Mathc complexes/08z
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c00e.c |
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/* ------------------------------------ */
/* Save as : c00e.c */
/* ------------------------------------ */
#include "w_a.h"
/* ------------------------------------ */
#define RAb R3
#define CA C4
#define Cb C1
#define Rv C4
/* ------------------------------------ */
# define FREEV C1
/* ------------------------------------ */
int main(void)
{
double ab[RAb*(CA+Cb)]={
+4, +3, -1, -2, -0, /* V1|0 See c00a.c */
-1, -1, +2, -3, -0, /* V2|0 */
-2, -5, +4, -3, -0 /* V3|0 */
};
double Projp[Rv*C1] ={
-12.06250000,
-13.75000000,
+15.37500000,
-15.75000000
};
double **Ab = ca_A_mRZ(ab,i_Abr_Ac_bc_mZ(RAb, CA, Cb));
double **Ab_free = i_Abr_Ac_bc_mZ(CA, CA, Cb + FREEV);
double **b_free = i_mZ(CA, Cb + FREEV);
double **A = c_Ab_A_mZ(Ab, i_mZ(RAb, CA));
double **At = transpose_mZ(A, i_mZ(CA, RAb));
double **vPp = ca_A_mRZ(Projp, i_mZ(Rv,C1));
double **v1 = c_c_mZ(At,(C1), i_mZ(Rv,C1),C1);
double **v2 = c_c_mZ(At,(C2), i_mZ(Rv,C1),C1);
double **v3 = c_c_mZ(At,(C3), i_mZ(Rv,C1),C1);
double **vQ = i_mZ(Rv,C1);
/* Compute the free variables (b_free) */
gj_PP_mZ(Ab);
put_zeroR_mZ(Ab,Ab_free);
put_freeV_mZ(Ab_free);
gj_mZ(Ab_free);
c_Ab_b_mZ(Ab_free,b_free);
c_c_mZ(b_free,C2, vQ,C1);
/* vQ is the free vector of the system Ab:
v1|0
A|b = v2|0
v3|0
vQ is orthogonal to the vectors v1,v2 and v3 */
clrscrn();
printf(" Compute the orthogonal vectors to: (See c00a .c)\n\n"
" v1:");
p_mRZ(v1, S10,P4, C1);
printf(" v2:");
p_mRZ(v2, S10,P4, C1);
printf(" v3:");
p_mRZ(v3, S10,P4, C1);
stop();
clrscrn();
printf(" vQ is orthogonal to the vectors v1,v2 and v3.\n\n"
" The equation of a hyperplan defined by v1, v2 and v3\n"
" is ax + by + cz + dt=0, where a,b,c,d are the coordinates\n"
" of the vector orthogonal (vQ).\n\n"
" vQ:");
p_mRZ(vQ, S10,P12, C1);
stop();
clrscrn();
printf(" vPp is the projection of a point P onto the hyperplan (See c00d.c)\n\n"
" vPp:");
p_mRZ(vPp, S10,P12, C1);
printf(" vQ is orthogonal to the hyperplan defined by the vectors v1,v2 and v3.\n\n"
" vQ:");
p_mRZ(vQ, S10,P12, C1);
printf(" dot(vPp,vQ) = ");p_Z(dot_Z(vPp,vQ), S4,P5, S5,P5);
printf("\n The dot product of vQ and vPp say that the vector vPp\n"
" is orthogonal to the vector vQ. Then the point p is in\n"
" the hyperplan v1, v2 and v3.\n\n");
stop();
f_mZ(Ab);
f_mZ(Ab_free);
f_mZ(b_free);
f_mZ(v1);
f_mZ(v2);
f_mZ(vQ);
return 0;
}
/* ------------------------------------ */
/* ------------------------------------ */
Vérifiez si la projection d'un point P appartient à l'hyperplan. Si la projection du point P, le vecteur vPp, est perpendiculaire au vecteur vQ, alors le produit scalaire de ces deux vecteurs est nul et le point Pp appartient bien à l'hyperplan.
Exemple de sortie écran :
Compute the orthogonal vectors to: (See c00a .c)
v1:
+4.0000
+3.0000
-1.0000
-2.0000
v2:
-1.0000
-1.0000
+2.0000
-3.0000
v3:
-2.0000
-5.0000
+4.0000
-3.0000
Press return to continue.
vQ is orthogonal to the vectors v1,v2 and v3.
The equation of a hyperplan defined by v1, v2 and v3
is ax + by + cz + dt=0, where a,b,c,d are the coordinates
of the vector orthogonal (vQ).
vQ:
+0.285714285714
+1.000000000000
+2.142857142857
+1.000000000000
Press return to continue.
vPp is the projection of a point P onto the hyperplan (See c00d.c)
vPp:
-12.062500000000
-13.750000000000
+15.375000000000
-15.750000000000
vQ is orthogonal to the hyperplan defined by the vectors v1,v2 and v3.
vQ:
+0.285714285714
+1.000000000000
+2.142857142857
+1.000000000000
dot(vPp,vQ) = -0.00000+0.00000i
The dot product of vQ and vPp say that the vector vPp
is orthogonal to the vector vQ. Then the point p is in
the hyperplan v1, v2 and v3.
Press return to continue.