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Question #41747

Difficulty: 8/10 | 🟢 Free

The Question

Let $L_\theta$ be a family of lines passing through the origin with the direction vector $\vec{d}(\theta) = (2\theta^2) \hat{i} + (4\theta - 2\sin \theta + 3 - 3\cos \theta) \hat{j} + (\theta^3) \hat{k}$ for $\theta \in (0, \pi/2)$ . Let $P_\theta$ be the orthogonal projection of the line $L_\theta$ onto the $xy$ -plane. As $\theta \to 0$ , the line $P_\theta$ approaches a limiting position denoted by the line $L_0$ . Find the shortest distance between the line $L_0$ and the point $Q(4, 3, 10)$ .

Options

A.
$2\sqrt{29}$
B.
10
C.
$\sqrt{109}$
D.
5

Check Your Answer

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

Orthogonal projection of a line|Line lying in a plane|The L'Hôpital's Rule Misuse|L'Hopital's Pitfall

Why This Question Is Tricky

Common Mistakes

This question tests the rigorous evaluation of limits in a geometric context, specifically targeting the mechanical misuse of L'Hôpital's Rule.

Students often blindly apply L'Hôpital's Rule multiple times when they see a $\theta^2$ in the denominator.

If they differentiate the numerator twice without checking if the first derivative evaluates to $0/0$ , they will incorrectly calculate the limit as $3/4$ .

This leads them to the wrong limiting line $y = \frac{3}{4}x$ and a trap distance of $10$ .

The problem forces students to verify the indeterminate form at every step of limit evaluation, penalizing rote algorithmic calculus.

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