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Question : 126 of 150
Marks:
+1,
-0
Solution:
=(4−)+λ(+2−3) .......(i)
and
=(−+2) +µ(2+4−5) ........(ii)
Shortest distance between lines (i) and (ii)
= Here,
1=4−;1=+2−3 and
2=−+2;2=2+4−5 Now,
1×2=|| =(−10+12)−(−5+6)+(4−4) =2− ∴|1×2|=√4+1=√5 Also,
(2−1)=(−3+2) ∴ Required shortest distance
=|| =||= units
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