1. Problem solving
To solve the problem of the intersection of the space line and the sphere, in order to solve the problem of the quadratic equation in two variables, the following is the solution process
Second, the code implementation
1. Code implementation
1 void cal_q ( double par_x, double par_y, double par_z, double par_r, double bpx, double bpy, double bpz, double spx, double spy, double spz ) { 2 double x_1, x_2, y_1, y_2, z_1, z_2; 3 4 if ( bpx != spx ) { 5 double k_xy = ( spy - bpy ) / ( spx - bpx ); 6 double b_xy = bpy - k_xy * bpx; 7 8 doublek_zx = ( spz - bpz ) / ( spx - bpx ); 9 double b_zx = bpz - k_zx * bpx; 10 11 double A = 1 + k_xy * k_xy + k_zx * k_zx; 12 double B = 2 * k_xy * ( b_xy - par_y ) + 2 * k_zx * ( b_zx - par_z ) - 2 * par_x; 13 double C = par_x * par_x + ( b_xy - par_y ) * ( b_xy - par_y ) + ( b_zx - par_z ) * ( b_zx - par_z ) - par_r * par_r; 14 15 x_1 = ( -B + sqrt ( B * B - 4 * A * C ) ) / ( 2 * A ); 16 y_1 = k_xy * x_1 + b_xy; 17 z_1 = k_zx * x_1 + b_zx; 18 19 x_2 = ( -B - sqrt ( B * B - 4 * A * C ) ) / ( 2 * A ); 20 y_2 = k_xy * x_2 + b_xy; 21 z_2 = k_zx * x_2 + b_zx; 22 23 printf ( "x_1: %10.5f y_1: %10.5f z_1: %10.5f \n", x_1, y_1, z_1 ); 24 printf ( "x_2: %10.5f y_2: %10.5f z_2: %10.5f \n", x_2, y_2, z_2 ); 25 26 printf( " bpx!=spx\n " ); 27 } else { 28 if ( bpy != spy ) { 29 double k_zy = ( spz - bpz ) / ( spy - bpy ); 30 double b_zy = spz - k_zy * spy; 31 32 double A = 1 + k_zy * k_zy; 33 double B = 2 * k_zy * ( b_zy - par_z ) - 2 * par_y; 34 double C = par_y * par_y + ( b_zy - par_z ) * ( b_zy - par_z ) + ( bpx - par_x ) * ( bpx - par_x ) - par_r * par_r; 35 36 x_1 = bpx; 37 y_1 = ( -B + sqrt ( B * B - 4 * A * C ) ) / ( 2 * A ); 38 z_1 = k_zy * y_1 + b_zy; 39 40 x_2 = bpx; 41 y_2 = ( -B - sqrt ( B * B - 4 * A * C ) ) / ( 2 * A ); 42 z_2 = k_zy * y_2 + b_zy; 43 44 printf ( "x_1: %10.5f y_1: %10.5f z_1: %10.5f \n", x_1, y_1, z_1 ); 45 printf ( "x_2: %10.5f y_2: %10.5f z_2: %10.5f \n", x_2, y_2, z_2 ); 46 47 printf ( "bpx == spx && bpy != spy \n" ); 48 } else { 49 if ( bpz != spz ) { 50 x_1 = bpx; 51 y_1 = bpy; 52 z_1 = par_z + sqrt ( par_r * par_r - ( bpx - par_x ) * ( bpx - par_x ) - ( bpy - par_y ) * ( bpy - par_y ) ); 53 54 x_2 = bpx; 55 y_2 = bpy; 56 z_2 = par_z - sqrt ( par_r * par_r - ( bpx - par_x ) * ( bpx - par_x ) - ( bpy - par_y ) * ( bpy - par_y ) ); 57 58 printf ( "x_1: %10.5f y_1: %10.5f z_1: %10.5f \n", x_1, y_1, z_1 ); 59 printf ( "x_2: %10.5f y_2: %10.5f z_2: %10.5f \n", x_2, y_2, z_2 ); 60 61 printf ( "bpx == spx && bpy == spy \n" ); 62 } else { 63 printf ( "Warning: two points overlap !!!! \n" ); 64 return 0; 65 } 66 } 67 } 68 }
2. Test
1 #include <cstdio> 2 #include <cmath> 3 4 void cal_q ( double, double, double, double, double, double, double, double, double, double ); 5 6 int main() { 7 double par_x = 20; 8 double par_y = 20; 9 double par_z = 20; 10 double par_r = 15; 11 12 double bpx = 20; 13 double bpy = 30; 14 double bpz = 40; 15 16 double spx = 20; 17 double spy = 20; 18 double spz = 20; 19 20 printf ( "bpx: %10.5f bpy: %10.5f bpz: %10.5f \n", bpx, bpy, bpz ); 21 printf ( "spx: %10.5f spy: %10.5f spz: %10.5f \n", spx, spy, spz ); 22 23 cal_q ( par_x, par_y, par_z, par_r, bpx, bpy, bpz, spx, spy, spz ); 24 25 return 0; 26 }
3. Graphical display of test results
