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Question #41654

Difficulty: 8/10 | 🟢 Free

The Question

Let $\Delta(t) = \begin{vmatrix} x & y & z \\ e^{-t} & 1-\cos t & \ln(1-t) \\ \sin t & e^{-2t} & -t \end{vmatrix}$ for $t < 1$ . A plane $P$ is defined by the equation $\left[ \frac{d}{dt} \Delta(t) \right]_{t=0} = 0$ . Let $L_1$ be the line $\frac{x-5}{1} = \frac{y-6}{2} = \frac{z}{1}$ and $L_1'$ be its orthogonal projection onto the plane $P$ . If $d$ is the shortest distance between $L_1'$ and the line $L_2: \frac{x}{2} = \frac{y-2}{-1} = \frac{z-7}{0}$ , then the value of $d^2$ is:

Options

A.
25
B.
30
C.
36
D.
42

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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

Orthogonal projection of a line|Distance from point to plane|Differentiation of Determinants

Why This Question Is Tricky

Common Mistakes

This question tests the differentiation of determinants and 3D geometry projections.

The primary trap lies in the determinant differentiation: students often incorrectly differentiate all elements simultaneously, yielding a null vector for the plane's normal, or they waste time expanding the determinant before differentiating, risking product rule errors.

A secondary geometric trap is failing to check the dot product between the line's direction and the plane's normal.

Recognizing that $L_1$ is parallel to $P$ bypasses the complex derivation of the projected line's direction vector, reducing the problem to a simple point projection.

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