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Question #74448
The Question
Let and be two mutually orthogonal unit vectors, and let . A point is the unique intersection of the three planes given by the equations: where . It is given that satisfies the following two conditions: (i) lies in the region containing the transverse axis bounded by the asymptotic planes of the hyperbolic cylinder . (ii) The scalar projection of along the vector is strictly greater than .
Find the set of all possible values of .
Options
- A.
- B.
- C.
- D.
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
Vector Triple Product Expansion|Intersection Point Accuracy|Sign Flipping Rule|Asymptote Geometry
Why This Question Is Tricky
Examiner Levers
Variable Manipulation
Algebraic solutions only (No numbers); tests pure symbolic logic.
Difficulty Boost : +4Common Mistakes
This question tests the Vector Triple Product expansion for solving linear systems of planes and spatial geometry constraints.
The primary trap is the 'Sign Flipping Rule' embedded in the Scalar Triple Product.
The basis vectors form a left-handed system (), introducing a negative denominator.
When solving the linear inequality for the scalar projection, students who ignore this negative sign will fail to reverse the inequality, leading to the incorrect interval .
The secondary constraint uses the asymptote geometry of a hyperbolic cylinder to restrict the domain, forcing students to correctly interpret the region containing the transverse axis as and intersect it with the projection condition.
Quick Facts About This Question
Question Information
| Subject | Mathematics |
| Chapter | Vector Algebra - Class 12 |
| Concept | Vector Triple Product Expansion |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |