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Question #41019

Difficulty: 8/10 | 🟢 Free

The Question

Let a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} be three non-coplanar vectors such that the volume of the parallelepiped formed by them is 325325. A point PP with position vector r⃗=xa⃗+yb⃗+zc⃗\vec{r} = x\vec{a} + y\vec{b} + z\vec{c} is chosen uniformly at random inside this parallelepiped. What is the volume of the region consisting of all such points PP for which line segments of lengths x,y,x, y, and zz can form an acute-angled triangle?

Options

  • A.

    325(1−π4)325 \left(1 - \frac{\pi}{4}\right)

  • B.

    325(1−π6)325 \left(1 - \frac{\pi}{6}\right)

  • C.

    3252\frac{325}{2}

  • D.

    325(1−π3)325 \left(1 - \frac{\pi}{3}\right)

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Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Equiprobability Fallacy|Scalar Triple Product Volume Logic|Triangle Inequality Mind Games

Why This Question Is Tricky

Common Mistakes

This question tests the 'Equiprobability Fallacy' by tempting students to analyze the geometric lengths and angles of the vector r⃗\vec{r} itself, rather than recognizing that the coefficients (x,y,z)(x,y,z) are uniformly distributed in a unit cube and mapped via a constant Jacobian.

Furthermore, the 'Triangle Inequality Mind Games' trap is deployed by asking for an acute triangle condition, which creates a non-convex domain bounded by conical surfaces.

Students often mistakenly assume linear boundaries, attempt to use Heron's formula, or fail to realize the three non-acute regions are mutually exclusive, leading to incorrect integration limits or overlapping volume subtractions.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterProbability - Class 11
ConceptEquiprobability Fallacy
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree