The solution of nonlinear equations.

[Acrobat]

Help solve the following system of equations:

(x-x1)^2 + (y-y1)^2 + (z-z1)^2 = (R - R11)^2
(x-x2)^2 + (y-y2)^2 + (z-z2)^2 = (R - R21)^2
(x-x3^2 + (y-y1)^2 + (z-z3)^2 = (R - R31)^2
(x+R)^2 + (y+R)^2 + (z+R)^2 = L^2

xi,yi,zi,Rij,L - consts

 

[Deleted member 4993]

Help solve the following system of equations:

(x-x1)^2 + (y-y1)^2 + (z-z1)^2 = (R - R11)^2
(x-x2)^2 + (y-y2)^2 + (z-z2)^2 = (R - R21)^2
(x-x3^2 + (y-y1)^2 + (z-z3)^2 = (R - R31)^2
(x+R)^2 + (y+R)^2 + (z+R)^2 = L^2

xi,yi,zi,Rij,L - consts

What are your thoughts?

Please share your work with us ...even if you know it is wrong

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[stapel]

[Please h]elp solve the following system of equations:

(x - x₁)² + (y - y₁)² + (z - z₁)² = (R - R₁₁)²
(x - x₂)² + (y - y₂)² + (z - z₂)² = (R - R₂₁)²
(x - x₃)² + (y - y₁)² + (z - z₃)² = (R - R₃₁)²
(x + R)² + (y + R)² + (z + R)² = L²

x_i, y_i, z_i, R_ij, and L are constants

You have four non-linear equations in terms of five variables. This cannot be solved, other than in a parameterized manner.

By the way, is the subscript on the second "y" in the third equation supposed to be "1", or is it actually "3"?

Note: Wolfram Alpha's widget can solve the system. Unfortunately, I can't figure out how to tell it which of the letters are the variables, so it solves for things you don't want. Either way, though, clearly the solution is quite complex. You may be on your own here... Sorry.

 

[Ishuda]

Help solve the following system of equations:

(x-x1)^2 + (y-y1)^2 + (z-z1)^2 = (R - R11)^2
(x-x2)^2 + (y-y2)^2 + (z-z2)^2 = (R - R21)^2
(x-x3)^2 + (y-y1)^2 + (z-z3)^2 = (R - R31)^2
(x+R)^2 + (y+R)^2 + (z+R)^2 = L^2

xi,yi,zi,Rij,L - consts

Assuming everything but the x, y, and z are constants, first get rid of the mess by letting
u = x - x1,
v = y - y1,
w = z - z1,
collecting constants and and rearranging to obtain
u² + v² + w² = a
for the first equation and use that on the other three equations to obtain
a₁₁ u + a₁₂ v + a₁₃ w = b₁
a₂₁ u + a₂₂ v + a₂₃ w = b₂
a₃₁ u + a₃₂ v + a₃₃ w = b₃
This is a simple linear 3X3 system which may or may not have a solution.

Oh, and see that other assumed correction in red above.

Hint: (x-x₂)² = [x - x₁ - (x₂-x₁)]² = [u - (x₂-x₁)]²