## Saturday, August 16, 2014

### Lines in Planes

(Click on the comic if you can't see the full image.)
(C)Copyright 2014, C. Burke.

You've been skewered.

Skew lines are neither parallel nor intersecting, occuring in three-dimensional space, not two. Imagine skew lines as an overpass and a roadway passing beneath. They don't intersect, but they're not parallel, and they aren't traveling in the same direction.

Or think of the front-right edge of a shoebox and the top-left edge. They each join two planes, but there isn't a common plane containing both of them.

Bill R said...

"They each join two planes,"

<quibble>

They each join an infinite number of (infinite) planes, two of which are realized (or distinguished) as (orthogonal) faces of the shoebox.

</quibble>

(x, why?) said...

Would it have been better, leaving the illustration aside, had I said that the edges were formed by two planes, or by the intersections of said planes?

Bill Ricker said...

in the pure world devoid of shoeboxes, yes, a Euclidean line is the unique intersection of ANY two of the infinite number of planes including the line.

Using consistent terminology (face, edge, vertex) in the realm of finite solids and (plane, line, point) in the realm of infinite space, bridging with 'face embedded in a plane', 'edge embedded in a line', may be of pedagogic value?

Why my pedantic pedagogy? Skewness is only interesting for the infinite line, since non-intersecting non-parallel edges aka line segments might be either collinear or intersecting in extension when their embedding in infinite rays or lines are considered.

All quibbling aside, using the shoebox's several pairs of non-intersecting edges as a familiar manifestation of Skewness is a good thing. It does make uniqueness and perpendicular-ness of the minimum approach appear obvious ... while proving those facts is more interesting. :-)

Fun combinatorial problem
How many pairs of skew lines are defined by the (extensions of the) 12 edges of the shoebox or a cube ?

I partition all pairings into intersecting, parallel, or skew (lema: that it IS a partition!) and so get

comb(12,2)
- 8*comb(3,2)
- 3*comb(4,2) ?
= 24 = 2*E = 2^3 * 3^1
where
12,8 are from cube's F,E,V=6,12,8;
one 3 is E-per-V (and only because it's a cube the number of dimensions), other 3 is number of dimensions = # (orthogonal) pencils of parallel edges.
and 4 is # edges per pencil,
2 is number of lines in a pair whether skew, intersecting or parallel.

But what is the direct counting calculation ? 12*(2+2)/2.