JEE Advanced 2023 — Paper 1 — MATH-II solution

JEE Advanced2023Paper 1MATH-II

Let $Q$ be the cube with the set of vertices ${(x_{1}, x_{2}, x_{3}) \in R^{3} : x_{1}, x_{2}, x_{3} \in {0, 1}}$ . 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 $ℓ_{1}$ and $ℓ_{2}$ , let $d (ℓ_{1}, ℓ_{2})$ denote the shortest distance between them. Then the maximum value of $d (ℓ_{1}, ℓ_{2})$ , as $ℓ_{1}$ varies over $F$ and $ℓ_{2}$ varies over $S$ , is

A
$\frac{1}{6}$
B
$\frac{1}{8}$
C
$\frac{1}{3}$
D
$\frac{1}{12}$

Solution

By symmetry, take the main diagonal $ℓ_{2}$ from $O (0, 0, 0)$ to $G (1, 1, 1)$ , with direction $d_{2} = (1, 1, 1)$ .

Consider a face diagonal that is skew to $ℓ_{2}$ , for example $ℓ_{1}$ joining $A (1, 0, 0)$ to $F (0, 1, 1)$ , with direction $d_{1} = (- 1, 1, 1)$ .

The common perpendicular direction is $d_{1} \times d_{2} = \hat{i} + \hat{j} - 2 \hat{k}$ (after computing the determinant), with magnitude $6$ .

Using $O A = \hat{i}$ , the shortest distance is

$d = \frac{∣ O A \cdot ( d _{1} \times d _{2} ) ∣}{∣ d _{1} \times d _{2} ∣} = \frac{∣1∣}{6} = \frac{1}{6}$ .

Checking the other configurations of face diagonal vs main diagonal (intersecting or with smaller separation) confirms that $\frac{1}{6}$ is the maximum.

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