Calculating the normal of a (3D) line

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

[Kaninika Pant]

How is this 3D line defined? ie. By a point and a direction vector? Or 2 points?

Answer 1

[Kaninika Pant]

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.

[Kees de Kapper]

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]

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.