JEE Main 20206 Sep, Morning shiftMathematicsMediumThree Dimensional Geometry249× since 2002The 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 :AOption A. 1BOption B. 12{1 \over 2}21COption C. 12{1 \over {\sqrt 2 }}21DOption D. 13{1 \over {\sqrt 3 }}31CheckSkip · show solution
