Expert: Paul Klarreich - 12/25/2009

Question
QUESTION: Hello!

I've the following two lines: L(1): z=y-c; L(2): z=c-x. My question is, how to find the shortest distance between these lines?

thank you

ANSWER: Are you sure these are lines?  They look like planes to me.

---------- FOLLOW-UP ----------

QUESTION: Oh yes, sorry, so L(1) line lies on the yz plan (x=0), and L(2) line lies on the xz plan (y=0). Hope it's correct now, and you can help me!

Thanks

Answer
Questioner: grull
Country: Hungary
Category: Advanced Math
Private: Yes
Subject: skew lines??
Question: QUESTION: Hello!

I've the following two lines: L(1): z=y-c; L(2): z=c-x. My question is, how to find the shortest distance between these lines?

thank you

ANSWER: Are you sure these are lines?  They look like planes to me.

---------- FOLLOW-UP ----------

QUESTION: Oh yes, sorry, so L(1) line lies on the yz plan (x=0), and L(2) line lies on the xz plan (y=0). Hope it's correct now, and you can help me!

Thanks


L1:  z = y - c,  x = 0, so:

<x,y,z> = <0,0,-c> + t<0,1,1>

L2:  z = c - x,  y = 0, so:

<x,y,z> = <0,0,c> + s<1,0,-1>

If P1 and P2 are points on their lines, their distance(squared) is:

M = (x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2

M = (s)^2 + (t)^2 + ((t-c)-(c-s)^2

M =  s^2 + t^2 + ((t + s - 2c)^2

M =  s^2 + t^2 + t^2 + s^2 - 4ct - 4cs + 2st + 4c^2

M =  2s^2 + 2t^2  - 4ct - 4cs + 2st + 4c^2

Now you want to minimize that;  find

dM/ds  and  dM/dt

and set them equal to zero.

I think you get :

t = 2c/3
s = 2c/3

That is it -- now plug in and find your distance.