MathematicsMedium249×since 2002Q2419The shortest distance between the lines x−10=y+1−1=z1{{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}0x−1=−1y+1=1z and x + y + z + 1 = 0, 2x – y + z + 3 = 0 is :A1B12{1 \over 2}21C12{1 \over {\sqrt 2 }}21D13{1 \over {\sqrt 3 }}31Check answerSkip