\documentclass[10pt]{article} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage[svgnames]{xcolor} \def\R{{\mathbb R}} \def\r{\color{Crimson}} \def\g{\color{SeaGreen}} \def\b{\color{Blue}} \def\a{\color{brown}} \def\x{\color{DarkViolet}} \begin{document} Vectors $\vec{KL} \equiv v$ and $\vec{QM} \equiv w$. And~let $\vec{KQ} = \vec l$. Linear spaces (straight lines) generated by~'em~-- $p\vec v$ and $\vec l + q\vec w$, with $p, q \in \R$. The square of~the~distance (generated by~$g_{ij} = \delta_{ij}$; or~whatever: invariants are~invariants): $$ \rho^2(p, q) = (pv_x - l_x - qw_x)^2 + (pv_y - l_y - qw_y)^2 + (pv_z - l_z - qw_z)^2 = $$ $$ = (p\vec v - \vec l - q\vec w, p\vec v - \vec l - q\vec w) = (p\vec v - q\vec w, p\vec v - q\vec w) + (\vec l, \vec l) - 2 (p\vec v - q\vec w, \vec l). $$ Geometry (in-plane~-- $\R^2$ and~in~$\R^3$) is~invariant under translations (at~least, the~geometry you're talking about), so~we may assume that one~of~the~vectors starts at~the~origin. Extremal (minimal or~maximal) distance~-- zero of~first derivative(s): $$ \frac{d\rho^2}{dp} = 2v_x(pv_x - l_x - qw_x) + 2v_y(pv_y - l_y - qw_y) + 2v_z(pv_z - l_z - qw_z) = 0, $$ $$ \frac{d\rho^2}{dq} = 2w_x(pv_x - l_x - qw_x) + 2w_y(pv_y - l_y - qw_y) + 2w_z(pv_z - l_z - qw_z) = 0. $$ $\delta_{ij}$ you~say? $g_{ij}$? Drop that: who cares? $$ p(v, v) - q(v, w) = (v, l); $$ $$ p(v, w) - q(w, w) = (w, l). $$ We~see the~(almost Gram-Schmidt) matrix $$ \hat A = \begin{bmatrix} (v, v) & -(v, w) \\ (w, v) & -(w, w) \\ \end{bmatrix}; \hat A^{-1} = \frac 1{(v, w)^2 - v^2w^2} \begin{bmatrix} -(w, w) & (v, w) \\ -(v, w) & (v, v) \\ \end{bmatrix} $$ By~the~way, we have the~square of~parallelogram area, $S^2(\vec v, \vec w) = v^2w^2 - (v, w)^2 \equiv |v * w|^2$ (and~$S^2 > 0$ due to~Kauchy-Buniakovsky inequality). Thus, $$ \begin{bmatrix} p \\ q \\ \end{bmatrix} = \frac 1{S^2(\vec v, \vec w)} \begin{bmatrix} (w, w) & -(v, w) \\ (v, w) & -(v, v) \\ \end{bmatrix} \begin{bmatrix} (v, l) \\ (w, l) \\ \end{bmatrix} = $$ $$ = \frac 1{S^2(v, w)} \begin{bmatrix} (w, w) (v, l) - (v, w) (w, l) \\ (w, v) (v, l) - (v, v) (w, l) \end{bmatrix} $$ Dig the~skew-symmetry! $$ \vec\gamma = p\vec v - q\vec w = \frac 1{S^2(v, w)} \cdot $$ $$ \cdot \left\{ \left[ (w, w) (v, l) - (v, w) (w, l) \right] \vec v + \left[ (v, v) (w, l) - (w, v) (v, l) \right] \vec w \right\}. $$ And ({\it leaping through time}) $$ \rho_{min} = \frac{}{|v * w|} = \frac{V(\vec l, \vec v, \vec w)}{|v * w|}. $$ Task: can you deduce $V(\vec l, \vec v, \vec w)$ from all possible scalar products of~$\vec v$,~$\vec w$~and~$\vec l$? \vskip 1em \centerline{{\Large\it Turn the page!}} \eject {\it The~leap through time} (reminder: $S^2 = v^2w^2 - (v, w)^2$), \vskip 1em $$ (\vec\gamma, \vec l) = \frac1{S^2} \left[ w^2 (v, l)^2 - (v, w) (w, l) (v, l) v^2 (w, l)^2 - (w, v) (v, l) (w, l) \right] = $$ $$ = \frac1{S^2} \left[ w^2 (v, l)^2 - 2 (v, w) (w, l) (l, v) + v^2 (w, l)^2 \right]. $$ \vskip 2em $$ (\vec\gamma, \vec\gamma) = \frac1{S^4}\left\{ v^2\left[ w^4{\g (v,l)^2} + (v, w)^2(w, l)^2 - 2 w^2{\r (v, l)(l, w)(w, v)} \right] \right. + $$ $$ +\left. w^2\left[ v^4{\b (w, l)^2} + (w, v)^2(v, l)^2 - 2 v^2{\r (w, l)(l, v)(v, w)} \right] \right. + $$ $$ \left. + 2(v, w)\left[ w^2v^2{\r (w, l)(l, v)} + (v, w)^2{\a (v, l)(w, l)} - w^2(w, v)^2(v, l) - v^2(v, w)^2(w, l) \right] \right\} = $$ $$ = \frac1{S^4}\left\{ {\x v^2w^4}{\g (v, l)^2} + {\x w^2v^4}{\b (w, l)^2} - v^2 (v, w)^2(w, l)^2 - w^2 (w, v)^2(v, l)^2 \right. - $$ $$ \left. - 2{\x v^2w^2}{\r (v, w)(w, l)(l, v)} + 2(v, w)^2{\a (w, v)(v, l)(l, w)} \right \} = $$ $$ \frac1{S^4}\left\{ {\x v^2w^2}\left[ w^2(v, l)^2 + v^2(w, l)^2 - 2(v, w)(w, l)(l, v) \right]\right. - $$ $$ \left. - (v, w)^2\left[ v^2(w, l)^2 + w^2(v, l)^2 - 2(v, w)(w, l)(l, v) \right] \right\} = $$ $$ = \frac{w^2(v, l)^2 + v^2(w, l)^2 - 2(v, w)(w, l)(l, w)}{S^2}. $$ Algebra is~{\it da~shit}, yeah~baby! Recall that $$ \rho^2 = (\vec\gamma, \vec\gamma) + (\vec l, \vec l) - 2 (\vec\gamma, \vec l), $$ so $$ S^2\rho^2 = l^2S^2 - \left[w^2(v, l)^2 + v^2(w, l)^2 - 2(v, w)(w, l)(l, v)\right] $$ $$ = \left[ l^2 v^2 w^2 - l^2(v, w)^2 - w^2(v, l)^2 - v^2(w, l)^2 + 2(v, w)(w, l)(l, v) \right] = $$ $$ = V^2(\vec l, \vec v, \vec w). $$ Why such an~identity for $V^2$? Since\footnote{$^t$~is~matrix transposition: $M^t_{ij} = M_{ji}$.} $$ V^2=\det\hat{V^2}= \det|v, w, l|\det|v, w, l|^t = \det \begin{bmatrix} (v, v) & (v, w) & (v, l) \\ (w, v) & (w, w) & (w, l) \\ (l, v) & (l, w) & (l, l) \\ \end{bmatrix} = $$ $$ = v^2w^2l^2 + {\r (v, w)(w, l)(l, v)} + {\r (v, l)(w, v)(l, w)} - v^2(w, l)(l, w) - (v, w)(w, v)l^2 - (v, l)w^2(l, v) = $$ $$ = v^2w^2l^2 - v^2 (w, l)^2 - w^2 (l, v)^2 - l^2 (v, w) + 2{\r (v, l)(l, w)(w, v)} $$ with $\hat{V^2}$ being the~real Gram-Schmidt matrix. \vskip 1em \centerline{{\Large\it That's about~it.}} \end{document}