Consider the line L1-x-13-y-21-z-12- L2-x-21-y-22-z-33The shortest distance between L1 and L2 is

The correct option is C 175√(3) L _ 1 : x+1 3 = y+2 1 = z+1 2 L _ 2 : x 2 1 = y+2 2 = z 3 3 (x_ 1 ,y_ 1 ,z_ 1 )≡ ( 1, 2, 1) ...

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Shortest Distance between Two Skew Lines

Consider the ...

Question

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

0

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\frac{17}{\sqrt{3}}

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\frac{41}{5 \sqrt{3}}

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\frac{17}{5 \sqrt{3}}

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Similar questions

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}

L_{1}

L_{2}

l_{1} : \frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 2}{4}

l_{2} : \frac{x + 2}{4} = \frac{y - 0}{- 3} = \frac{z + 1}{1}

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

L_{2} : \frac{x}{1} = \frac{y}{- 1} = \frac{z - 5}{3}

L_{1}

L_{2}

L_{1}

L_{2}

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}

(1, 1, 1)

(- 1, - 2, - 1)

L_{1}

L_{2}

L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}

L_{2} : \frac{x - 2}{3} = \frac{y - 4}{2} = \frac{z - 5}{5}

L_{1}

L_{2}

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