A) \[9\]
B) \[\frac{19}{3}\]
C) \[\frac{16}{3}\]
D) \[4\]
E) None of these
Correct Answer: E
Solution :
Given lines can be rewritten as \[\frac{x-6}{1}=\frac{y-2}{-2}=\frac{z-2}{2}\] and \[\frac{x+4}{3}=\frac{y}{-2}=\frac{z-1}{-2}\] Here, \[{{x}_{1}}=6,\,\,{{y}_{1}}=2,\,{{z}_{1}}=2\] \[{{x}_{2}}=-4,{{y}_{2}}=0,\,{{z}_{2}}=1\] \[{{l}_{1}}=1,\,{{m}_{1}}=-2,\,{{n}_{1}}=2\] and \[{{l}_{2}}=3,\,{{m}_{2}}=-2,\,{{n}_{2}}=-2\] Now, \[\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} -10 & -2 & -1 \\ 1 & -2 & 2 \\ 3 & -2 & -2 \\ \end{matrix} \right|\] \[=-10(4+4)+2(-2-6)-1(-2+6)\] \[=-100\] and \[\sqrt{\begin{align} & {{({{m}_{1}}{{n}_{2}}-{{m}_{2}}{{n}_{1}})}^{2}}+{{({{n}_{2}}{{l}_{2}}-{{l}_{1}}{{n}_{2}})}^{2}} \\ & +{{({{l}_{1}}{{m}_{2}}-{{l}_{2}}{{m}_{1}})}^{2}} \\ \end{align}}\] \[=\sqrt{{{(4+4)}^{2}}+{{(6+2)}^{2}}+{{(-2+6)}^{2}}}\] \[=\sqrt{64+64+16}=12\] \[\therefore \] Required shortest distance \[=\frac{-100}{12}=\frac{25}{3}\] [neglect \[(-ve)\] sign]
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