How to find the orientation of the line of the intersection between two planes?
1 view (last 30 days)
Show older comments
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 Comments
Accepted Answer
Matt J
on 5 Feb 2023
Edited: Matt J
on 5 Feb 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 Comments
More Answers (1)
Torsten
on 5 Feb 2023
Edited: Torsten
on 5 Feb 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 Comments
Torsten
on 6 Feb 2023
Edited: Torsten
on 6 Feb 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.
See Also
Categories
Find more on Interpolating Gridded Data in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!