   Ray to plane intersection
  Hi all. I am trying to solve intersection between a ray and plane but not getting through.I got plane at  = c = ( 2,0 ).plane normal = n =  ( -1, 0 )ray start po  = 0.5, -1.5ray end pe = 2.5 , -0.5;dir = pe - po = 2.0, 1.0dirUnit = 0.63, 0.316.plane distance from origin = d = 2.0calculating for  tt = d - ( p0.n ) / dirUnit.nt = -3.968p0 + ( dir * t ) => intersection point = (-2.0, -2.75)which does'nt sounds right. Expected intersection point = (2.0 , -0.75)Just to be more clear I have uploaded the graph pic at http://i44.tinypic.com/x5nkzm.jpgany help would be appreciated.RegardsKazz   
  Ok I'm not sure of your method after you get the distance from the origin...how I did it was...get p0c = p0 - c = 1.5, 1.5p0c.n = (1.5, 1.5).(-1,0) = -1.5pec = pe - c = (-0.5, 0.5)pec.n = (-0.5, 0.5).(-1,0) = 0.5these two scalars define the distance from p0, pe to the plane.  Get the absolute sum (2.0), and divide the absolute value of the distance for p0 by it, to get the 'time' value t of the point on the line that lies on the plane.t = 1.5/2.0 = 0.75p = (1-t) p0 + t pep = (1 - 0.75) (0.5, -1.5) + (0.75) (2.5, -0.5)p = 0.25 (0.5, -1.5) + (0.75 (2.5, -0.5)...p = (2, -0.75);  Hi. Pardon me for any silly misunderstanding, but looking at the solution , you have used ray end point. What if we only have ray start point and normalized direction and no end point. Also  I think I mis calculated value for d.I am using the equation for plane "p.n = d", where p is any point on plane, n is normal and d is distance of plane from origin. So for my case its C which is c=[2,0], so I took d as magnitude of cd = 2.Seems quite straight forward maths, but some how I am not getting though this one.Please helpRegardsKazz