Let Q be the cube with the set of vertices $\{(x_1, x_2, x_3) \in \mathbb{R}^3 : x_1, x_2, x_3 \in \{0,1\}\}$. Let F be the set of all twelve lines containing the diagonals of the six faces of the cube Q. Let S be the set of all four lines containing the main diagonals of the cube Q; for instance, the line passing through the vertices $(0, 0, 0)$ and $(1, 1, 1)$ is in S. For lines $\ell_1$ and $\ell_2$, let $d(\ell_1, \ell_2)$ denote the shortest distance between them. Then the maximum value of $d(\ell_1, \ell_2)$, as $\ell_1$ varies over F and $\ell_2$ varies over S, is

Let Q be the cube with the set of vertices {(x₁, x₂, x₃) ∈ R³ : x₁, x₂, x₃ ∈ {0,1}}. Let F be the set of all twelve lines containing the diagonals of the six faces of the cube Q. Let S be the set of all four lines containing the main diagonals of the cube Q; for instance, the line passing through the vertices (0, 0, 0) and (1, 1, 1) is in S. For lines ell₁ and ell₂, let d(ell₁, ell₂) denote the shortest distance between them. Then the maximum value of d(ell₁, ell₂), as el

$\text{Given: } Q = \{(x_1, x_2, x_3) : x_1, x_2, x_3 \in \{0,1\}\}$

$F = \{\text{12 lines containing the face diagonals}\}$

$S = \{\text{4 lines containing the main diagonals}\}$

$\text{To find: } \max_{\ell_1 \in F,\ \ell_2 \in S} d(\ell_1, \ell_2)$

$Q: \text{cube with vertices at } \{0,1\}^3$ $M = OB: \text{ through } (0,0,0) \text{ and } (1,1,1)$ $\vec{u} = (1,1,1)$ $F: \text{12 face-diagonal lines (2 per face)}$ $\text{Each face contains exactly one endpoint of } M$ $\text{Diagonal through it meets } M \Rightarrow d = 0$ $\text{The other 6 face diagonals are skew to } M$ $D: (0,1,0)\ \text{to}\ (0,0,1);\quad \vec{v} = (0,-1,1)$ $\text{Plan: fix } M \text{ by cube symmetry}$ $6 \text{ meet } M \Rightarrow d = 0; \quad 6 \text{ skew}$ $\text{compute } d \text{ for one skew}; \ \text{symmetry} \Rightarrow \max$ $d = \dfrac{|(P_2 - P_1) \cdot (\vec{u} \times \vec{v})|}{|\vec{u} \times \vec{v}|}$ $P_1 = O = (0,0,0), \quad \vec{u} = (1,1,1)$ $P_2 = (0,1,0), \quad \vec{v} = (0,-1,1)$ $\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & -1 & 1 \end{vmatrix}$ $= \hat{i}(1+1) - \hat{j}(1-0) + \hat{k}(-1-0)$ $\vec{u} \times \vec{v} = (2,-1,-1)$ $|\vec{u} \times \vec{v}| = \sqrt{4+1+1} = \sqrt{6}$ $P_2 - P_1 = (0,1,0)$ $(P_2 - P_1)\cdot(\vec{u}\times\vec{v}) = 0 + (-1) + 0 = -1$ $d = \dfrac{|-1|}{\sqrt{6}} = \dfrac{1}{\sqrt{6}}$ $P\left(\tfrac13,\tfrac13,\tfrac13\right), \quad Q\left(0,\tfrac12,\tfrac12\right)$ $\vec{PQ} = \left(-\tfrac13,\tfrac16,\tfrac16\right)$ $|PQ| = \sqrt{\tfrac19 + \tfrac1{36} + \tfrac1{36}} = \sqrt{\tfrac16} = \dfrac{1}{\sqrt{6}}$ $\text{4 main diagonals equivalent by rotation of the cube}$ $\text{6 skew face diagonals equivalent (fixing } M) \Rightarrow \text{same } d$ $\max d(\ell_1,\ell_2) = \dfrac{1}{\sqrt{6}} \ \Rightarrow\ \text{option } (1)$