Perpendicular Cubes - Possible in Three-Space?

[Hypercube]

I was just wondering; is it physically possible for two cubes to be perpendicular other than in more than three dimensions, or without having to have them bisect each other?

[Country Boy]

First thing you will have to do is DEFINE "perpendicular" for cubes. As far as I know "perpendicular" is only defined for lines (or, more generally, curves, but always one dimensional) in any dimension space.

 

Since you don't mention squares, what would it mean for two squares to be "perpendicular"?

[Hypercube]

Where the heck did you learn that HallsofIvy? In a cube; the top face is perpendicular to the side faces. In one dimension, only points can be perpendicular, in two dimensions; lines are perpendicular, in three dimensions; two dimensional shapes are perpendicular. That would suggest that in four dimensions; three dimensional shapes can be perpendicular, and so on.

[Country Boy]

Then I'll have to ask again what definition of "perpendicular" you are using.

 

I've never heard of points being "perpendicular". Yes, we can say that planes are perpendicular if their normal lines are perpendicular. Whether or not you can use the term "perpendicular" in higher dimension depends upon the precise definition of "perpendicular" you are using.

[Bignose]

Well, starting from a purely mathematical point of view, one way to define perpendicular is to define the angle between two vectors as [math]\cos \theta = \mathbf{n}_1 \cdot \mathbf{n}_2 [/math] and thusly when [math]\theta = \frac{\pi}{2}[/math] then [math]\cos \theta = 0[/math] and the vectors are considered perpendicular.

 

In 2-D, to see if two lines are perpendicular, you just computed the dot product of vectors that go along the line.

 

In 3-D, you usually used a vector that was normal to the plane and computed the dot product between those normals. (You could do exactly the same thing in 2-D.)

 

For example, consider two planes in 3-D:

[math]a_1 x + b_1 y + c_1 z + d_1 = 0[/math]

[math]a_2 x + b_2 y + c_2 z + d_2 = 0[/math]

 

Then the normals are [math]\mathbf{n}_1 = [a_1, b_1, c_1][/math] and [math]\mathbf{n}_2 = [a_2, b_2, c_2][/math]

 

There is no reason that this same thing cannot be extended to 4-D or more. Certainly dot products can be extended to 4-vectors and higher, and normals to planes in higher dimensions can be calculated in just the same way.

 

Finally, though, I want to come back to "In one dimension, only points can be perpendicular" This statement is essentially meaningless. Especially when you look at the definition above. The normal of every number in 1_D is "1" Then, you take the dot product between two "1"s? It pretty quickly becomes meaningless.