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Question #76198

Difficulty: 8/10 | 🟢 Free

The Question

Let $P$ be a plane passing through the origin with normal vector $\vec{n} = (\sin \theta) \hat{i} + (\cos \theta) \hat{j} + (\tan \theta) \hat{k}$ . The angle $\theta \in \left(0, \frac{\pi}{4}\right)$ satisfies the equation $\tan^2 \theta - \left(a + \frac{1}{a}\right)\tan \theta + 1 = 0$ , where $a > 1$ . Let $L$ be a line passing through the origin and parallel to the vector $\vec{d} = a \hat{i} + \hat{j}$ . If $\phi$ is the acute angle between the line $L$ and the plane $P$ , which of the following represents $\sin \phi$ ?

Options

A.
$\frac{2a^2}{(a^2+1)^{3/2}}$
B.
$\frac{1}{\sqrt{a^2+1}}$
C.
$\frac{2a}{a^2+1}$
D.
$\frac{a^2}{(a^2+1)^{3/2}}$

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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

Unit Vector Normalization|Angle between line and plane|Extraneous Solution Rejection

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 synthesis of 3D geometry and trigonometric constraints, specifically targeting the 'Extraneous Solution Rejection' trap.

Students must determine the plane's normal vector by solving a quadratic equation in $\tan \theta$ .

The trap lies in the roots $a$ and $1/a$ ; ignoring the domain constraint $\theta \in (0, \pi/4)$ leads to selecting the extraneous root $\tan \theta = a$ (since $a > 1$ ), which yields the incorrect option B.

Furthermore, the problem tests 'Unit Vector Normalization' by defining a normal vector whose magnitude is $\sec \theta \neq 1$ .

Students who forget to divide by $|\vec{n}|$ will incorrectly arrive at option C.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Vector Algebra - Class 12
Concept	Unit Vector Normalization
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free