Shortest distance of a tetrahedron

Question

Show that the shortest distance between the opposite edges of a regular tetrahedron is $ \frac{1}{2} \sqrt{2} \alpha $ when the length of each side is $ \alpha $.

What I've done so far:

Take the vertices to be $ O=(0,0,0), A =(\alpha,0,0), B=(\frac{\alpha}{2},\alpha\frac{\sqrt3}{2},0). $

Take $ \hat i $ in the direction of $ OA $, $ OBA\ $ lying in the $ Oxy $ plane with $ \hat{j} $ in the direction of $ Oy $ and $ \hat{k} $ perpendicular to $ Oxy $ plane.

Then the other vertex will have coordinates $ C=(\frac{\alpha}{2}, \frac{\alpha}{2 \sqrt{3}},\alpha) $.

The vector equation of $ BC= \frac{\alpha}{2}\hat{i}+\frac{\alpha \sqrt{3}}{2} \hat{j}+\lambda (-\frac{\sqrt{3}\alpha}{2} \hat{j} + \alpha\hat{k}). $ $ OA = \alpha \hat{i}. $

$ OA \land BC = -\alpha^2\hat{j}+ \frac{\alpha^2}{2 \sqrt{3}} \hat{j}. $ Taking the dot product of $ \frac{\alpha}{2}\hat{i}+ \frac{\alpha \sqrt{3}}{2} \hat{j} $ and this vector would not give the required answer ( $ \land $ denotes the cross product. )

Please help. Thank you in advance.

Look at the bottom animation in this old answer of mine. Placing the tetrahedron inside a cube makes answering this question easy. The opposite edges of a tetrahedron are diagonals of the opposite faces of a cube. The ratio of their lengths and distance is immediate from the geometry of the surrounding cube. — Jyrki Lahtonen, Apr 18 '20 at 05:41

score 3 · Answer 1 · answered Apr 18 '20 at 04:56

3

Your coordinates for the tetrahedron are somewhat convoluted. If we instead take $(0,0,0),(1,1,0),(1,0,1),(0,1,1)$ as coordinates, opposite sides lie in opposite faces of a cube, and the distance between those faces is $1$. This tetrahedron's side length is $\sqrt2$, so scaling down gives the requisite answer of $\frac12\sqrt2\alpha$.

answered Apr 18 '20 at 04:56

Parcly Taxel

103,344

There was a hint to one vertex at the origin and to take one of the coordinates as (0,0,0) – pseudogeek Apr 18 '20 at 05:29
@pseudogeek I did that, right? – Parcly Taxel Apr 18 '20 at 05:30
I'm still having trouble. Why do the opposite sides have to lie on opposite faces of a cube? Also, the side length of the tetrahedron is alpha. – pseudogeek Apr 18 '20 at 05:35
@pseudogeek Draw the tetrahedron inscribed in the cube and it will be clear. We can take any side length we want for the tetrahedron and then generalise to any $\alpha$ by scaling. – Parcly Taxel Apr 18 '20 at 05:36
I had the same idea. This is surely the simplest way. – Jyrki Lahtonen Apr 18 '20 at 05:43
The problem that I'm having is that sides of a tetrahedron have an angle of 60 degrees whereas the sides of a cube is perpendicular – pseudogeek Apr 18 '20 at 05:43
@pseudogeek Did you look at the animation? – Jyrki Lahtonen Apr 18 '20 at 05:49
Thank you. Makes sense now. – pseudogeek Apr 18 '20 at 05:51

Shortest distance of a tetrahedron

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