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Question #41201

Difficulty: 8/10 | 🟢 Free

The Question

Let a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} be three non-coplanar vectors representing the position vectors of points A,B,CA, B, C with respect to an origin OO. A new origin DD is chosen with position vector d⃗=xa⃗+yb⃗+zc⃗\vec{d} = x\vec{a} + y\vec{b} + z\vec{c} such that x+y+z≠1x+y+z \neq 1. Let r⃗A,r⃗B,r⃗C\vec{r}_A, \vec{r}_B, \vec{r}_C be the position vectors of A,B,CA, B, C with respect to DD. A 3×33 \times 3 matrix MM is formed such that its rows are the scalar components of the vectors u⃗,v⃗,w⃗\vec{u}, \vec{v}, \vec{w} respectively, where: u⃗=7(r⃗A×r⃗B)+3(r⃗B×r⃗C)\vec{u} = 7(\vec{r}_A \times \vec{r}_B) + 3(\vec{r}_B \times \vec{r}_C) v⃗=7(r⃗B×r⃗C)+7(r⃗C×r⃗A)\vec{v} = 7(\vec{r}_B \times \vec{r}_C) + 7(\vec{r}_C \times \vec{r}_A) w⃗=3(r⃗C×r⃗A)+λ(r⃗B×r⃗A)\vec{w} = 3(\vec{r}_C \times \vec{r}_A) + \lambda(\vec{r}_B \times \vec{r}_A) If MM is a singular matrix for all such choices of DD, then the value of λ\lambda is:

Options

  • A.

    7

  • B.

    -7

  • C.

    21

  • D.

    -21

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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

Singular vs Non-Singular Matrices|Vector Product Anti-Commutativity|Position Vector Origin Dependence|Collinearity and Coplanarity

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of matrix singularity, scalar triple products, and vector anti-commutativity under a shifted origin.

The primary trap is the anti-commutativity of the cross product: students rushing under pressure often hastily substitute r⃗B×r⃗A\vec{r}_B \times \vec{r}_A as X⃗\vec{X} instead of −X⃗-\vec{X}, leading to the incorrect determinant equation 147+21λ=0147 + 21\lambda = 0 and the trap answer −7-7.

The secondary conceptual layer is the origin dependence; students must recognize that shifting the origin by d⃗\vec{d} scales the volume by (1−x−y−z)(1-x-y-z), which is explicitly constrained to be non-zero, ensuring the base vectors X⃗,Y⃗,Z⃗\vec{X}, \vec{Y}, \vec{Z} remain linearly independent and the determinant condition is strictly required.

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Question Information

SubjectMathematics
ChapterMatrices - Class 12
ConceptSingular vs Non-Singular Matrices
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree