How can we find the intersection between two planes in higher dimensions (4d space and above)?

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How can we find the intersection between two planes in higher dimensions (4d space and above)? For example we have the following 2 planes in 4d:
Plane 1
P1 =[252716585.970010 -136769230.769231 0 0];
P2 =[ -136769230.769231 252716585.970010 -136769230.769231 0];
P3= [0 -136769230.769231 252716585.970010 -136769230.769231];
P4 = [0 0 -136769230.769231 126358292.985005];
Plane 2
P11= [191269260.712188 -136769230.769231 0 0];
P22=[ -136769230.769231 259653876.096803 -136769230.769231 0];
P33= [0 -136769230.769231 259653876.096803 -136769230.769231];
P44=[0 0 -136769230.769231 129826938.048402];
  2 commentaires
Jan
Jan le 6 Fév 2023
What are your inputs? Du you mean 2D planes in 4D space? 4 Points? 1 Point and 2 lines in the plane?
Matt J
Matt J le 6 Fév 2023
It is more compact to describe the planes in equation form Aeq*x=beq. For plane 1, this would be
P1 =[252716585.970010 -136769230.769231 0 0];
P2 =[ -136769230.769231 252716585.970010 -136769230.769231 0];
P3= [0 -136769230.769231 252716585.970010 -136769230.769231];
P4 = [0 0 -136769230.769231 126358292.985005];
Aeq=null([P2;P3;P4]-P1)'
Aeq = 1×4
0.2420 0.4472 0.5843 0.6325
beq=Aeq*P1'
beq = -3.2783e-07
and similarly for plane 2.

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Réponse acceptée

Matt J
Matt J le 6 Fév 2023
Modifié(e) : Matt J le 6 Fév 2023
In general, intersections of two hyperplanes would be expressed algebraically by a 2xN set of linear equations Aeq*x=beq. A geometric description can be made in terms of an origin vector, which gives the position of some point in the intersection space, and a set of direction vectors which span the linear space parallel to it. Example:
Aeq=[1,2,3,4;
5,6,7,8];
beq=[5;7];
assert( rank([Aeq,beq])==rank(Aeq) , 'Hyperplanes do not intersect')
origin = pinv(Aeq)*beq
origin = 4×1
-1.0000 -0.2500 0.5000 1.2500
directions = null(Aeq)
directions = 4×2
-0.4001 -0.3741 0.2546 0.7970 0.6910 -0.4717 -0.5455 0.0488
  9 commentaires
M
M le 17 Juin 2023
Hello @Matt J, could you please tell me based on what you programed these two lines in your code:
origin = pinv(Aeq)*beq
directions = null(Aeq)
Could you please elaborate them?
Thanks
Torsten
Torsten le 17 Juin 2023
Modifié(e) : Torsten le 17 Juin 2023
The complete set of solutions of a linear systems of equations (interpreted as the intersection of the hyperplanes) is given by one solution of the inhomogeneous system (origin) + the solutions of the homogeneous system (directions).
Read the last two paragraphs (Homogeneous solution set, Relation to nonhomogeneous systems) under

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