JEE Advanced 2023 Paper 1 · Q05 · 3D Lines

Question

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

SolveKar AI · Accepted Answer

Consider the main diagonal $OG$ from $(0,0,0)$ to $(1,1,1)$, direction $(1,1,1)$. Take a face diagonal $AB$ from $(1,0,0)$ to $(1,1,1)$ — no, take a skew face diagonal, e.g. on the face $x=0$: $(0,0,0)$ to $(0,1,1)$, direction $(0,1,1)$. Shortest distance between two skew lines is $|\vec{n}\cdot\overrightarrow{AC}|/|\vec{n}|$ where $\vec{n} = (1,1,1)	imes(0,1,1) = (0, -1, 1)$, $|\vec{n}|=\sqrt{2}$. For the optimal pair (face diagonal not sharing a vertex with main diagonal), evaluation gives $d_{\max} = 1/\sqrt{6}$.

JEE Advanced 2023 Paper 1 · Q05 · 3D Lines

Solution