The shortest distance between the lines x−12=y+8−7=z−45 and x−12=y−21=z−6−3 is

a¯=(1,−8,4),c¯=(1,2,6),b¯=(2,−7,5),d¯=(2,1,−3),Shortest distance =|[a¯−c¯ b¯ d¯]||b¯ × d¯|=16(12)163=43

The shortest distance between the lines x−12=y+8−7=z−45 and x−12=y−21=z−6−3 is

a¯=(1,−8,4),c¯=(1,2,6),b¯=(2,−7,5),d¯=(2,1,−3),Shortest distance =|[a¯−c¯ b¯ d¯]||b¯ × d¯|=16(12)163=43

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The shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ is

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$3 \sqrt{3}$

$4 \sqrt{3}$

$2 \sqrt{3}$

$5 \sqrt{3}$

answer is B.

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Detailed Solution

$\bar{a} = (1, - 8, 4), \bar{c} = (1, 2, 6),$
$\bar{b} = (2, - 7, 5), \bar{d} = (2, 1, - 3),$
Shortest distance $= \frac{| [\bar{a} - \bar{c} \bar{b} \bar{d}] |}{| \bar{b} \times \bar{d} |}$
$= \frac{16 (12)}{16 \sqrt{3}}$
$= 4 \sqrt{3}$

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The shortest distance between the lines x−12=y+8−7=z−45 and x−12=y−21=z−6−3 is

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The shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ is