Before You Begin

Question #40695

Difficulty: 8/10 | 🟢 Free

The Question

A variable plane $P$ has non-zero intercepts on the $x, y,$ and $z$ axes, which respectively form the first, second, and third terms of a geometric progression with common ratio $r$ ( $r \neq 0, 1$ ). The plane $P$ passes through the point $(1, 1, 1)$ . Let $L$ be the line passing through the origin with direction ratios $(1, -r, 0)$ . The orthogonal projection of the line $L$ onto the plane $P$ is the line $L'$ . If $L'$ lies in the plane $z = c$ , then the maximum value of $c$ as $r$ varies is:

Options

A.
$\frac{4}{3}$
B.
$\frac{3}{4}$
C.
1
D.
$\frac{2}{3}$

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

Selection of GP Terms|Variable plane intercepts|Line-Plane intersection logic|Orthogonal projection of a line

Why This Question Is Tricky

Common Mistakes

The problem tests the student's ability to handle variable plane intercepts and orthogonal projections.

A common trap is attempting to find a finite intersection point between the line and the plane, failing to realize that the line's direction ratios $(1, -r, 0)$ and the plane's normal $(r^2, r, 1)$ are orthogonal, making them parallel.

This leads to a degenerate system of equations.

Furthermore, incorrectly assigning the GP terms to the axes destroys this parallelism, resulting in a variable $z$ -coordinate and a breakdown of the locus condition.

Recognizing the parallelism simplifies the projection to finding the perpendicular distance of the origin, elegantly reducing the problem to maximizing a rational function.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sequences and Series - Class 11
Concept	Selection of GP Terms
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free