Point that lies on the angle bisector.

Hey all. This is my problem, I hope somebody is gonna solve it. :p Let's suppose to work in the three dimensional space. We have three different point: 'P1', 'P2' and 'P3'. For example ..

    % P1 = [0 10000 0];
    % P2 = [0 0 1];
    % P3 = [100 0 0];

Of course, we can obtain the angle of the triangle given by the three points. Let's focus on the angle of the endpoint 'P1':

    % seg2 = [(P2(1)-P1(1)),(P2(2)-P1(2)),(P2(3)-P1(3))];
    % seg3 = [(P3(1)-P1(1)),(P3(2)-P1(2)),(P3(3)-P1(3))];
    % u2 = seg2/norm(seg2);
    % u3 = seg3/norm(seg3);
    % Ang = acos(dot(u2,u3));

What I need is to get the coordinates of the point that lies on the angle bisector. We want that the distance between this point and 'P1' is '(R2+R3)/2' where ..

    % R2 = norm(P2-P1);
    % R3 = norm(P3-P1);

I don't know how to get this coordinates. I mean, I can do it in the two dimensional space .. but not in three dim. Anybody? :D

[Roger Stafford]

By the way,

 Ang = atan2(norm(cross(seg2,seg3)),dot(seg2,seg3));

is more accurate than acos(dot(u2,u3)) for seg2 and seg3 which are nearly parallel or nearly in opposite directions.

Answer 1

[Emiliano]

Nobody ..? :p

Answer 2

[Roger Stafford]

 R2 = norm(P2-P1);
 R3 = norm(P3-P1);
 T = (P2-P1)/R2+(P3-P1)/R3;
 P0 = P1+T/norm(T)*(R2+R3)/2; % <-- P0 is the desired point

This works unless P2-P1 and P3-P1 point in opposite directions, in which case there is no unique solution.