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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 greater than or equal to

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$\frac{1}{\sqrt{2}}$

$\frac{1}{\sqrt{8}}$

$\frac{1}{\sqrt{12}}$

$\frac{1}{\sqrt{6}}$

answer is A, B, D.

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Detailed Solution

DR’s of $\vec{O G} = (1, 1, 1)$
DR’s of $\vec{A C} = (- 1, 1, 0)$
Equation of $\vec{O G} = \frac{x}{1} = \frac{y}{1} = \frac{z}{1}$
Equation of $\vec{A C} = \frac{x - 1}{- 1} = \frac{y}{1} = \frac{z}{0}$
$\vec{O A} = \hat{i}$

Normal of $\vec{O G}$ and $\vec{A C}$

$= | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ - 1 & 1 & 0 \end{matrix} | = (- \hat{i} - \hat{j} + 2 \hat{k})$

$S . D . = \frac{| \hat{i} (- \hat{i} - \hat{j} + 2 \hat{k}) |}{| - \hat{i} - \hat{j} + 2 \hat{k} |} = \frac{1}{\sqrt{6}}$

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