Determining the coordinates of a node between two known nodes and knowing the distances between the known nodes

Hi! This may seem like a rather trivial question. Do you know how to determine the coordinates of a node C knowing node A, node B, distance AC and distance CB?

At the moment I can use these two equations:

C = [x, y, z]; %unknown

A = [x_A, y_A, z_A];
d1 = (x-x_A)^2;
d2 = (y-y_A)^2;
d3 = (z-z_a)^2;
d_AC = sqrt(d1 + d2 + d3);

B = [x_B, y_B, z_B];
d4 = (x_B-x)^2;
d5 = (y_B-y)^2;
d6 = (z_B-z)^2;
d_CB = sqrt(d4 + d5 + d6);

A third equation is missing. At the moment I have no idea, and I don't know if it is possible to determine the coordinates directly on matlab.

Answer 1

[Matt J]

Assuming the nodes form a straight line,

C = A + (B-A)*d_AC/(d_AC+dCB)

[Torsten]

Assuming the nodes form a straight line

As said: this is a very special case, namely if dAC + dCB = dAB.

Answer 2

[Torsten]

There is only a unique solution for this problem if AC + CB = AB. In all other cases, you either get no solution (AC + CB < AB) or infinitly many solutions (AC + CB > AB).

[Matt J]

or infinitly many solutions (AC + CB > AB).

Wouldn't there be 2 solutions, assuming A~=B? The intersection of the circles of radii AC and BC?

[Torsten]

Not in 3d where spheres intersect. And this is reflected by having two equations for three unknowns.

Answer 3

[Hassaan]

function C = findPointC(A, B, d_AC, d_CB)
    % Define the system of equations
    function F = distanceEquations(C)
        F(1) = (C(1) - A(1))^2 + (C(2) - A(2))^2 + (C(3) - A(3))^2 - d_AC^2;
        F(2) = (C(1) - B(1))^2 + (C(2) - B(2))^2 + (C(3) - B(3))^2 - d_CB^2;
    end

    % Initial guess (can be changed based on the problem context)
    initialGuess = (A + B) / 2;

    % Solve the equations
    C = fsolve(@distanceEquations, initialGuess, optimoptions('fsolve', 'Display', 'off'));
end

You can use this function by passing the coordinates of points A and B, and the distances dAC​ and dCB​. Keep in mind that this approach might find one of the possible solutions or may fail if the distances provided are not physically consistent with the positions of A and B.

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[Matt J]

Why a numerical solution? Why not use solve()?