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Question #40860

Difficulty: 8/10 | 🟢 Free

The Question

Let $M = \begin{pmatrix} 0 & 0 & -1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{pmatrix}$ . The characteristic equation of $M$ is given by $\det(\lambda I - M) = \lambda^3 + c_1\lambda^2 + c_2\lambda + c_3 = 0$ . Consider the vectors $\vec{u} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$ and $\vec{v} = -2\hat{j} - \hat{k}$ . Let $A(3, 0, 0)$ , $B(0, 3, 0)$ , and $C(0, 0, 3)$ be three fixed points in space. A point $P(x,y,z)$ moves such that the vector $\vec{w} = PA^2\hat{i} + PB^2\hat{j} + PC^2\hat{k}$ is linearly dependent on $\vec{u}$ and $\vec{v}$ . Which of the following best describes the locus of $P$ ?

Options

A.
A sphere of radius $3$
B.
A plane passing through $(2, -1, 2)$
C.
A single point
D.
An empty set

Check Your Answer

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

Locus via Distance Constraints|Linear Independence vs Dependence|Characteristic Equation

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of linear algebra (characteristic equations, linear dependence) and 3D coordinate geometry (locus).

The specific trap is the 'singularity collapse'.

Students often mechanically expand the locus equation, see the non-canceling quadratic terms ( $x^2+y^2+z^2$ ), and immediately assume it represents a sphere.

They fail to check the radius condition $R^2 \ge 0$ .

Here, the coefficients derived from the characteristic equation perfectly balance the constants to yield $R^2 = 0$ , collapsing the expected spherical surface into a single point.

This forces students to follow through the algebraic geometry to its absolute conclusion rather than relying on surface-level pattern recognition.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Introduction to Three-dimensional Geometry - Class 11
Concept	Locus via Distance Constraints
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free