 -36.0000000002 ]

-So the sphere's equation is x² + y² + z² - 4 x/3 + 50 y/3 - 14 z - 36 = 0
-The exact radius is sqrt(1394)/3
-The center is ( 2/3 , -25/3 , 7 )

12°) Heron Formula

-Let a triangle ABC with 3 known sides a , b , c

-Heron's formula is Area = [ p(p-a)(p-b)(p-c) ]^1/2 where p = (a+b+c)/2 = semiperimeter

STACK	INPUTS	OUTPUTS
Level 3	a	/
Level 2	b	/
Level 1	c	Area

Example: a = 2 b = 3 c = 4

   2   ENTER
   3   ENTER
   4   HERON -> Area = 2.90473750966

13°) Brahmagupta Formula

-We can use Brahmagupta formula to calculate the area of a cyclic quadrilateral

STACK	INPUTS	OUTPUTS
Level 4	a	/
Level 3	b	/
Level 2	c	/
Level 1	d	Area

Example: a = 4 , b = 5 , c = 6 , d = 7

   4 ENTER
   5 ENTER
   6 ENTER
   7 BRAHM -> Area = 28.9827534925

14°) Tetrahedron Volume

-We use the following conventions:
-The edges d , e , f are respectively opposite to a , b , c ( non co-planar )
-The edges a , b , c intersect at the same vertex S

                                        C
                                         *
                                     *      *
                                  *             *
                               *         d *     *
                        c *                           *    e
                         *                      B         *
                      *                         *              *
                   *               b *                *         *
               *            *                             f *      *
             *      *                                                * *
       S *     *    *    *    *    *    *    *    *    *    *     * A
                                         a

-Given the 6 sides of a tetrahedron a , b , c , d , e , f , this program computes the volume V of the tetrahedron,
and the radii r , R , r₁ , r₂ , r₃ , r₄ of the insphere, the circumsphere and the 4 exspheres.

STACK	INPUTS	OUTPUTS
Level 7	/	r₄
Level 6	a	r₃
Level 5	b	r₂
Level 4	c	r₁
Level 3	d	r
Level 2	e	R
Level 1	f	Volume

Example: a = 3   b = 5   c = 7   d = 6   e = 8   f = 4

3 ENTER
5 ENTER
7 ENTER
6 ENTER
8 ENTER
4 THVR -> V = 8.42614977317 and

R = 4.35040875528    ( circumsphere ) r₁ = 1.29830860673 r₃ = 1.15302542336 ( the radii of the
r = 0.591887582737    ( insphere )    r₂ = 1.89859159932 r₄ = 0.823182563658    4 exspheres )

15°) Generating Integer Tetrahedra

-Here, you specify a , b , c and ITH searches for integers d , e , f that produce a tetrahedron with an integer volume V
-Then, it returns - if any - the integer volume V corresponding to the edges lengths ( a , b , c, d , e , f )

STACK	INPUTS	OUTPUTS
Level 3	a	/
Level 2	b	V
Level 1	c	{ a b c d e f }

Example: a = 4 b = 5 c = 6

4 ENTER
5 ENTER
6 ITH -> ( HALT ) { 4 5 6 2 8 7 } and V = 6

-Press CONT there is no other solution

Note:

-This program is very slow...

16°) Quadratic Hypersurface

QHS reduces the cartesian equation of a quadratic hypersurface in a n-dimensional space ( n > 1 )

-Given (QHS): a₁₁ x₁² + a₂₂ x₂² + .............. + a_nn x_n² + Sum_{i < j} a_{i j} x_i x_j + b₁ x₁ + b₂ x₂ + ............. + b_n x_n + c = 0

the elements a_{i j} ( i < j ) are gradually zeroed by the Jacobi's iterative method.

The reduced equation is: (QHS): a'₁₁ X₁² + a'₂₂ X₂² + .............. + a'_nn X_n² + b'₁ X₁ + b'₂ X₂ + ............. + b'_n X_n + c' = 0

where b'_i = 0 if a'_ii # 0 and c' = 0 or 1

STACK	INPUT	OUTPUT
Level 1	cartesian equation	reduced equation

Example: (QHS): 3 x² + 4 y² + 7 z² + 6 t² + 9 x.y + 3 x.z + 7 x.t + 4 y.z + 8 y.t + 2 z.t + 9 x + 2 y + 5 z + 4 t - 10 = 0 in a 4-dimensional space.

[ 3 4 7 6 9 3 7 4 8 2 9 2 5 4 -10 ] QHS -> [ 0.209131158075 -2.71616606068 -1.2354670922 -0.297006168891 0 0 0 0 1 ]

-So, the reduced equation is 0.209131158075 X² - 2.71616606068 Y² - 1.2354670922 Z² - 0.297006168891 T² + 1 = 0 a "hyper-hyperboloid"?

Note:

QHS calls the following routine JCB to find the eigenvalues & eigenvectors of the symmetric matrix defined by the cartesian equation.

Remark:

-The variable 'eps' contains a small number ( E-10) to define "tiny" elements that are replaced by 0.
-Change this value if you prefer another test...

17°) Jacobi's Method

JCB uses a variant of the Jacobi's algorithm:

*** The eigenvalues of a matrix A are the diagonal-elements of an upper triangular matrix T equal to the infinite product ...U_k^-1.U_k-1^-1....U₂^-1.U₁^-1.A.U₁.U₂....U_k-1.U_k....
where the U_k are unitary matrices. The eigenvectors are the columns of U = U₁.U₂....U_k-1.U_k.... if A is Hermitian ( i-e if A equals its transconjugate )
Actually, T is diagonal if A is Hermitian.

-"JCB" finds the greatest element a_{i j} below the main diagonal
-Then, U₁ is determined so that it places a zero in position ( i , j ) in U₁^-1.A.U₁

U₁ has the same elements as the Identity matrix, except that u_{i i} = u_{j j} = x and u_{i j} = y + i.z , u_{j i} = -y + i.z

with y + i.z = C.x , x = ( 1 + | C |² )^-1/2 and C = 2.a_{i
j} / [ a_{i i} - a_{i j} + ( ( a_{i i} - a_{i j} )² + 4 a_{i j} a_{j i} )^1/2 ]

-The process is repeated until the greatest sub-diagonal element is smaller than eps

-The successive greatest | a_{i j} | ( i > j ) are displayed when the routine is running.

STACK	INPUTS	OUTPUTS
Level 2	/	V
Level 1	A	M

Where A is a square matrix ( symmetric or hermitian ) , V contains the eigenvalues, the columns of M are the eigenvectors

Example1: Let's compute all the eigenvalues and the eigenvectors of the matrix

            1 2 4
    A = 2 7 3 JCB returns the 3-vector in level 2 [ -0.730676198691 4.91074121335 12.8199349853 ]
            4 3 9

-So, the eigenvalues are:

                 k₁ = -0.730676198691
                 k₂ = 4.91074121335
                 k₃ = 12.8199349853

and the 3 corresponding eigenvectors are the 3 columns of the matrix in level 1

             V₁ ( 0.930757325639 ; -0.104865823188 ; -0.350276975967 )
             V₂ ( -0.101146468234 ; 0.846760700436 ; -0.522269765696 )
             V₃ ( 0.351369026422 ; 0.521535689482 ; 0.777521917338 )

-All the eigenvectors are unit vectors.

Note:

-This program only works well if the matrix is symmetric ( or more generally hermitian ).

References:

[1] Dunkl & Ramirez - "Computing Hyperelliptic Integrals for Surface Measure of Ellipsoids"
[2] Abramowitz and Stegun - "Handbook of Mathematical Functions" - Dover Publications - ISBN 0-486-61272-4