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Question #41745

Difficulty: 8/10 | 🟢 Free

The Question

Consider the family of lines $L_\alpha$ defined by $\frac{x - 1}{\alpha^2 - 4} = \frac{y + 1}{|\alpha - 2|(\alpha + 2)} = \frac{z - 2}{\alpha^2 - 4}$ for $\alpha \in (0, 2)$ . Let $L'$ be the orthogonal projection of the line $L_\alpha$ onto the plane $\Pi: x + y + z = 3$ . If $L_2$ is the line given by $\frac{x - 2}{1} = \frac{y}{-1} = \frac{z - 1}{2}$ , then the shortest distance between $L'$ and $L_2$ is

Options

A.
$\frac{4}{\sqrt{11}}$
B.
$\frac{2\sqrt{30}}{9}$
C.
$\sqrt{2}$
D.
$\frac{4}{\sqrt{14}}$

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

Orthogonal projection of a line|Shortest distance between skew lines|The Zero-Factor Cancellation Trap|The Modulus Trap

Why This Question Is Tricky

Common Mistakes

This question tests the 'Modulus Trap' combined with orthogonal projections.

Students often blindly cancel the terms $(\alpha^2 - 4)$ and $|\alpha - 2|(\alpha + 2)$ in the direction ratios without considering the domain $\alpha \in (0, 2)$ .

This algebraic oversight leads to an incorrect direction vector $\vec{d} = \hat{i} + \hat{j} + \hat{k}$ .

Because this incorrect vector is parallel to the normal of the plane $\Pi$ , its orthogonal projection degenerates into a single point rather than a line.

Calculating the distance from this degenerated point to $L_2$ yields the false positive option $\frac{2\sqrt{30}}{9}$ , heavily punishing students who fail to evaluate the signum discontinuity.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Three-Dimensional Geometry - Class 12
Concept	Orthogonal projection of a line
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free