Show that the shortest distance between two opposite edges of a tetrahedron is $\frac{2a}{\sqrt{6}}$.

Question

The tetrahedron is formed by the planes
$ y+z=0$,
$z+x=0$,
$x+y=0$, and
$x+y+z=a$

I'm not able to visualise the sides that will constitute the tetrahedron, so not able to figure out which edges will lie opposite to each other. Please help me understand the formation of the tetrahedron.

Answer 1

The "easy" way to visualize the tetrahedron is to do a coordinate change: $$X=z+y\\Y=z+x\\Z=x+y\\x+y+z=(X+Y+Z)/2$$ In this new coordinate system the sides are the $X=0$, $Y=0$, $Z=0$, and the plane that intersects each axis at $2a$. Your original tetrahedron is similar to this one, rotated $45^\circ$ two times

Answer 2

At the start, it can be analyzed mostly algebraically.

Assume $a > 0$.

First find the vertices . . .

To find a vertex, choose $3$ of the $4$ equations, and solve the system.

Do that for each choice of $3$ equations from the $4$ given ones.

If you do it correctly, you should get vertices $$A=(0,0,0),\;B=(a,a,-a),\;C = (a,-a,a),D = (-a,a,a)$$

Next, consider pairs of disjoint edges.

There are $3$ pairs $$(AB,CD),\;(AC,BD),\;(AD,BC)$$ of disjoint edges.

Noting the algebraic symmetry of the points $B,C,D$, it follows that all $3$ pairs of disjoint edges have the same minimum distance between them, so only one pair, $(AB,CD)$, say, needs to be considered.

Note that face $BCD$ is an equilateral triangle with edge length $2a$, and the edges from $A$ have length $a\sqrt{3}$.

A little visualization is needed here . . .

Draw a tetrahedron with an equilateral base of edge length $2a$, and draw the remaining vertex centered "above" the base, equidistant, with distance $a\sqrt{3}$, from the vertices of the base.

By symmetry, the point on edge $CD$ which is closest to edge $AB$ is the midpoint, $P$ say, of edge $CD$.

So we want the minimum distance from $P$ to edge $AB$.

By the midpoint formula, $P = (0,0,a)$, hence, by the distance formula, triangle $ABP$ has edge lengths $$AB = a\sqrt{3},\;AP = a,\;BP=a\sqrt{6}$$

By the law of cosines, angle $APB$ is obtuse, hence the altitude from $P$ to edge $AB$ hits line $AB$ at a point outside of edge $AB$.

It follows the minimum distance from $P$ to a point on edge $AB$ is the distance from $P$ to $A$, which is $a$.

So if minimum distance between two edges means the minimum distance between two points which are actually on the respective edges, then the answer is $a$.

Answer 3

Without having to draw a sketch, you can still deduce the equation of the lines that contains each of the edges. Each edge comes from letting a pair of sides intersect; therefore, there are exactly $\binom{4}{2} = 6$ edges.

For instance, if you consider the faces $x+y = 0$ and $y+z=0$ and call the edge $AB$ when you let those faces intersect, then the equation of the line containing $AB$ will be $$\begin{gather} x+y=0 \\ y+z = 0 \end{gather}$$ When you solve this, you will get the parametric equation $AB: (t,-t,t), t\in\mathbb{R}.$

We also want the opposite side, so the opposite edge $CD$ will have the equation: $$\begin{gather} x+z = 0 \\ x+y+z = a \end{gather}$$ This is parametrically $CD: (p, a, -p), p\in\mathbb{R}.$ Finally, if we connect compute the distance between two points on $AB$ and $CD$, it will be: $$d^2(t,p) = (t-p)^2+(a+t)^2+(t+p)^2=3t^2+2at+2p^2+a^2$$

and we want to minimize it over $(t,p)\in\mathbb{R}^2.$ Solving for the critical points, we get $6t+2a = 0$ and $p=0.$ Hence, the minimum is attained at $(t^*, p^*) = (-a/3, 0)$, where the distance would be $$d(-a/3, 0) = \sqrt{a^2/9 + 4a^2/9+a^2/9} = \dfrac{2a}{\sqrt{6}}.$$