Linear Algebra and the C Language/a06c

__NOTOC__ The coefficients of the equation of a sphere. Presentation : A homogeneous linear system with as many equations as unknowns has a nontrivial solution if and only if the determinant of the matrix is zero. Let us calculate the equation of the sphere passing through points P, Q, R and S : c1(x^2 +y^2 +z^2) +c2x +c3y +c4z +c5 = 0 This same equation with the points P(x1,y1,z1) Q(x2,y2,z2) R(x3,y3,z3) and S(x4,y4,z4): c1(x1^2+y1^2+z1^2)+c2x1+c3y1+c4z1+c5 = 0 c1(x2^2+y2^2+z2^2)+c2x2+c3y2+c4z2+c5 = 0 c1(x3^2+y3^2+z3^2)+c2x3+c3y3+c4z3+c5 = 0 c1(x4^2+y4^2+z4^2)+c2x4+c3y4+c4z4+c5 = 0 The system of five equations: c1(x^2 +y^2 +z^2) +c2x +c3y +c4z +c5 = 0 c1(x1^2+y1^2+z1^2)+c2x1+c3y1+c4z1+c5 = 0 c1(x2^2+y2^2+z2^2)+c2x2+c3y2+c4z2+c5 = 0 c1(x3^2+y3^2+z3^2)+c2x3+c3y3+c4z3+c5 = 0 c1(x4^2+y4^2+z4^2)+c2x4+c3y4+c4z4+c5 = 0 The determinant of the system: |(x ^2+y ^2+z ^2) x y z 1| |(x1^2+y1^2+z1^2) x1 y1 z1 1| |(x2^2+y2^2+z1^2) x2 y2 z2 1| = 0 |(x3^2+y3^2+z3^2) x3 y3 z3 1| |(x4^2+y4^2+z4^2) x4 y4 z4 1| The determinant in C language: | 1 1 1 1 1| |(x1^2+y1^2+z1^2) x1 y1 z1 1| |(x2^2+y2^2+z1^2) x2 y2 z2 1| = 0 |(x3^2+y3^2+z3^2) x3 y3 z3 1| |(x4^2+y4^2+z4^2) x4 y4 z4 1| c1 (x^2 +y^2 +z^2) + c2 x + c3 y + c4 z + c5 = 0 To calculate the coefficients of the equation of the sphere, we use the cofactor expansion along the first row. cof(R1,C1) (x ^2+y ^2+z ^2)+ cof(R1,C2) x+ cof(R1,C3) y+ cof(R1,C4) z+ cof(R1,C5) = 0 This equation gives us the equation of the sphere that passes through the four points P, Q, R and S. Application dsphere.h .............. Temporary h file Delete this file at the end of the session. c00a.c c00b.c