   Perpendicular line
  Say I have a point and a plane. How could I determine the point on the plane that, when paired with my first point, creates a line segment perpendicular to the plane? 
  You need to visualize the plane as a line with a slope to make it simpler to work with. If you're talking about a totally flat plane then that would work of course.The plane has a slope of m  .In order for a line to be perpendicular to that line or plane it needs to have a alope of -(1/m) for this condition to yield true. Testing that and dealing with the result, simple as that.EDIT: If you are looking for a 3d solution then mine obviously will have no bearing and the post below should put you on track.   [edited by - nervo on August 3, 2003 10:54:18 AM];  If you have the plane defined asAx+By+Cz +D = 0where[A,B,C] is the normal to the planethe line that passes through your point having the coordinates(x0,y0,z0)is defined by three equations x = x0 + Aty = y0 + Btz = z0 + Ctwhere t is an unknown.Hope it helpsCiehoo ;  plane normal is N, and has parameter D such as D = (PointOnPlane * N)point Q on Plane is the projection of P on the plane, sothe distance of point P to the plane isfloat dist = (P * Plane.N) - Plane.Dthen Q is Q = P - Plane.N * dist;  I assume you have an equation of the plane and a normalized normal to the plane.  If not, get them.The known off-plane point is PThe normal to the plane is N = (A, B, C)  (in 2D, C = 0)The plane equation is A*x + B*y + C*z = DYou can make a vector V from P to some point Q on the plane by arbitrarily picking a point on the plane and subtracting P''s components from Q''s components.  V = Q - PFind the vector projection J of V onto N.  Since N is normalized, J = (V.N)*N where . is dot product and * is scalar mulitplication with a vector (N).P + J now gives you the location of the point K on the plane that is the closest point on the plane to the point P, and J is a vector projection onto N, so it''s also normal to the plane.  K = P + J ;  Thanks for the replies, I got it now. 