JEE Math Practice Question
The shortest distance between the lines $${{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}$$ and x + y + z + 1 = 0, 2x – y + z + 3 = 0 is :
- A.1
- B.$${1 \over 2}$$
- C.$${1 \over {\sqrt 2 }}$$
- D.$${1 \over {\sqrt 3 }}$$
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