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Question #75897

Difficulty: 8/10 | 🟢 Free

The Question

Let PP be a 3×33 \times 3 matrix with det⁡(P)=2024\det(P) = 2024. Let Q=(sin⁡xcos⁡xsin⁡(x−λx∘)sin⁡ycos⁡ysin⁡(y−λy∘)sin⁡zcos⁡zsin⁡(z−λz∘))Q = \begin{pmatrix} \sin x & \cos x & \sin(x - \lambda x^\circ) \\ \sin y & \cos y & \sin(y - \lambda y^\circ) \\ \sin z & \cos z & \sin(z - \lambda z^\circ) \end{pmatrix} where x,y,z∈Rx, y, z \in \mathbb{R} are measured in radians and x∘x^\circ denotes the equivalent angle in degrees. Let M=PQTM = P Q^T. If the row vectors of MM are coplanar for all x,y,z∈Rx, y, z \in \mathbb{R}, let SS be the sum of all possible strictly positive values of λ\lambda. Find SS.

Options

  • A.

    540π\frac{540}{\pi}

  • B.

    360π\frac{360}{\pi}

  • C.

    180π\frac{180}{\pi}

  • D.

    720π\frac{720}{\pi}

Check Your Answer

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

Coplanarity Condition|Determinant Product Rule|Transpose of Product|Radian-Degree Conversion Identity

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

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

Difficulty Boost : +4

Common Mistakes

This question tests the integration of the Determinant Product Rule with vector coplanarity, while embedding a rigorous Radian-Degree Conversion trap.

Students must recognize that det⁡(PQT)=det⁡(P)det⁡(Q)=0\det(PQ^T) = \det(P)\det(Q) = 0.

The primary trap lies in the term sin⁡(x−λx∘)\sin(x - \lambda x^\circ).

Students often incorrectly treat x∘x^\circ as a constant angle shift, assuming the third column is trivially a linear combination of the first two via standard trigonometric expansion.

However, x∘x^\circ is a variable dependent on xx (x∘=π180xx^\circ = \frac{\pi}{180}x).

Recognizing this unit mismatch forces the student to solve the functional identity sin⁡(kx)=Asin⁡x+Bcos⁡x\sin(kx) = A\sin x + B\cos x, rigorously restricting kk to {−1,0,1}\{-1, 0, 1\} and yielding the correct values for λ\lambda.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterVector Algebra - Class 12
ConceptCoplanarity Condition
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree