Question

# The shortest distance between the lines x−y=0=2x+z and x+y−2=0=3x−y+z−1 is

A
111
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B
123
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C
12
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D
1
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Solution

## The correct option is D 1√11L1:x−y=0=2x+z and L2:x+y−2=0=3x−y+z−1It can also be write asL1:x1=y1=z−2L2:x−1=y−21=z−34A(0,0,0) be the point on L1 and →B=(^i+^j−2^k) is a direction of L1C(0,2,3) be the point on L2 and →D=(−^i+^j+4^k) is a direction of L2→B×→D=6^i−2^j+2^k∣∣→B×→D∣∣=√62+22+22=2√11→C−→A=2^j+3^k(→B×→D).(→C−→A)=(6^i−2^j+2^k)⋅(2^j+3^k)=2The shortest distance between line L1 and L2 can bee found using the following formula.Distance =∣∣ ∣∣→B×→D|→B×→D|.(→C−→A)∣∣ ∣∣Distance =1√11Hence, option A.

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