How to find the closest point to three vector lines? Regarding the image below, I know the angles $A$ and $B$ for each individual, and their positions in longitude and latitude (assuming height of person $z =0$). 
Am I correct in thinking that for any position in the z plane it would be $z = x\tan(B)$ and in the $y$ plane $y = x\tan(A)$? 
If so this would mean I could form a  $3$ dimensional vector straight line from each person. 
The problem is I am highly doubting all three straight lines will intersect, so my question is how do I calculate the closest point to all three vector lines mathematically? 
Cheers

 A: If you can convert the line information that you currently have into $(x,y,z)$-coordinates then you can calculate the closest point between two straight, skew lines using the following:
Line 1:
\begin{equation}
\vec{r}_{1} = \vec{a}+\lambda\vec{b}
\end{equation}
Line 2:
\begin{equation}
\vec{r}_{2} = \vec{c}+\mu\vec{d}
\end{equation}
where $\vec{a}$, $\vec{c}$ are the position vectors of the line equations, and $\vec{b}$, $\vec{d}$ are the direction vectors of the line equations.
The linking line passing through the shortest distance between $\vec{r}_{1}$ and $\vec{r}_{2}$ is perpendicular to both $\vec{r}_{1}$ and $\vec{r}_{2}$:
\begin{equation}
\vec{n} = \vec{b}\times\vec{d}.
\end{equation}
The component of $\vec{a}-\vec{c}$ (or $\vec{c}-\vec{a}$) in the direction of $\vec{n}$ is then needed:
\begin{equation}
\rm{Distance} = |(\vec{a}-\vec{c})\cdot\hat{n}|
\end{equation}
or
\begin{equation}
\rm{Distance} = |(\vec{c}-\vec{a})\cdot\hat{n}|.
\end{equation}
Where $\hat{n} = \vec{n}/|\vec{n}|$.
If you repeat this process to calculate the distance from line 1 to line 2, line 1 to line 3, line 2 to line 3, then you can take the average to find the shortest distance between all three lines.
