SOLUTION: Find the shortest distance between the two lines: (x,y,z) = (2,-1,2) + t(-3,-2,-2) (x,y,z) = (-3,4,1) + s(0,2,0) Thanks sooo much!

Question 62451: Find the shortest distance between the two lines:
(x,y,z) = (2,-1,2) + t(-3,-2,-2)
(x,y,z) = (-3,4,1) + s(0,2,0)

Thanks sooo much!

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Find the shortest distance between the two lines:
L1 SAY..(x,y,z) = P SAY(2,-1,2) + t(-3,-2,-2)
L2 SAY (x,y,z) = Q SAY(-3,4,1) + s(0,2,0)
IT IS PROJECTION OF PQ ON DIRECTION(L,M,N SAY) PERPENDICULAR TO L1 AND L2
-3L-2M-2N=0........1
2M=0
3L=-2N
N=-1.5L
LET L=2...M=0....N=-3
PQ IS (-3-2,4+1,1-2)=(-5,5,-1)
S.D.= PROJECTION OF PQ ON (2,0,-3) DIRECTION
[-5*2+1*3)/SQRT(2^2+3^2)=-7/SQRT(13)
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