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Question #41754

Difficulty: 8/10 | 🟢 Free

The Question

Let P1P_1 be a plane with normal vector n⃗1=ai^+bj^+ck^\vec{n}_1 = a\hat{i} + b\hat{j} + c\hat{k}, where a,b,a, b, and cc are the coefficients of x2,x3,x^2, x^3, and x4x^4 respectively in the expansion of (1−x)−3(1-x)^{-3} for ∣x∣<1|x| < 1. Let P2P_2 be another plane with a unit normal vector n⃗2=pi^+qj^+rk^\vec{n}_2 = p\hat{i} + q\hat{j} + r\hat{k}, where p,q,r∈Rp, q, r \in \mathbb{R} satisfy the equation sin⁡−1(p)+cos⁡−1(q)+csc⁡−1(r)=π\sin^{-1}(p) + \cos^{-1}(q) + \csc^{-1}(r) = \pi. If θ\theta is the acute angle between the planes P1P_1 and P2P_2, then the value of 19cos⁡θ19 \cos \theta is equal to

Options

  • A.

    10

  • B.

    15

  • C.

    19

  • D.

    6

Check Your Answer

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

Angle between two planes|Coordinate Axis Projections|Domain Restrictions|Selection from Identical Items

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of binomial expansion for negative indices with 3D geometry, while embedding a critical domain restriction trap.

Students often attempt to solve the inverse trigonometric equation algebraically or assume standard values.

However, recognizing that n⃗2\vec{n}_2 is a unit vector (p2+q2+r2=1p^2+q^2+r^2=1) alongside the domain requirement of csc⁡−1(r)\csc^{-1}(r) (∣r∣≥1|r| \ge 1) forces a boundary condition ∣r∣=1|r|=1.

This elegantly collapses the vector to a single coordinate axis projection, rewarding students who check domain constraints before algebraic manipulation.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterThree-Dimensional Geometry - Class 12
ConceptAngle between two planes
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree