A) \[\frac{1}{6}\]
B) \[\frac{1}{\sqrt{6}}\]
C) \[\frac{1}{\sqrt{3}}\]
D) \[\frac{1}{3}\]
E) 6
Correct Answer: B
Solution :
The given lines are \[\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\] and \[\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}\] Now, \[\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 2-1 & 4-2 & 5-3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 1 & 2 & 2 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \\ \end{matrix} \right|=1\] and\[\sqrt{\begin{align} & {{({{m}_{1}}{{n}_{2}}-{{m}_{2}}{{n}_{1}})}^{2}}+{{({{n}_{1}}{{l}_{2}}-{{l}_{1}}{{n}_{2}})}^{2}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+{{({{l}_{1}}{{m}_{2}}-{{l}_{2}}{{m}_{1}})}^{2}} \\ \end{align}}\] \[=\sqrt{{{(15-16)}^{2}}+{{(12-10)}^{2}}+{{(8-9)}^{2}}}\] \[=\sqrt{6}\] \[\therefore \]Required shortest distance \[=\frac{\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ \end{matrix} \right|}{\sqrt{\begin{align} & {{({{m}_{1}}{{n}_{2}}-{{m}_{2}}{{n}_{1}})}^{2}}+{{({{n}_{1}}{{l}_{2}}-{{l}_{1}}{{n}_{2}})}^{2}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+{{({{l}_{1}}{{m}_{2}}-{{l}_{2}}{{m}_{1}})}^{2}} \\ \end{align}}}\] \[=\frac{1}{\sqrt{6}}\]
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