# reverse 3D euclidean distance

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Ferdinand Grosse-Dunker on 25 Jul 2019
Commented: Star Strider on 25 Jul 2019
Hi there,
I am standing at an unknown point U(x,y,z) in the room. I can measure 3 (euclidean) distances D to 3 known points P in the room. I try to find the point, where I am at. My equation system looks like this:
(x-3)²+(y-1)²+(z-4)²=D1²=81
(x-12)²+(y-1)²+(z-4)²=D2²=36
(x-34)²+(y-2)²+(z-4)²=D3²=601
I can put the known points into a matrix P, the measured distance in a vector D:
P=[3 1 4;12 1 4; 34 2 4]
P =
3 1 4
12 1 4
34 2 4
D=[81 36 601]
Do you know how I can find U(x,y,z) ?
I am not sure if I can use
D = pdist(X,'euclidean');

Star Strider on 25 Jul 2019
Try this:
P=[3 1 4;12 1 4; 34 2 4];
D=[81 36 601];
fcn = @(b,x) (b(1)-x(:,1)).^2 + (b(2)-x(:,2)).^2 + (b(3)-x(:,3)).^2;
B = fminsearch(@(b) norm(D(:) - fcn(b,P)), [1; 1; 1])
producing:
B =
10.0000
5.0000
-0.0000
that are the (x,y,z) coordinates, as best fminsearch can calculate them.

Ferdinand Grosse-Dunker on 25 Jul 2019
Works perfekt, thank you. Can you comment on the function you use?
Star Strider on 25 Jul 2019
As always, my pleasure.
The documentation for the fminsearch function is at the link. It is an unconstrained optimiser that uses a derivative-free method to find the minimum.
The code I use for my objective function ‘fcn’ and as an argument to fminsearch are Anonymous Functions. They are quite useful for coding short functions, although they have their limitations.
I use the norm function so that the fminsearch function finds the minimum value that satisfies the sum-of-squares criterion (since this is essentially a curve-fiting problem).

Akira Agata on 25 Jul 2019
Here is my try.
P = [3 1 4;12 1 4; 34 2 4];
D = [81 36 601];
func = @(x) (vecnorm(x - P(1,:))-sqrt(D(1)))^2+...
(vecnorm(x - P(2,:))-sqrt(D(2)))^2+...
(vecnorm(x - P(3,:))-sqrt(D(3)))^2;
x1 = fminsearch(func,[1 1 1]);
x2 = fminsearch(func,[10 10 10]);
>> x0
x0 =
10.0000 5.0000 -0.0000
>> x1
x1 =
10.0000 5.0000 8.0000