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Question #40299

Difficulty: 8/10 | 🟢 Free

The Question

Let the universal set be $U = \left\{ \vec{r} \in \mathbb{R}^3 : |\vec{r}|^2 \le \frac{11}{4} \right\}$ . Two sets $S_1$ and $S_2$ are defined as: $S_1 = \left\{ \vec{r} \in U : \vec{r} \times (\hat{i} + \hat{j} + \hat{k}) = (\hat{i} - \hat{j}) \times (\hat{i} + \hat{j} + \hat{k}) \right\}$ $S_2 = \left\{ \vec{r} \in U : \vec{r} \times (\hat{i} - \hat{j} + \hat{k}) = (2\hat{i} + \hat{k}) \times (\hat{i} - \hat{j} + \hat{k}) \right\}$ If $d$ is the shortest distance between any point in $S_1$ and any point in $S_2$ , then the value of $d$ is:

Options

A.
0
B.
1
C.
$\frac{1}{2}$
D.
$\frac{\sqrt{3}}{2}$

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

Universal Set Relativity|Collinearity vs Coplanarity|Shortest distance between skew lines|Bounded vs Unbounded Regions

Why This Question Is Tricky

Common Mistakes

This question tests the concept of universal set relativity applied to 3D geometry, specifically targeting the trap of blindly applying the shortest distance formula for skew or intersecting lines.

The lines containing $S_1$ and $S_2$ actually intersect at $(2, 0, 1)$ , which would yield a shortest distance of $0$ (Option A).

However, the problem restricts the sets to a bounded spherical region $U$ .

Students must recognize that the intersection point lies outside this universal set.

By parameterizing the lines and applying the boundary constraints, the problem transforms into minimizing a two-variable quadratic function over a closed square domain, forcing students to evaluate boundary extrema rather than relying on unbounded geometric formulas.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sets - Class 11
Concept	Universal Set Relativity
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free