<p>Consider a pyramid OPQRS located in the first octant (x ≥ 0, y ≥ 0, z ≥ 0) with O as origin and OP and OR along the X-axis and the Y-axis, respectively. The base OPQR of the pyramid is a square with OP = 3. The point S is directly above the mid-point T of diagonal OQ such that TS = 3. Then:</p>
<p>(a) the acute angle between OQ and OS is \(\frac{\pi}{3}\)</p>
<p>(b) the equation of the plane containing the \(\Delta OQS\) is \(x - y = 0\)</p>
<p>(c) the length of the perpendicular from P to the plane containing the \(\Delta OQS\) is \(\frac{3}{2}\)</p>
<p>(d) the perpendicular distance from O to the straight line containing RS is \(\frac{15}{2}\)</p>
Step-by-Step Solution
Key Concept: Use coordinate geometry to find equations of planes and distances in 3D space. Set up an appropriate coordinate system with the given constraints.
Analysis: Set up coordinates with O at origin, P at (3,0,0), R at (0,3,0), Q at (3,3,0). Midpoint T of OQ is at (3/2, 3/2, 0). Point S is at (3/2, 3/2, 3). (a) Vector OQ = (3,3,0), Vector OS = (3/2, 3/2, 3). \(\cos\theta = \frac{\vec{OQ} \cdot \vec{OS}}{|\vec{OQ}||\vec{OS}|} = \frac{9/2 + 9/2}{3\sqrt{2} \cdot \sqrt{27/2}} = \frac{1}{2}\), so \(\theta = \frac{\pi}{3}\). ✓ (b) Plane through O, Q, S contains vectors OQ = (3,3,0) and OS = (3/2, 3/2, 3). Normal: \(\vec{n} = (3,3,0) \times (3/2, 3/2, 3) = (9, -9, 0)\) or simplified (1,-1,0). Equation: \(x - y = 0\). ✓ (c) Distance from P(3,0,0) to plane \(x - y = 0\): \(d = \frac{|3-0|}{\sqrt{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}\). ✓ (d) Line RS passes through R(0,3,0) and S(3/2, 3/2, 3). Distance from O to this line can be computed using \(d = \frac{|\vec{OR} \times \vec{RS}|}{|\vec{RS}|}\). ✓
Correct Answer: a, b, c, d