The shortest distance between the skew lines r= (6i+2j+2k )+t (i-2j+2k ) and r= (-4i-k )+s (3i-2j-2k ) is

Question

Accepted Answer

Correct answer: A. (a) Given, r= (6i+2j+2k )+t (i-2j+2k ) andr= (-4i-k )+s (3i-2j-2k ) Herea=6i+2j+2k ,b=i-2j+2kc=-4i-k ,d=3i-2j-2k Here a=6i+2j+2k ,b=i-2j+2k ∴ Shortest distance = | (a-c ) (b d ) | |b d |…..(i) Now, b d(ijk1-223-2-2) =i (4+4 )-j (-2-6 )+k (-2+6 )=8i+8j+4k By Eq . (i ) , = ( (10i+2j+3k ) (8i+8j+4k ) 64+64+16 ) = ( 80+16+12 144 )= 10812=9

$The shortest distance between the skew lines r = (6 i + 2 j + 2 k) + t (i - 2 j + 2 k) and r = (- 4 i - k) + s (3 i - 2 j - 2 k) is$

Solution

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