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3D Figures and Intersections


Date: 03/04/99 at 21:08:18
From: Matt
Subject: 3D Figures and Intersections

I have been working with some 3D problems lately, mainly in the context 
of projecting 3D objects into a 2D plane.

One particular problem I ran across is determining whether a line and 
plane intersect and where they will intersect, given a line 
perpendicular to the plane (one point is on the plane, another is out 
in space), and 2 points P and Q out in space that determine a line.

I have already browsed a bit through your archives (I found the 
information on 4D figures very interesting, and I found a lot of useful 
information on dot products. I think the problem works something like 
this:

Get the dot product of the vector describing the line between P and Q 
(the change in x, y, and z) and the vector describing the line normal 
to the plane. If the dot product is 0 (perpendicular) then the line is 
definitely parallel to the plane, otherwise the line intersects the 
plane somewhere.

I am not 100% sure what I have done so far is correct. The other thing 
I am wondering is, once I have this, how would I obtain the point of 
intersection between the line and the plane?


Date: 03/06/99 at 14:35:45
From: Doctor Peterson
Subject: Re: 3D Figures and Intersections

What you have done so far is fine. We can just take it a little further 
to find the intersection.

Let us say the plane is defined by a point A in the plane and a normal 
vector N, and the line is defined by point P on the line and vector V 
(= Q - P) in the direction of the line. Then we can write vector 
equations for a point X on the plane and on the line (using "*" for 
the dot product):

    plane: N * (X - A) = 0
    line:  X - P = tV      for real t

                ^N         /line
                |         /
               E+--------+Q
                |       /V
        A      D|      /
    ----+-------+-----+-------plane
         \      |    / X
          \     |   /
           \   C+--+P
            \   | /
             \  |
              \ |
               \|
                +O

One way to approach this is to think in terms of projections. The 
component of vector V (PQ) in the direction of N (that is, the length 
of the projection of V on N, CE) is given by

    V * N
    -----
     |N|

The point of intersection is a point X on the line where the component 
CD of PX on N the same as the projection of PA on N:

    (X-P)*N   (A-P)*N
    ------- = -------
      |N|       |N|

But since X-P = tV, this just means that

    tV*N = (A-P)*N

which we can solve to get

        (A-P)*N
    t = -------
          V*N

So the point of intersection is

            (A-P)*N
    X = P + ------- V
              V*N

This uses your two vectors V and N, and the two points A and P; it is 
not as complex as it looks.

Notice what happens if V*N = 0: there will be no point of intersection, 
because the line is parallel to the plane, as you said.

There is a lot to learn about vectors, and they are extremely useful-- 
if only because they let me write equations like this without ever 
mentioning individual coordinates.  Have fun learning!

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Higher-Dimensional Geometry
High School Linear Algebra

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