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Q.

Let PQRS be a unit Square folded along the Diagonal $\bar{P R},$ so that ‘Q’ Reaches at $Q^{1}$ and planes $P R Q^{1}, P R S$ are perpendicular to one other then which of the following options is/are correct?

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a

Shortest distance between $\bar{P Q^{1}}, \bar{R S} = \frac{2}{\sqrt{3}}$

b

$| Q Q^{1} | = \frac{1}{\sqrt{2}}$

c

Shortest distance between $\bar{P Q^{1}}, \bar{R S} = \sqrt{\frac{2}{3}}$

d

$| Q Q^{1} | = 1$

answer is B, C.

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Detailed Solution

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Take P =(0 0 0)

$\begin{array}{l} \bar{P Q} = i, \bar{P R} = i + j, \bar{P S} = j \\ Q' = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}) \\ Q = (1 0 0) \end{array} \begin{array}{l} \overset{\leftrightarrow}{P Q'} \equiv \bar{r} = \bar{o} + t (i + j + \sqrt{2} k) \\ \overset{\leftrightarrow}{R S} \equiv \bar{r} = i + j + α (i) \\ S . D = \frac{\sqrt{2}}{\sqrt{3}} \end{array}$

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