The shortest distance between the lines x+7−6=y−67=Z and 7−x2=y−2=z−6 is

L1:x+7−6=y−67=z−01 ⇒ a1¯=(−7,6,0), b1¯=−6i¯+7j¯+k¯ L2:x−7−2=y−21=z−61 ⇒ a2¯=(7,2,6), b1¯=−2i¯+j¯+k¯Shortest distance between L1 and L2 is =|[a2→−a1→b1→b2→]b1→ ×b2→ |=229

The shortest distance between the lines x+7−6=y−67=Z and 7−x2=y−2=z−6 is

L1:x+7−6=y−67=z−01 ⇒ a1¯=(−7,6,0), b1¯=−6i¯+7j¯+k¯ L2:x−7−2=y−21=z−61 ⇒ a2¯=(7,2,6), b1¯=−2i¯+j¯+k¯Shortest distance between L1 and L2 is =|[a2→−a1→b1→b2→]b1→ ×b2→ |=229

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The shortest distance between the lines $\frac{x + 7}{- 6} = \frac{y - 6}{7} = Z$ and $\frac{7 - x}{2} = y - 2 = z - 6$ is

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$\frac{\sqrt{42}}{2}$

$2 \sqrt{29}$

$\frac{\sqrt{34}}{2}$

$1$

answer is A.

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Detailed Solution

$L_{1} : \frac{x + 7}{- 6} = \frac{y - 6}{7} = \frac{z - 0}{1} \Rightarrow \bar{a_{1}} = (- 7, 6, 0), \bar{b_{1}} = - 6 \bar{i} + 7 \bar{j} + \bar{k}$
$L_{2} : \frac{x - 7}{- 2} = \frac{y - 2}{1} = \frac{z - 6}{1} \Rightarrow \bar{a_{2}} = (7, 2, 6), \bar{b_{1}} = - 2 \bar{i} + \bar{j} + \bar{k}$
Shortest distance between $L_{1}$ and $L_{2}$ is
$= | \frac{[\begin{matrix} \vec{a_{2}} - \vec{a_{1}} & \vec{b_{1}} & \vec{b_{2}} \end{matrix}]}{\vec{b_{1}} \times \vec{b_{2}}} | = 2 \sqrt{29}$