Let $Q$ be the cube with the set of vertices $\left\{(x_{1}, x_{2}, x_{3}) \in \mathbb{R}^{3} : x_{1}, x_{2}, x_{3} \in \{0, 1\}\right\}$. 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, e

$Q = \{0,1\}^3$

$F = \text{12 lines through face diagonals}$

$S = \text{4 main-diagonal lines}$

$d(\ell_1,\ell_2) = \text{shortest distance}$

$\max d(\ell_1,\ell_2),\quad \ell_1 \in F,\ \ell_2 \in S$

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

$\text{8 vertices, edge length } 1$

$S:\ \text{4 body diagonals}$

$\text{dirs } (1,1,1),\ (1,1,-1),\ (1,-1,1),\ (1,-1,-1)$

$F:\ 12 \text{ face diagonals }(2 \text{ per face})$

$\text{face diagonal shares vertices with 2 body diagonals} \Rightarrow d=0$

$\text{skew to the other 2} \Rightarrow \text{max comes from these}$

$d = \dfrac{|(\vec B - \vec A)\cdot(\vec u \times \vec v)|}{|\vec u \times \vec v|}$

$A=(0,0,0),\ \vec u=(1,1,1);\quad B=(1,0,0),\ \vec v=(-1,0,1)$

u⃗ × v⃗ = î ĵ k̂ 1 1 1 -1 0 1

$= \hat{i}(1) - \hat{j}(2) + \hat{k}(1) = (1,-2,1)$

$|\vec u \times \vec v| = \sqrt{1+4+1} = \sqrt{6}$

$\vec B - \vec A = (1,0,0)$

$(\vec B-\vec A)\cdot(\vec u\times\vec v) = 1(1)+0(-2)+0(1) = 1$

$d = \dfrac{|1|}{\sqrt{6}} = \dfrac{1}{\sqrt{6}}$

$\max d = \dfrac{1}{\sqrt{6}}$

$\boxed{\text{option (1)}\quad \dfrac{1}{\sqrt{6}}}$