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Question #74448

Difficulty: 8/10 | 🟢 Free

The Question

Let a⃗\vec{a} and b⃗\vec{b} be two mutually orthogonal unit vectors, and let c⃗=a⃗+b⃗−a⃗×b⃗\vec{c} = \vec{a} + \vec{b} - \vec{a} \times \vec{b}. A point r⃗0\vec{r}_0 is the unique intersection of the three planes given by the equations: r⃗⋅a⃗=t,r⃗⋅b⃗=2t−1,r⃗⋅c⃗=2t2+3t−2\vec{r} \cdot \vec{a} = t, \quad \vec{r} \cdot \vec{b} = 2t - 1, \quad \vec{r} \cdot \vec{c} = 2t^2 + 3t - 2 where t∈Rt \in \mathbb{R}. It is given that r⃗0\vec{r}_0 satisfies the following two conditions: (i) r⃗0\vec{r}_0 lies in the region containing the transverse axis bounded by the asymptotic planes of the hyperbolic cylinder (r⃗⋅a⃗)2−(r⃗⋅b⃗)2=1(\vec{r} \cdot \vec{a})^2 - (\vec{r} \cdot \vec{b})^2 = 1. (ii) The scalar projection of r⃗0\vec{r}_0 along the vector a⃗×b⃗\vec{a} \times \vec{b} is strictly greater than 12\frac{1}{2}.

Find the set of all possible values of tt.

Options

  • A.

    (13,12)\left( \frac{1}{3}, \frac{1}{2} \right)

  • B.

    (32,1)\left( \frac{\sqrt{3}}{2}, 1 \right)

  • C.

    (−12,13)\left( -\frac{1}{2}, \frac{1}{3} \right)

  • D.

    (13,32)\left( \frac{1}{3}, \frac{\sqrt{3}}{2} \right)

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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

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 : +4

Common 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 ([a⃗b⃗c⃗]=−1[\vec{a}\vec{b}\vec{c}] = -1), 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 (3/2,1)(\sqrt{3}/2, 1).

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 (r⃗⋅a⃗)2−(r⃗⋅b⃗)2>0(\vec{r}\cdot\vec{a})^2 - (\vec{r}\cdot\vec{b})^2 > 0 and intersect it with the projection condition.

Quick Facts About This Question

Question Information

SubjectMathematics
ChapterVector Algebra - Class 12
ConceptVector Triple Product Expansion
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree