JEE Advanced · Mathematics · 31. 3D Geometry

Consider a pyramid $O P Q R S$ located in the first octant $(x \geq 0, y \geq 0, z \geq 0)$ with $O$ , as origin, and $O P$ and, $O R$ along the $x ‐ a x i s$ and the $y ‐ a x i s$ , respectively. The base $O P Q R$ of the pyramid is a square with $O P = 3$ . The point S is directly above the mid-point T of diagonal, $O Q$ , such that, $T S = 3$ . Then

A The acute angle between OQ and OS is $\frac{π}{3}$
B The equation of the plane containing the triangle OQS is $x - y = 0$
C The length of the perpendicular from P to the plane containing the triangle OQS is $\frac{3}{\sqrt{2}}$
D The perpendicular distance from O to the straight line containing RS is $\sqrt{\frac{15}{2}}$

Verified Solution

Answer & Solution

Correct Answer

(B) The equation of the plane containing the triangle OQS is $x - y = 0$

Step-by-step Solution

Detailed explanation

O (0,0, 0) \to Origin

P (3,0, 0) \to on x axis

R (0,3, 0) \to on y axis

Q (3,3, 0)

T (\frac{3}{2}, \frac{3}{2}, 0), 5 (\frac{3}{2}, \frac{3}{2}, 3)

Given OP = OR = 3 and OPQR is a square

\Rightarrow

O Q = 3 \sqrt{2} \Rightarrow O T = \frac{3}{\sqrt{2}}

and

S T = 3

Let θ be a angle between OQ & OS
Using

Δ S O T, \tan θ = \frac{S T}{O T} = \sqrt{2} \Rightarrow θ = \tan^{- 1} \sqrt{2}

Clearly, equation of plane containing triangle OQS is x - y = 0 as

O (0,0, 0), Q (3,3, 0), S (\frac{3}{2}, \frac{3}{2}, 3) lies on it

let ax + by + cz = d

(0,0, 0) \Rightarrow d = 0

(3,3, 0) \Rightarrow a + b = 0

(\frac{3}{2}, \frac{3}{2}, 3) \Rightarrow \frac{3 a}{2} + \frac{3 b}{2} + 3 c = 0

\Rightarrow c = 0

\Rightarrow b = - a

x - y = 0

Also, length of perpendicular from P to the plane containing the triangle OQS is PT

= \frac{3}{\sqrt{2}}

.
Also equation of RS is

\vec{r} = 3 \hat{j} + t (\frac{3}{2} \hat{i} - \frac{3}{2} \hat{j} + 3 \hat{k})

= (\frac{3 t}{2}, 3 - \frac{3 t}{2}, 3 t)

Let M be foot of

⊥

from origin on line passing through R,S
Let co - ordinates of

M = (\frac{3 t}{2}, 3 - \frac{3 t}{2}, 3 t)

∵

\vec{O M} . \vec{R S} = 0

\Rightarrow

\frac{9}{4} t - \frac{3}{2} (3 - \frac{3 t}{2}) + 9 t = 0

\Rightarrow

\frac{9 t}{2} + 9 t = \frac{9}{2}

\Rightarrow

t = \frac{1}{3}

∴

M = (\frac{1}{2}, \frac{5}{2}, 1)

\Rightarrow

O M = \sqrt{\frac{1}{4} + \frac{25}{4} + 1} = \sqrt{\frac{30}{4}} = \sqrt{\frac{15}{2}}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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D)	$d^{2} s p^{3}$	S)	${[F e C l_{4}]}^{2 -}$
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Consider a pyramidOPQRS located in the first octant x ≥0, y ≥0, z ≥0 withO, as origin, andOP and,OR along the x‐axis and the y‐axis, respectively. The baseOPQR of the pyramid is a square withOP = 3. The point S is directly above the mid-point T of diagonal, OQ, such that, TS = 3. Then

Step-by-step Solution