How do I make a vector exactly reach a line?

Question

Let's say a vector starts at Point P and points in the direction of v. How do I make sure that this vector exactly reaches the line? Its end should be exactly on the line, and not cross it.

Answer 1

You should calculate the intersection point (IP) between the two lines.

      (X - Ax)     (Y - Ay)
EQ1 = --------  =  --------   =>   (X-Ax) * (By-Ay) - (Y - Ay) * (Bx - Ax) = 0;
      (Bx - Ax)    (By - Ay)

      (X - Px)     (Y - Py)
EQ2 = --------  =  --------   =>   (X-Px) * (Vx) - (Y - Py) * (Vy) = 0;
         (Vx)        (Vy)

Solve the two equations and you get IP (X,Y).

Maybe you'd need to check that IP is between A and B... before continue...

Then you build a new line with this equation:

  (P + V) = P + t*(IP-P)   => t = V / (IP - P) 

   t = (IP.X-P.X != 0) 
        ? (V.X / (IP.X - P.X)) 
        : (V.Y / (IP.Y - P.Y));

  if t>1 then (P+V) has overpassed the line

EDIT:

Remember:

1) You have to calculate IP. the intersection point
2) You should check that IP is between A and B
3) You had the vector V, and now have the vector (IP-P),
4) You can choose betwwen two options:
   a) Now you can compare the vector lenghts to now if V is greater that (IP-P)
   b) Calculate t as I described before...
           if t==0 => V=(0,0);  
           if t=1 => V=(IP-P)
           if (t>0 && t<1) length(V) < length(IP-P)
           if (t>1) length(V) > length(IP-P)

Answer 2

You should:

1. Find an (X, Y) point of intersection of your line and a support line for the vector.
2. Extend your vector by the distance from its end to than point you've found.

Example:

1. Find the (X, Y) point: solve the

   (X - Ax) * (By - Ay) - (Y - Ay) * (Bx - Ax) = 0

   and

   (X - Px) * Vx - (Y - Py) * Vy = 0

   equations.

2. Your new vector will be Your old vector (both components) plus YVN * Dist;

   • where YVN is Your old vector normalized (one-unit length);
   • where Dist is distance between N and V;
   • where N is your newfoundland point.