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Question : 14 of 90
Marks:
+1,
-0
Solution:
Let
L1:=(−)+λ(2+)‌L2:=(2−)+µ(−+)‌=||‌=−−2
Equation of line along shortest distance of
L1 and
L2‌‌=‌=‌=r‌⇒(x,y,z)≡(r−1,2−r,3−2r)‌⇒(r−1)−2(2−r)+3(3−2r)=10‌⇒r=−2‌⇒Q(x,y,z)≡(−3,4,7)‌⇒PQ=√4+4+16=2√6
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