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Question #40678

Difficulty: 8/10 | 🟢 Free

The Question

Let a⃗\vec{a} and b⃗\vec{b} be two unit vectors such that the angle between them is π6\frac{\pi}{6}. A sequence of position vectors is defined for k∈Nk \in \mathbb{N} as: r⃗k=Ω⃗+∑n=1∞(kk+1)na⃗+∑n=1∞(k2k2+1)nb⃗\vec{r}_k = \vec{\Omega} + \sum_{n=1}^{\infty} \left( \frac{k}{k+1} \right)^n \vec{a} + \sum_{n=1}^{\infty} \left( \frac{k^2}{k^2+1} \right)^n \vec{b} where Ω⃗=(lim⁡x→0sin⁡x−xx3)a⃗+(∫0π/2ln⁡(tan⁡x)dx)b⃗+(∑n=1∞(−1)n−1n)(a⃗×b⃗)\vec{\Omega} = \left( \lim_{x \to 0} \frac{\sin x - x}{x^3} \right) \vec{a} + \left( \int_{0}^{\pi/2} \ln(\tan x) dx \right) \vec{b} + \left( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \right) (\vec{a} \times \vec{b}). Find the area of the triangle formed by the points with position vectors r⃗1,r⃗2,\vec{r}_1, \vec{r}_2, and r⃗3\vec{r}_3.

Options

  • A.

    12\frac{1}{2}

  • B.

    1

  • C.

    34\frac{\sqrt{3}}{4}

  • D.

    14\frac{1}{4}

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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

Infinite GP Convergence|Geometric Interpretation of Cross Product|Position Vector Origin Dependence

Why This Question Is Tricky

Common Mistakes

This question tests the student's understanding of translational invariance in vector geometry, specifically the 'Position Vector Origin Dependence' trap.

The position vectors r⃗k\vec{r}_k are defined with a highly complex constant vector Ω⃗\vec{\Omega} and infinite geometric progressions.

A student who relies on brute-force procedural calculation will waste significant time evaluating the limits, integrals, and series in Ω⃗\vec{\Omega}, and then attempt to compute the area using the origin-dependent formula 12∣r⃗1×r⃗2+r⃗2×r⃗3+r⃗3×r⃗1∣\frac{1}{2}|\vec{r}_1 \times \vec{r}_2 + \vec{r}_2 \times \vec{r}_3 + \vec{r}_3 \times \vec{r}_1|.

The trap is failing to recognize that the area of a triangle is invariant under origin shifts.

By defining a new origin at Ω⃗\vec{\Omega}, the constant vector completely vanishes, and the infinite GPs simplify to basic polynomials kk and k2k^2, reducing a seemingly monstrous problem to a trivial cross product.

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Question Information

SubjectMathematics
ChapterSequences and Series - Class 11
ConceptInfinite GP Convergence
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree