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Question #41656

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of all real values of $\lambda$ such that the orthogonal projection of the line $L: \frac{x - 1}{\lambda} = \frac{y - 2}{\lambda^2 - 4} = \frac{z - 1}{\lambda - 1}$ on the plane $\Pi: x + 3y + 2z = 0$ is coplanar with the z-axis. If $z \in \mathbb{C}$ satisfies $|z - \lambda| \le 4$ for all $\lambda \in S$ , then the maximum value of $|z - 2|$ is:

Options

A.
$\sqrt{13}$
B.
5
C.
4
D.
$\sqrt{15}$

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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|Collinearity vs Coplanarity|Triangle Inequality Mind Games

Why This Question Is Tricky

Common Mistakes

The problem integrates 3D geometry with complex numbers.

The core trap lies in the cross product $\vec{n} = \vec{d} \times \vec{n}_\Pi$ .

Students often divide by the common factor $(\lambda + 1)$ assuming $\vec{n} \neq \vec{0}$ , which causes them to lose the boundary solution $\lambda = -1$ .

At $\lambda = -1$ , the line is normal to the plane, and its projection degenerates to a point.

Since a point is always coplanar with a line, this is a valid solution.

Missing this degenerate case leads to $S = \{3\}$ , resulting in the incorrect maximum distance of 5.

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