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Question #74996

Difficulty: 8/10 | 🟢 Free

The Question

Let $L$ be a line with direction cosines $(l, m, n)$ satisfying the equation: $(b-c)l + (c-a)m + (a-b)n = p \sqrt{(b-c)^2 + (c-a)^2 + (a-b)^2}$ where $a, b, c$ are distinct real numbers and $p \in (0, 1)$ . Let $\Pi$ be the plane given by the equation: $(b+c-2a)x + (c+a-2b)y + (a+b-2c)z = 0$ If $\theta$ is the angle between the line $L$ and the plane $\Pi$ , what is the maximum possible value of $\theta$ ?

Options

A.
$\cos^{-1}(p)$
B.
$\sin^{-1}(p)$
C.
$\cos^{-1}\left(\frac{p}{\sqrt{2}}\right)$
D.
$\frac{\pi}{2} - \sin^{-1}(p^2)$

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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

Direction Cosines constraint|Angle between line and plane|Auxiliary Angle Method

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 3D geometry with algebraic constraints.

The primary trap is failing to recognize that the direction cosines $(l,m,n)$ inherently satisfy $l^2+m^2+n^2=1$ , which acts as a bounding sphere.

Students often attempt to use Lagrange multipliers or auxiliary angle methods directly on the coordinates without noticing the hidden orthogonality of the constraint's normal vector and the target plane's normal vector.

By transforming the problem into an orthonormal basis, the Cauchy-Schwarz inequality (or Pythagorean identity) naturally reveals the maximum projection, bypassing tedious calculus.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Direction Cosines constraint
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free