3D Figures and IntersectionsDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/