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

Question

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.

Answer 1

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 $\vec{OP_2}$.

Answer 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.

Answer 3

I presume, since you have the lines, you also have their direction vectors. Now, we know that the addition of two vectors, by the parallelogram law of addition, is the diagonal of the parallelogram which forms when you take those two vectors as adjacent sides.

Since you’re trying to find the angle bisector, let’s think of a parallelogram whose diagonal bisects the angle between its sides - which happens to be (at least) a rhombus. So, we want to convert that parallelogram pertaining to the direction vectors, into a rhombus. The best way to do that, is to consider unit vectors in the direction of the direction vectors of the lines. (Since, then the adjacent sides of the parallelogram will be equal, making it a rhombus.)

Simply adding the unit vectors will give you the diagonal of the rhombus, and the diagonal’s direction vectors is the direction vector of the angle bisector. If you want the line, all you have to do is find the point of intersection of the two lines in question.