5 ms·

> dot product of two vectors is just cosine of the angle between them, multiplied by their lengths How do you define the "angle" between two n-dimensional vect

by adwn 5y ago

> dot product of two vectors is just cosine of the angle between them, multiplied by their lengths How do you define the "angle" between two n-dimensional vectors? Most likely using the dot-product and the arccos. "cos(angle) times lengths" might give a good intuition for 2D or 3D space, but it doesn't help in higher-dimensional vectors.

Nevermark 5y ago
It generalizes perfectly. The angle between two lines in any dimensions is the same concept. Two (non-collinear) lines share a plane. The angle on that plane is just the ordinary angle, no matter how many dimensions the two lines are embedded in. In the case they are collinear, the angle between them is zero on any plane that intersects them. So that corner case works too, regardless of numbers of dimensions.
- adwn 5y ago
  > Two (non-collinear) lines share a plane. The angle on that plane is just the ordinary angle, no matter how many dimensions the two lines are embedded in. Okay, but now you've got a plane in n-dimensional space. How do you define/calculate the angle between the two vectors without falling back on the dot product? You could say: The angle between the two vectors A and B is defined as the smallest rotation around their common normal vector so that the rotated first vector points in the same direction as the second vector. But what is the normal vector? It's a third vector C which is 90° to each of A and B. Now your definition is cyclic. Okay, then: C is a non-zero vector so that dot(A,C)=0 and dot(B,C)=0. Now you're back to using dot-products for defining angles in higher-dimensional space.
  - Nevermark 5y ago
    The normal vector is the cross product of the two vectors. u = [u1 u2 u3] v = [v1 v2 v3] in dimensions i, j, k u x v = determinate of this matrix: = | i j k| |u1 u2 u3| |v1 v2 v3| = (u2v3-v2y3)i - (u1v3 - v1y3)j + (u1v2 - v2u2)k
    - adwn 5y ago
      That works for 3D space. What about n-dimensional vectors?
      - Nevermark 5y ago
        You just do the cross product for n dimensions which is the determinant with n dimensions. [i j k m ...], with vector u, v, w, s, .... Its the same for all dimensions.
BongoMcCat 5y ago
take the first line/vector, then just call that line/vector "first-new-dimension". Then take the second line/vector, and decide that this vector can be written as a 2-dimensional vector made out of "first-new-dimension" and "second-new-dimension", now you just have to figure out what "second-new-dimension" is. A simple trick would be to measure the length of the line, and then add/remove a multiple of the first dimension until the length of the line becomes as short as possible, this new line is your "second-new-dimension". Now, even if you are working with a 10-dimensional space, you have two lines that only exist in the first two (new) dimensions, so, you can treat them as two-dimensional objects and find an angle using your old 2-dimensional methods.

by adwn 5y ago

Nevermark 5y ago

It generalizes perfectly. The angle between two lines in any dimensions is the same concept. Two (non-collinear) lines share a plane. The angle on that plane is just the ordinary angle, no matter how many dimensions the two lines are embedded in. In the case they are collinear, the angle between them is zero on any plane that intersects them. So that corner case works too, regardless of numbers of dimensions.

adwn 5y ago
> Two (non-collinear) lines share a plane. The angle on that plane is just the ordinary angle, no matter how many dimensions the two lines are embedded in. Okay, but now you've got a plane in n-dimensional space. How do you define/calculate the angle between the two vectors without falling back on the dot product? You could say: The angle between the two vectors A and B is defined as the smallest rotation around their common normal vector so that the rotated first vector points in the same direction as the second vector. But what is the normal vector? It's a third vector C which is 90° to each of A and B. Now your definition is cyclic. Okay, then: C is a non-zero vector so that dot(A,C)=0 and dot(B,C)=0. Now you're back to using dot-products for defining angles in higher-dimensional space.
- Nevermark 5y ago
  The normal vector is the cross product of the two vectors. u = [u1 u2 u3] v = [v1 v2 v3] in dimensions i, j, k u x v = determinate of this matrix: = | i j k| |u1 u2 u3| |v1 v2 v3| = (u2v3-v2y3)i - (u1v3 - v1y3)j + (u1v2 - v2u2)k
  - adwn 5y ago
    That works for 3D space. What about n-dimensional vectors?
    - Nevermark 5y ago
      You just do the cross product for n dimensions which is the determinant with n dimensions. [i j k m ...], with vector u, v, w, s, .... Its the same for all dimensions.

BongoMcCat 5y ago

take the first line/vector, then just call that line/vector "first-new-dimension". Then take the second line/vector, and decide that this vector can be written as a 2-dimensional vector made out of "first-new-dimension" and "second-new-dimension", now you just have to figure out what "second-new-dimension" is. A simple trick would be to measure the length of the line, and then add/remove a multiple of the first dimension until the length of the line becomes as short as possible, this new line is your "second-new-dimension". Now, even if you are working with a 10-dimensional space, you have two lines that only exist in the first two (new) dimensions, so, you can treat them as two-dimensional objects and find an angle using your old 2-dimensional methods.