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Question #40773

Difficulty: 8/10 | 🟢 Free

The Question

Let $O$ be the origin. Consider a tetrahedron $OPQR$ where the position vectors of the vertices $P, Q,$ and $R$ are given by: $\vec{p} = \left( \sum_{k=1}^{89} \sin^2(k^\circ) \right) \hat{i} + \left( \sum_{k=1}^{89} \cos^2(k^\circ) \right) \hat{j} + 89 \hat{k}$ $\vec{q} = \left( \int_{0}^{\pi/4} \tan^2 x \, dx \right) \hat{i} + \frac{\pi}{4} \hat{j} + \left( \int_{0}^{\pi/4} \sec^2 x \, dx \right) \hat{k}$ $\vec{r} = (e^\pi - \pi^e) \hat{i} + (\pi^e + \sqrt{2}) \hat{j} + (e^\pi + \sqrt{2}) \hat{k}$ The volume of the tetrahedron $OPQR$ is:

Options

A.
0
B.
$\frac{89}{6} \left(1 - \frac{\pi}{4}\right) (e^\pi - \pi^e)$
C.
$\frac{1}{6}$
D.
$\frac{89}{6}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Area of Triangle via Determinants|Scalar Triple Product Volume Logic|Linear Independence vs Dependence

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize linear dependence (coplanarity) before blindly executing laborious calculations for the scalar triple product.

The trap is set by using intimidating expressions (trigonometric series, definite integrals, and transcendental numbers) for the vector components, which tempts the student into either giving up or wasting significant time evaluating each term to compute a $3 \times 3$ determinant.

The core JEE Advanced logic here is that if the columns of a determinant satisfy a linear relation (in this case, $C_1 + C_2 = C_3$ ), the determinant vanishes identically, rendering the volume of the tetrahedron zero.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Straight Lines - Class 11
Concept	Area of Triangle via Determinants
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free