Properties of Pyramid OPQRS in First Octant with Square Base

To analyze the properties of the pyramid $OPQRS$, we first establish the coordinates of its vertices in three-dimensional space based on the given information:

1. Origin: $O = (0, 0, 0)$.

2. Base $OPQR$: The base is a square of side length $OP = 3$ situated in the $xy$-plane ($z = 0$) within the first octant.

   • Since $OP$ lies along the positive $x$-axis: $P = (3, 0, 0)$.
   • Since $OR$ lies along the positive $y$-axis: $R = (0, 3, 0)$.
   • $Q$ is the fourth vertex of the square base: $Q = (3, 3, 0)$.

3. Mid-point $T$ of diagonal $OQ$: $T = \left(\frac{0+3}{2}, \frac{0+3}{2}, 0\right) = \left(\frac{3}{2}, \frac{3}{2}, 0\right)$

4. Apex $S$: $S$ is directly above $T$ along the positive $z$-axis at a distance of $TS = 3$: $S = \left(\frac{3}{2}, \frac{3}{2}, 3\right)$

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Analysis of Options:

Option (A): Angle between $OQ$ and $OS$

The position vectors of $Q$ and $S$ relative to $O$ are: $\vec{OQ} = 3\hat{i} + 3\hat{j}$ $\vec{OS} = \frac{3}{2}\hat{i} + \frac{3}{2}\hat{j} + 3\hat{k}$

The dot product of $\vec{OQ}$ and $\vec{OS}$ is: $\vec{OQ} \cdot \vec{OS} = 3\left(\frac{3}{2}\right) + 3\left(\frac{3}{2}\right) + 0(3) = \frac{9}{2} + \frac{9}{2} = 9$

The magnitudes of $\vec{OQ}$ and $\vec{OS}$ are: $|\vec{OQ}| = \sqrt{3^2 + 3^2 + 0^2} = 3\sqrt{2}$ $|\vec{OS}| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{3}{2}\right)^2 + 3^2} = \sqrt{\frac{9}{4} + \frac{9}{4} + 9} = \sqrt{\frac{27}{2}} = \frac{3\sqrt{6}}{2}$

Let $\theta$ be the acute angle between $OQ$ and $OS$: $\cos \theta = \frac{\vec{OQ} \cdot \vec{OS}}{|\vec{OQ}||\vec{OS}|} = \frac{9}{(3\sqrt{2}) \cdot \left(\frac{3\sqrt{6}}{2}\right)} = \frac{9}{9\sqrt{3}} = \frac{1}{\sqrt{3}}$

Since $\cos \theta = \frac{1}{\sqrt{3}} \neq \frac{1}{2}$, $\theta \neq \frac{\pi}{3}$. Thus, Option (A) is incorrect.

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Option (B): Equation of the plane containing triangle $OQS$

A normal vector $\vec{n}$ to the plane containing $O(0,0,0)$, $Q(3,3,0)$, and $S\left(\frac{3}{2}, \frac{3}{2}, 3\right)$ is given by:

$$\vec{n} = \vec{OQ} \times \vec{OS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 3 & 0 \\ \frac{3}{2} & \frac{3}{2} & 3 \end{vmatrix} = \hat{i}(9 - 0) - \hat{j}(9 - 0) + \hat{k}\left(\frac{9}{2} - \frac{9}{2}\right) = 9\hat{i} - 9\hat{j}$$

Simplifying the normal vector gives $\vec{n} = \hat{i} - \hat{j}$. Since the plane passes through the origin $O(0,0,0)$, its equation is: $1(x - 0) - 1(y - 0) + 0(z - 0) = 0 \implies x - y = 0$

Thus, Option (B) is correct.

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Option (C): Perpendicular distance from $P$ to the plane $OQS$

The equation of the plane containing $\triangle OQS$ is $x - y = 0$. The perpendicular distance $d$ from the point $P(3, 0, 0)$ to this plane is: $d = \frac{|3 - 0|}{\sqrt{1^2 + (-1)^2 + 0^2}} = \frac{3}{\sqrt{2}}$

Thus, Option (C) is correct.

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Option (D): Perpendicular distance from $O$ to the line containing $RS$

The line containing $RS$ passes through $R(0, 3, 0)$ and $S\left(\frac{3}{2}, \frac{3}{2}, 3\right)$. The direction vector $\vec{d}_{RS}$ of the line $RS$ is: $\vec{d}_{RS} = \vec{S} - \vec{R} = \left(\frac{3}{2} - 0\right)\hat{i} + \left(\frac{3}{2} - 3\right)\hat{j} + (3 - 0)\hat{k} = \frac{3}{2}\hat{i} - \frac{3}{2}\hat{j} + 3\hat{k}$

We can take a parallel direction vector $\vec{u} = \hat{i} - \hat{j} + 2\hat{k}$.

The perpendicular distance $D$ from origin $O(0,0,0)$ to the line passing through point $R$ with direction vector $\vec{u}$ is: $D = \frac{|\vec{OR} \times \vec{u}|}{|\vec{u}|}$

Evaluating the vector cross product: $\vec{OR} \times \vec{u} = (3\hat{j}) \times (\hat{i} - \hat{j} + 2\hat{k}) = 3(\hat{j} \times \hat{i}) + 6(\hat{j} \times \hat{k}) = -3\hat{k} + 6\hat{i} = 6\hat{i} - 3\hat{k}$

Magnitudes: $|\vec{OR} \times \vec{u}| = \sqrt{6^2 + 0^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}$ $|\vec{u}| = \sqrt{1^2 + (-1)^2 + 2^2} = \sqrt{6}$

Therefore, the perpendicular distance is: $D = \frac{3\sqrt{5}}{\sqrt{6}} = \sqrt{\frac{9 \times 5}{6}} = \sqrt{\frac{15}{2}}$

Thus, Option (D) is correct.

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Conclusion:

The correct options are B, C, and D.