Consider a pyramid OPQRS located in the first octant \left(\right. x \geq 0 , \textrm{ } y \geq 0 , \textrm{ } z \geq 0 \left.\right) 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

Engineering

Mathematics

Scalar Triple Product

Question

Consider a pyramid OPQRS located in the first octant $(x \geq 0, y \geq 0, z \geq 0)$ 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

The length of the perpendicular from P to the plane containing the triangle OQS is $\frac{3}{\sqrt{2}}$

The perpendicular distance from O to the straight line containing RS is $\sqrt{\frac{15}{2}}$

The equation of the plane containing the triangle OQS is x – y = 0

The acute angle between OQ and OS is $\frac{π}{3}$

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(A) $OT = \sqrt{\frac{9}{4} + \frac{9}{4}} = \frac{3}{\sqrt{2}}$

$OS = \sqrt{\frac{9}{4} + \frac{9}{4} + 9} = \frac{3 \sqrt{3}}{\sqrt{2}}$

$\cos θ = \frac{\frac{3}{\sqrt{2}}}{\frac{3 \sqrt{3}}{\sqrt{2}}} = \frac{1}{\sqrt{3}}$

(B) Equation of plane OQS

$| \begin{matrix} x & y & z \\ 3 & 3 & 0 \\ 3 / 2 & 3 / 2 & 3 \end{matrix} | = 0 \Rightarrow x - y = 0$

(D) Straight line RS

$\frac{x - 0}{\frac{3}{2}} = \frac{y - 3}{\frac{- 3}{2}} = \frac{z}{3} \Rightarrow \frac{x}{1} = \frac{y - 3}{- 1} = \frac{z}{2} = λ$

Let M º (λ, – λ + 3, 2λ)

OM perpendicular to RS Þ λ + λ – 3 + 4λ = 0 ⇒ $λ = \frac{1}{2}$

$∴ M \equiv (\frac{1}{2}, \frac{5}{2}, 1), OM = \sqrt{\frac{1}{4} + \frac{25}{4} + 1} = \sqrt{\frac{15}{2}} .$