Start Learning Free

Before You Begin

Question #73946

Difficulty: 8/10 | 🟢 Free

The Question

Let α=sin⁡−1(sin⁡10π9)\alpha = \sin^{-1}\left(\sin \frac{10\pi}{9}\right) and β=cos⁡−1(cos⁡10π9)\beta = \cos^{-1}\left(\cos \frac{10\pi}{9}\right). Let A1,A2,…,A8A_1, A_2, \dots, A_8 be 88 arithmetic means inserted between α\alpha and β\beta. Let dd be the common difference of this arithmetic progression, and define k=πdk = \frac{\pi}{d}. Consider three vectors u⃗,v⃗,w⃗\vec{u}, \vec{v}, \vec{w} defined as: u⃗=sin⁡−1(sin⁡Ak−5)i^+cos⁡−1(cos⁡Ak−3)j^+tan⁡−1(tan⁡Ak−1)k^\vec{u} = \sin^{-1}(\sin A_{k-5})\hat{i} + \cos^{-1}(\cos A_{k-3})\hat{j} + \tan^{-1}(\tan A_{k-1})\hat{k} v⃗=cos⁡−1(cos⁡Ak−1)i^+tan⁡−1(tan⁡Ak−5)j^+sin⁡−1(sin⁡Ak−3)k^\vec{v} = \cos^{-1}(\cos A_{k-1})\hat{i} + \tan^{-1}(\tan A_{k-5})\hat{j} + \sin^{-1}(\sin A_{k-3})\hat{k} w⃗=tan⁡−1(tan⁡Ak−3)i^+sin⁡−1(sin⁡Ak−1)j^+cos⁡−1(cos⁡Ak−5)k^\vec{w} = \tan^{-1}(\tan A_{k-3})\hat{i} + \sin^{-1}(\sin A_{k-1})\hat{j} + \cos^{-1}(\cos A_{k-5})\hat{k} Find the volume of the parallelepiped whose coterminous edges are represented by u⃗,v⃗\vec{u}, \vec{v}, and w⃗\vec{w}.

Options

  • A.

    26π381\frac{26\pi^3}{81}

  • B.

    20π381\frac{20\pi^3}{81}

  • C.

    5π39\frac{5\pi^3}{9}

  • D.

    π381\frac{\pi^3}{81}

Check Your Answer

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

Scalar Triple Product Volume Logic|Insertion of Means|Principal Value Branch Boundaries|The Quadrant Adjustment Logic

Why This Question Is Tricky

Examiner Levers

Recursive Logic

The answer to part A is required to solve part B (Step-wise failure test).

Difficulty Boost : +5

Common Mistakes

This question tests the integration of arithmetic progressions with vector scalar triple products, heavily gated by inverse trigonometric principal value branch (PVB) traps.

Students often incorrectly assume f−1(f(x))=xf^{-1}(f(x)) = x universally.

Here, the arithmetic means A6=5π/9A_6 = 5\pi/9 and A8=7π/9A_8 = 7\pi/9 fall outside the principal domains of sin⁡−1x\sin^{-1}x [−π/2,π/2][-\pi/2, \pi/2] and tan⁡−1x\tan^{-1}x (−π/2,π/2)(-\pi/2, \pi/2), requiring quadrant adjustments (e.g., tan⁡−1(tan⁡(5π/9))=−4π/9\tan^{-1}(\tan(5\pi/9)) = -4\pi/9).

Failing to apply these PVB mappings results in sign errors in the vector components, leading to incorrect determinant evaluations for the parallelepiped's volume (yielding the trap answers like 20π3/8120\pi^3/81 or 5π3/95\pi^3/9).

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterVector Algebra - Class 12
ConceptScalar Triple Product Volume Logic
Difficulty8
Cognitive Load7
Target Solve Time170 sec
AccessFree