How do show points on plane going through 3d plot

Hello,

I have 3d plot of points and curves which are connected (picture attached). ("o" and "*" in the graph are my points in matrix and "-" & "--" connect the points)

Then I make a new "plane" (gray color) going through the graph. I need to show points of curves going through the plane on the plane.

Is there a way make a plane goint through curves and then show and write points of the curves?

Answer 1

[albara]

Yes, you can create a plane that intersects the 3D curves and then display the intersection points on the plane. I will provide a general outline of the steps you can take to achieve this in MATLAB.

1. Define your 3D points and curves.
2. Define the plane equation (e.g., Ax + By + Cz + D = 0).
3. Find intersection points between the plane and the curves.
4. Plot the points and curves in a 3D plot.
5. Plot the intersection points on the plane.

Here's an example to help you get started:

% Define your 3D points and curves (dummy example)
curve1 = [1 2 3; 2 3 4; 3 4 5];
curve2 = [5 4 3; 4 3 2; 3 2 1];

% Define the plane equation: Ax + By + Cz + D = 0
A = 1;
B = 1;
C = 1;
D = -6;

% Find intersection points between the plane and the curves
intersect1 = intersectPlaneCurve(A, B, C, D, curve1);
intersect2 = intersectPlaneCurve(A, B, C, D, curve2);

% Plot the points and curves in a 3D plot
figure
hold on
plot3(curve1(:,1), curve1(:,2), curve1(:,3), 'o-')
plot3(curve2(:,1), curve2(:,2), curve2(:,3), 'o--')
plotIntersectionPlane(A, B, C, D)

% Plot the intersection points on the plane
scatter3(intersect1(:,1), intersect1(:,2), intersect1(:,3), 'r*')
scatter3(intersect2(:,1), intersect2(:,2), intersect2(:,3), 'g*')
hold off

function intersection = intersectPlaneCurve(A, B, C, D, curve)
    intersection = [];
    for i = 1:size(curve, 1)-1
        p1 = curve(i,:);
        p2 = curve(i+1,:);
        [is_intersect, point] = linePlaneIntersection(A, B, C, D, p1, p2);
        if is_intersect
            intersection = [intersection; point];
        end
    end
end

function [is_intersect, point] = linePlaneIntersection(A, B, C, D, p1, p2)
    direction = p2 - p1;
    denominator = A*direction(1) + B*direction(2) + C*direction(3);
    
    if abs(denominator) < 1e-6
        is_intersect = false;
        point = [];
        return
    end
    
    t = -(A*p1(1) + B*p1(2) + C*p1(3) + D) / denominator;
    point = p1 + t * direction;
    
    if t >= 0 && t <= 1
        is_intersect = true;
    else
        is_intersect = false;
        point = [];
    end
end

function plotIntersectionPlane(A, B, C, D)
    [x, y] = meshgrid(linspace(-10, 10, 100));
    z = (-D - A*x - B*y) / C;
    surf(x, y, z, 'FaceAlpha', 0.5, 'EdgeColor', 'none', 'FaceColor', [0.5 0.5 0.5]);
end

This example shows two 3D curves and a plane that intersects them.

Important: There may be some mistakes in this answer Experts can tell if there are any mistakes

[Simon Spielmann]

Yes thank you very much for Your advice and quick respond, this may be solution for my problem!

Maybe dummy quastion but is there a way to determine a position of the plane?

I understand that a,b,c determine a normal vector of the plane. So if I determine normal vector n=[1; 0; 0] it would be parallel to x axis. But for example if I wanted plane that be x = -10 (as seen on in the picture) what should I change?

PS: sorry for late response

[albara]

You are welcome, Yes, you can determine the position of the plane by adjusting the coefficients A, B, C, and D of the plane equation Ax + By + Cz + D = 0. If you have a normal vector n = [A; B; C] and a point on the plane p = [x0; y0; z0], you can find the value of D as follows:

D = -A * x0 - B * y0 - C * z0

In your example, you want a plane parallel to the YZ plane (normal vector n = [1; 0; 0]) and positioned at x = -10. To find D, you can use any point on the plane, for instance, p = [-10; 0; 0]:

D = -1 * (-10) - 0 * 0 - 0 * 0 = 10

So, your plane equation is x + 10 = 0, or equivalently, 1x + 0y + 0z + 10 = 0.

In the provided MATLAB code, you can set A, B, C, and D as follows:

A = 1; B = 0; C = 0; D = 10;

This will create a plane parallel to the YZ plane at x = -10.

Important: There may be some mistakes in this answer Experts can tell if there are any mistakes

[Simon Spielmann]

Thank You very much again for explanation!

Helped me a lot

Answer 2

[chicken vector]

Not sure if I understood what you need.

You can have plane transparency using 'FaceAlpha'.

Is this helpful? Otherwise I ask you for more details.

nCurves = 10;
nPoints = 15;
xPlane = 5;

x = 0 : nPoints-1;
y = (1 : nCurves)' + rand(nCurves,nPoints)/2;
z = flip(x) + rand(nCurves,nPoints);

lineStyles = {'-o','--*'};

figure;
grid on;
hold on;
for line = 1 : nCurves
    plot3(x,y(line,:),z(line,:),lineStyles{round((line-1)/nCurves)+1});
end
fill3(xPlane*ones(4,1), [-2, nPoints*2, nPoints*2, -2], [-2, -2, nPoints*2, nPoints*2], ... 
      [.85 .85 .85],'FaceAlpha',0.85)
hold off;
view([2 1 1]);
xlabel('x'); ylabel('y'); zlabel('z');
xlim([-1 nPoints-1]); ylim([-1 nCurves+1]); zlim([0 nPoints*1.5]);

[Simon Spielmann]

Thank you for your respond.

My problem is I would like to determine plane in the 3d plot and then show on the plane intersect point of curves going through the plane.

albara understood my problem and gave me solution to my problem