3D line plane intersection, with simple plane

i have two points in 3D space which have X-coordinates with different signum. so one of them lies definitely on one side of the X-plane and one on the other.

now i want to find the int开发者_JAVA技巧ersection of this plane and the line made up by the two points in the most simple and optimized way.

i know how to do general line plane intersection, but since in this case the plane is just the x-plane, i think there should be some shortcuts i can take.

thanks!

Connect the two points and get the equation of line using two-point form (the 3D generalization is simple).

Then solve the equation for x = 0.

After you've got the solutions, translate them into your programming language.

P1 = (x1,y1,z1)
P2 = (x2,y2,z2)

k1 = -x2/(x1-x2)
k2 = 1-k1

Intersection = k1*P1 + k2*P2    or:
Ix = 0              - we know this one
Iy = k1*y1 + k2*y2
Iz = k1*z1 + k2*z2

I'm assuming P1 is to the right and P2 to the left. It may work with them reversed.

Try this I am still calculating :) improving... Let me know if it works.

A = (x1,y1,z1)
B = (x2,y2,z2)
C = (x,y,z)

C will divide line joining A and B in ratio x1/x2.

So by similarity (y,z) will also divide line joining (y1,z1) and (y2,z2) in the same ratio.

As the point C lies in Y-Z plane

x = 0 

by Section Formula

y = (r*y2 + y1) / (r+1)

z = (r*z2 + z1) / (r+1)

where r = |x1| / |x2|

Simple example:

Let A = (1,2,2) and B = (-2,2,2) then C should clearly be (0,2,2).

x = 0
r = 1 / 2 = 0.5
y = (0.5*2 + 2)/(0.5+1) = 2
z = (0.5*2 + 2)/(0.5+1) = 2

CODE C#:

 public class Point
    {
        public double x { get; set; }
        public double y { get; set; }
        public double z { get; set; }

        public Point(double X, double Y, double Z)
        {
            x = X;
            y = Y;
            z = Z;
        }

        public override string ToString()
        {
            return string.Format("({0},{1},{2})",x,y,z);
        }
    }

    public class Program
    {
        public static void Main()
        {
            Point a = new Point(-10, 0, 10);
            Point b = new Point(10, 0, 0);

            Console.WriteLine(GetIntersectionWithYZ(a,b));
        }

        public static Point GetIntersectionWithYZ(Point A, Point B)
        {
            double r = - A.x / B.x;

            double y = (r * B.y + A.y) / (r + 1);
            double z = (r * B.z + A.z) / (r + 1);

            return new Point(0, y, z);
        }
    }

This question is old but since there is such a much more convenient solution I figured it might help someone.

The general case is very nearly exactly as fast in practice if correctly implemented.

Plane and line intersections are quite elegant when expressed in homogeneous coordinates but lets assume you just want the solution:

There is a vector 4x1 p which describes the plane such that p^Tx =0 for any homogeneous point on the plane. Next compute the plucker coordinates for the line L=ab^T - ba^T where a = {point_1; 1}, b={point_2;1}, both 4x1 on the line compute: x=Lp = {x0,x1,x2,x3} x_intersect=({x0,x1,x2}/x3)

For higher performance use of expressions templates will allow the compiler to collapse the solution to the minimal case.