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Question

Find the shortest distance between the pair of parallel lines
r=i+2j+3k+λ(ij+k)andr=2ijk+μ(i+jk)is

A
773
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B
1114
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C
783
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D
683
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Solution

The correct option is B 1114
Shortest distance between lines with vector equations
r=a1+λb1 and r=a2+μb2 is
∣ ∣ ∣(b1×b2).(a2a1)b1×b2∣ ∣ ∣

Given: r=^i+2^j+3^k+λ(^i^j+^k) and r=2^i^j^k+μ(^i+^j^k)

a1=^i+2^j+3^k,a2=2^i^j^k and b1=2^i^j^k,b2=^i+^j^k

b1×b2=∣ ∣ ∣^i^j^k211111∣ ∣ ∣

b1×b2=(1+1)^i(21)^j+(21)^k

b1×b2=2^i+3^j+^k

b1×b2=22+32+12=4+9+1=14

a2a1=2^i^j^k^i2^j3^k=^i3^j4^k

Now (b1×b2).(a2a1)=(2^i+3^j+^k)(^i3^j4^k)=294=11

Now,Shortest distance between lines=∣ ∣ ∣(b1×b2).(a2a1)b1×b2∣ ∣ ∣

=1114=1114units by substituting the above values

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