Skew lines/R^3/Distance/Extremal problem/Example

Let

G = P + ℝ v

and

H = Q + ℝ w

be skew lines. We want to understand the distance problem between the two lines as an extremal problem in the sense of higher-dimensional analysis. Let

P = (\begin{matrix} a_{1} \\ a_{2} \\ a_{3} \end{matrix})

and

Q = (\begin{matrix} b_{1} \\ b_{2} \\ b_{3} \end{matrix}) .

The square of the distance between the two points

P^{'} = (\begin{matrix} a_{1} \\ a_{2} \\ a_{3} \end{matrix}) + s (\begin{matrix} v_{1} \\ v_{2} \\ v_{3} \end{matrix})

and

Q^{'} = (\begin{matrix} b_{1} \\ b_{2} \\ b_{3} \end{matrix}) + t (\begin{matrix} w_{1} \\ w_{2} \\ w_{3} \end{matrix})

is (setting $c_{i} = a_{i} - b_{i}$ )

\begin{matrix} d (P^{'}, Q^{'})^{2} & = {(c_{1} + s v_{1} - t w_{1})}^{2} + {(c_{2} + s v_{2} - t w_{2})}^{2} + {(c_{3} + s v_{3} - t w_{3})}^{2} \\ = c_{1}^{2} + s^{2} v_{1}^{2} + t^{2} w_{1}^{2} + 2 s c_{1} v_{1} - 2 t c_{1} w_{1} - 2 s t v_{1} w_{1} + c_{2}^{2} + s^{2} v_{2}^{2} + t^{2} w_{2}^{2} + 2 s a_{2} v_{2} - 2 t c_{2} w_{2} - 2 s t v_{2} w_{2} + c_{3}^{2} + s^{2} v_{3}^{2} + t^{2} w_{3}^{2} + 2 s c_{3} v_{3} - 2 t c_{3} w_{3} - 2 s t v_{3} w_{3} \\ = c_{1}^{2} + c_{2}^{2} + c_{3}^{2} + 2 s (c_{1} v_{1} + c_{2} v_{2} + c_{3} v_{3}) - 2 t (c_{1} w_{1} + c_{2} w_{2} + c_{3} w_{3}) + s^{2} (v_{1}^{2} + v_{2}^{2} + v_{3}^{2}) + t^{2} (w_{1}^{2} + w_{2}^{2} + w_{3}^{2}) - 2 s t (v_{1} w_{1} + v_{2} w_{2} + v_{3} w_{3}) . \end{matrix}

We interpret this expression with methods of Analysis 2. We consider the data given by the lines as fixed parameters, so that we have a real-valued expression $f (s, t)$ in the two real variables $s$ and $t$ , and we want to determine its extrema. The partial derivatives are

\frac{\partial f}{\partial s} = 2 (c_{1} v_{1} + c_{2} v_{2} + c_{3} v_{3}) + 2 s (v_{1}^{2} + v_{2}^{2} + v_{3}^{2}) - 2 t (v_{1} w_{1} + v_{2} w_{2} + v_{3} w_{3})

and

\frac{\partial f}{\partial t} = 2 (c_{1} w_{1} + c_{2} w_{2} + c_{3} w_{3}) + 2 t (w_{1}^{2} + w_{2}^{2} + w_{3}^{2}) - 2 s (v_{1} w_{1} + v_{2} w_{2} + v_{3} w_{3}) .

If we equate this with $0$ , then we obtain an inhomogeneous linear system of equations with two equations in the variables $s$ and $t$ . Using Cramer's rule, we get

s = \frac{det (\begin{matrix} - c_{1} v_{1} - c_{2} v_{2} - c_{3} v_{3} & - v_{1} w_{1} - v_{2} w_{2} - v_{3} w_{3} \\ - c_{1} w_{1} - c_{2} w_{2} - c_{3} w_{3} & w_{1}^{2} + w_{2}^{2} + w_{3}^{2} \end{matrix})}{det (\begin{matrix} v_{1}^{2} + v_{2}^{2} + v_{3}^{2} & - v_{1} w_{1} - v_{2} w_{2} - v_{3} w_{3} \\ - v_{1} w_{1} - v_{2} w_{2} - v_{3} w_{3} & w_{1}^{2} + w_{2}^{2} + w_{3}^{2} \end{matrix})}

and

t = \frac{det (\begin{matrix} v_{1}^{2} + v_{2}^{2} + v_{3}^{2} & - c_{1} v_{1} - c_{2} v_{2} - c_{3} v_{3} \\ - v_{1} w_{1} - v_{2} w_{2} - v_{3} w_{3} & - c_{1} w_{1} - c_{2} w_{2} - c_{3} w_{3} \end{matrix})}{det (\begin{matrix} v_{1}^{2} + v_{2}^{2} + v_{3}^{2} & - v_{1} w_{1} - v_{2} w_{2} - v_{3} w_{3} \\ - v_{1} w_{1} - v_{2} w_{2} - v_{3} w_{3} & w_{1}^{2} + w_{2}^{2} + w_{3}^{2} \end{matrix})} .

If $v$ and $w$ are normed, then these expressions can be simplified to

s = \frac{- ⟨ P - Q, v ⟩ - ⟨ P - Q, w ⟩ ⟨ v, w ⟩}{1 - {⟨ v, w ⟩}^{2}}

and

t = \frac{- ⟨ P - Q, w ⟩ - ⟨ P - Q, v ⟩ ⟨ v, w ⟩}{1 - {⟨ v, w ⟩}^{2}} .