MaplePrimes - answers and comments on Question, Equation of a plane (3)

The idea of the solution

Kitonum — Sun, 22 Jan 2012 02:47:33 Z

If N (x0, y0, z0) is the point of tangency, then the three conditions:

1) The distance from N to the center of the sphere (point O) is equal to the radius.

2) ON is perpendicular to AN.

3) ON is perpendicular to BN.

I used your code. restart:with(LinearAlgebra

toandhsp — Sun, 22 Jan 2012 04:41:44 Z

I used your code.

restart:
with(LinearAlgebra):A:=<-1,3,6>: B:=<2,2,-0>: N:=<x,y,z>: o:=<1,-1,7>:
{Norm(o-N,2)=3, DotProduct(N - o, N-A) = 0, DotProduct(N - o, N-B) = 0}:
solve({Norm(o-N,2)=3, DotProduct(N - o, N-A) = 0,DotProduct(N - o, N-B) = 0}): assign(%): N:
'N'=[seq(N[i],i=1..3)];

and i got Warning, solutions may have been lost.

Please check my code. Thank you.

This is my idea: Let (P) be finding

toandhsp — Mon, 23 Jan 2012 06:10:24 Z

This is my idea: Let (P) be finding plane. Put n = (a, b, c) is a normal vector of (P). Then, the equation of (P) has the form

a*(x - 1) + b*(y -3) + c(z + 6) = 0. We find a, b, c from following conditions:

1) a^2 + b^2 +c^2 = 1;

2) vector n is perpendicular to vector AB;

3) The distance form C(1; -1; 7) (C is center of the given sphere) to the plane (P) equal to 3.

But i assume(a^2 + b^2 + c^2 > 0) and use command

plane(P,[A, n],[x, y, z]) can not write the Equation of (P). Plese help me. Thank you very much.

Solution

Kitonum — Mon, 23 Jan 2012 20:24:53 Z

You have 2 planes: ABN1 and ABN2. Find the equation ABN1. The simplest way to use the determinant. The code (the continuation of the previous code):

N1:=convert(N1,Vector):

T:=<x,y,z> - N1: N1-A: N1-B:

P1:=collect(LinearAlgebra[Determinant](< T | N1-A | N1-B >),[x,y,z]):

k:=simplify(sqrt(coeff(P1,x)^2+coeff(P1,y)^2+coeff(P1,z)^2)):

P2:=simplify(collect(P1/k,[x,y,z])):

P:=collect(P2,[x,y,z])=0; # Equation of the plane P

simplify(coeff(lhs(P),x)^2+coeff(lhs(P),y)^2+coeff(lhs(P),z)^2); # Checking

Similarly, we find a second plane.

Rewrited code

Kitonum — Mon, 23 Jan 2012 23:54:41 Z

Once posted the previous message, noticed in it one essential disadvantage. The code will not work if the points A, B, N1 are collinear. Therefore, rewrote the code using your idea. The code is simpler, and the result , of course, is the same:

with(LinearAlgebra):

N1:=convert(N1,Vector):

n:=o-N1: e:=n/simplify(Norm(n,2)):

T:=<x,y,z>-N1:

collect(expand(DotProduct(e,T,conjugate=false)),[x,y,z])=0; # Equation of the plane P

Correction

Kitonum — Sun, 22 Jan 2012 20:21:14 Z

Your code does not work because:

1) For the command DotProduct the option conjugate=false is required.

2) The system has not one but two solutions.

The corrected code:

restart:

with(LinearAlgebra):

A:=<-1,3,6>: B:=<2,2,-0>: N:=<x,y,z>: o:=<1,-1,7>:

Sys:={Norm(o-N,2)^2=9, DotProduct(N - o, N-A,conjugate=false) = 0, DotProduct(N - o, N-B,conjugate=false) = 0}:

Sol:=[solve(Sys)]:

N1:=[seq(rhs(Sol[1,i]),i=1..3)];

N2:=[seq(rhs(Sol[2,i]),i=1..3)];