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View Full Version : How is a line perpendicular to a space?


nykkyo
Jul 20, 2011, 09:47 PM
In rectagular coordintes, all dimension lines are mutually perpendicular.
In 1D - a line is perpendiular to a point.
In 2D - a line is perpendicular to a line.
In 3D - a line is perpendicular to a plane. Each point on the 3D line, belongs to a plane.
It follows that in nD, a line is perpendicular to an (n-1)D space. Again, each point on the nD line belongs to an (n-1)D space.
Aside from the above question: How do an infinite number of dimensionless points make a line with length? That is like, adding an infinite numer of zeroes and their sum is not zero. So, does the mean sub-spaces are also dimensionless.

ebaines
Jul 21, 2011, 05:50 AM
To be ,more specific:

In 2D there is a line that is perpendicular to any point on any line.

In 3D there is a line that is perpendicular to any point on any plane. Each point on any 3D line belongs to one perpendicular plane.

It follows that in nD there is a line that is perpendicular to any point in any (n-1)D space. Each point on the nD line belongs to one perpendicular (n-1)D space.

As to how an infinite number of zero dimension points can ad up to have physical length - recall that the concept of a point is purely an abstract mathematical concept. It's impossible to "count" the number of infinitely small points along a line, no matter how short that line is. Similarly a line has dimension only in one direction, and is infinitely small in the perpendicular direction. Again, a purely mathemetacal concept. It follows that in nD space a line, or plane, or cube or... up through (n-1)D space has 0 dimension in the perpendicular direction in nD space.

nykkyo
Jul 21, 2011, 05:38 PM
If the _nD line is \bot \ to every point in _(_n_-_1_)D spcem then an arbitrary point in space (3D) that is equadistant from each axial plane, lines normal to the axial planes (3) , form a cube. In 6D, there are _6C_2= 15 axial planes. Normals to the planes form a parallelapilepiped. If the point is equadistant from all planes, it is a 6D cube. The Tetrahedon is a projection onto a plane, \bot \ to the ray to the point, and rotated so the cube shows a cube within a cube.