Given two points and two vectors, find point of intersection [duplicate]

Question

Possible Duplicate:
How do you detect where two line segments intersect?

Given two points a and b plus two vectors v and u I want to find a third point c, which is the point of intersection in the following manner:

vector2 intersection(vector2 a, vector2 v, vector2 b, vector2 u)
{
    float r, s;

    a + r * v = b + s * u;
    r * v - s * u = b - a

    r * v.x - s * u.x = b.x - a.x
    r * v.y - s * u.y = b.y - a.y
}

Is there any other way than using gaussian elimination to solve this system? Or is this the best (or at least an acceptable) way to handle this?

EDIT: Definition of vector2

typedef union vector2
{
    float v[2];
    struct { float x, y; };
} vector2;

a and b are also of type vector2, because the only difference between a point and a vector is in the the way it is transformed by an affine transformation.

Answer 1

It's simple math.

But, first, check that you have intersection. If both vector are parallel you will fail to solve that:

// Edit this line to protect from division by 0 
if (Vy == 0 && Uy == 0) || ((Vy != 0 && Uy != 0 && (Vx/Vy == Ux/Uy)) // => Fail.

Then (I want show the calculation cause they long but):

R = (AxUy - AyUx + ByUx - BxUy) / (VyUx - VxUy)  
S = (Ax - Bx + RVx) / Ux  

Hope that helped you.

Answer 2

Looks like an assignment problem to me. Here is the logic that will help you write the code.

Let us call the first Ray as R0.
Locus of a point on R0 is defined as P:

P = P0 + alpha x V0

For the second ray R1:

P = P1 + beta x V1

Since they should intersect:

P0 + alpha x V0 = P1 + beta x V1

alpha and beta are unknowns and we have two equations in x any y.

Solve for the unknowns and get back the point of intersection.

i.e.,

P0.x + alpha * V0.x = P1.x + beta * V1.x
P0.y + alpha * V0.y = P1.y + beta * V1.y

solve for alpha and beta.

If there is a real positive solution for both alpha and beta, rays intersect.
If there is a real but at least one negative solution for both alpha and beta, extended rays intersect.