What are the significant angles between intersecting planes in 4D space?

[Hornbein]

TL;DR Summary

Wondering how to find the angles between planes in a four dimensional space.

Suppose I have two intersecting planes in a four dimensional space. It seems to me that there are two angles between these planes. If the two planes intersect in a line then one of those angles is zero. If the two angles are non-zero then the planes intersect in a point. If one plane is the WX plane and the other the YZ plane then the two angles are both 90 degrees.

The best I can do is this. Each plane is defined by two orthonormal basis vectors. Project a basis vector onto the other plane. Then the angle is the arccos of the length of the projection. Do this with each basis vector to get the two angles. However I believe that the result depends on the choice of basis vectors. What I need is an eigenvector or maximum or minimum or something like that, but I don't know what.

 

[Mark44]

Suppose I have two intersecting planes in a four dimensional space. It seems to me that there are two angles between these planes. If the two planes intersect in a line then one of those angles is zero. If the two angles are non-zero then the planes intersect in a point. If one plane is the WX plane and the other the YZ plane then the two angles are both 90 degrees.

There are two angles, but they aren't oriented as you describe above. If two planes intersect in a line, the angle between them is the angle made by the normals of the two planes. One angle is the acute angle between the planes, and the other is the supplement of the angle (180° minus the other angle).

The best I can do is this. Each plane is defined by two orthonormal basis vectors.
The vectors don't have to be orthogonal, and they don't have to have a length of 1. Any two nonparallel vectors in the plane will do.

Project a basis vector onto the other plane. Then the angle is the arccos of the length of the projection. Do this with each basis vector to get the two angles. However I believe that the result depends on the choice of basis vectors. What I need is an eigenvector or maximum or minimum or something like that, but I don't know what.

This seems like the hard way, if indeed it actually works.
Assuming you have the equations of the two planes,

1. Find a normal to each plane
2. Find the angle between the two planes, using this definition of the dot product:
   $\cos(\theta) = \frac{\vec N_1 \cdot \vec N_2}{| \vec N_1 | |\vec N_2 |}$

$\cos(\theta) > 0$ if $\theta$ is between 0° and 90°, and is negative if $\theta$ is between 90° and 180°.

 

[Hornbein]

There are two angles, but they aren't oriented as you describe above. If two planes intersect in a line, the angle between them is the angle made by the normals of the two planes. One angle is the acute angle between the planes, and the other is the supplement of the angle (180° minus the other angle).

This seems like the hard way, if indeed it actually works.
Assuming you have the equations of the two planes,

1. Find a normal to each plane
2. Find the angle between the two planes, using this definition of the dot product:
   $\cos(\theta) = \frac{\vec N_1 \cdot \vec N_2}{| \vec N_1 | |\vec N_2 |}$

$\cos(\theta) > 0$ if $\theta$ is between 0° and 90°, and is negative if $\theta$ is between 90° and 180°.

In 4D there is no unique normal to a 2D plane. All vectors in the complementary plane are normal to the first.

 

[fresh_42]

In 4D there is no unique normal to a 2D plane. All vectors in the complementary plane are normal to the first.

Do the planes intersect at all? If so, then you can find a 3D subspace that contains both of them and that allows you to determine the normals.

 

[Hornbein]

Do the planes intersect at all? If so, then you can find a 3D subspace that contains both of them and that allows you to determine the normals.

In 4D, two 2D planes may intersect in a point and span the entire space.

 

[fresh_42]

In 4D, two 2D planes may intersect in a point and span the entire space.

How do you even define the angle in such a case?

If they intersect in a line, then proceed as described.

 

[Mark44]

In 4D there is no unique normal to a 2D plane.

I didn't say it was unique. What I said was "Find a normal to each plane." Any normal will do.

All vectors in the complementary plane are normal to the first.

I don't know what you mean by "complementary plane." In any case, all vectors in the plane are perpendicular to the normal.

In 4D, two 2D planes may intersect in a point and span the entire space.

It's impossible for two planes to intersect at exactly one point.
Edit: My intuition was incorrect on that one.
Two planes, in 3D, 4D, whatever, can intersect in a line (the normals aren't parallel), at all points (the planes are identical, and the normals are parallel), or nowhere (the normals are parallel, but the planes share no points).

If you are given the equations of the two planes, it's a simple matter to find the normals to each. For example, the equation $x + y + z + w = 2$ is a hyperplane in 4D. A normal to it is N = <1, 1, 1, 1>. Any nonzero multiple of that vector will also be a normal. You can verify this by noticing that A(1, 1, 0, 0), B(0, 0, 1, 1), and C(0, 1, 0, 1) are points in that hyperplane. Construct vectors $\overline {AB}$ and $\overline {AC}$ and calculate the dot product of each with N. Each dot product is zero.

If you aren't given the equations of the planes, but are given points in the planes, it's a little trickier than it is for planes in 3D, since you don't have a cross-product to work with. Assuming you are given three points in the plane, A, B, and C, construct vectors $\overline {AB}$, $\overline {AC}$, and $\overline {BC}$, and dot each of them with the assumed normal N = <a, b, c, d>. From the resulting system of equations you can determine coefficients for N.

 

[fresh_42]

It's impossible for two planes to intersect at exactly one point.

$$\left\{\alpha \begin{bmatrix}
1\\0\\0\\0
\end{bmatrix}+\beta
\begin{bmatrix}
0\\1\\0\\0
\end{bmatrix}
\right\} \bigcap
\left\{\gamma \begin{bmatrix}
0\\0\\1\\0
\end{bmatrix}+\delta
\begin{bmatrix}
0\\0\\0\\1
\end{bmatrix}
\right\}=\begin{bmatrix}
0\\0\\0\\0
\end{bmatrix}$$

 

[Mark44]

I stand corrected.

 

[Hornbein]

I got an answer but don't have the vocabulary to understand it.

https://math.stackexchange.com/questions/4487760/finding-angles-between-planes-in-4d-space

Could someone tell me what area of math this is so that I can find a textbook?

 

[Mark44]

I got an answer but don't have the vocabulary to understand it.

The first paragraph in the link is a long-winded way of saying about what I said back in post #2, which is to find two normal vectors and then use the cosine definition of the dot product to find the angle between the two planes.

Could someone tell me what area of math this is so that I can find a textbook?

Some linear algebra textbooks would have this sort of problem.

 

[fresh_42]

I read this as: If the planes intersect in a line, then take the product formula for two vectors.

Hot air.

 

[fresh_42]

For example, I don't know what the height of an ideal is.

This is algebraic geometry, or commutative algebra. It is the length of a chain of ideals, the Krull dimension of an ideal. It has nothing to do with your problem, and I also did not read it anywhere on the site you linked to.

 

[Hornbein]

I'm going to show that given two 2D planes in 4D we can define two angles between them. If both angles are small then the planes are close to being coplanar. If both angles are large then the planes are close to being perpendicular. If one is small and the other large then the planes are close together in one sense but not in the other. Given two proper angles we can construct two planes that have that relationship.

So what is my question? What I would like to do is the inverse of that. Given two planes, find the two angles.

First we will show that these two angles are always well-defined. Consider four free variables w,x,y,z. Then the sets [w,x,0,0] and [0,0,y,z] are planes that intersect only at the origin. Take any vector in one plane and project it into the other. You get a vector of length zero. We can say that the angle between the two planes is arccos(0)=pi/2. The two planes are perpendicular.

The sets [w,x,w,x] and [0,0,y,z] are also planes that intersect only at the origin. The projection of any vector in the first plane into the second is [0,0,w,x]. For any unit vector in the first plane we have 2(w^2+x^2) = 1. The length of the projection is 1/2. We have a well-defined angle of arccos(1/2) = pi/3.

The sets [w,x,w,0] and [0,0,y,z] are also planes that intersect only at the origin. The projection of any vector in the first plane into the second is [0,0,w,0]. For any unit vector in the first plane the length of the projection can be any number from 0 to 1/s, where s=sqrt(2). While not well-defined, the minimum and maximum of that length are well-defined. We get a major angle of arccos(0) = pi/2 and a minor of arccos(1/s) = pi/4.

Next have a constant a. The sets [w,x,aw,0] and [0,0,y,z] are also planes that intersect only at the origin. The projection of any vector in the first plane into the second is [0,0,aw,0]. We get a a major angle that is pi/2 and minor angle that depending on the value of a can be anything between 0 and pi/2.

Next have a constant b. Sets [w,x,aw,bx] and [0,0,y,z] are also planes that intersect only at the origin. The projection of any vector in the first plane into the second is [0,0,aw,bx]. Using the same process we get two angles. The point is that either angle could be anything from pi/2 to a limit of zero. We can make whatever we like, from perpendicular planes to a pair that is nearly coplanar but still intersect at a point.

We have shown that given two proper angles we can construct two planes that have that relationship. So what is my question? What I would like to do is the inverse of that. Given two planes, find the two angles. I can do it numerically by starting with a random vector in each plane and searching for the maximum and minimum, but I would like to think there is something more elegant.

 

[robphy]

Is this problem of finding two angles part of a larger problem?
(What would you do with these two angles once you have found them?)

Could the larger problem be solved using a different approach?

 

[Hornbein]

Is this problem of finding two angles part of a larger problem?
(What would you do with these two angles once you have found them?)

Could the larger problem be solved using a different approach?

I needed two planes in 4D such that the minor angle was small and the major angle large. I was able to construct such planes. I was able to figure out this numerical method of computing the angles, but it would not be much use to actually do it. I would like to get a closed formula instead. This might help me find out whether the two vectors associated with the major and minor angles are always perpendicular, which seems like basic useful general knowledge.

Given a plane and the two angles the possible second planes constructed this way form a class. Maybe there are more such classes. If I had the closed formula I might be able to find such.

Perhaps the main reason is if I knew how to calculate the angles easily then I might be able to apply the method elsewhere.

Once I've got the answer in 4D it should work in N number of dimensions and any combination of m-plane and n-plane, with the number of angles of interest varying depending on N,m,and n.

 

[Mark44]

I'm going to show that given two 2D planes in 4D we can define two angles between them. If both angles are small then the planes are close to being coplanar. If both angles are large then the planes are close to being perpendicular. If one is small and the other large then the planes are close together in one sense but not in the other. Given two proper angles we can construct two planes that have that relationship.

This business about the two angles is something you mentioned in post #1. From what you said in that post and in what I've quoted, I think you have a misunderstanding about planes. The usual way of determining the angle between planes is to look at the normals to the planes, and use the cosine definition of the dot product. This is what I said before, and what the stackexchange answer said, in essence.

So what is my question? What I would like to do is the inverse of that. Given two planes, find the two angles.

Here's a simple example, given the equations of two planes in $\mathbb R^4$.

$P_1: x + y + z + w = 2$
$P_2: x - y + z + w = 6$

The normals and their magnitudes:
$\vec N_1 = <1, 1, 1, 1>$
$\vec N_2 = <1, -1, 1, 1>$
$| \vec N_1| = \sqrt{1^2 + 1^2 + 1^2 + 1^2} = \sqrt 4 = 2$
$| \vec N_2| = \sqrt{1^2 + (-1)^2 + 1^2 + 1^2} = \sqrt 4 = 2$
Cosine def. of the dot product (rearranged):
$\cos(\theta) = \frac{\vec N_1 \cdot \vec N_2}{|\vec N_1| |\vec N_2 |}$
$= \frac{2 } { 2 \cdot 2} = \frac 1 2$
$\cos(\theta) = \frac 1 2 \Rightarrow \theta = \frac \pi 3$ or 60°

The main difference between what I did and what the stackexchange answer described was that my normals aren't of unit length.

Here's a simple drawing of the two planes of my example, looking straight down the line of intersection of the planes. One of the points of intersection is (1, -2, 1, 2). Another is (2, -2, 1, 1).

 

[Hornbein]

This will not work in 4D. In 4D planes have normal planes, not normal vectors.

 

[Mark44]

This will not work in 4D. In 4D planes have normal planes, not normal vectors.

This makes no sense. I gave you an example with two planes (hyperplanes actually), and showed you the normals to each one.

The hyperplane of the first equation, $P_1$ has the equation x + y + z + 2 = 2, and has a normal of <1, 1, 1, 1>. Certainly you could find a plane that is normal to the plane of this equation, but you don't need to.

Some points on $P_1$: A(2, 0, 0, 0), B(0, 1, 1, 0), and C(0, 0, 0, 2), found by observation.
Construct vectors $\overline{AB}$ = <-2, 1, 1, 0> and $\overline{AC}$ = <-2, 1, 1, 0>

If you calculate the dot product of each of these two vectors with $N_1$, you get zero, which says that the normal is perpendicular to each vector.
$\overline{AB} \cdot \vec N_1 = <-2, 1, 1, 0> \cdot <1, 1, 1, 1> = -2 + 1 + 1 = 0$
$\overline{AC} \cdot \vec N_1 = <-2, 0, 0, 2> \cdot <1, 1, 1, 1> = -2 + 2 = 0$

The equations I gave are of hyperplanes, 3-dimensional subsets of $\mathbb R^4$. If you're asking about actual planes (i.e., 2-dimensional subsets of $\mathbb R^4$, please give an example of the type of problem you're working on.

 

[Hornbein]

Aha. By planes I mean 2D planes. (In self defense I did give the example of "If one plane is the WX plane and the other the YZ plane"). Thank you for putting all that effort into your reply.

I have gottena a partial but nevertheless satisfactory solution from an expert in geometric algebra. Using this it is very basic to calculate the product of the sines of the major and minor angles, sin a₁ sin a₂. It is common to have another constraint on the angles (such as maybe we know that a₁=a₂) that allows a solution to be found. Even when we can't solve for the angles it gives a useful sense of what is going on. So I'm happy with this.

On the off chance that anyone is interested, this answer is at http://hi.gher.space/forum/viewtopic.php?f=32

 

[Hornbein]

Aha. By planes I mean 2D planes. (In self defense I did give the example of "If one plane is the WX plane and the other the YZ plane"). Thank you for putting all that effort into your reply.

I have gottena a partial but nevertheless satisfactory solution from an expert in geometric algebra. Using this it is very basic to calculate the product of the sines of the major and minor angles, sin a₁ sin a₂. It is common to have another constraint on the angles (such as maybe we know that a₁=a₂) that allows a solution to be found. Even when we can't solve for the angles it gives a useful sense of what is going on. So I'm happy with this.

On the off chance that anyone is interested, this answer is at http://hi.gher.space/forum/viewtopic.php?f=32

I didn't try geo

 

[Hornbein]

To my surprise there is a closed form solution for the two angles between the 2D planes in 4D which may be viewed at
http://hi.gher.space/forum/viewtopic.php?f=32

 

[robphy]

http://hi.gher.space/forum/viewtopic.php?f=32

“The requested topic does not exist.”

Is the url correct?

 

[Hornbein]

“The requested topic does not exist.”

Is the url correct?

How about this one.

http://hi.gher.space/forum/viewtopic.php?f=32&t=2620

 

[Hornbein]

Let's look at the familiar case of two 2D planes here in 3D space. Using these methods we calculate two angles between the planes, but one of the angles is constrained to be always zero so it is of no interest.

 

[WWGD]

Maybe you can use the fact that $2D$-planes in general position intersect at a point. in general, if you have spaces $S_1, S_2$ (Manifolds, I think) of dimensions D1 , intersecting in an ambient space $S_3$of dimension $D3 ; D3> D1, D2$, then, in general position, the dimension of $Dim( S_1 \cap S_2)= D3-(D_1 +D_2)$. So, in dimension $4$ ( I assume Euclidean $4$ -space) , the dimension of the intersection would be :$4-(2+2)=0$. And if you have an intersection at a point, how do you define the angle. If the intersection is higher-dimensional, you have a degenerate case.

My intuition would be: for each plane, choose a basis, extend it to a basis for the ambient space. Then find the dot product between the two planes. Just intuition, haven't thought it through.

edit: for books, other sources, maybe look up Intersection Theory, Intersection Homology.

 

[Hornbein]

Not surprisingly all this was figured out many years ago. They are known as "significant angles."