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Question #41143

Difficulty: 8/10 | 🟢 Free

The Question

Let $B = \{ \vec{v} \in \mathbb{R}^3 : |\vec{v}| \le 1 \}$ be the closed unit ball in $\mathbb{R}^3$ . A binary operation $\star$ is defined on $B$ by $\vec{u} \star \vec{v} = \lambda (\vec{u} \cdot \vec{v}) (\vec{u} \times \vec{v})$ where $\lambda > 0$ is a real constant. What is the maximum value of $\lambda$ such that $B$ is closed under the operation $\star$ ?

Options

A.
1
B.
$\sqrt{2}$
C.
2
D.
4

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

Binary Operation Closure|Geometric Interpretation of Cross Product|Lagrange Identity

Why This Question Is Tricky

Common Mistakes

This question tests the closure of a binary operation on a restricted vector space (the unit ball).

The primary trap is the assumption that since the dot and cross products of unit vectors are individually bounded by 1, their product is maximized when both are 1, leading students to incorrectly guess $\lambda = 1$ .

By applying the Lagrange Identity ( $|\vec{u} \times \vec{v}|^2 = |\vec{u}|^2|\vec{v}|^2 - (\vec{u} \cdot \vec{v})^2$ ), students must rigorously model the geometric constraint.

The identity reveals that the magnitudes of the dot and cross products are coupled; they cannot be simultaneously maximized.

The true maximum occurs at $\theta = \pi/4$ , yielding a bound of $1/4$ for the squared terms, which allows the scalar $\lambda$ to be as large as 2 without exiting the restricted set.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Relations and Functions - Class 12
Concept	Binary Operation Closure
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free