JEE Advanced 2023 · Paper 1 · Mathematics Q5

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MathematicsSingle correct+3 / -1Question 5 of 51

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

\dfrac{1}{\sqrt6}

\dfrac{1}{\sqrt8}

\dfrac{1}{\sqrt3}

\dfrac{1}{\sqrt{12}}

Answer and worked solution

Official answer: A

By symmetry, take the main diagonal $\ell_2$ through $O$ with direction $(1,1,1)$ .

Which face diagonals matter. Every face has $O$ or $(1,1,1)$ as a vertex.

The diagonal through that vertex meets $\ell_2$ , so its distance is 0.
The other diagonal of each face (6 lines in all) is skew to $\ell_2$ , e.g. the line through $(1,0,0)$ and $(0,1,0)$ .

Distance. Take the line through $(1,0,0)$ with direction $(-1,1,0)$ :

$(1,1,1)\times(-1,1,0)=(-1,-1,2)$ , with modulus $\sqrt6$ ;
the vector between the lines' points is $(1,0,0)$ .

$d=\frac{|(1,0,0)\cdot(-1,-1,2)|}{\sqrt6}=\frac1{\sqrt6}.$

The symmetries of the cube fixing $\ell_2$ carry these 6 lines into each other, and all main diagonals are alike, so the maximum is $\dfrac1{\sqrt6}$ .

Topics:Skew lines Cross product

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JEE Advanced 2023 · Paper 1, Mathematics question 5