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Question #66794

Difficulty: 8/10 | 🟢 Free

The Question

Let a⃗=i^+sin⁡θj^+cos⁡θk^\vec{a} = \hat{i} + \sin \theta \hat{j} + \cos \theta \hat{k}, b⃗=i^+sin⁡2θj^+cos⁡2θk^\vec{b} = \hat{i} + \sin 2\theta \hat{j} + \cos 2\theta \hat{k}, and c⃗=i^+sin⁡3θj^+cos⁡3θk^\vec{c} = \hat{i} + \sin 3\theta \hat{j} + \cos 3\theta \hat{k}. Let MM be the 3×33 \times 3 matrix whose rows are the components of a⃗,b⃗,\vec{a}, \vec{b}, and c⃗\vec{c}. Let SS be the set of all θ∈[0,2π]\theta \in [0, 2\pi] such that the points with position vectors a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} and i^+j^+k^\hat{i} + \hat{j} + \hat{k} are coplanar, and the matrix MM is invertible. What is the minimum volume of the parallelepiped formed by the row vectors of M−1M^{-1} for θ∈S\theta \in S?

Options

  • A.

    239\frac{2\sqrt{3}}{9}

  • B.

    332\frac{3\sqrt{3}}{2}

  • C.

    439\frac{4\sqrt{3}}{9}

  • D.

    00

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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

Inverse Existence Criterion|Scalar Triple Product Volume Logic|Collinearity and Coplanarity|Collinearity vs Coplanarity

Why This Question Is Tricky

Examiner Levers

Logical Assertion

Tests the causal relationship between two true statements.

Difficulty Boost : +4

Boundary Condition

Tests behavior at limits like 0, 1, infinity, or empty sets.

Difficulty Boost : +4

Common Mistakes

This question exploits the lethal collision between set-theoretic coplanarity of points and the linear dependency of their position vectors.

Students reflexively equate 'coplanar' with 'determinant = 0'.

If they apply this to the position vectors a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}, they will incorrectly assume det⁡(M)=0\det(M) = 0, leading to a contradiction with the invertibility condition and concluding the set SS is empty.

The trap forces them to realize that points can be coplanar without their position vectors being linearly dependent (here, they all lie on the affine plane x=1x=1, which does not pass through the origin).

Once past the trap, it rigorously tests the Inverse Existence Criterion and the geometric interpretation of determinants (volume of the parallelepiped of inverse matrix rows).

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterMatrices - Class 12
ConceptInverse Existence Criterion
Difficulty8
Cognitive Load8
Target Solve Time180 sec
AccessFree