Question

I have one line segment formed by two vectors, let’s say v1 and v2, a vector v3 and an angle a. How do I write a method in Java (I’m also using Apache Commons Math to represent a vector) which gives me a vector v4, so that the line segments v1-v2 and v3-v4 are at angle a? There are infinite v4 elements, it would even be better if I could give a size to that method so that the line segment v3-v4 has that size. (all in 2d space, angle can be radians or degrees, doesn’t matter)

Edit: as promised I have included an image of the problem I’m trying to solve. I have a line segment defined by 2 vectors (the line is a bit longer but that doesn’t matter), an angle, and a third point. I need to draw the second line which intersects the first one at angle a. Since all lines in Javafx (which I’m using here) are drawn by defining two points, I needed to find the red point (or any of the possible ones).

Edit: Using Ali’s answer I got the following method which does what I need:

public Pair<Vector2D, Vector2D> calculateFourthPoint(Vector2D v1, Vector2D v2, Vector2D v3, double angleInDegrees) {
    Vector2D r = v1.subtract(v2);
    double rx = r.getX();
    double ry = r.getY();
    double angle = toRadians(angleInDegrees);

    double a = pow(rx, 2) + pow(ry, 2);
    double b = 2 * sqrt(pow(rx, 2) + pow(ry, 2)) * cos(angle) * rx;
    double c = pow(rx, 2) * pow(cos(angle), 2) + pow(ry, 2) * pow(cos(angle), 2) - pow(ry, 2);
    double discriminant = sqrt(pow(b, 2) - (4 * a * c));

    double sx1 = (-b + discriminant) / (2 * a);
    double sx2 = (-b - discriminant) / (2 * a);

    double sy1 = sqrt(1 - pow(sx1, 2));
    double sy2 = sqrt(1 - pow(sx2, 2));

    Vector2D s1 = new Vector2D(sx1, sy1);
    Vector2D s2 = new Vector2D(sx2, sy2);

    Vector2D v4_1 = v3.subtract(s1);
    Vector2D v4_2 = v3.subtract(s2);

    return new Pair<Vector2D, Vector2D>(v4_1, v4_2);
}

Answer 1

I don’t know Apache Commons Math so I am writing in pseudo code. Let vx and vy denote the x and y components of vector v, respectively.

Let r=v1-v2 and s=v3-v4. You have 2 unknowns (namely sx and sy; and v4=v3-s) so you need 2 equations. These should be:

dot_product(r,s)=length(r)*cos a // forces the desired angle

dot_product(s,s)=1 // just sets the length of s to 1

To spell it out, the above equations are:

(1)    rx*sx + ry*sy = sqrt(rx^2+ry^2)*cos a

(2)    sx^2 + sy^2 = 1

The first equation is linear in both sx and sy. Let’s eliminate sy from the first equation (assuming that ry is not zero)

 sy = (1/ry)*(sqrt(rx^2+ry^2)*cos a - rx*sx)

and substitute this sy into the second equation. You get a quadratic equation in sy (I don’t want to write it here because it is complicated) and that has 2 solutions. You get the corresponding sx by substituting the sy values into (assuming rx is not zero):

 sx = (1/rx)*(sqrt(rx^2+ry^2)*cos a - ry*sy).

Finally, v4=v3-s. You get 2 solutions for v4, one for each of the solutions to the quadratic equation. (Degenerate cases, such as r being a null vector, are ignored in my answer.)