How far apart are the diagonals on two adjacent faces (not intersecting each other) of a unit cube?
Solution. Let the vertices of the cube be $A, B, C, D, E, F, G, H$ as shown in the diagram. Due to the symmetry of the cube, it is sufficient to determine the distance between any pair of non-intersecting face diagonals on adjacent faces. Such a pair of face diagonals could be $A F$ and $B G$.

The distance between two skew lines is equal to the distance between two parallel planes, one of which contains one of the lines, and the other contains the other line. In our case, these planes are $A F H$ and $B D G$, because $A H$ is parallel to $B G$ and $D G$ is parallel to $A F$ (since the corresponding face diagonals on opposite faces of the cube are parallel). Therefore, our task is to determine the distance between the planes $A F H$ and $B D G$. We will show that both planes are perpendicular to the space diagonal $E C$ and trisect it.

![](https://cdn.mathpix.com/cropped/2024_05_02_4370f0514331e96b860bg-1.jpg?height=401&width=420&top_left_y=518&top_left_x=842)

The triangles $A F H$ and $B D G$ are equilateral because each of their sides is a face diagonal of the cube. $E A = E F = E H = 1 = C B = C D = C G$, because each of these is an edge of the cube. Therefore, if we connect $E$ to the centroid $S$ of the triangle $A F H$ (which is also the center of the circumscribed circle), then the line $E S$ will be perpendicular to the plane $A F H$, and similarly, if the centroid of the triangle $B D G$ is $T$, then $C T$ will be perpendicular to the plane $B D G$. Consider the vectors $\mathbf{a} = E A$, $\mathbf{f} = E F$, and $\mathbf{h} = E H$. Due to the properties of the cube, $\mathbf{a} + \mathbf{f} + \mathbf{h} = \overrightarrow{E C}$, and by the centroid formula, $\frac{\mathbf{a} + \mathbf{f} + \mathbf{h}}{3} = \overrightarrow{E S}$. This means that $S$ is one of the trisection points of the space diagonal $E C$. Similarly, we find that $T$ is the other trisection point of the space diagonal.

The distance between the two planes is therefore exactly one-third of the space diagonal. The distance between the two face diagonals is thus one-third of the space diagonal of the unit cube, which is $\frac{\sqrt{3}}{3}$.