Linear algebar!!!distance between two lines,help for part(b)by tengxiaona Tags: algebardistance, linear, lines, partb 

#1
Apr609, 09:55 PM

P: 8

1. The problem statement, all variables and given/known data
(a)use cross product to find the distance d from the line K determined by the two points(4,4,4) and (1,2,5) to the line L determined by the parametric equations x=3,y=1,z=55t (b)find two points P and Q on the lines K and L respectively in part(a) such that the idstance from P to Q is d. 2. Relevant equations projection formula, corss product 3. The attempt at a solution i slove part(a) i set A as (4,4,4) and B as (1,2,5), then we can get >AB=(3,6,9) from x=3,y=1,z=55t , we get C as (3,1,5),then >AC=(1,5,1) and we know l(distance between L and K)=area of parallelogram/length of AB which is l= >AB*>AC / >AB now i use cross product >AB*>AC=(3,6,9)*(1,5,1)=(39,12,21) ................... ................... FINALLY i get l= sqrt(117/7) i think i did right way for part(a)  but for part(b) i find the projection of >AC first proj(AC)={AC*AB/AB*AB}AB =(3+309)/126 (3,6,9) =4/21(3,6,9) ok, till here, can i use (4,4,4)as point P then use C(3,1,5) minus (4/21)(3,6,9) to get point Q? if i did wrong way ,plz give me a hand thanks!!! 



#2
Apr709, 12:41 AM

P: 8

i get some ideas from another topic
http://www.physicsforums.com/showthread.php?t=274638 i use the eqution wut Dick offered X_p(s)X_q(t)=u*W then i get (4,4,4)+t[3,6,9]  (3,1,5)+s[0,0,5] = u[30,15,0]<cross product by deriction vectors i solve it and get t=11/15 s=38/25 u=1/25 wuts next? PLZ HELP 



#3
Apr809, 12:06 AM

P: 8

any1 help me !! plz!!




#4
Apr809, 10:15 AM

P: 394

Linear algebar!!!distance between two lines,help for part(b)
Your original posted solution to (a) gives the dist from C to K, but how do you know C is the closest pt to K? Your original (a) is wrong because K and L don't lie in the same plane.
Look at your second post. The skew lines K and L lie on two parallel planes P1 and P2. The distance between K and L is the distance between the two planes P1 and P2. Your vector [30,15,0] is perpendicular to both planes P1 and P2. (4,4,4)+t[3,6,9] when t=11/15 is pt M on line K (3,1,5)+s[0,0,5] when s=38/25 is pt N on line L Vector MN is parallel to [30,15,0]. You are ready to write down the answer to both parts now, do you see it? 


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