Consider the line L1:3x+1=1y+2=2z+1,L2:1x−2=2y+2=3z−3 The shortest distance between L1andL2 is

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Consider the line {L}_{{1}}:{{{x}+{1}}}{{3}}={{{y}+{2}}}{{1}}=

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Consider the line $L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}, L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}$ The shortest distance between $L_{1} and L_{2}$ is

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Emma Discussed

Consider the line

L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}, L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}

The shortest distance between

L_{1} and L_{2}

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Question Text

Consider the line

L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}, L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}

The shortest distance between

L_{1} and L_{2}