Please Scroll Down to See Forums Below 
Discuss.
Can two skew lines in parallel planes ever intersect?
- Thread starter the_alcatraz
- Start date
*ebony*
New member
I thought skew lines never intersect. It's the definition, two lines that are not parallel but are not in the same plane, such that they never intersect. also parallel lines never intersect, unless they are the same line (in which case they intersect everywhere, but this is a fairly trivial case, no?)
IDK...if two planes never intersect, their intersection is never a plane, unless they are the same plane (think of two pieces of paper on top of each other)
IDK...if two planes never intersect, their intersection is never a plane, unless they are the same plane (think of two pieces of paper on top of each other)
Lets hear Dr.Sketchs answer!
*ebony*
New member
Doctors are smart!!
I don't beleive that, that is why i quit going to mine. Well not really, he died at like 45yo.
That was a great answere thou ebony, when I was reading I just kept thinking of your body as one plain and my body as the other.
Except my tangent kept crossing deep into your line....
lostinfitness
New member
ebony has it mos def right
chris302001
Banned
its gotta be, why else would he ask it? They do have google in canada right?
It all depends on the geometry. If you're talking about a differentiable manifold or something, then the geometry is intrinsic and locally defined by a metric. Curvature and topology are some of the most weird and difficult areas in mathematics.
If you're assuming a well-behaved system with infinitely long planes in euclidean 2-space, then I believe it's true. (I think it might always be true if you reduce the system enough.)
The biggest thing in pure mathematics is defining the system. I would think it would be relatively straightforward to formulate a layman's "proof" in E2 using a few postulates from high school geometry. You could probably extend the proof to higher dimensions, too, since they're all orthogonal.
If you're assuming a well-behaved system with infinitely long planes in euclidean 2-space, then I believe it's true. (I think it might always be true if you reduce the system enough.)
The biggest thing in pure mathematics is defining the system. I would think it would be relatively straightforward to formulate a layman's "proof" in E2 using a few postulates from high school geometry. You could probably extend the proof to higher dimensions, too, since they're all orthogonal.
*ebony*
New member
LOL...samoth if anyone can figure out a way i would bet on you!lol
i forgot about euclideans..... well the definition for parallel lines is different is it not when it's discussed with respect to non-euclidean gemotries. if i remember correctly it calls parallel lines those which, at some point, are both at right angles to a third line? the normal parallel lines have this property too. arn't the lines in non-euclidean geometries with this property are also called parallel, and these types of lines could be shown to meet in some situations.
In normal cases, we call parallel lines those that do not intersect, and they happen to be mutually at right angles to another line, in fact infinitely many lines........so they kind of took that aspect of parallel lines and made it part of a non-euclidean gemotrical definition for "parallel." dont ya think? this is just the impression I got anyways.
i forgot about euclideans..... well the definition for parallel lines is different is it not when it's discussed with respect to non-euclidean gemotries. if i remember correctly it calls parallel lines those which, at some point, are both at right angles to a third line? the normal parallel lines have this property too. arn't the lines in non-euclidean geometries with this property are also called parallel, and these types of lines could be shown to meet in some situations.
In normal cases, we call parallel lines those that do not intersect, and they happen to be mutually at right angles to another line, in fact infinitely many lines........so they kind of took that aspect of parallel lines and made it part of a non-euclidean gemotrical definition for "parallel." dont ya think? this is just the impression I got anyways.
*ebony*
New member
So many improper things to say I just can't settle on one
lay 'em out there, I could use a smile lol.
I'll behave, sometimes I forget about the b/f and don't want to cross any lines. I'm working on being a better person, yay!
chris302001
Banned
chris302001
Banned
That's just something stupid parents tell their kids when they dont have a clue...
It all depends on the geometry. If you're talking about a differentiable manifold or something, then the geometry is intrinsic and locally defined by a metric. Curvature and topology are some of the most weird and difficult areas in mathematics.
If you're assuming a well-behaved system with infinitely long planes in euclidean 2-space, then I believe it's true. (I think it might always be true if you reduce the system enough.)
The biggest thing in pure mathematics is defining the system. I would think it would be relatively straightforward to formulate a layman's "proof" in E2 using a few postulates from high school geometry. You could probably extend the proof to higher dimensions, too, since they're all orthogonal.
Samoth, I'm dealing with solid geometry and Desargues' theorem. Trying to uderstand how rules of geometry change in projective space.
Care to elaborate?
JR76
New member
TITCR! Skewness of lines is incidental. It's all about "da planes" as they might have said on "Fantasy Island" and the fact that they are parallel.
Creepusmaximus
New member
It all depends on the geometry. If you're talking about a differentiable manifold or something, then the geometry is intrinsic and locally defined by a metric. Curvature and topology are some of the most weird and difficult areas in mathematics.
If you're assuming a well-behaved system with infinitely long planes in euclidean 2-space, then I believe it's true. (I think it might always be true if you reduce the system enough.)
The biggest thing in pure mathematics is defining the system. I would think it would be relatively straightforward to formulate a layman's "proof" in E2 using a few postulates from high school geometry. You could probably extend the proof to higher dimensions, too, since they're all orthogonal.
Unless you were breaking the speed of light at the time there by warping the planes.
chris302001
Banned
ololol @ you and your light speed breaker...... olololol!
And your third dimension too.....its so cute.
Really, I got nothin........
U
UA_Iron
Guest
They intersect at infinity.
Creepusmaximus
New member
Hey Chris had a linkin log today, want to hear about that??
So in other words, they don't?
also is it at positive infinity or negative infinity?
Actually, mapping euclidean space to a moving frame near c would not make any difference here (see special relativity and Lorentz contraction).
However, near a gravitational field it would, which is the whole idea behind differential geometry and general relativity -- incorporating spacetime geometry into everything.
The thing is, since gravity goes as the inverse square of the distance, it never technically goes to zero, and thus the familiar geometries we're used to are actually only a special case of something much more complicated.
Actually, mapping euclidean space to a moving frame near c would not make any difference here (see special relativity and Lorentz contraction).
However, near a gravitational field it would, which is the whole idea behind differential geometry and general relativity -- incorporating spacetime geometry into everything.
The thing is, since gravity goes as the inverse square of the distance, it never technically goes to zero, and thus the familiar geometries we're used to are actually only a special case of something much more complicated.
The symmetry of the problem in question would not make such a thing matter.
By definition, the derivative of a straight line is zero, so the only way they could intersect would be for the geometry of the system to change from the everywhere-smooth and infinitely flat orthogonal planes that was implied.
The symmetry of the problem in question would not make such a thing matter.
By definition, the derivative of a straight line is zero, so the only way they could intersect would be for the geometry of the system to change from the everywhere-smooth and infinitely flat orthogonal planes that was implied.
I'm impressed. Really.
I'm curious why you started this thread. Is this something you were just pondering one night?
I'm curious why you started this thread. Is this something you were just pondering one night?
I love math. I always read theories. Geometry, trig, calculus. was a math wiz as a kid, sucked at everything else in school...but math came to me naturally...I loved it. It must be cause I play chess.
waiting for Samoth's update on this.
Euclid's parallelism postulate must hold for all Euclidean space. Since that is what we approximate ourselfs to exist in, then I take the original post to imply that, and thus the answer is no, they cannot.
Interstingly enough, if you take away this fifth axiom of "normal" euclidean space, you get non-euclidean, or curved space. In four dimensions, this is spacetime.
Euclid's parallelism postulate must hold for all Euclidean space. Since that is what we approximate ourselfs to exist in, then I take the original post to imply that, and thus the answer is no, they cannot.
Interstingly enough, if you take away this fifth axiom of "normal" euclidean space, you get non-euclidean, or curved space. In four dimensions, this is spacetime.
if we were to talk about curved space, can the rules of normal euclidean space be bent?
chris302001
Banned
How bout we drop dead cats outta airplanes and see make calculations as to how far away they fall from the point at which they dropped? Or we can show how the refridgerator in your house has more gravitational pull (and thereby more of an "astrological effect") than the nearest star outside our solar system (alpha centauri), or play with something like time dilation and the twin paradox..... That all involves math......
LOL, I havent the slightest idea about this geometry though........
LOL, I havent the slightest idea about this geometry though........
Well-behaved curved geometries obey the other four axioms as far as I know, because they are either well-defined spaces (sphere, torus, etc.) or, more abstractly, because they are differential (Riemannean) manifolds with a defined metric and stuff that works with the other four postulates of Euclid. Basically, on an everywhere-connected n-dimensional differential manifold, you work in the (n-1)-dimensional tangent vector space (or bundle) of the manifold, which is a locally flat place to work (like the infintesimal distance from differential calculus).
U
UA_Iron
Guest
Ah I am glad I'm not in school anymore. Though I do miss the higher thinking part. Now I just design medical devices.
Well-behaved curved geometries obey the other four axioms as far as I know, because they are either well-defined spaces (sphere, torus, etc.) or, more abstractly, because they are differential (Riemannean) manifolds with a defined metric and stuff that works with the other four postulates of Euclid. Basically, on an everywhere-connected n-dimensional differential manifold, you work in the (n-1)-dimensional tangent vector space (or bundle) of the manifold, which is a locally flat place to work (like the infintesimal distance from differential calculus).
I'm specifically interested in the 5th postulate of Euclid and how it contrasts with non-Euclidean geometry
That's an area I should look into besides pharm dev.
How limited is it to just engineers?
5th postulate failing = non-Euclidean.
Degree to which fails = curvature tensor
Curvature tensor = allows metric and one to define curvature
U
UA_Iron
Guest
That's an area I should look into besides pharm dev.
How limited is it to just engineers?
Just engineers do the designing. We have people with science degrees design experiments, validate, verify, test and do all sorts of things with the instruments. I think you're overqualified for that outfit.
It's been my experience that groups of scientists together in the professional world harbor a lot of resentment towards others and especially engineers (engineers get paid more at my work). They like to hoard ideas and are kind of secretive about things. This is especially true in the chemistry field - something about them makes it a nasty work environment. My ex will agree 100% - she worked for a defense contractor and endured it throughout school. Often times the scientist is not concerned with making it work and short deadlines, but more interested in just testing and finding new things out - which is also important to product development, but not as much to a competitive marketplace.
We have a Physics Major with a Master's Degree. He's a great guy. I think he was definitely over qualified for the position. He's going back to school for a master's in systems engineering, something our company recently has felt the need for.
That being said, I dont think a physics degree necessarily black balls you from the field. If you have enough experience with design programs (solidworks, Pro/E, Catia) then that would easily supplement a lot of the schooling. With a Master's Degree then management positions come a lot faster than with a Bachelor's (if that was your goal), and a technical manager is a highly regarded position at our work. No d-bag with an MBA or general business degree would understand the dynamics well.
Awesome post.
The med devices area of biotech is the one that's pretty unfamiliar to me. I've been looking into it more and debating if I should add it to my potential list of job interests. I really need to do more research, though.
chris302001
Banned
Can we pllllllllllllllease drop a dead cat out of an airplane??????
needtogetaas
New member
hmmmmmm
needtogetaas
New member
(The answer is no.)
Yes Yes it is. Or is it? Maybe
5th postulate failing = non-Euclidean.
Degree to which fails = curvature tensor
Curvature tensor = allows metric and one to define curvature
talk about oversimplifying things
