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Question

The shortest distance between the lines xy=0=2x+z and x+y2=0=3xy+z1 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 111
L1:xy=0=2x+z and
L2:x+y2=0=3xy+z1
It can also be write as
L1:x1=y1=z2
L2:x1=y21=z34

A(0,0,0) be the point on L1 and B=(^i+^j2^k) is a direction of L1
C(0,2,3) be the point on L2 and D=(^i+^j+4^k) is a direction of L2
B×D=6^i2^j+2^k

B×D=62+22+22=211

CA=2^j+3^k

(B×D).(CA)=(6^i2^j+2^k)(2^j+3^k)=2

The shortest distance between line L1 and L2 can bee found using the following formula.
Distance =∣ ∣B×D|B×D|.(CA)∣ ∣
Distance =111

Hence, option A.

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