CODEWITHSUNDEEP
javamathtrigonometrylwjgl
CyanPrime
CyanPrime

Reputation: 5193

getting the distance of a point in between two lines

I have 4 points, three of the points make two lines like a V or a < or a >, you get the idea, now I got a point that's in that cone (the V) and I can get the dist from the top to the bottom left, and bottom right, but not where the bottom pos is.

Maybe this will help. https://dl.dropbox.com/u/28109593/examplevec.png

And I have code to go along with this problem:

public float GetDist(Vector3f one, Vector3f two, Vector3f three, Vector3f p){


          Vector3f one_to_point = new Vector3f(0,0,0);
          Vector3f.sub(p,one,one_to_point);            //Storing vector A->P



          Vector3f one_to_two = new Vector3f(0,0,0);
          Vector3f.sub(two, one, one_to_two);            //Storing vector A->B

          Vector3f one_to_three = new Vector3f(0,0,0);
          Vector3f.sub(three, one, one_to_three);            //Storing vector A->C

          float q1 =  Vector3f.dot(one_to_point,  one_to_two) / one_to_two.lengthSquared();            // The normalized "distance" from a to b
          float q2 =  Vector3f.dot(one_to_point,  one_to_three) / one_to_three.lengthSquared();            // The normalized "distance" from a to c

Now I already know that the pos vector is in the cone, so what do I need to do to get the pos as shown by the green circle's pos in the image?

Upvotes: 2

Views: 807

Answers (2)

mathematician1975
mathematician1975

Reputation: 21351

You have 4 points in the plane : A,B,C and D (where D is what you label as pos in your diagram). A unique straight line can be drawn between any 2 distinct points so find the straight line that links A and D and get its equation in the form

    y = m_1 * x + c_1

Do the same for line the points B and C to get 

    y = m_2 * x + c_2

Now you know the 2 lines, you can solve this pair of simultaneous equations to get the point (x,y) that lies at the green circle on your diagram - I shall call this E. Given E, calculate the length of vector BE and divide by the length of vector BC. This value is the value of X that you are looking for in your question.

If you do not know how to find the equation for a line that goes through 2 points look at this link for details http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/zoo/eqline.html

I do not doubt that there is a simpler more elegant way to do this but if you get no other answer here, this approach will serve your purpose.

Upvotes: 2

autonymous
autonymous

Reputation: 303

It's been a while since I've done vector algebra, but let me see if I got this right. You're looking for the green point, which is where a line from A to Pos intersects a line from B to C.

I believe if you knew the ratio of the angle that B-A-Pos forms to the angle that B-A-C forms, that ratio would be the same as the ratio of the distance from B to Green to the distance from B to C. From B, the direction of Green is the same as the direction of C, so the vector that represents Green's location is

VectorGreen = VectorB + (x1 / x2)(VectorC - VectorB) //The vector to B plus a fraction of the vector from B to C

x1 = arccos ( Normalize(VectorP - VectorA ) * Normalize(VectorB - VectorA) )//Angle between A to B and A to Pos

x2 = arccos ( Normalize(VectorB - VectorA ) * Normalize(VectorC - VectorA) ) //Angle between A to B and A to C

Upvotes: 1

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