# Two vectors and two perpendicular lines

• I
• LCDF
In summary,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 ##\mathbb{R}^2## from the origin corresponding to points ##(x_1, y_1)## and ##(x_2, y_2)##. The angle between these vectors is ##\theta \in (0, 2\pi)##
LCDF
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 ##\mathbb{R}^2## from the origin corresponding to points ##(x_1, y_1)## and ##(x_2, y_2)##. The angle between these vectors is ##\theta \in (0, 2\pi)## measured in anticlockwise direction, and has the probability density function ##f_{\theta}##. I want to calculate the probability that none of the perpendicular lines lies between the two vectors.

My attempt: I notice that this is equivalent to calculating the probability that the two points ##(x_1, y_1)## and ##(x_2, y_2)## lie on the same side of each of the perpendicular lines. In other words, a line does not lie between the two vectors iff ##(a_1x_1 + b_1y_1 +c_1)(a_1x_2 + b_1y_2 +c_2) > 0## AND ##(a_2x_1 + b_2y_1 +c_1)(a_2x_2 + b_2y_2 +c_2) > 0##, where ##a_1x + b_1y +c = 0## and ##a_2x + b_2y +c = 0## are the equations of the lines. But I am unable to obtain an expression for the probability.

For the case of only one line passing through the origin (ignore the second line), the probability that the line does not lie between two vectors is
##p = \frac{1}{\pi}\int_{0}^{\pi} (\pi -\rho) f_{\theta}(\rho)\mathrm{d}\rho + \frac{1}{\pi}\int_{\pi}^{2\pi} (\rho-\pi) f_{\theta}(\rho)\mathrm{d}\rho.##

How are you selecting the points ##(x_1, y_1)## and ##(x_2, y_2)##?

Introducing ##\psi## may be unnecessary. Why not take the two lines to be the ##x## and ##y## axes?

In any case,the problem reduces to the two points being in the same quadrant, with respect to your axes.

PeroK said:
How are you selecting the points ##(x_1, y_1)## and ##(x_2, y_2)##?

Introducing ##\psi## may be unnecessary. Why not take the two lines to be the ##x## and ##y## axes?

In any case,the problem reduces to the two points being in the same quadrant, with respect to your axes.
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 them.

LCDF said:
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, and I know the distribution of the angle between them.
The angle between them doesn't specify the relationship between, say, ##(x_1, y_1)## and your axes. You need to specify how ##(x_1, y_1)## is chosen and then the distribution for ##\theta##.

Are you assuming that ##(x_1, y_1)## is uniformly distributed in the plane?

PeroK said:
The angle between them doesn't specify the relationship between, say, ##(x_1, y_1)## and your axes. You need to specify how ##(x_1, y_1)## is chosen and then the distribution for ##\theta##.

Are you assuming that ##(x_1, y_1)## is uniformly distributed in the plane?
Yes, ##(x_1, y_1)## is uniformly distributed in the plane.

PeroK said:
The angle between them doesn't specify the relationship between, say, ##(x_1, y_1)## and your axes. You need to specify how ##(x_1, y_1)## is chosen and then the distribution for ##\theta##.

Are you assuming that ##(x_1, y_1)## is uniformly distributed in the plane?
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, \pi]##. Then one averages over the distribution of ##\theta##. Am I missing something here?

LCDF said:
Yes, ##(x_1, y_1)## is uniformly distributed in the plane.
Then ##\psi## is irrelevant. Take the ##x## and ##y## axes. Assume, without loss of generality, that ##(x_1, y_1)## is in the first quadrant, with an angle ##\phi## uniform between ##0## and ##\pi/2##.

For each ##\phi## you need ##\theta## in the ranges ##[0, \frac \pi 2 - \phi)## or ##(2\pi - \phi, 2\pi)##.

Draw a diagram to see this.

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PeroK said:
Then ##\psi## is irrelevant. Take the ##x## and ##y## axes. Assume, without loss of generality, that ##(x_1, x_2)## is in the first quadrant, with an angle ##\phi## uniform between ##0## and ##\pi/2##.

For each ##\phi## you need ##\theta## in the ranges ##[0, \frac \pi 2 - \phi)## or ##(2\pi - \phi, 2\pi)##.

Draw a diagram to see this.
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 that the axis will not lie within the angle ##\theta## is simply ##(\pi - \theta)/\pi## for ##\theta \in (0, \pi)## as the axis is oriented uniformly between ##[0, \pi]##. Then one averages over the distribution of ##\theta##. One need not say anything else about ##(x_1, y_1)##. Am I missing something here?

LCDF said:
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 that the axis will not lie within the angle ##\theta## is simply ##(\pi - \theta)/\pi## for ##\theta \in (0, \pi)## as the axis is oriented uniformly between ##[0, \pi]##. Then one averages over the distribution of ##\theta##. One need not say anything else about ##(x_1, y_1)##. Am I missing something here?
That's a different problem.

LCDF said:
But I am not sure why one needs to specify the location of ##(x_1, y_1)##.

You don't. It's the same in every quadrant. Which is why we can take it to be the first, for simplicity.

It's a form of decluttering the problem.

LCDF
LCDF said:
##p = \frac{1}{\pi}\int_{0}^{\pi} (\pi -\rho) f_{\theta}(\rho)\mathrm{d}\rho + \frac{1}{\pi}\int_{\pi}^{2\pi} (\rho-\pi) f_{\theta}(\rho)\mathrm{d}\rho.##
You should check this when ##f_\theta## is uniformly distributed. It doesn't look right.

If I understand the problem, you have a single line amd a pair of lines separated by an angle ##\theta##. As above, we take the line to be the x-axis, ##(x_1, y_1)## to be above the x-axis at an angle ##\phi##. Then the probability that the second point is also above the x-axis is:
$$p = \int_0^{\pi - \phi} f(\theta) d\theta + \int_{2\pi - \phi}^{2\pi} f(\theta) d\theta$$
And if you check with a uniform ##f(\theta)## you get ##p = \frac 1 2##, as expected.

PeroK said:
Then ##\psi## is irrelevant. Take the ##x## and ##y## axes. Assume, without loss of generality, that ##(x_1, x_2)## is in the first quadrant, with an angle ##\phi## uniform between ##0## and ##\pi/2##.

For each ##\phi## you need ##\theta## in the ranges ##[0, \frac \pi 2 - \phi)## or ##(2\pi - \phi, 2\pi)##.

Draw a diagram to see this.
You mean ##(x_1,y_1)## instead of ##(x_1, x_2)##?

LCDF said:
You mean ##(x_1,y_1)## instead of ##(x_1, x_2)##?
Yes, fixed.

LCDF said:
Yes, ##(x_1, y_1)## is uniformly distributed in the plane.
There is no such thing as a uniform distribution in the plane.

LCDF and PeroK
jbriggs444 said:
There is no such thing as a uniform distribution in the plane.
That's a good point. I guess we could limit it to points at a fixed distance; or, have an arbitrary radial distribution.

jbriggs444
jbriggs444 said:
There is no such thing as a uniform distribution in the plane.
Right. I think saying that the angle is uniform would be accurate.

PeroK said:
Then ##\psi## is irrelevant. Take the ##x## and ##y## axes. Assume, without loss of generality, that ##(x_1, y_1)## is in the first quadrant, with an angle ##\phi## uniform between ##0## and ##\pi/2##.

For each ##\phi## you need ##\theta## in the ranges ##[0, \frac \pi 2 - \phi)## or ##(2\pi - \phi, 2\pi)##.

Draw a diagram to see this.

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 uniformly distributed ##\psi## in ##[0, \pi]##.

LCDF said:
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 uniformly distributed ##\psi## in ##[0, \pi]##.
If ##\psi## is relevant, then ##p## will be a function of ##\psi##.

If you take ##\psi = 0## and ##\psi = \frac \pi 4##, say, and do the calculations, you would get a different answer. Do you expect that? If so, why?

If we haven't understood the problem and you get a different scenario for different ##\psi## then it is relevant. From what I understand the probability should be independent of ##\psi##.

PeroK said:
If ##\psi## is relevant, then ##p## will be a function of ##\psi##.

If you take ##\psi = 0## and ##\psi = \frac \pi 4##, say, and do the calculations, you would get a different answer. Do you expect that? If so, why?

If we haven't understood the problem and you get a different scenario for different ##\psi## then it is relevant. From what I understand the probability should be independent of ##\psi##.
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, y_1)## is equivalent to the case of considering a fixed line as the x-axis and a point whose angle with the x-axis is uniformly distributed in ##[0, \pi]##. Isn't it?

If I am not misunderstood you earlier suggestion, you were exactly implying the second interpretation, while I the first. And I get the desired answer by putting things in your interpretation that matches my simulation while considering a fixed point and a randomly oriented line!

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LCDF said:
If I am not misunderstood you earlier suggestion, you were exactly implying the second interpretation, while I the first. And I get the desired answer by putting things in your interpretation that matches my simulation while considering a fixed point and a randomly oriented line!

I don't understand that. The calculations should be independent of ##\psi##.

PeroK said:
I don't understand that. The calculations should be independent of ##\psi##.
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!

## 1. What is the definition of a vector?

A vector is a mathematical object that has both magnitude (size) and direction. It is typically represented by an arrow pointing in the direction of the vector, with the length of the arrow representing the magnitude.

## 2. How are two vectors added together?

To add two vectors, you can use the parallelogram law. This means placing the tail of one vector at the head of the other vector, and then drawing a line from the tail of the first vector to the head of the second vector. The resulting vector, from the tail of the first vector to the head of the second vector, is the sum of the two vectors.

## 3. What does it mean for two vectors to be perpendicular?

Two vectors are perpendicular if they form a 90 degree angle (also known as a right angle) when placed tail-to-tail or head-to-head. This means that their dot product, or the product of their magnitudes and the cosine of the angle between them, is equal to zero.

## 4. How do you determine if two lines are perpendicular?

To determine if two lines are perpendicular, you can use the slope-intercept form of the lines (y = mx + b). If the slopes of the two lines are negative reciprocals of each other (meaning one is the negative inverse of the other), then the lines are perpendicular.

## 5. Can two vectors be perpendicular if they have the same magnitude?

Yes, two vectors can be perpendicular even if they have the same magnitude. The magnitude of a vector only represents its size, while the direction of the vector is what determines if it is perpendicular to another vector.

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