   Planes and intersections
  I have a few questions relating to planes (the geometrical ones). 1. What does ''d'' represent in the equation of a plane ''ax + by + cz = d''. I keep reading it''s the distance from the origin of the co-ordinate system to the the normal of the plane but how can this be as the normal of a plane is not actually a point on the plane?I know I''m probably missing something simple here so hopefully someone can enlighten me.2. What would the equations of the three planes that cross the x, y, and z axes at 0,0,0 be?   I.e the plane that contains the x axis and runs perpendicular/ parallel to itwould have the equation "?x + ?y + ?z =?"3. If an object had a position of <0,-10,0> and a veloctiy of <1,2,0> it would obviously cross the intersect the y-axis plane at <5,0,0>.How could this be shown mathematically  using the equation of a plane and the objects position and velocity vectors?    POI<0,0,0>------<5,0,0>------------------  |  |  |  |  |  |  |  |  |  O   <-object <0,-10,0>Thanks. 
  1) d is a measure of how far the plane is from the origin.  The location of the normal of the plane is not involved in this distance (directly), as the normal is not located at any specific place... the normal is a vector perpendicular to the plane, which can be located anywhere.  The normal is often shown or thought of as starting somewhere on the plane, but it''s a vector, and doesn''t really have a ''start'' position...  (If it did, it would be a ray or line)Anyway, consider:ax + by + cz = dYou can rearrange this to determine how far the plane is from the origin on various axes.  For example, set x = z = 0by = dy = d/bThis tells you how far from the origin the plane meets the y-axis.  The same can be done for the other axes, or any vector that''s not parallel to the plane.  It''s often easiest to think of this as indicating how far ''above'' the origin the plane is located, however.2) x = 0, y = 0, z = 0more generally, set d = 0ax + by + cz = 0Now any real a, b, c will work.  Above used 1, 0, 0; 0, 1, 0; and 0, 0, 1 respectively3) let P = (0,-10,0) -> point on the linelet V = (1, 2, 0)   -> vector parallel to direction of motion/line"y-axis plane" is ambiguous.  Do you mean plane with normal parallel to the y-axis?  This is commonly the "xz plane".  It has equation y = 0.You can parameterize the motion like soS(t) = P + t*Vwhich expands to(X(t), Y(t), z(t)) = (0, -10, 0) + (t, 2*t, 0)orX(t) = tY(t) = -10 + 2*tZ(t) = 0so, solve for Y(t) = 0  (which you want for intersection with th eplane y = 0)0 = -10 + 2*tt=5thus, you know X(t) = 5so, the intersectin point is(t, -10 + 2*t, 0) = (5, 0, 0) ;  Thanks for the answers, they were really helpful.