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Question #40437

Difficulty: 8/10 | 🟢 Free

The Question

Let $L$ be a line passing through the point $P(3, 4, 5)$ with direction cosines $(l, m, n)$ . The line $L$ intersects the line $L_1: x-1 = y-2 = z-3$ at exactly one point, and also intersects the line $L_2: \frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ at exactly one point. If $S$ is the set of all possible values of $m$ , then $S$ is equal to:

Options

A.
$\left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$
B.
$\left[-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right] \setminus \left\{-\sqrt{\frac{2}{7}}, \sqrt{\frac{2}{7}}\right\}$
C.
$\left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \setminus \left\{-\sqrt{\frac{2}{7}}, \sqrt{\frac{2}{7}}\right\}$
D.
$\left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \setminus \{0\}$

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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

Auxiliary Angle Method|Direction Cosines constraint|Collinearity and Coplanarity

Why This Question Is Tricky

Common Mistakes

This question tests the 'Collinearity and Coplanarity' traps in 3D geometry combined with the Auxiliary Angle Method.

Students often correctly identify the coplanarity condition $l-2m+n=0$ and use trigonometric parameterization to find the absolute bounds $m \in [-1/\sqrt{3}, 1/\sqrt{3}]$ .

However, the lethal trap lies in the intersection constraints: if $m = \pm 1/\sqrt{3}$ , $L$ becomes collinear with $L_1$ (intersecting at infinitely many points), and if $m = \pm\sqrt{2/7}$ , $L$ becomes strictly parallel to $L_2$ (intersecting nowhere).

Failing to check the cross product of direction vectors for these boundary and internal points leads to incorrect interval closures.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Trigonometric Functions - Class 11
Concept	Auxiliary Angle Method
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free