Finding coordinates of closest approach

Question

Given two lines $l_1=\mathbf E_1+k\mathbf E'_1$ and $l_2=\mathbf E_2+\mu\mathbf E'_2$ in 3D, there exists a shortest distance between the two lines. How does one find the coordinates of the points $P$ on $l_1$ and $Q$ on $l_2$, such that $P$ and $Q$ are the points where the distance between the two lines is the shortest?

Answer 1

$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$

${\bf d} = {\bf l}_{1} - {\bf l}_{2}$ $$ {\bf d}^{2} = \pars{{\bf E}_{1} + k{\bf E}_{1}' - {\bf E}_{2} - \mu{\bf E}_{2}'}^{2} = \pars{{\bf E} + k{\bf E}_{1}' - \mu{\bf E}_{2}'}^{2}\,, \quad {\bf E} \equiv {\bf E}_{1} - {\bf E}_{2} $$

$$ {\bf d}^{2} = {\bf E}^{2} + {\bf E}_{1}'^{2}k^{2} + {\bf E}_{2}'^{2}\mu^{2} + 2{\bf E}\cdot{\bf E}_{1}'k - 2{\bf E}\cdot{\bf E}_{2}'\mu - 2{\bf E}_{1}'\cdot{\bf E}_{2}'k\mu $$

Minimize ${\bf d}^{2}$ respect of $k$ and $\mu$: \begin{align} \partiald{{\bf d}^{2}}{k} = 0 &\imp\quad 2{\bf E}_{1}'^{2}k + 2{\bf E}\cdot{\bf E}_{1}' - 2{\bf E}_{1}'\cdot{\bf E}_{2}'\mu = 0 \\[3mm] \partiald{{\bf d}^{2}}{\mu} = 0 &\imp\quad 2{\bf E}_{2}'^{2}\mu - 2{\bf E}\cdot{\bf E}_{2}' - 2{\bf E}_{1}'\cdot{\bf E}_{2}'k = 0 \end{align} Now, we have two equations for $k$ and $\mu$: $$ \left\lbrace% \begin{array}{rcrcl} {\bf E}_{1}'^{2}\,k & - & {\bf E}_{1}'\cdot{\bf E}_{2}'\,\mu & = & -{\bf E}\cdot{\bf E}_{1}' \\ {\bf E}_{1}'\cdot{\bf E}_{2}'\,k & - & {\bf E}_{2}'^{2}\,\mu & = & -{\bf E}\cdot{\bf E}_{2}' \end{array}\right. $$

Find the values of $k$ and $\mu$. They determine the points we are looking for on the lines ${\bf l}_{1}$ and ${\bf l}_{2}$, respectively.

Answer 2

Since we can take any value on two parameters K and μ,

let k=0 then we have a point on the line, which has coordinate same as $\overrightarrow{E1}$. (note: you can let k be other numbers but zero makes question easier.)

Now, this question becomes finding the shortest distance from a point to a line.

Let's call the point we just got point P, the shortest distance would be the length of the linesegment formed by point P and a point on the other line, we then assume the point lies on line2 is point Q.

vector: $\overrightarrow {PQ}$= $\overrightarrow {OQ}$-$\overrightarrow {OP}$ = ($\overrightarrow{E2}$-$\overrightarrow{E1}$)+μ$\overrightarrow{E'2}$ (note: I used bracket for the sake of calculating and to emphasize that E2&E1 doesn't have parameter) will be the shortest distance when it's perpendicular to the parallel vector of Line2. so the dot product of $\overrightarrow {AX}$ and E'2 will be zero

$\overrightarrow {PQ}$. $\overrightarrow{E'2}$=0

[($\overrightarrow{E2}$-$\overrightarrow{E1}$)+μ$\overrightarrow{E′2}$].$\overrightarrow{E}$=0

Then you can solve μ as a constant and bring it back into $\overrightarrow{PQ}$ and find the magnitude of $\overrightarrow{PQ}$ which is the shortest distance you want.