Maximum Shortest Distance Between Face Diagonals and Main Diagonals of a Cube

To find the maximum shortest distance $d(\ell_1, \ell_2)$ between any line $\ell_1 \in F$ containing a face diagonal and any line $\ell_2 \in S$ containing a main diagonal of the unit cube $Q$, we can analyze the geometric configuration using vector algebra.

Step 1: Coordinates and Line Equations

Let the vertices of the cube $Q$ be $(x_1, x_2, x_3)$ where $x_1, x_2, x_3 \in \{0, 1\}$.

By symmetry, all four main diagonals in $S$ are congruent under the symmetry group of the cube. Thus, we can choose one specific main diagonal $\ell_2 \in S$:

• Let $\ell_2$ be the main diagonal passing through the origin $A(0, 0, 0)$ and the opposite vertex $B(1, 1, 1)$.
• A vector along $\ell_2$ is given by $\vec{d}_2 = (1, 1, 1)$.
• The parametric equation of $\ell_2$ is $\vec{r}_2(t) = (0,0,0) + t(1,1,1) = (t, t, t)$ for $t \in \mathbb{R}$.

Step 2: Categorizing the Face Diagonals $\ell_1 \in F$

The cube has 6 faces and each face has 2 diagonals, yielding 12 face diagonals in total. Relative to our chosen main diagonal $\ell_2$:

1. Intersecting Face Diagonals: Six of the face diagonals pass through either $(0,0,0)$ or $(1,1,1)$.

   • For example, the face diagonal joining $(0,0,0)$ to $(1,1,0)$ intersects $\ell_2$ at $(0,0,0)$.
   • For these 6 lines, the shortest distance to $\ell_2$ is $d(\ell_1, \ell_2) = 0$.

2. Skew Face Diagonals: The remaining 6 face diagonals do not share any endpoint with $\ell_2$ and are skew to it. Let us calculate the shortest distance for one such skew face diagonal $\ell_1$, for example, the diagonal on the face $z = 0$ joining the vertices $P(1,0,0)$ and $Q(0,1,0)$:

   • Point on $\ell_1$: $\vec{a}_1 = (1, 0, 0)$
   • Direction vector of $\ell_1$: $\vec{d}_1 = (0, 1, 0) - (1, 0, 0) = (-1, 1, 0)$
   • Point on $\ell_2$: $\vec{a}_2 = (0, 0, 0)$
   • Direction vector of $\ell_2$: $\vec{d}_2 = (1, 1, 1)$

Step 3: Distance Calculation

The shortest distance $d(\ell_1, \ell_2)$ between two skew lines defined by points $\vec{a}_1, \vec{a}_2$ and direction vectors $\vec{d}_1, \vec{d}_2$ is given by: $d(\ell_1, \ell_2) = \frac{|(\vec{a}_1 - \vec{a}_2) \cdot (\vec{d}_1 \times \vec{d}_2)|}{|\vec{d}_1 \times \vec{d}_2|}$

First, compute the cross product $\vec{d}_1 \times \vec{d}_2$: $\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ 1 & 1 & 1 \end{vmatrix} = (1 - 0)\hat{i} - (-1 - 0)\hat{j} + (-1 - 1)\hat{k} = (1, 1, -2)$

Next, find the magnitude of the cross product: $|\vec{d}_1 \times \vec{d}_2| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$

Now, compute the scalar triple product: $\vec{a}_1 - \vec{a}_2 = (1, 0, 0) - (0, 0, 0) = (1, 0, 0)$ $(\vec{a}_1 - \vec{a}_2) \cdot (\vec{d}_1 \times \vec{d}_2) = (1, 0, 0) \cdot (1, 1, -2) = 1$

Therefore, the shortest distance is: $d(\ell_1, \ell_2) = \frac{|1|}{\sqrt{6}} = \frac{1}{\sqrt{6}}$

By symmetry, the shortest distance between any of the remaining skew face diagonals and any main diagonal yields the exact same value of $\frac{1}{\sqrt{6}}$.

Conclusion

The possible values for the shortest distance $d(\ell_1, \ell_2)$ between any face diagonal $\ell_1 \in F$ and main diagonal $\ell_2 \in S$ are $0$ and $\frac{1}{\sqrt{6}}$.

Thus, the maximum value of $d(\ell_1, \ell_2)$ is $\frac{1}{\sqrt{6}}$, which corresponds to Option A.