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Question : 35 of 160
Marks:
+1,
-0
Solution:
L1:r=(3+4−2)+λ(−+2+) L2:r=(−7−2)+µ(+3+2‌) Shortest distance will be along
(n1×n2) So,
n1×n2=|| ,,;−1,2,1;1,3,2|=+3−5 |
unit vector along
n1×n2=‌(+3−5 Shortest distance is the projection of position vector passing through
L1 and
L2 along
n1×n2 So,
[(3−1)+(4+7)+(−2+2)]⋅‌(+3−5 =‌=√35 units
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