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Question

Find the shortest distance between lines: x11=y23=z32 and x42=y53=z61

A
6
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B
5
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C
3
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D
6
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Solution

The correct option is C 3
Given lines
l1:x11=y23=z32
l2:x42=y53=z61
position vector of line l1
a=^i+2^j+3^k
position vector of line l2
c=4^i+5^j+6^k
normal vector of line l1
n1=^i+3^j+2^k
normal vector of line l2
n2=2^i+3^j+^k
so line are skews line
SD=(ca)(n1×n2)|(n1×n2)|

ca=4^i+5^j+6^k^i2^j3^k
ca=3^i+3^j+3^k
n1×n2=∣ ∣ ∣^i^j^k132231∣ ∣ ∣
n1×n2=^i(36)^j(14)+^k(36)
n1×n2=3^i+3^j3^k

|n1×n2|=(32)+32+(3)2

|n1×n2|=27

putting ca,n1×n2,|n1×n2| in formula

SD=∣ ∣(3^i+3^j+3^k)(3^i+3^j3^k27∣ ∣

SD=9+9927

SD=927

SD=933

SD=3


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