Plot a plane from 3 points

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Alberto Acri
Alberto Acri 2023년 3월 15일
댓글: Mathieu NOE 2023년 3월 16일
Hi, I want to create a plane that goes through points A, B, C. I am using the "plot_line" function found on "file exchange" below:
%% ====== FUNCTION ======
function [normal, d, X, Y, Z] = plot_line_deviazione(p1, p2, p3, x_min, x_max, y_min, y_max)
% This function plots a line from three points.
% I/P arguments:
% p1, p2, p3 eg, p1 = [x y z]
%
%
% O/P is:
% normal: it contains a,b,c coeff , normal = [a b c]
% d : coeff
normal = cross(p1 - p2, p1 - p3);
d = p1(1)*normal(1) + p1(2)*normal(2) + p1(3)*normal(3);
d = -d;
x = x_min:x_max; y = y_min:y_max;
[X,Y] = meshgrid(x,y);
Z = (-d - (normal(1)*X) - (normal(2)*Y))/normal(3);
mesh(X,Y,Z)
Can anyone tell me why it is not shown in the plotting? I have this code:
A = [91.0049, 222.2614, -12.2419];
B = [12.8, 48.2, -5.6631];
C = [12.8, 48.2, -53.7279];
figure
plot3(A(:,1),A(:,2),A(:,3),'r.','Markersize',25)
hold on
plot3(B(:,1),B(:,2),B(:,3),'r.','Markersize',25)
plot3(C(:,1),C(:,2),C(:,3),'r.','Markersize',25)
[normal0, d0, X0, Y0, Z0] = plot_line(A, B, C, -100, 100, -100, 100); % new plane
hold off
grid off
xlabel('x')
ylabel('y')
zlabel('z')
view([15,50,30])
axis equal
NOTE: using these three points shows the plane
A = [-62.6412, 32.6849, -195.9077];
B = [-36.5, -33.1755, -2];
C = [-36.5, 34.2726, -2];

채택된 답변

Mathieu NOE
Mathieu NOE 2023년 3월 16일
hello Alberto
welcome back
I was first a bit puzzled because you speak of "line" whereas we are talking here "plane" that goes through 3 points. just a minor comment
also I don't see wich Fex function you refer to (missing link) but again this is not important
more important :
  • 1 in your code you don't call the function plot_line_deviazione that you posted but this one : plot_line.
[normal0, d0, X0, Y0, Z0] = plot_line(A, B, C, -100, 100, -100, 100); % new plane
I supposed that it's actually plot_line_deviazione that you wanted to use in your main code
  • now let's see what happen when you test your code with the following data (representing a triangle in the vertical plane)
A = [91.0049, 222.2614, -12.2419];
B = [12.8, 48.2, -5.6631];
C = [12.8, 48.2, -53.7279];
the cross product is a vector in the horizontal plane , so it's z coordinate is zero
normal = 1.0e+03 *( 8.3662 -3.7589 0)
therefore in this line you are making a division by zero resulting in Z being Inf :
Z = (-d - (normal(1)*X) - (normal(2)*Y))/normal(3);
so you have to check if your cross product has one zero value and make sure you are not making a division by zero when you generate the mesh.
for this specific case you have to create first the x and z plan mesh and then compute the Y result (see below new code of your function)
all the best
% 3 points on vertical plane
A = [91.0049, 222.2614, -12.2419];
B = [12.8, 48.2, -5.6631];
C = [12.8, 48.2, -53.7279];
% % 3 points on inclined plane
% A = [-62.6412, 32.6849, -195.9077];
% B = [-36.5, -33.1755, -2];
% C = [-36.5, 34.2726, -2];
figure(1)
plot3(A(:,1),A(:,2),A(:,3),'r.','Markersize',25)
hold on
plot3(B(:,1),B(:,2),B(:,3),'r.','Markersize',25)
plot3(C(:,1),C(:,2),C(:,3),'r.','Markersize',25)
[normal0, d0, X, Y, Z] = plot_line_deviazione(A, B, C, -100, 100, -100, 100); % new plane
mesh(X,Y,Z)
hold off
grid off
xlabel('x')
ylabel('y')
zlabel('z')
view([15,50,30])
axis equal
%% ====== FUNCTION ======
function [normal, d, X, Y, Z] = plot_line_deviazione(p1, p2, p3, x_min, x_max, y_min, y_max)
% This function plots a line from three points.
% I/P arguments:
% p1, p2, p3 eg, p1 = [x y z]
%
%
% O/P is:
% normal: it contains a,b,c coeff , normal = [a b c]
% d : coeff
normal = cross(p1 - p2, p1 - p3);
d = p1(1)*normal(1) + p1(2)*normal(2) + p1(3)*normal(3);
d = -d;
if abs(normal(3))>eps % cross product is non zero in Z direction ( your original code)
x = x_min:x_max;
y = y_min:y_max;
[X,Y] = meshgrid(x,y);
Z = (-d - (normal(1)*X) - (normal(2)*Y))/(normal(3));
else % cross product is zero in Z direction => modified code
x = x_min:x_max;
z = min([p1(3) p2(3) p3(3)]):max([p1(3) p2(3) p3(3)]);
[X,Z] = meshgrid(x,z);
Y = (-d - (normal(1)*X) - (normal(3)*Z))/(normal(2));
end
end
  댓글 수: 6
Alberto Acri
Alberto Acri 2023년 3월 16일
Thank you!
Mathieu NOE
Mathieu NOE 2023년 3월 16일
again, my pleasure !

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추가 답변 (1개)

Luca Ferro
Luca Ferro 2023년 3월 16일
편집: Luca Ferro 2023년 3월 16일
basically after debugging i got this results:
mesh(X,Y,Z) fails because Z is a 201x201 whose elements are ALL Inf.
This is caused by its declaration Z = (-d - (normal(1)*X) - (normal(2)*Y))/normal(3) in which the last term normale(3) is 0.
The last term is 0 since in cross(p1 - p2, p1 - p3), the two differences inside are 78.2049 174.0614 41.4860 and 78.2049 174.0614 -6.5788. As you can see, using cross product definition of ||a|| ||b|| sin(theta), the theta is 0 due to their alignment, resulting in the 0.
The latter 3 points work because the normal vector has a non zero third element, 0.1763 to be exact, so using it to divide is not problematic.

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