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Question #41661

Difficulty: 8/10 | 🟢 Free

The Question

For a parameter $\theta \in [0, 2\pi]$ , let $P_1$ and $P_2$ be two planes with normal vectors $\vec{n_1} = \cos 3\theta \hat{i} + \sin 2\theta \hat{j} + \cos \theta \hat{k}$ and $\vec{n_2} = \cos \theta \hat{i} + \sin 2\theta \hat{j} + \cos 3\theta \hat{k}$ respectively. The sum of all values of $\theta$ for which $P_1$ and $P_2$ are perpendicular is:

Options

A.
$4\pi$
B.
$6\pi$
C.
$2\pi$
D.
$8\pi$

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

Angle between two planes|Trigonometric Identical Substitutions|Double and Triple Angle Ratios

Why This Question Is Tricky

Common Mistakes

The question tests the intersection of 3D geometry and trigonometric equations.

The primary trap lies in the mechanical solving of the dot product equation $\vec{n_1} \cdot \vec{n_2} = 0$ .

Students will correctly reduce the trigonometric identity to $\cos 2\theta(\cos 2\theta + 1) = 0$ and find all six roots in the interval $[0, 2\pi]$ , yielding a sum of $6\pi$ .

However, they often fail to verify the geometric existence condition: a plane is only defined if its normal vector is non-zero ( $|\vec{n}| \neq 0$ ).

At the roots $\theta = \pi/2$ and $3\pi/2$ (where $\cos 2\theta = -1$ ), the normal vectors collapse to the null vector $\vec{0}$ , rendering the planes undefined.

Recognizing this domain constraint is a hallmark of JEE Advanced problems.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Angle between two planes
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free