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Question #40890

Difficulty: 8/10 | 🟢 Free

The Question

Let $P_1(0,0,0)$ , $P_2(2,0,0)$ , $P_3(0,2,0)$ and $P_4(x,y,z)$ be four points in $\mathbb{R}^3$ . Let $Q_1, Q_2, Q_3, Q_4$ be any four points in $\mathbb{R}^3$ . Define a $4 \times 4$ matrix $M$ such that its entries are given by $M_{ij} = [d(P_i, Q_j)]^2$ , where $d(A,B)$ is the Euclidean distance between $A$ and $B$ . If $\det(M) = 0$ for every possible choice of $Q_1, Q_2, Q_3, Q_4$ , then the locus of $P_4$ is a curve $C$ . The length of $C$ is:

Options

A.
$2\pi$
B.
$2\sqrt{2}\pi$
C.
$4\pi$
D.
$4\sqrt{2}\pi$

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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

Distance Formula in 3D|Collinearity vs Coplanarity|Determinant Product Rule

Why This Question Is Tricky

Common Mistakes

This question tests the integration of 3D geometry, determinants, and matrix factorization.

The primary trap is assuming that $\det(M) = 0$ implies the points $P_i$ are merely coplanar (which would yield a 2D plane as the locus, leading to an infinite length).

By factoring $M = AB$ into $4 \times 5$ and $5 \times 4$ matrices, students must apply the Cauchy-Binet theorem.

The condition that $\det(M) = 0$ for all $Q_j$ forces all $4 \times 4$ minors of $A$ to vanish, meaning the rank of $A$ is less than 4.

This simultaneously enforces coplanarity ( $z=0$ ) and a spherical locus ( $x^2+y^2+z^2-2x-2y=0$ ), restricting the locus to a circumcircle.

It filters out students who memorize the scalar triple product for coplanarity but cannot generalize determinant properties to distance matrices.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Distance Formula in 3D
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free