A square $A B C D$ of diagonal $2 a$ is folded along the diagonal $A C$ so that the planes $D A C$ and $B A C$ are at right angle. The shortest distance between $D C$ and $A B$ is

\sqrt{2} a

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\frac{2 a}{\sqrt{3}}

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\frac{2 a}{\sqrt{5}}

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\frac{\sqrt{3} a}{2}

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Solution

The correct option is A $\frac{2 a}{\sqrt{3}}$
When folded coordinates will be $D (0, 0, a); C (a, 0, 0); A (- a, 0, 0); B (0, - a, 0)$
Equation of $D C$ is, $\frac{x}{a} = \frac{y}{b} = \frac{z - a}{a}$
Equation of $A B$ is, $\frac{x + a}{a} = \frac{y}{- a} = \frac{z}{0}$
$∴$ Shortest distance $= \frac{2 a}{\sqrt{3}} .$

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