10-03-2003 12:58									#1
	Tough maths problem - 3D Vectors Support Apolyton buy from Amazon

well i think its tough

Two lines in space are skew if they are not parallel and do not interscet. Let l be a line in space with direction vector v passing through the point P. Let m be a line in space with direction vector w passing through the point Q. Suppose that l and m are skew. Determine the minimum distance between l and m.

stumped on where to start...i did it all with x's and y's..ended up with 14 variables..lol..new approach needed!

	10-03-2003 13:09									#2
	Got spare money?

Here's a hint: what is the cross-product of v and w?

	10-03-2003 14:41									#3
	Support Apolyton, buy Civilization: The Boardgame

Geometrically? The area formed by the parallelogram with sides v and w...

I like the idea of a hint...gets me started but allows me to make progress. Thanks...gotta get thinking about this now!

	10-03-2003 14:44									#4
	Help yourself to an AD-FREE life

Think about its direction.

	10-03-2003 14:48									#5
	Inflate your Upload Space

perpendicular to both v and w...

	10-03-2003 15:07									#6
	Put an end to popups!

That's right. And what do you know about the direction of the shortest vector between the lines (the answer should be obvious given the context in which I'm writing the question, but visualize it - play around with your fingers and arms and whatnot)?

	10-03-2003 15:11									#7
	Increase Your PM Length

Ok it is a line that is perpendicular to both v and w.

Here's whats running though my head
Doesnt the cross product inolve vectors that are tail to tail, which these cant be?

	10-03-2003 15:16									#8
	Increase Your PM Length

thanks guys!

i think ive got something down

	10-03-2003 16:16									#9
	Avatar Enlargement: We've got the solution

agghh, that was tough

here is basicaly what i did

Find the cross-product which is V x W

The point with position vector P lies on a line L
The point with position vector Q lies on a line M

Shifting a copy of M until it interstets with L gives the plane A(1) with the point P lying on it

Shifting a copy of L until it interstets with M gives the plane A(2) with the point Q lying on it

The cross product give a common normal to both planes

Call this unit normal 'n'

Then I found A(1) = n.p
Then I found A(2) = n.q

The distance to the origin of each is the modulus of the dot product

therefore for my answer i get

|p.n-q.n|

wow...am i even close? tell me i am

[edit...correct variables put in!]

Last edited by heardie on 10-03-2003 at 16:24

	10-03-2003 16:36									#10
	Support Apolyton, buy Call to Power 2

There is a easier way yeah?
ok

Call the normal vector n same as above

Would the distance b/w the 2 skew lines be the modulus of the scalar projection of PQ onto n???

	10-03-2003 16:46									#11
	Increase Your PM Length

talking to myself here....

from my post above

=|(PQ.n)|
=|(p-q).n|
=|p.n-q.n|

holy crap thats the same as above...so does this mean im right? or just totally dumb, and co-incidently get the same (wrong) answer twice

	10-03-2003 18:17									#12
	Support Apolyton, buy Civilization 2

quote: Would the distance b/w the 2 skew lines be the modulus of the scalar projection of PQ onto n???

yes