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Question #74678

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The Question

Let $A$ and $B$ be two points with position vectors $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 4\hat{i} + 5\hat{j} + 6\hat{k}$ respectively. A point $P$ with position vector $\vec{p}$ divides the line segment $AB$ internally in the ratio $\lambda : 1$ . Let $M$ be a $3 \times 3$ matrix defined as: $M = \begin{pmatrix} 0 & \lambda^3 - 216 & \lambda^2 - 36 \\ 216 - \lambda^3 & 0 & \lambda - 6 \\ 36 - \lambda^2 & 6 - \lambda & 0 \end{pmatrix}$ If $M$ is an involutory matrix, then the position vector $\vec{p}$ is:

Options

A.
$\frac{25}{7}\hat{i} + \frac{32}{7}\hat{j} + \frac{39}{7}\hat{k}$
B.
3\hat{i} + 4\hat{j} + 5\hat{k}
C.
6\hat{i} + 6\hat{j} + 6\hat{k}
D.
\text{No such vector } \vec{p} \text{ exists}

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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

Section Formula Internals|Skew-Symmetric Determinant|Idempotent and Involutory Properties

Why This Question Is Tricky

Examiner Levers

Logical Assertion

Tests the causal relationship between two true statements.

Difficulty Boost : +4

Boundary Condition

Tests behavior at limits like 0, 1, infinity, or empty sets.

Difficulty Boost : +4

Common Mistakes

This question tests the conceptual intersection of linear algebra and coordinate geometry, specifically targeting the properties of skew-symmetric and involutory matrices.

The primary trap lies in the matrix entries: students will likely notice the roots at $\lambda = 6$ (from $\lambda^3 - 216 = 0$ ) and incorrectly assume this value satisfies the matrix condition, confusing an involutory matrix ( $M^2 = I$ ) with an idempotent one ( $M^2 = M$ , which holds for the zero matrix when $\lambda = 6$ ).

The rigorous mathematical bottleneck is that $M$ is a $3 \times 3$ (odd-order) skew-symmetric matrix, meaning its determinant is strictly zero.

An involutory matrix requires $\det(M^2) = \det(I) \implies (\det(M))^2 = 1$ .

Equating $0 = 1$ yields a fundamental contradiction, proving that no such scalar ratio $\lambda$ can exist, rendering the section formula calculation a deliberate dead-end.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Vector Algebra - Class 12
Concept	Section Formula Internals
Difficulty	8
Cognitive Load	8
Target Solve Time	180 sec
Access	Free