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Question #67774

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ be a $2 \times 2$ symmetric matrix such that $A^3 = A$ , $\det(A) \neq 0$ , and $A \neq \pm I$ . A family of curves in the $xy$ -plane is given by the position vector $\mathbf{r}(t)$ which satisfies the differential equation $\frac{d^2\mathbf{r}}{dt^2} = \mathbf{r}$ , subject to the initial conditions $\mathbf{r}(0) = \mathbf{c}$ and $\mathbf{r}'(0) = A\mathbf{c}$ , where $\mathbf{c} \in \mathbb{R}^2$ is an arbitrary non-zero constant vector. Which of the following represents the family of orthogonal trajectories to these curves? (Here $K$ is an arbitrary real constant).

Options

A.
\mathbf{r}^T A \mathbf{r} = K
B.
\mathbf{r}^T \mathbf{r} = K
C.
\mathbf{r}^T (I - A) \mathbf{r} = K
D.
\mathbf{r}^T (I + A) \mathbf{r} = K

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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

Idempotent and Involutory Properties|Orthogonal Trajectories|Geometric Interpretation

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question tests the student's ability to avoid a massive algebraic trap by recognizing geometric invariants and matrix properties.

By hiding the first-order ODE $\mathbf{r}' = A\mathbf{r}$ behind a second-order IVP and the condition $A^3=A$ , students are baited into parameterizing $\mathbf{r}(t)$ into $x(t)$ and $y(t)$ components using a general symmetric matrix.

This leads to a 'Lethal Collision' of hyperbolic and trigonometric terms that is nearly impossible to eliminate algebraically to find $dy/dx$ .

The elegant geometric interpretation relies on recognizing the involutory property ( $A^2=I$ ) to reduce the system to $\mathbf{r}' = A\mathbf{r}$ , and then using matrix transposes and symmetry to directly integrate the orthogonal condition $(A\mathbf{r}) \cdot d\mathbf{r} = 0$ into the quadratic form $\mathbf{r}^T A \mathbf{r} = K$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Matrices - Class 12
Concept	Idempotent and Involutory Properties
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free