You can input either a 2D or a 3D line and 2D or 3D points. This function is an extended version of the line below, with some input checking:
Rik (2020). point to line distance (https://www.mathworks.com/matlabcentral/fileexchange/64396-point-to-line-distance), MATLAB Central File Exchange. Retrieved .
Thank you very much for your effort in the file.
Could you please improve the code a little more to add two optional outputs: (1) the coordinates of the projection points for all points on the line and (2) a flag if the projection point is inside or outside of the line segment for each point?
@Nicola, thanks for pointing out this issue. I now uploaded a version where this should be fixed. The function now properly accepts 2D inputs.
The input point "pt" shouldn't be checked against the case in which it's of size 2?
@Kaleesh, I'll try to help you your question thread https://www.mathworks.com/matlabcentral/answers/425347
lemme put this way
curvexy1 = [ 0 20 ];
curvexy2 =[20 50];
curvexy = (curvexy1 & curvexy2 );
[x,y] = ginput(1);
h1 = text(x,y,'o', ...
'Color', [1 0 0], ...
%% let curvexy1 ,curvexy2 be v1,v2 and ginput(1) be pt - I tried using your code to identify the perpendicular distance but ??
You can find the solution here: https://gamedev.stackexchange.com/questions/72528/how-can-i-project-a-3d-point-onto-a-3d-line
A non-vectorized solution in the terms of the input to this function:
ap = pt-v1;ab = v2-v1;
result = v1 + dot(ap,ab)/dot(ab,ab) * ab;
Hi! Thanks, works great. I was wondering: how do I find the POINT of intersection between the original line and the projection of the point to this line? Thank you!
fixed issue with 2D input
added compatibility for R13 (MATLAB 6.5)
uploaded incorrect version, so now a bug in the orientation is fixed