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Question #41612

Difficulty: 8/10 | 🟢 Free

The Question

Let the quadratic equations x2−ax+6=0x^2 - ax + 6 = 0 and x2−bx+12=0x^2 - bx + 12 = 0 have a common root α\alpha. Let their other roots be β\beta and γ\gamma respectively. It is given that β\beta and γ\gamma are positive integers such that β+γ=6\beta + \gamma = 6.

Let PP be a point whose position vector with respect to the origin O(0,0,0)O(0,0,0) is p⃗=αi^+βj^+γk^\vec{p} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}. The origin is then shifted to the point Q(2,0,2)Q(2, 0, 2). Let v⃗=v1i^+v2j^+v3k^\vec{v} = v_1 \hat{i} + v_2 \hat{j} + v_3 \hat{k} be the position vector of PP with respect to the new origin.

If a matrix MM is defined as M=(2v1v2−v12v3−v2−v32)M = \begin{pmatrix} 2 & v_1 & v_2 \\ -v_1 & 2 & v_3 \\ -v_2 & -v_3 & 2 \end{pmatrix}, then the value of det⁡(adj(M))\det(\text{adj}(M)) is:

Options

  • A.

    26

  • B.

    676

  • C.

    4356

  • D.

    5184

Check Your Answer

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

Position Vector Origin Dependence|Adjoint-Determinant Relation|Skew-Symmetric Determinant|Condition for Common Roots

Why This Question Is Tricky

Common Mistakes

This question integrates the theory of equations, vector origin dependence, and determinant properties of skew-symmetric matrices.

The primary trap lies in the 'Origin Dependence' of position vectors.

Students often calculate the initial position vector p⃗\vec{p} using the common root conditions and immediately substitute its components into the matrix MM, ignoring the coordinate transformation (origin shift to QQ).

This leads to the highly plausible distractor 4356.

Furthermore, the matrix MM is structured as λI+S\lambda I + S (where SS is skew-symmetric), testing the student's ability to bypass tedious Sarrus rule expansions by recognizing that det⁡(M)=λ3+λ(v12+v22+v32)\det(M) = \lambda^3 + \lambda(v_1^2 + v_2^2 + v_3^2).

Finally, the adjoint-determinant relation ∣adj(M)∣=∣M∣2|adj(M)| = |M|^2 acts as the final computational lock.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterVector Algebra - Class 12
ConceptPosition Vector Origin Dependence
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree