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Question #41035

Difficulty: 8/10 | 🟢 Free

The Question

A unit vector $\hat{u}$ is chosen uniformly at random in $\mathbb{R}^3$ . Let $\pi$ be a plane passing through the origin with a fixed normal vector $\hat{n}$ . Let $\vec{v}$ be the orthogonal projection of $\hat{u}$ onto the plane $\pi$ . Consider the triangle $T$ whose vertices are the origin $O$ , and the points $U$ and $V$ such that $\vec{OU} = \hat{u}$ and $\vec{OV} = \vec{v}$ . Let $E_1$ be the event that the area of triangle $T$ is at most $\frac{\sqrt{3}}{8}$ , and $E_2$ be the event that $|\vec{v}| \le \frac{1}{2}$ . Find the conditional probability $P(E_2 \mid E_1)$ .

Options

A.
$\frac{3 - \sqrt{3}}{6}$
B.
$\frac{1}{2}$
C.
$\frac{3 + \sqrt{3}}{6}$
D.
$\frac{\sqrt{3}}{3}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Conditional Probability Foundations|Orthogonal projection of a line|Unit Vector Normalization

Why This Question Is Tricky

Common Mistakes

This problem tests the student's understanding of probability measures on a sphere versus its orthogonal projection.

The core trap is a 'double naive measure' fallacy.

Students often incorrectly assume that either the projected vector $\vec{v}$ is uniformly distributed on the unit disk, or that the polar angle $\theta$ is uniformly distributed in $[0, \pi/2]$ .

Remarkably, both of these incorrect assumptions lead to the exact same erroneous probability of $1/2$ .

To solve correctly, students must recognize that a uniform distribution of unit vectors in $\mathbb{R}^3$ implies the polar angle $\theta$ has a probability density function of $\sin\theta$ , which acts as the Jacobian for the spherical measure, fundamentally altering the probability space of the geometric inequalities.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Conditional Probability Foundations
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free