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Question

Find the shortest distance between the following pairs of parallel lines.
r=(^i+2^j+3^k)+λ(^i^j+^k) and r=(2^i^j^k)+μ(^i+^j^k).

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Solution

The equations of the given lines are
r=(^i+2^j+3^k)+λ(^i+^j+^k)
r=(2^i^j^k)+μ(^i+^j^k)
It is known that the shortest distance between the lines r=a1+λb1 and r=a2+μb2 is given by
d=∣ ∣ ∣(b1×b2).(a2a1)b1×b2∣ ∣ ∣ ......(1)
Comparing the equations,
a1=^i+2^j+3^k
a2=2^i^j^k
b1=^i+^j+^k
b2=^i+^j^k
a2a1=2^i^j^k^i2^j3^k=^i3^j4^k
b1×b2=∣ ∣ ∣^i^j^k111111∣ ∣ ∣
=^i(11)^j(1+1)+^k(1+1)
=2^i+2^k
b1×b2=(2)2+22=22
d=∣ ∣ ∣(b1×b2).(a2a1)b1×b2∣ ∣ ∣
=∣ ∣ ∣(2^i+2^k).(^i3^j4^k)22∣ ∣ ∣
=2822
=52
the shortest distance between the two lines is 52

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