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Question #65475

Difficulty: 8/10 | 🟢 Free

The Question

Let v⃗(x)=sin⁡−1(sin⁡x)i^+cos⁡−1(cos⁡x)j^+tan⁡−1(tan⁡x)k^\vec{v}(x) = \sin^{-1}(\sin x)\hat{i} + \cos^{-1}(\cos x)\hat{j} + \tan^{-1}(\tan x)\hat{k}. Let a,b,c∈(π,3π2)a, b, c \in \left(\pi, \frac{3\pi}{2}\right) be distinct real numbers, and AA be the 3×33 \times 3 matrix whose rows are v⃗(a),v⃗(b),v⃗(c)\vec{v}(a), \vec{v}(b), \vec{v}(c) respectively. Let n^\hat{n} be the unit normal vector to the row space of AA such that its i^\hat{i} component is strictly positive. If B=ATA+n^n^TB = A^T A + \hat{n}\hat{n}^T, find the value of det⁡(B)\det(B).

Options

  • A.

    2π2((a−b)2+(b−c)2+(c−a)2)2\pi^2 \left( (a-b)^2 + (b-c)^2 + (c-a)^2 \right)

  • B.

    π2((a−b)2+(b−c)2+(c−a)2)\pi^2 \left( (a-b)^2 + (b-c)^2 + (c-a)^2 \right)

  • C.

    00

  • D.

    4π2((a−b)2+(b−c)2+(c−a)2)4\pi^2 \left( (a-b)^2 + (b-c)^2 + (c-a)^2 \right)

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

The Quadrant Adjustment Logic|Linear Independence vs Dependence|Unit Vector Normalization|Properties of Transpose Determinant

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This problem tests the 'Quadrant Adjustment Logic' combined with the spectral properties of the Gramian matrix.

The primary trap lies in ignoring the quadrant constraints: if a student incorrectly assumes sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x, the vectors become collinear, making ATAA^T A rank 1 and det⁡(B)=0\det(B) = 0.

Furthermore, the 'Properties of Transpose Determinant' trap tempts students to incorrectly distribute the determinant as ∣ATA∣+∣n^n^T∣=0+0=0|A^T A| + |\hat{n}\hat{n}^T| = 0 + 0 = 0.

The correct path requires recognizing that n^n^T\hat{n}\hat{n}^T acts as a rank-1 update that perfectly fills the null space of ATAA^T A, shifting its zero eigenvalue to 1, making det⁡(B)\det(B) the product of the non-zero eigenvalues of ATAA^T A (computable via the Cauchy-Binet formula).

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterInverse Trigonometric Functions - Class 12
ConceptThe Quadrant Adjustment Logic
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree