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Question

Find the shortest distance and the vector equation of the line of shortest distance between the lines given by:
r=(3i+8j+3k)+λ(3ij+k)
r=(3i7j+6k)+μ(3i+2j+4k)

A
330 units
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B
90 units
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C
230 units
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D
310 units
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Solution

The correct option is A 330 units
We have a1=3^i+8^j+3^k b1=3^i^j+^k
a2=3^i7^j+6^k and b2=3^i+2^j+4^k
Now, b1×b2=∣ ∣ ∣^i^j^k311324∣ ∣ ∣=6^i15^j+3^k
b1×b2=62+152+32=330
a1a2=6^i+15^j3^k
Therefore shortest distance between the given lines is

d=(a1a2)×(b1×b2)b1×b2=62+152+32330=330

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