Calculating the normal of a (3D) line

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Kees de Kapper
Kees de Kapper 2018-6-1
评论： Kees de Kapper 2018-6-3
Hi all,
I would like to calculate the normal vector of a 3D line. This normal vector should cross the origin (0,0,0), otherwise the normal vector is undefined.
How can I do this easily?
Many thanks in advance.
/Kees
  1 个评论
Kaninika Pant
Kaninika Pant 2018-6-1
How is this 3D line defined? ie. By a point and a direction vector? Or 2 points?

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Kaninika Pant
Kaninika Pant 2018-6-1
Note that the normal vector is only a direction (cannot pass through a point). I think what you are looking for is a line passing through the origin with a direction vector normal to the previous line.
Also, given a line in any form it is always possible to find the direction vector and a point on the line. So I will answer assuming that you have found those Consider a point (x,y,z) on this normal line. Let the direction vector of the original 3D line is (l,m,n) and a point on the line be (x1,y1,z1). To find this line you need to solve the following:
1. (x-0)l+(y-0)m+(z-0)n=0 (ie. direction vectors of both lines must be perpendicular) 2. (x-x1)/l + (y-y1)/m + (z-z1)/n = k(ie. (x,y,z) must lie on the original 3D line) So, x=l*k + x1; y=m*k + y1; z=n*k + z1
Thus you only have one variable k. Put this in the first equation and simplify to get: k=-(l*x1+m*y1+n*z1)/(l^2+m^2+n^2);
After finding k plug it back to get x,y,z. Note that (x,y,z) is a point on your normal line and also the direction vector. Now you can form the equation of your normal line.
  2 个评论
Kees de Kapper
Kees de Kapper 2018-6-2
Dear Kaninika,
thank you very much for your help and extensive answer. It sounds to be the solution for my problem. I'll try to implement this.
All the best, Kees
Kees de Kapper
Kees de Kapper 2018-6-3
C = [1,2,3;4,4,4];
NC=(C(2,:)-C(1,:))./norm(C(2,:)-C(1,:));
k = -(NC(1)*C(1,1)+NC(2)*C(1,2)+NC(3)*C(1,3))/(NC(1)^2+NC(2)^2+NC(3)^2);
P = [0,0,0; NC(1)*k + C(1,1), NC(2)*k + C(1,2), NC(3)*k + C(1,3)];
C2 = [C(1,:) - 20*NC; C(1,:) + 20*NC];
P2 = [[0,0,0] - 20*P(2,:); [0,0,0] + 20*P(2,:)];
figure(10); plot3(C2(:,1), C2(:,2), C2(:,3), 'k', C(:,1), C(:,2), C(:,3), 'r', P2(:,1), P2(:,2), P2(:,3), 'k', P(:,1), P(:,2), P(:,3), 'b'); axis equal;
I think it works! Great. Thanks again.

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