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Question #41164

Difficulty: 8/10 | 🟢 Free

The Question

Let u⃗=sin⁡−1(sin⁡10)i^+cos⁡−1(cos⁡10)j^+tan⁡−1(tan⁡10)k^\vec{u} = \sin^{-1}(\sin 10)\hat{i} + \cos^{-1}(\cos 10)\hat{j} + \tan^{-1}(\tan 10)\hat{k}. If the vectors u⃗\vec{u}, v⃗=2i^+j^+3k^\vec{v} = 2\hat{i} + \hat{j} + 3\hat{k}, and w⃗=i^+λj^+2k^\vec{w} = \hat{i} + \lambda\hat{j} + 2\hat{k} are coplanar, then the value of λ\lambda is:

Options

  • A.

    π−43π−10\frac{\pi - 4}{3\pi - 10}

  • B.

    π10−3π\frac{\pi}{10 - 3\pi}

  • C.

    45\frac{4}{5}

  • D.

    25\frac{2}{5}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

The Quadrant Adjustment Logic|Collinearity and Coplanarity|Coplanarity Condition

Why This Question Is Tricky

Common Mistakes

This question tests the 'Quadrant Adjustment Logic' intertwined with vector coplanarity.

The primary trap lies in the evaluation of the inverse trigonometric functions.

Students frequently assume sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x or incorrectly map the angle to the wrong principal branch, failing to recognize that 10 radians falls in the 4th quadrant relative to 3π3\pi for sine and tangent, but requires referencing 4π4\pi for cosine to stay within [0,π][0, \pi].

This leads to incorrect signs or offsets in the vector components (e.g., assuming all components are identical or merely differ by a sign).

When these flawed components are substituted into the scalar triple product determinant, it yields an incorrect value for λ\lambda, perfectly matching the distractor options.

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