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Question

Find the shortest distance between the lines,
r=^i+2^j+3^k+t(2^i+3^j+4^k) and r=2^i+4^j+5^k+s(3^i+4^j+5^k)

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Solution

Equation of given lines:
r=^i+2^j+3^k+t(2^i+3^j+4^k)
and r=2^i+4^j+5^k+s(3^i+4^j+5^k)
a1=^i+2^j+3^k,b1=2^i+3^j+4^k
and a2=2^i+4^j+5^k,b2=3^i+4^j+5^k
a2a1=(2^i+4^j+5^k)(^i+2^j+3^k)
a2a1=^i+2^j+2^k
b1×b2=∣ ∣ ∣^i^j^k234345∣ ∣ ∣
=^i(1516)^j(1012)+^k(89)
=^i+2^j^k
|b1×b2|=|^i+2^j^k|
=(1)2+22+(1)2
=1+4+1=6
shortest distance
=(a2a1)(b1×b2)|b1×b2|
=(^i+2^j+2^k).(^i+2^j^k)6
=1+426=16.

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