A square ABCD of diagonal 2a is folded along the diagonal AC so that the planes DAC and BAC are at right angle. The shortest distance between DC and AB 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}}{2}) a

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Solution

The correct option is B $\frac{2 a}{\sqrt{3}}$
When folded co-ordinates will be D(0,0, a); C(a, 0, 0);
A(-a, 0, 0); B(0, -a, 0)

Equation DC is, $\frac{x}{a} = \frac{y}{0} = \frac{z - a}{- a}$
Equation AB is, $\frac{x + a}{a} = \frac{y}{- a} = \frac{z}{0}$
$∴$ Shortest distance = $\frac{2 a}{\sqrt{3}}$ ,(By formula).

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A square ABCD of diagonal 2a is folded along the diagonal AC so that the planes DAC and BAC are at right angle. The shortest distance between DC and AB is

The correct option is B 2a√3When folded co-ordinates will be D(0,0, a); C(a, 0, 0); A(-a, 0, 0); B(0, -a, 0) Equation DC is, xa=y0=z−a−a Equation AB is, x+aa=y−a=z0 ∴ Shortest distance =2a√3,(By formula).