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Question

The shortest distance betwwen lines ¯¯¯r=2¯i¯j+λ(2¯i+¯j3¯¯¯k) and ¯¯¯r=¯i¯j+2¯¯¯k+μ(2¯i+¯j5¯¯¯k)

A
6
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B
66
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C
55
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D
1
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Solution

The correct option is C 55
the given line (1) has error the eq should be
r=2^i^j+λ(2^i+^j5^k)
in place of
r=2^i^j+λ(2^i+^j3^k)
Given lines
r=2^i^j+λ(2^i+^j5^k)------(1)
r=^i^j+2^k+μ(2^i+^j5^k)------(2)
the line sare parallel with normal vector
b=2^i+^j5^k
position vectors of line(1) and (2) respectively
a=2^i^j
and
c=^i^j+2^k
ca=^i^j+2^k(2^i^j)
ca=^i^j+2^k2^i+^j
ca=^i+2^k
shortest distance

SD=(ca)×bb

SD=(^i+2^k)×(2^i+^j5^k)22+12+(5)2

SD=∣ ∣ ∣∣ ∣ ∣^i^j^k102215∣ ∣ ∣∣ ∣ ∣22+12+(5)2

SD=^i(02)^j(54)+^k(10)4+1+25

SD=2^i^j^k30

SD=(2)2+(1)2+(1)230

SD=630

SD=15
on multiplying and dividing by 5 in RHS

SD=1×55×5

SD=55


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