If the foot of the perpendicular from point (4,3,8) on the line L1:x−al=y−23=z−b4, l≠0 is (3,5,7), then the shortest distance between the line L1 and line L2:x−23=y−44=z−55 is equal to :
A
√23
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B
1√3
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C
12
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D
1√6
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Solution
The correct option is D1√6 (3,5,7) lies on given line L1.
Then 3−al=33=7−b4 7−b4=1⇒b=3 3−al=1⇒3−a=l A(4,3,8) and B(3,5,7)
DR's of AB is (1,−2,1)
AB ⊥ line L1 ⇒(1)(l)+(−2)(−3)+4(1)=0 ⇒l=2 ∴a=1