How to find the orientation of the line of the intersection between two planes?
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Is there any method/indiacator that i can use to know the orientation of the the intersection line between two planes( using Dual Plucker Matrix )?
I used the follwoing code get the line:
P1 =[177668442.453315 ,-102576923.076923, 0];
P2 =[ -102576923.076923 ,177668442.453315 ,-102576923.076923];
P3= [0, -102576923.076923, 88834221.2266576];
P11= [152763459.308716 , -102576923.076923, 0];
P22=[ -102576923.076923, 183536536.231793 , -102576923.076923];
P33= [0, -102576923.076923, 91768268.1158967];
A=null([[P1;P2;P3],ones(3,1)]); %plane 1
B=null([[P11;P22;P33],ones(3,1)]); %plane 2
L=A*B.' - B*A.' %line of intersection
2 commentaires
Réponse acceptée
Matt J
le 5 Fév 2023
Modifié(e) : Matt J
le 5 Fév 2023
P1 =[177668442.453315 ,-102576923.076923, 0];
P2 =[ -102576923.076923 ,177668442.453315 ,-102576923.076923];
P3= [0, -102576923.076923, 88834221.2266576];
P11= [152763459.308716 , -102576923.076923, 0];
P22=[ -102576923.076923, 183536536.231793 , -102576923.076923];
P33= [0, -102576923.076923, 91768268.1158967];
A=null([[P1;P2;P3],ones(3,1)]); %plane 1
B=null([[P11;P22;P33],ones(3,1)]); %plane 2
L=A*B.' - B*A.' %line of intersection;
N=null(L);
a=N(:,1)+N(:,2);
b=N(:,1)+2*N(:,2);
direction=normalize( b(1:3)/b(4)-a(1:3)/a(4) ,'n')
or,
direction=normalize( cross(A(1:3),B(1:3)) ,'n')
6 commentaires
Plus de réponses (1)
Torsten
le 5 Fév 2023
Modifié(e) : Torsten
le 5 Fév 2023
If you look at the next lines in Matt's code, he creates 100 points on the line.
Thus in the modified code below, Pstart could be taken as a point on the line and d as a direction vector for the line emanating from Pstart.
Did you mean something like this ?
P1 =[177668442.453315 ,-102576923.076923, 0];
P2 =[ -102576923.076923 ,177668442.453315 ,-102576923.076923];
P3= [0, -102576923.076923, 88834221.2266576];
P11= [152763459.308716 , -102576923.076923, 0];
P22=[ -102576923.076923, 183536536.231793 , -102576923.076923];
P33= [0, -102576923.076923, 91768268.1158967];
A=null([[P1;P2;P3],ones(3,1)]); %plane 1
B=null([[P11;P22;P33],ones(3,1)]); %plane 2
L=A*B.' - B*A.'; %line of intersection
N=null(L);
t=linspace(-.1,.1);
xyz=N(:,1) + N(:,2)*t;
xyz = xyz(1:3,:)./xyz(4,:);
P1 = xyz(:,1);
P2 = xyz(:,2);
Pstart = P1
d = (P2-P1)/norm(P2-P1)
13 commentaires
Torsten
le 6 Fév 2023
Modifié(e) : Torsten
le 6 Fév 2023
The usual representation of the plane of intersection is given by all solutions x of a linear system of the form
A*x = b
Here, A is a 2x4 matrix and b is a 2x1 vector.
The rows of A are the normal vectors of 2 hyperplanes in the 4d space.
The vector b somehow represents the distance of these hyperplances to the origin.
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