Question

I don’t know how to calculate the angle bisector of two vectors (that defined by 3 points) in 3D.
example:
my points are:

P1=[0 0 0];
p2=[26.94,-6.68,42.76];
P3=[7.2804  -11.1231   46.6817];

p2 is the start point of the angle bisector .

p1=[0 0 0];
p2=[26.94,-6.68,42.76];
p3=[7.2804  -11.1231   46.6817];


V1=[(p1(1)-p2(1)) (p1(2)-p2(2)) (p1(3)-p2(3))];
V2=[(p2(1)-p3(1)) (p2(2)-p3(2)) (p2(3)-p3(3))];

V1=V1/norm(V1);
V2=V2/norm(V2);

Bisector=V1+V2;
figure
hold on

plot3([p3(3),p2(3)],[p3(2),p2(2)],[p3(1),p2(1)],'Color','r','LineWidth',2)
plot3([p1(3),p2(3)],[p1(2),p2(2)],[p1(1),p2(1)],'Color','r','LineWidth',2)

plot3([p2(3),Bisector(3)],[p2(2),Bisector(2)],[p2(1),Bisector(1)],'Color','y','LineWidth',2)

Answer 1

You have some errors in your code above. This should fix it.

V1=[(p1(1)-p2(1)) (p1(2)-p2(2)) (p1(3)-p2(3))];
V2=[(p3(1)-p2(1)) (p3(2)-p2(2)) (p3(3)-p2(3))];

V1=V1/norm(V1);
V2=V2/norm(V2);

Bisector=V1+V2;
figure
hold on

plot3([p3(3),p2(3)],[p3(2),p2(2)],[p3(1),p2(1)],'Color','r','LineWidth',2)
plot3([p1(3),p2(3)],[p1(2),p2(2)],[p1(1),p2(1)],'Color','r','LineWidth',2)

plot3([p2(3),Bisector(3)+p2(3)],[p2(2),Bisector(2)+p2(2)],[p2(1),Bisector(1)+p2(1)],'Color','y','LineWidth',2)

Note the two lines in boldface.

The first issue is essentially a sign issue. You need to translate your vectors so p2 is at the origin. So, to get V1 and V2, you subtract p2 from p1 and p3 respectively. You do this when computing V1 but your are computing V2=p2-p3 rather than V2=p3-p2.

The second issue is that your vector Bisector is translated relative to the an origin placed at p2. Thus, you need to translate it back to the original coordinate system. This is done simply by adding p2 to your Bisector vector for display purposes. This is what is now done in the last line.