How to solve parametric system of vector equations?

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Dániel Széplaki 2020-3-21

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评论： Christopher Creutzig 2020-3-25

I have two parameters: P1 and P2, they are vectors. I am looking for a C vector and r scalar.

I have constraints: (P1-C)^2 == (P2-C)^2 == r^2 ; (C-(P1+P2)/2)*(P1-P2) == 0 ; 1+r^2 == C^2

Now I have tried creating symbolic variables, such as a, b and then P1 = (a,b). But somehow the dot product of two vectors becomes some complex vector.

So the question is, can I solve a system of equations such as this, using vectors?

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Christopher Creutzig 2020-3-25

If you do not say differently, symbolic variables are complex (and scalar). The dot product therefore follows the rules in the complex plane. Please try syms a b real, and if you still run into problems, please post a minimal, but complete code snippet, i.e., something others can copy and run.

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Answer 1

David Goodmanson 2020-3-22

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编辑：David Goodmanson 2020-3-22

在 MATLAB Online 中打开

Hi Daniel,

interesting problem. does this correspond to a particular physical situation?

% P1 --> a,   P2 --> b
% solution is for c^2 = r^2 + z^2
a = 2*rand(3,1)-1;
b = 2*rand(3,1)-1;
z = 1;             % specific case 
p = (a+b)/2;
q = (b-a)/2;
u = cross(p,q);     % perpendicular to plane defined by a and b
w = cross(u,q);
w = w/norm(w);      % unit vector in ab plane, perpendicular to (b-a)
lambda = (dot(q,q)+z^2-dot(p,p))/(2*dot(p,w));
c = p+lambda*w;
r = sqrt(dot(q,q)+lambda^2);
% checks, should be small
dot(c-a,c-a) - r^2 
dot(c-b,c-b) - r^2 
dot(c -((a+b)/2),b-a)
dot(c,c) - (r^2+z^2)

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Dániel Széplaki 2020-3-22

编辑：Dániel Széplaki 2020-3-22

It corresponds to a geometric problem. You are given two points inside a unit circle, and you have to find a another circle, that fits on those points, and is perpendicular to the first circle.

I see how your answer works now. Very elegant, thank you!

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