Procedure

Kitonum — Wed, 25 Jan 2012 14:16:54 Z

Your problem is no different from the previous plane (3). The procedure P solves the problem for any points A, B, M and numeric d (points should not be collinear).

Procedure code:

restart:

P:=proc(A,B,M,d)

local P,Sol,L,f,L1;

uses RealDomain, LinearAlgebra, ListTools;

if Equal(simplify(CrossProduct(convert(B,Vector)-convert(A,Vector),convert(M,Vector)-

convert(A,Vector))),<0,0,0>) then error `Points A, B, M should not be collinear`; fi;

P:=a*(x-A[1])+b*(y-A[2])+c*(z-A[3]);

Sol:=[solve({subs(x=B[1],y=B[2],z=B[3],P)=0,a^2+b^2+c^2=1,

abs(subs(x=M[1],y=M[2],z=M[3],P))=d},{a,b,c})];

L:=[seq(<rhs(Sol[i,1]),rhs(Sol[i,2]),rhs(Sol[i,3])>,i=1..nops(Sol))];

f:=(x,y)->Equal(simplify(CrossProduct(x,y)),<0,0,0>);

L1:=[Categorize(f, L)];

if nops(L1)=0 then print(`The problem has no solutions`); fi;

if nops(L1)=1 then print(`The problem has 1 solution`); print(collect(subs(a=L1[1][1][1],b=L1[1][1][2],c=L1[1][1][3],P),[x,y,z])=0); fi;

if nops(L1)=2 then print(`The problem has 2 solutions`); print(collect(subs(a=L1[1][1][1],b=L1[1][1][2],c=L1[1][1][3],P),[x,y,z])=0); print(collect(subs(a=L1[2][1][1],b=L1[2][1][2],c=L1[2][1][3],P),[x,y,z])=0); fi;

end proc;

As an exapmle see solution the problem plane (3)

P([-1,3,-6],[2,2,-10],[1,-1,7],3);

Similarly, you can find equation of a plane (4)

Another way

toandhsp — Fri, 17 Feb 2012 14:53:08 Z

This is my code. Please comment to me.

>restart;f:=(x,y,z)->a*x + b*y +c*z + d:

A:=f(2,-2,3):

B:=f(4,-5,6):

M:=f(1,2,3):

sol:=solve([A = 0, B=0, abs(M) = 2*sqrt(2), a^2 + b^2 + c^2 = 1],[a, b, c, d]):

for i to nops(sol) do subs(sol[i],f(x,y,z)=0) end do;

Thank you very much.

toandhsp — Wed, 25 Jan 2012 18:33:07 Z

Thank you very much.

MaplePrimes - answers and comments on Question, Equation of a plane (4)

Procedure

Another way

Thank you very much.