Finding the vector between 2 vectors (2D)

Question

Lets say for example I have the following 2 vectors:

       *B





*A

The vector I would want would be C

*C       *B





*A

What I'm trying to do is generate square outlines. I use 2d slurp: where v0 would be A and v2 would be B. Right now I use sslerp2D to make round edges but I also want regular square edges hence why I want to modify this.

POINTFLOAT slerp2d( const POINTFLOAT &v0, 
                   const POINTFLOAT &v1, float t )
{
    float dot = (v0.x * v1.x + v0.y * v1.y);

    if( dot < -1.0f ) dot = -1.0f;
    if( dot > 1.0f ) dot = 1.0f;

    float theta_0 = acos( dot );
    float theta = theta_0 * t;

    POINTFLOAT v2;
    v2.x = -v0.y;
    v2.y = v0.x;

    POINTFLOAT result;
    result.x = v0.x * cos(theta) + v2.x * sin(theta);
    result.y = v0.y * cos(theta) + v2.y * sin(theta);

    return result;
}

Thanks

Edit:

I generate outlines like this:

void OGLSHAPE::GenerateLinePoly(std::vector<DOUBLEPOINT> &input, int width)
{
    OutlineVec.clear();
    if(input.size() < 2)
    {
        return;
    }


    if(connected)
    {
        input.push_back(input[0]);
        input.push_back(input[1]);
    }


    float w = width / 2.0f;

    //glBegin(GL_TRIANGLES);
    for( size_t i = 0; i < input.size()-1; ++i )
    {
        POINTFLOAT cur;
        cur.x = input[i].point[0];
        cur.y = input[i].point[1];


        POINTFLOAT nxt;


        nxt.x = input[i+1].point[0];
        nxt.y = input[i+1].point[1];

        POINTFLOAT b;
        b.x = nxt.x - cur.x;
        b.y = nxt.y - cur.y;

        b = normalize(b);



        POINTFLOAT b_perp;
        b_perp.x = -b.y;
        b_perp.y = b.x;


        POINTFLOAT p0;
        POINTFLOAT p1;
        POINTFLOAT p2;
        POINTFLOAT p3;

        p0.x = cur.x + b_perp.x * w;
        p0.y = cur.y + b_perp.y * w;

        p1.x = cur.x - b_perp.x * w;
        p1.y = cur.y - b_perp.y * w;

        p2.x = nxt.x + b_perp.x * w;
        p2.y = nxt.y + b_perp.y * w;

        p3.x = nxt.x - b_perp.x * w;
        p3.y = nxt.y - b_perp.y * w;

        OutlineVec.push_back(p0.x);
        OutlineVec.push_back(p0.y);
        OutlineVec.push_back(p1.x);
        OutlineVec.push_back(p1.y);
        OutlineVec.push_back(p2.x);
        OutlineVec.push_back(p2.y);

        OutlineVec.push_back(p2.x);
        OutlineVec.push_back(p2.y);
        OutlineVec.push_back(p1.x);
        OutlineVec.push_back(p1.y);
        OutlineVec.push_back(p3.x);
        OutlineVec.push_back(p3.y);



        // only do joins when we have a prv
        if( i == 0 ) continue;


        POINTFLOAT prv;
        prv.x = input[i-1].point[0];
        prv.y = input[i-1].point[1];

        POINTFLOAT a;
        a.x = prv.x - cur.x;
        a.y = prv.y - cur.y;

        a = normalize(a);

        POINTFLOAT a_perp;
        a_perp.x = a.y;
        a_perp.y = -a.x;

        float det = a.x * b.y  - b.x * a.y;
        if( det > 0 )
        {
            a_perp.x = -a_perp.x;
            a_perp.y = -a_perp.y;

            b_perp.x = -b_perp.x;
            b_perp.y = -b_perp.y;
        }

        // TODO: do inner miter calculation

        // flip around normals and calculate round join points
        a_perp.x = -a_perp.x;
        a_perp.y = -a_perp.y;

        b_perp.x = -b_perp.x;
        b_perp.y = -b_perp.y;

        size_t num_pts = 4;

        std::vector< POINTFLOAT> round( 1 + num_pts + 1 );
        POINTFLOAT nc;
        nc.x = cur.x + (a_perp.x * w);
        nc.y = cur.y + (a_perp.y * w);

        round.front() = nc;

        nc.x = cur.x + (b_perp.x * w);
        nc.y = cur.y + (b_perp.y * w);

        round.back() = nc;

        for( size_t j = 1; j < num_pts+1; ++j )
        {
            float t = (float)j/(float)(num_pts+1);
            if( det > 0 )
         {
             POINTFLOAT nin;
             nin = slerp2d( b_perp, a_perp, 1.0f-t );
             nin.x *= w;
             nin.y *= w;

             nin.x += cur.x;
             nin.y += cur.y;

             round[j] = nin;
         }
            else
         {
             POINTFLOAT nin;
             nin = slerp2d( a_perp, b_perp, t );
             nin.x *= w;
             nin.y *= w;

             nin.x += cur.x;
             nin.y += cur.y;

             round[j] = nin;
         }
        }

        for( size_t j = 0; j < round.size()-1; ++j )
        {

            OutlineVec.push_back(cur.x);
            OutlineVec.push_back(cur.y);


            if( det > 0 )
         {
             OutlineVec.push_back(round[j + 1].x);
             OutlineVec.push_back(round[j + 1].y);
             OutlineVec.push_back(round[j].x);
             OutlineVec.push_back(round[j].y);
         }
            else
         {

             OutlineVec.push_back(round[j].x);
             OutlineVec.push_back(round[j].y);

             OutlineVec.push_back(round[j + 1].x);
             OutlineVec.push_back(round[j + 1].y);
         }
        }
    }

}

Answer 1

You seem to be working with points, not vectors. If you have two points (X₁, Y₁), (X₂, Y₂), then (X₁, Y₂) and (X₂, Y₁) should both form right triangles with the first two points [Edit: provided, of course, that the first two points don't form a vertical or horizontal line]. For the inputs you gave, one would be the point you've marked C, and other will give the mirror-image triangle on the bottom-right.

Further Editing: I'm not quite sure what you mean about an angle that's not exactly 90 degrees. As for the situation mentioned above where the original line is vertical or horizontal, it's trivial to form an infinite variety of right triangles from one of these: pick an arbitrary length for a second side and extend it at 90 degrees from one of the two points. Your third side (the hypotenuse) will extend from that new point to the other of the two original points. You can also construct an infinite variety of other right triangles with the original pair forming the hypotenuse, if you prefer (though it involves minutely more difficult math -- A² = C²-B², where C is the length of the original line and B is an arbitrary length you select in the range 0..C.