# How do you find a point on a line bisecting an angle in three-dimensional space?

Given the x, y, z coordinates of three points P1, P2, P3 with the angle between them being $$\angle$$P1P2P3, how do you find a point, say at a distance of 1 from P2, on the line that bisects the angle?

I know from the angle bisector theorem that the point must be equidistant from the the vector (P3-P2) and (P1-P2), but I can't seem to figure out how to find such a point in 3-space.

• Forget the distance for a moment. You know that the sides of the angle are directed along the mentioned vectors (P3-P2) and (P1-P2). Now what about the bisector? Oct 4, 2020 at 22:15

The sum of the two unit vectors $${\bf v} =\vec{P_2P_1}/ |P_2P_1|\: + \;\vec{P_2P_3}/ |P_2P_3|$$ is a vector lying on the bisector of the angle between them.
Make $$\bf v$$ unitary, multiply it by the distance $$d$$ you want from $$P_2$$ and add to $$\vec{OP_2}$$.
Assume $$P_1,P_2, \text{ and } P_3$$ aren't collinear. Put $$v_1=P_1-P_2$$ and $$v_2=P_3-P_2$$. You need to find $$v\in\text{span}\{v_1,v_2\}$$ satisfying the relationship $$\frac{v_1\cdot v}{||v_1||}=\frac{v_2 \cdot v}{||v_2||}$$. If you write $$v=c_1v_1+c_2v_2$$ you'll immediately recognize that $$\frac{v_1\cdot v}{||v_1||}=\frac{v_2 \cdot v}{||v_2||} \iff c_2=c_1\Bigg(\frac{||v_1||-\frac{v_1 \cdot v_2}{||v_2||}}{||v_2||-\frac{v_1 \cdot v_2}{||v_1||}}\Bigg)$$ So if we assign $$c_1=1$$ and $$c_2=\frac{||v_1||-\frac{v_1 \cdot v_2}{||v_2||}}{||v_2||-\frac{v_1 \cdot v_2}{||v_1||}}$$ we see $$v=v_1+\Bigg(\frac{||v_1||-\frac{v_1 \cdot v_2}{||v_2||}}{||v_2||-\frac{v_1 \cdot v_2}{||v_1||}}\Bigg)v_2$$ and the line $$l(t)=P_2+tv$$ bisects $$\angle{P_1P_2P_3}$$. Notice how $$l\Big(\frac{1}{||v||}\Big)$$ is a point on this bisector that lands one unit away from $$P_2$$. You can see this by clicking here.