JEE Math Practice Question
The shortest distance between the lines $${x \over 2} = {y \over 2} = {z \over 1}$$ and $${{x + 2} \over { - 1}} = {{y - 4} \over 8} = {{z - 5} \over 4}$$ lies in the interval :
- A.[0, 1)
- B.[1, 2)
- C.(2, 3]
- D.(3, 4]
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