# Connect Two Points in 3D Space with Three Lines

I have two points in 3D space, $$P_1$$ and $$P_4$$.

From either point, I have a line extended ($$L_1$$ intersecting $$P_1$$ and $$L_3$$ intersecting $$P_4$$). I want to join these two lines together with another line, $$L_2$$.

I'm trying to determine the lengths of Lines $$L_1$$, $$L_2$$, and $$L_3$$, but only the direction of lines $$L_1$$ and $$L_3$$ are known and given by the unit vectors $$t_1$$ and $$t_3$$ respectively. The direction of line $$L_2$$, or the unit vector $$t_2$$, is unknown. I'm not all that great at linear algebra, vector geometry, etc. so correct me if I'm wrong here but I'm guessing I still am missing a variable to define this problem? So let's also say that the angle between $$t_1$$ and $$t_2$$ is known as $$\alpha_1$$ OR that the angle between $$t_2$$ and $$t_3$$ is known as $$\alpha_2$$.

Does all of these parameters make this problem solvable?

EDIT: Thinking about this further it looks like the parameters listed above still leave things under-defined (?) This is a somewhat open ended problem and I can play around with what values are known and unknown. But I'm trying to determine the minimum number required. Values that can be considered known or known: $$L_1$$, $$L_2$$, $$L_3$$, $$\alpha_1$$, $$\alpha_2$$

Values that must be considered as "known": $$t_1$$, $$t_3$$

• where is this question from? is it a part of homework in some class? it's really unclear what you are trying to find and what is given. – GuySa Nov 23 '18 at 7:56
• How about $P_1$ and $P_4$, are they known? – Alex Vong Nov 23 '18 at 9:35
• Yes P1 and P4 are known. This is for designing three dimensional trajectories and the way this problem was originally presented to me this was solved through iteration. However I want to try and solve this more or less analytically. – CLH Nov 23 '18 at 14:55
• For "designing" this trajectory the user will always specify the position of the start and end points and the direction of the lines at those points. Then the user will specify some other combination of variables. – CLH Nov 23 '18 at 15:11

Firstly, let $$\vec{p_1}, \vec{p_2}, \vec{p_3}, \vec{p_4}$$ be the position vector of point $$P_1, P_2, P_3, P_4$$ respectively. Notice that $$P_1$$ and $$\vec{t_1}$$ determines line $$L_1$$ by $$\vec{r_1}(s) = \vec{p_1} + s \vec{t_1}$$ where $$s \in \mathbb{R}$$. Similarly, $$P_4$$ and $$\vec{t_3}$$ determines line $$L_3$$ by $$\vec{r_3}(s) = \vec{p_4} + s \vec{t_3}$$ where $$s \in \mathbb{R}$$.
Following the pattern, line $$L_2$$ can be parameterized as $$\vec{r_2}(s) = \vec{p_2} + s \vec{t_2}$$ or $$\vec{r_2}(s) = \vec{p_3} + s \vec{t_2}$$ where $$s \in \mathbb{R}$$, which one to use is up to you.
So we need at most $$2$$ more additional parameters, $$\{P_2, \vec{t_2}\}$$ or $$\{P_3, \vec{t_2}\}$$. Finally, you can draw some pictures to convince yourself that adding only one of the parameters $$P_2, P_3, \vec{t_2}, \alpha_1, \alpha_2$$ doesn't determine $$L_2$$. So $$2$$ is the minimal number of additional parameters required.
EDIT: In the above, we do not consider $$L_1, L_2$$ and $$L_3$$ as parameters, but if we do so, then the minimal number of additional parameters required drops down to $$1$$, because in the argument above, we have already shown that $$L_1$$ and $$L_3$$ are in fact determined, thus we only need $$L_2$$ as an additional parameter to determine all lines.