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In every geometry textbook it states that the shortest distance between two skew lines (lines that are not co-planar) is given by the unique line that runs perpendicular to both skew lines. This is fairly simple to prove given a shortest line exists (see my comments below). However, how can we prove that between any two skew lines there exists a shortest straight path?

I have tried using calculus to show that for lines $L_1 = \mathbf{p}+s\mathbf{u}$ and $L_2 = \mathbf{q} + t\mathbf{v}$ the equation:

$$R(s,t) = \Vert \mathbf{p}+s\mathbf{u}-(\mathbf{q}+t\mathbf{v})\Vert^2$$

  • Has a local minimum

However I am not quite able to show that (without getting into page and pages of calculations) and what is more, even after I have shown this, it only proves a local minimum exists.

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    $\begingroup$ Note, once this has been proven, then we are done as: First we prove that a line perpendicular to both L1 and L2 exists. This can be done by finding the general line equation perpendicular to L1 and also the general line equation perpendicular to L2 (use dot product = 0) and then finding the subset where these two general lines parallel to each other. $\endgroup$ – smotala1 Jul 9 at 16:22
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    $\begingroup$ Finally, after proving that a perpendicular always exist and using that a shortest line always exists you can use the fact that the shortest route from a point to L1 is perpendicular to L1 (let the set of these lines be A) and likewise let B equal the set of lines perpendicular to L2. We know that that the two must intersect (at the line perpendicular to both) and by definition of A and B this is the shortest distance between L1 and L2. $\endgroup$ – smotala1 Jul 9 at 16:30
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If you have two skew lines $a$ and $b$ it is easy to construct a line perpendicular to both, hence proving that it exists.

  1. Construct the plane $\beta$ containing $b$ and parallel to $a$.

  2. Construct the plane $\alpha$ containing $a$ and perpendicular to $\beta$.

  3. If $B$ is the intersection of $\alpha$ with $b$, then the line $AB$ passing through $B$ and perpendicular to $a$ is also perpendicular to $b$ and is thus the solution.

enter image description here

It is then immediate to show that $AB$ is the line of minimum distance: given any two points $P\in a$ and $Q\in b$, if $H$ is the projection of $P$ on $\beta$ we have:

$$ PQ^2=PH^2+HQ^2\ge PH^2=AB^2. $$

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  • $\begingroup$ Fantastic diagram! What software did you use to make it? $\endgroup$ – Ben Grossmann Jul 9 at 21:28
  • $\begingroup$ @Omnomnomnom The shading / fading out along the edges looks a lot like GeoGebra to me. $\endgroup$ – Xander Henderson Jul 10 at 2:51
  • $\begingroup$ @Omnomnomnom I used GeoGebra 5. $\endgroup$ – Intelligenti pauca Jul 10 at 8:32
  • $\begingroup$ @Aretino Thanks $\endgroup$ – Ben Grossmann Jul 10 at 9:02
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It can be done. We have $$\frac{\partial R}{\partial s} (s,t)= 2(\mathbf{p}-\mathbf{q}+s\mathbf{u}+t\mathbf{v})\cdot \mathbf{u}, \quad \frac{\partial R}{\partial t} (s,t)= 2(\mathbf{p}-\mathbf{q}+s\mathbf{u}+t\mathbf{v})\cdot \mathbf{v}.$$

The condition for local extremum is $$\frac{\partial R}{\partial s} (s_0,t_0) = \frac{\partial R}{\partial t} (s_0,t_0)=0$$ so such $s_0,t_0$ satisfy $\mathbf{p}-\mathbf{q}+s_0\mathbf{u}+t_0\mathbf{v} \perp \mathbf{u},\mathbf{v}$. Therefore, there is a scalar $\alpha$ such that $$\mathbf{p}-\mathbf{q}+s_0\mathbf{u}+t_0\mathbf{v} = \alpha(\mathbf{u} \times \mathbf{v}).$$ Note that $\mathbf{u}$ and $\mathbf{v}$ are linearly independent (since the lines are skew) so $\{\mathbf{u}, \mathbf{v}, \mathbf{u} \times \mathbf{v}\}$ is a basis for $\Bbb{R}^3$ and hence $s_0, t_0, \alpha$ exist and are unique. Hence, if we know that this is a local minimum (e.g. by calculating the Hessian), it has to be a global minimum.

Scalar multiplying the above relation by $\mathbf{u} \times \mathbf{v}$, we get $$(\mathbf{p}-\mathbf{q})\cdot (\mathbf{u} \times \mathbf{v}) = (\mathbf{p}-\mathbf{q}+s_0\mathbf{u}+t_0\mathbf{v}) \cdot (\mathbf{u} \times \mathbf{v})= \alpha \|\mathbf{u} \times \mathbf{v}\|^2$$ so $$\alpha = \frac{(\mathbf{p}-\mathbf{q})\cdot (\mathbf{u} \times \mathbf{v})}{\|\mathbf{u} \times \mathbf{v}\|^2}.$$ The minimal distance is now given by $$\|\mathbf{p}-\mathbf{q}+s_0\mathbf{u}+t_0\mathbf{v}\| = \alpha \|\mathbf{u} \times \mathbf{v}\| = \frac{(\mathbf{p}-\mathbf{q})\cdot (\mathbf{u} \times \mathbf{v})}{\|\mathbf{u} \times \mathbf{v}\|}.$$

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  • $\begingroup$ You can prove it is a munimum using $u \cdot u>0$ and $v \cdot v >0$ $\endgroup$ – Rd Basha Jul 9 at 17:10
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    $\begingroup$ @RdBasha I've found an elegant argument for minimality. For $s,t \in \Bbb{R}$ we have \begin{align*}\|\mathbf{p}-\mathbf{q}-s\mathbf{u}+t\mathbf{v}\|^2 &= \|\overbrace{\mathbf{p}-\mathbf{q}-s_0\mathbf{u}+t_0\mathbf{v}}^{\perp \mathbf{u}, \mathbf{v}} + (s-s_0)\mathbf{u} + (t-t_0)\mathbf{v}\|^2 \\ &= \|\mathbf{p}-\mathbf{q}-s_0\mathbf{u}+t_0\mathbf{v}\|^2 + \|(s-s_0)\mathbf{u} + (t-t_0)\mathbf{v}\|^2\\ &= \|\mathbf{p}-\mathbf{q}-s_0\mathbf{u}+t_0\mathbf{v}\|^2 + |s-s_0|^2\|\mathbf{u}\|^2 + |t-t_0|\|\mathbf{v}\|^2 + 2(s-s_0)\mathbf{u}\cdot(t-t_0)\mathbf{v} \end{align*} $\endgroup$ – mechanodroid Jul 9 at 20:14
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    $\begingroup$ \begin{align*}&\ge \|\mathbf{p}-\mathbf{q}-s_0\mathbf{u}+t_0\mathbf{v}\|^2 + |s-s_0|^2\|\mathbf{u}\|^2 + |t-t_0|\|\mathbf{v}\|^2 - 2|(s-s_0)\mathbf{u}\cdot(t-t_0)\mathbf{v}|\\ &\ge \|\mathbf{p}-\mathbf{q}-s_0\mathbf{u}+t_0\mathbf{v}\|^2 \end{align*} using Cauchy-Schwartz and AM-GM $$|(s-s_0)\mathbf{u}\cdot(t-t_0)\mathbf{v}| \le \|(s-s_0)\mathbf{u}\|\|(t-t_0)\mathbf{v}\| \le \frac12(|s-s_0|^2\|\mathbf{u}\|^2 + |t-t_0|\|\mathbf{v}\|^2).$$ Moreover, the inequality above is strict if $|s-s_0|, |t-t_0| > 0$ since $\mathbf{u}$ and $\mathbf{v}$ are linearly independent. $\endgroup$ – mechanodroid Jul 9 at 20:14
  • $\begingroup$ This looks like an expansion if you take $s-s_0$ and $t-t_0$ to be small. The first order vanishes, as should for an extremum, and the second order is related to the Hessian. $\endgroup$ – Rd Basha Jul 9 at 21:38
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Here is quick proof, making heavy use of matrix analysis. Let $A$ denote the matrix whose columns are $\mathbf u, -\mathbf v$, let $\mathbf x$ denote the column vector $\mathbf x = (s,t)$, and let $\mathbf b = \mathbf q - \mathbf p$.

The function that we are trying to minimize is $$ R(\mathbf x) = \left\|A \mathbf x - \mathbf b \right\|^2. $$ In other words, we are looking for the least-squares solution to the equation $A\mathbf x = \mathbf b$. There are many derivations/justifications of the solution $\mathbf x = (A^TA)^{-1}A^T\mathbf b$, one of which is given here.


Here is another proof: I claim (without proof) that because distances are fixed under rotation and translation, we can assume without loss of generality that $\mathbf q = 0$ and $\mathbf v = (0,0,1)$. With that, we find $$ (s\mathbf u + \mathbf p) - (t \mathbf v + \mathbf q) = \\ (su_1 + p_1, su_2 + p_2, su_3 - t + p_3 - q_3). $$ With the substitution $k = su_3 - t + (p_3 - q_3)$, this is simply the vector $$ (su_1 + p_1, su_2 + p_2, k). $$ Of course, we can rearrange $$ k = u_3\,s - t + (p_3 - q_3) \implies t = u_3\,s - k + (p_3 - q_3). $$ In other words, the change of coordinates $(s,t) \mapsto (s,k)$ is bijective. So, minimizing $R(s,t)$ is equivalent to minimizing $R(s,k)$.

Now, it is easy to see that $R(s,k)$ attains a minimum, since $$ R(s,k) = \|(su_1 + p_1, su_2 + p_2, k)\|^2 = (su_1 + p_1)^2 + (s u_2 + p_2)^2 + k^2, $$ which means that $R$ is minimized at $s = s_0, k=0,$ where $s_0$ is the value of $s$ that minimizes $(su_1 + p_1)^2 + (s u_2 + p_2)^2$.


Here is a proof by "completing the square." Expand the inner product $$ | s \mathbf u - t\mathbf v + (\mathbf p- \mathbf q) |^2 = \\ ( s \mathbf u - t\mathbf v + (\mathbf p- \mathbf q))\cdot ( s \mathbf u - t\mathbf v + (\mathbf p- \mathbf q)) =\\ s^2 \| \mathbf u\|^2 - 2st (\mathbf u \cdot \mathbf v) + t^2 \|\mathbf v\|^2 + s \mathbf u \cdot (\mathbf p - \mathbf q) - t\mathbf v \cdot (\mathbf p - \mathbf q) + |\mathbf p - \mathbf q|^2. $$ The constant term plays no roll, which is to say that it suffices to minimize the function $$ s,t \mapsto s^2 \| \mathbf u\|^2 - 2st (\mathbf u \cdot \mathbf v) + t^2 \|\mathbf v\|^2 + s [\mathbf u \cdot (\mathbf p - \mathbf q)] - t[\mathbf v \cdot (\mathbf p - \mathbf q)]. $$ To simplify things, rewrite our function $$ R(s,t) = s^2 \| \mathbf u\|^2 - 2st (\mathbf u \cdot \mathbf v) + t^2 \|\mathbf v\|^2 + cs + dt + C, $$ where $C$ is some constant and we simply note that $c = \mathbf u \cdot (\mathbf p - \mathbf q)$ and $d = \mathbf v \cdot (\mathbf p - \mathbf q)$ are real numbers.

Take out a perfect square $(\|\mathbf u\|s - \frac{\mathbf u\cdot \mathbf v}{\|\mathbf u\|}t)^2$ to get $$ R(s,t) = (\|\mathbf u\|s - \frac{\mathbf u\cdot \mathbf v}{\|\mathbf u\|}t)^2 + (\|\mathbf v\|^2 - \frac{(\mathbf u \cdot \mathbf v)}{\|\mathbf u\|^2})t^2 + cs + dt + C. $$ Importantly, we note that $\|\mathbf v\|^2 - \frac{(\mathbf u \cdot \mathbf v)}{\|\mathbf u\|^2} > 0$ as a consequence of the Cauchy-Schwarz inequality, i.e. that for non-parallel $\mathbf u,\mathbf v$, we have $$ |\mathbf u \cdot \mathbf v| = \|\mathbf u\|\,\|\mathbf v\| \cdot |\cos \theta| < \|\mathbf u\| \|\mathbf v\|. $$ Thus, we have written $R(s,t)$ in the form $$ R(s,t) = a (s - kt)^2 + bt^2 + cs + dt + C, $$ with $a,b > 0$ and $c,d,k \in \Bbb R$. Noting that $cs = c(s - kt) + ckt + C$, we have $$ R(s,t) = a (s - kt)^2 + bt^2 + c(s-kt) + \bar d t + C\\ = [a (s - kt)^2 + c(s - kt)] + [bt^2 + \bar d t] + C. $$ With that, it suffices to note that the functions $$ f(x) = ax^2 + cx, \quad g(x) = bx^2 + dx $$ both attain minimums.


Here is a proof along the lines of a typical "real analysis" inequality. Note that $\inf_{s,t \in \Bbb R} R(s,t)$ refers to the greatest lower bound ("infimum") of $R(s,t)$ over all real $s,t$. This lower bound must exist because $R(s,t)$ is always non-negative.

First, note that we necessarily have $$ \inf_{s,t \in \Bbb R} R(s,t) \leq R(0,0) = |\mathbf p - \mathbf q|^2. $$ We note that there is a shortest distance between a point and a line. Because the lines are not parallel, $\mathbf u \neq \mathbf v$. Thus, there exist a $m_1,m_2 > 0$ such that for all $t$, $|\mathbf u - t \mathbf v| \geq m_1$ and $|\mathbf v - t \mathbf u| \geq m_2$.

Now note that for $|s| > c_1 = 2|\mathbf p - \mathbf q|/m_1$, we have $$ R(s,t) = |\mathbf p + s \mathbf u - \mathbf q - t\mathbf v|^2 \geq (|s\mathbf u - t \mathbf v| - |\mathbf p - \mathbf q|)^2\\ = (s| \mathbf u - (t/s) \mathbf v | - |\mathbf p - \mathbf q| )^2\\ \geq (sm_1 - |\mathbf p - \mathbf q| )^2\\ > (2|\mathbf p - \mathbf q| - |\mathbf p - \mathbf q|) = |\mathbf p - \mathbf q|. $$ Similarly, if $|t| > c_2 = |\mathbf p - \mathbf q|/m_2$, then $R(s,t) > |\mathbf p - \mathbf q|$.

It follows that $$ \inf_{s,t \in \Bbb R} R(s,t) = \inf_{|s|\leq c_1,|t|\leq c_2} R(s,t). $$ In other words, it suffices to consider $R(s,t)$ over the closed and bounded set of values $[-c_1,c_1]\times[-c_2,c_2]$. However, any real-valued function over a compact domain must attain its maximum and minimum. So, the lower bound over $[-c_1,c_1]\times[-c_2,c_2]$ (which is necessarily the lower bound over $\Bbb R \times \Bbb R$) is necessarily attained.

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