Geometry quickie: volume of a tetrahedron

I saw this formula on Wikipedia for the volume of tetrahedron and was intrigued.

Any two opposite edges of a tetrahedron lie on two skew lines, and the distance between the edges is defined as the distance between the two skew lines. Let $d$ be the distance between the skew lines formed by opposite edges... 

then it gave a volume formula. I want to derive it. This post gives my thought process, where I keep using the trick of reducing a general problem to a good example. This is a general strategy for solving problems, often written as "wlog".
(I write it as "wolog" and pronounce as "volog", as if it's a German adverb. So that I can write "We assume wolog that...", "... and thus wolog we have...")

How would a true geometer approach this problem? She would first do a dimension analysis. The volume has dimension $meter^3$, and the lengths have dimension $meter$, so it ought to look like
$$Volume = k \cdot AB \cdot CD \cdot h$$
where $h$ is the distance between the two skew lines of $AB$ and $CD$, and $k$ is some dimensionless number yet to be determined.

She would then figure out whether this problem is well-formed. Given the tetrahedron's opposing skewed sides, is it enough information to derive its volume?

If we double $AB$ to $AB'$, then the volume is doubled too. This is easily seen by noting that
$$Volume \propto Area(ABC) \cdot h_{D-ABC}$$
where $h_{D-ABC}$ is the height of point $D$ relative to the plane spanned by $ABC$.

Similarly, scaling $CD$ would also scale the volume.

Thus, we can "slide" the sides $AB$ and $CD$ along the lines they are on, since to slide $AB$ to $A'B'$, we only need to first lengthen $AB$ to $AB'$, then shorten it to $A'B'$. The volume is first scaled up then scaled down, unchanged in the end.

Now we can set up coordinates $(x, y, z)$, so that the z-axis is the common perpendicular of $AB$, $CD$.  $AB$ is in the planes $z = 0$, and $CD$ in $z = h$. Also, slide $AB$, $CD$, so that A, C are on the z-axis. Set $y$-axis parallel to $CD$.

Then notice that if we stretch the space in the $z$-direction so that it becomes double in length, it would double $h$ and $Volume$.

Thus, our original analysis
$$Volume = k \cdot AB \cdot CD \cdot h$$
is fully proven. It remains to figure out what $k$ is.

To find $k$, we do a shearing along the $y$-direction until $AB$ becomes perpendicular to $CD$. This shearing preserves $Volume$, $l$, $CD$, and the shadow of $AB$ on the $x$-axis, which is in general equal to $AB \sin\theta$, where $\theta$ is the angle between $AB$ and $CD$.

The only possible way for this shearing to fail is if $AB$ is also parallel to the $y$-axis. In this case, the tetrahedron degenerates into a flat trapezoid with $0$ area, and so our volume formula would still work.

Thus we only need to determine the number $k_0$. which can be found by calculate the volume for a prototype/archetype/canonical example/etc.

What is the tetrahedron with the most basic skew-line geometry? Of course, it's this tetrahedron inscribed in a unit cube!

Its volume is $1/6$ of the unit cube, by pure dissection:

Figure from catlikecoding

Thus, we obtain the final formula
$$Volume = \frac 1 6 \cdot l \cdot (AB \cdot CD\cdot \sin\theta)$$  If we use the vector cross-product, it is
$$Volume = \frac 1 6 \cdot l (\vec{AB} \times \vec{CD})$$

Conclusion

We solved the problem by successively transforming a general problem to more special examples. In a hand-wavy way, we can say that we give up symmetries in the problem and gain power in calculating the answer, and as long as we knew how to regain the symmetrize the answers, we can climb back the tower of symmetries to obtain the answer to the general problem we started with.

We can prove that the determinant gives the volume, using this strategy. A great exposition at a higher level is in What is a Gauge? (Terence Tao, 2008).