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Question #41730

Difficulty: 8/10 | 🟢 Free

The Question

Let t0∈(π,2π)t_0 \in (\pi, 2\pi) be the unique solution to the equation sin⁡−1(sin⁡t)+cos⁡−1(cos⁡t)=π2\sin^{-1}(\sin t) + \cos^{-1}(\cos t) = \frac{\pi}{2}. A line LL passes through the point P(2,−1,3)P(2, -1, 3) and is parallel to the vector v⃗=sin⁡−1(sin⁡t0)i^+cos⁡−1(cos⁡t0)j^+tan⁡−1(tan⁡t0)k^\vec{v} = \sin^{-1}(\sin t_0) \hat{i} + \cos^{-1}(\cos t_0) \hat{j} + \tan^{-1}(\tan t_0) \hat{k}. If the line LL intersects the plane x+y+z=10x + y + z = 10 at the point QQ, then the value of PQ2PQ^2 is:

Options

  • A.

    44

  • B.

    12

  • C.

    108

  • D.

    56

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

Parametric form pitfalls|The Sum-Difference Formula Constraints|Principal Value Branch Boundaries

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly identify the principal value branches of inverse trigonometric functions.

A common lethal mistake is assuming f−1(f(x))=xf^{-1}(f(x)) = x universally, which leads to the incorrect direction vector v⃗∝⟨1,1,1⟩\vec{v} \propto \langle 1, 1, 1 \rangle and a 'ghost' intersection point yielding PQ2=12PQ^2 = 12.

By forcing the parameter t0t_0 into the third quadrant, the problem demands rigorous application of domain constraints for arcsin⁡\arcsin, arccos⁡\arccos, and arctan⁡\arctan to find the true parametric direction of the line.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterThree-Dimensional Geometry - Class 12
ConceptParametric form pitfalls
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree