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Question #41284

Difficulty: 8/10 | 🟢 Free

The Question

Let $A = [a_{ij}]$ be a $3 \times 3$ skew-symmetric matrix with real entries such that $A \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix}$ . For $i=1,2,3$ , consider the lines $L_i: a_{i1}x + a_{i2}y + a_{i3} = 0$ . Assuming $a_{12} \neq 0$ , the lines $L_1, L_2, L_3$ are concurrent at a point $P$ . If the locus of $P$ as the entries of $A$ vary is a curve $S$ , and $d$ is the perpendicular distance from the origin to $S$ , find the value of $10d^2$ .

Options

A.
2
B.
4
C.
5
D.
8

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Skew-Symmetric Determinant|Concurrent Lines Condition|Shifting Origin|Point-Line Relative Position

Why This Question Is Tricky

Common Mistakes

This question tests the geometric properties of lines derived from a skew-symmetric matrix.

The primary trap is the concurrency condition: students often attempt to find the locus by setting the determinant of the coefficient matrix to zero.

However, since the matrix is $3 \times 3$ skew-symmetric, its determinant is identically zero, yielding $0=0$ and trapping the student in an algebraic dead end.

The correct logical path requires bypassing the determinant entirely, explicitly solving for the intersection of the first two lines, and recognizing that the skew-symmetry guarantees this point always lies on the third line.

This shifts the problem from a matrix determinant trap to parametric locus identification.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Determinants - Class 12
Concept	Skew-Symmetric Determinant
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free