the projected distances between the line and edge A

I have rectangular 3D surface with coordinates of 4 sides A=[1.6 0.8441 1.53]; B=[1.6 2.5441 1.53]; C=[1.6 2.5441 1.23]; D=[1.6 0.8441 1.23];

and ray passing throug it with ends p1= [-8.4153 6.5982 2.6492] , p2=[ 2.5736 1.6997 1.3800].

The line is intersecting the surface at point I=[ 1.6000 1.6977 1.4930]

i want to find projected distances between the line and edge A.

for that i have to find projection point of line. (like gram smith )

how can i find projection of line on surface and then calculate its distance to edge A

[Torsten]

What is edge A ? Or do you mean point A ?

[Mariam Ali]

Side of surface is edge A and so point A is edge A

[Torsten]

??

An edge is a line connecting two points. A is a point, thus it cannot be an edge.

Do we only have problems communicating in English or do you mean something else with "edge" ?

[Mariam Ali]

I am sorry i mean point A. ABCD makes a rectangular surface.

So i need Distance between projection of line on surface and point A

Answer 1

[Torsten]

You will have to check for correctness:

% Project p1 on plane ABC (gives point p1-proj)
Mat = [(B-A)*(B-A).' , (B-A)*(C-A).';(B-A)*(C-A).' ,(C-A)*(C-A).'];
rhs = [(B-A)*(p1-A).';(C-A)*(p1-A).'];
coord = Mat\rhs;
p1_proj = A + coord(1)*(B-A) + coord(2)*(C-A);

% Project A on line connecting p1_proj and I
P = p1_proj + ( (I-p1_proj).'\(A-p1_proj).' )*(I-p1_proj);

% Calculate distance
distance = norm(P-A)

[Mariam Ali]

can you tell or link the basic maths that you followed so i can check ?

[Torsten]

Solve

minimize: (p1-(A+lambda*(B-A)+mu*(C-A)))*(p1-(A+lambda*(B-A)+mu*(C-A)))'

for lambda and mu. Call the solution lambda_sol and mu_sol.

Then

p1_proj = A+lambda_sol*(B-A)+mu_sol*(C-A) 

is the projection of p1 on the plane ABCD.

We know that I is in the plane ABCD. Thus the line through p1_proj and I is the projection of the ray

through p1 and p2 on ABCD. To find the point P on this line with minimum distance to A,

solve

minimize: (A-(I+nu*(p1_proj-I)))*(A-(I+nu*(p1_proj-I)))'

for nu. Call the solution nu_sol.

Then

P = I+nu_sol*(p1_proj-I) 

is the point on the line through p1_proj and I with minimum distance to A.

You know how to solve the two unconstrained minimization problems ?

Building partial derivatives with respect to the unknowns, setting the partial derivatives to 0 and solving for the unknowns ?

[Torsten]

In your special case, everything is of course much simpler since the points A,B,C and D share the same x-coordinate.

This means that p1 and p2 projected on the plane spanned by A,B,C and D are

p1_proj = [1.6 6.5982 2.6492]

p2_proj = [1.6 1.6997 1.3800]

Now you can work in two instead of three dimensions and calculate the distance of A' = [0.8441 1.53] to the line passing through p1_proj' = [6.5982 2.6492] and p2_proj' = [1.6997 1.3800]. This gives you the distance of A to the projected ray through p1 and p2.

I guess that you should have a formula for this in your lecture notes.