A) \[\sqrt{6}\]
B) \[6\]
C) \[2\sqrt{3}\]
D) \[3\sqrt{2}\]
Correct Answer: A
Solution :
Given, lines are \[\frac{x}{-1}=\frac{y}{1}=\frac{z}{1}\] and \[\frac{x-3}{0}=\frac{y+3}{1}=\frac{z-3}{-1}\] Here, \[{{x}_{1}}=0,\,{{y}_{1}}=0,\,{{z}_{1}}=0\] \[{{l}_{1}}=-1,{{m}_{1}},=1,{{n}_{1}}=1\] and \[{{x}_{2}}=3,\,{{y}_{2}}=-3,\,{{z}_{2}}=3\] \[{{l}_{2}}=0,{{m}_{2}}=1,{{n}_{2}}=-1\] Now, shortest distance \[d=\left| \frac{\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{I}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{I}_{2}} & {{m}_{2}} & {{n}_{2}} \\ \end{matrix} \right|}{\sqrt{{{({{m}_{1}}{{n}_{2}}-{{m}_{2}}{{n}_{1}})}^{2}}+{{({{n}_{1}}{{I}_{2}}-{{n}_{2}}{{I}_{1}})}^{2}}+{{({{I}_{1}}{{m}_{2}}-{{I}_{2}}{{m}_{1}})}^{2}}}} \right|\] \[=\left| \frac{\left| \begin{matrix} 3 & -3 & 3 \\ -1 & 1 & 1 \\ 0 & 1 & -1 \\ \end{matrix} \right|}{\sqrt{{{(-1-1)}^{2}}+{{(0-1)}^{2}}+{{(-1-0)}^{2}}}} \right|\] \[=\left| \frac{3(-1-1)+3(1-0)+3(-1-0)}{\sqrt{4+1+1}} \right|\] \[=\left| \frac{-6+3-3}{\sqrt{6}} \right|=|-\sqrt{6}|=\sqrt{6}\]
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