Recent content by LCDF

Certainly, the calculations should be independent of ##\psi## and my answer reflects that. But I now see that it is nature of the randomness in points than the randomness in the orientation. Thanks!

Your suggestion was very useful, and I obtained the answer based on it. And I agree that the probability should not be dependent on ##\psi##. What I was thinking is that the case of a random line with orientation ##\psi## that is uniformly distributed in ##[0, \pi]## and a fixed point at ##(x_1...

I do not see why ##\psi## is irrelevant. A uniformly distributed ##\psi## in ##[0, \pi]## is equivalent to fixing the line as the x-axis and one of the points makes an angle ##\psi## that is uniformly distributed in ##[0, \pi]##. But this is possible only if the orientation of the line is...

Right. I think saying that the angle is uniform would be accurate.

You mean ##(x_1,y_1)## instead of ##(x_1, x_2)##?

I get it in the case when ##(x_1, y_1)## is uniformly distributed. But I am not sure why one needs to specify the location of ##(x_1, y_1)##. So, please let me repeat my previous comment in the case you missed it. For instance, when there is only one line (or, only one axis), the probability...

But I am not sure why one needs to specify the location of ##(x_1, y_1)##. For instance, when there is only one line (or, only one axis), the probability that the axis will not lie within the angle ##\theta## is simply ##(\pi - \theta)/\pi## as the axis is oriented uniformly between ##[0...

Yes, ##(x_1, y_1)## is uniformly distributed in the plane.

You are right that the problem reduces to the points being in the same quadrant. But I am unsure how to calculate the probability of it. The points ##(x_1, y_1)## and ##(x_2, y_2)## are the points of the Poisson point process (so they are random), and I know the distribution of the angle between...

In ##\mathbb{R}^2##, there are two lines passing through the origin that are perpendicular to each other. The orientation of one of the lines with respect to ##x##-axis is ##\psi \in [0, \pi]##, where ##\psi## is uniformly distributed in ##[0, \pi]##. Also, there are two vectors in...

Thanks for suggestions. I am new to Physics Forum and its protocols. I was following the typical Latex notation. Taking angles in clockwise is due to the setup in the bigger problem that I am trying to solve. ##\rho## is just a dummy variable representing the random variable ##\theta##. I think...

What if we assume the angle $\psi$ is uniformly distributed between $[0, \pi]$? I understand measuring angle clockwise is against the convention, but I need it due to the setup of a bigger problem.

I want to find the probability that the two points ($x_1, y_1$) and ($x_2, y_2$) lie on the opposite sides of a line passing through the origin $o = (0, 0)$ and makes an angle $\psi$ that is uniformly distributed in $ [0, \pi]$ with the $x$ axis when the angle is measured in clockwise direction...

Thanks. Is there a mathematical proof for your observation? A hint would also work.

Is the intersection of a 4D line segment and a 3D polyhedron in 4D a point in 4D, if they at all intersect? Intuitively, it looks like so. But I am not sure about it and how to prove it.