How to solve parametric system of vector equations?
3 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Dániel Széplaki
le 21 Mar 2020
Commenté : Christopher Creutzig
le 25 Mar 2020
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?
1 commentaire
Christopher Creutzig
le 25 Mar 2020
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.
Réponse acceptée
David Goodmanson
le 22 Mar 2020
Modifié(e) : David Goodmanson
le 22 Mar 2020
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)
1 commentaire
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Assumptions dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!