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Question #41216

Difficulty: 8/10 | 🟢 Free

The Question

Let $P(x_0, y_0, z_0)$ be a point in 3D space where $x_0 = \sin^{-1}\left(\sin \frac{17\pi}{5}\right)$ , $y_0 = \cos^{-1}\left(\cos \frac{17\pi}{5}\right)$ , and $z_0 = \tan^{-1}\left(\tan \frac{17\pi}{5}\right)$ . If $Q(\alpha, \beta, \gamma)$ is the image of $P$ in the plane $x + y - z = 0$ , then the coordinates of $Q$ are:

Options

A.
$\left( -\frac{4\pi}{15}, \frac{11\pi}{15}, \frac{4\pi}{15} \right)$
B.
$\left( \frac{17\pi}{15}, \frac{17\pi}{15}, \frac{17\pi}{3} \right)$
C.
$\left( \frac{2\pi}{15}, \frac{2\pi}{15}, \frac{2\pi}{3} \right)$
D.
$\left( -\frac{\pi}{3}, \frac{2\pi}{3}, \frac{\pi}{3} \right)$

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|Image of a point in a plane|Octant Sign Convention|Parametric form pitfalls

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to correctly evaluate inverse trigonometric functions outside their principal domains (Quadrant Adjustment Logic) before applying 3D geometry transformations.

A common trap is assuming $f^{-1}(f(x)) = x$ universally, which leads to incorrect initial coordinates (Option B).

Furthermore, students might confuse the image formula with the foot of the perpendicular formula by missing the factor of 2 (Option D), or misapply the reference angles and octant sign conventions when dealing with the negative coordinates generated by the inverse sine function (Option C).

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Inverse Trigonometric Functions - Class 12
Concept	The Quadrant Adjustment Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free