Let's say I have: the set of 3D data point (xi,yi,zi) contains around 10,000 points located in plane L
I also have : plane M which is specified by normal vector n , and point A(xA,yA,zA). How to find the intersection point between 2 plane L & M?
MATLAB: Find the intersection point between 2 plane in matlab
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This would be an easy implementation of a direct geometric solution.
1) First you need the line in a parametric form by calculating L0 and u:
say L(t) = L0 + t*u, where t is any real scalar
u is the line direction vector given by (R-Q). So, u = (x2-x1, y2-y1, z2-z1)
and L0 is Q = (x1, y1, z1).
Thus any point on the line can be generate by using some real value of t. In particular, here, L(0) gives you point Q, and L(1) gives you point R.
2) Check if line is parallel to plane (either never intersecting or completely lying in plane) i.e. if the plane normal vector n and line vector are perpendicular
if dot(n, u) == 0, that is the case
If the line is on the plane, then check if point Q (or R) is on the plane i.e it satisfies:
dot(n, Q-P) = 0 %the line between any two points on the plane is perpendicular to the normal vector.
Otherwise, the line never intersects the plane.
3) If the line is not parallel, then it intersects the plane at one point, which can be solved for:
t_int = dot(n, P - Q) / dot (n, R - Q )
and the point is simply L(t) = L0 + t_int* u
Here's some sample code you can use to define a function:
% define:
n = [1, 1, 1];P = [2, 3, 0];Q = [2, .51, 1];R = [4, 1, 1];% calculate L0 and U for line:
L0 = Q;u = R- Q;%check if parallel
if dot(n,u)==0 %line parallel to plane
if dot( n, Q-P)==0 disp('Line lies in plane') else disp('Line never intersects plane') end else %find point of intersection
t = dot(n, P - Q)/dot(n, u); POI = L0 + t*u; fprintf('Point of intersection is [%f %f %f]', POI(1), POI(2), POI(3) )endTry this:
% Create sample data.
numCircles = 10;xy = rand(numCircles, 2);radii = 0.1 * rand(numCircles, 1) + 0.1;viscircles(xy, radii);% Label them so we know what circle is what.
for k = 1 : numCircles text(xy(k, 1), xy(k, 2), num2str(k));endaxis square;% See if any intersect
for k = 1 : numCircles % Compute the distance between centers.
distances = sqrt((xy(k, 1) - xy(:, 1)).^2 + (xy(k, 2) - xy(:, 2)) .^2); sumOfRadii = radii(k) + radii(:); intersectingIndexes = distances < sumOfRadii; % Don't compare circle to itself. Of course that one intersects!
intersectingIndexes(k) = false; % Convert to linear indexes
intersectingIndexes = find(intersectingIndexes); for k2 = 1 : length(intersectingIndexes) fprintf('Circle %d intersects with circle #%d.\n', k, intersectingIndexes(k2)); endend
You see
Circle 1 intersects with circle #2.
Circle 1 intersects with circle #5.
Circle 1 intersects with circle #7.
Circle 2 intersects with circle #1.
Circle 2 intersects with circle #5.
Circle 3 intersects with circle #7.
Circle 3 intersects with circle #9.
Circle 5 intersects with circle #1.
Circle 5 intersects with circle #2.
Circle 6 intersects with circle #10.
Circle 7 intersects with circle #1.
Circle 7 intersects with circle #3.
Circle 8 intersects with circle #9.
Circle 9 intersects with circle #3.
Circle 9 intersects with circle #8.
Circle 10 intersects with circle #6.
and it's trivial to find out if one circle is completely contained within another. See if you can do it.
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