Randomly choose vector b in range such that $\vec{a} \cdot \vec{b} = 1$

Question

Given

Problem

I want to randomly and unbiasedly choose a vector $\vec{b}$ such that:

How to do that?

Attempt

I thought about the following:

Choose $\vec{c}$, a random direction perpendicular to $\vec{b}$.
Find range of $\theta$ such that $\vec{b} = \frac{\vec{a}}{\lVert \vec{a} \rVert} + \theta\vec{c}$ is within bounds.
Randoly choose a $\theta$

The problem is that this is not unbiased. Some directions have a larger range of legal $\theta$, and those directions should be chosen more often.

Is that equivalent to choosing a random point on a simplex with $\vec{a}$ as the normal vector? — R zu, Nov 7 '18 at 18:16
Before going to $n$-dimensional, have you made a detailed analysis on 2-dimensional? How do you define unbiased? Are you able to formalize your requirement, "Some directions have a larger range of legal θ, and those directions should be chosen more often" into probability terms? — Apass.Jack, Nov 7 '18 at 18:31
Unbiased: Every $\vec{c}$ that satisfies the conditions in the problem has equal chance to be chosen. "Some directions..." is not a requirement in the problem. — R zu, Nov 7 '18 at 19:17

R zu · Accepted Answer · 2018-11-07 19:40:58Z

1

Uniformly sampling from a simplex with equation $\vec{a} \cdot \vec{b} = 1$, and reject samples out of bounds.

Several answers are posted here for uniform sampling over a simplex: Uniform sampling from a simplex

answered Nov 7 '18 at 19:35

R zu

11310 bronze badges

$\begingroup$ found another method that doesn't need rejection. $\endgroup$ – R zu Nov 7 '18 at 19:46

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