A) \[1\]
B) \[2\]
C) \[3\]
D) \[4\]
Correct Answer: B
Solution :
The shortest distance, between two lines \[\frac{x-{{x}_{1}}}{{{l}_{1}}}=\frac{y-{{y}_{1}}}{{{m}_{1}}}=\frac{z-{{z}_{1}}}{{{n}_{1}}}\] and \[\frac{x-{{x}_{2}}}{{{l}_{2}}}=\frac{y-{{y}_{2}}}{{{m}_{2}}}=\frac{z-{{z}_{2}}}{{{n}_{2}}}\] is \[d=\frac{\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ \end{matrix} \right|}{\sqrt{\begin{align} & {{({{m}_{1}}{{n}_{2}}-{{m}_{2}}{{n}_{1}})}^{2}}+{{({{n}_{1}}{{l}_{2}}-{{l}_{1}}{{n}_{2}})}^{2}} \\ & +{{({{l}_{1}}{{m}_{2}}-{{m}_{1}}{{l}_{2}})}^{2}} \\ \end{align}}}\] Given lines are \[\frac{x+1}{12}=\frac{y}{6}=\frac{z}{-1}\] and \[\frac{x}{6}=\frac{y+2}{6}=\frac{z-1}{1}\] \[\therefore \]\[d=\frac{\left| \begin{matrix} 0+1 & -2-0 & 1-0 \\ 12 & 6 & -1 \\ 6 & 6 & 1 \\ \end{matrix} \right|}{\sqrt{{{(6+6)}^{2}}+{{(-6-12)}^{2}}+{{(72-36)}^{2}}}}\] \[=\frac{\left| \begin{matrix} 1 & -2 & 1 \\ 12 & 6 & -1 \\ 6 & 6 & 1 \\ \end{matrix} \right|}{\sqrt{144+324+1296}}\] \[=\frac{1(6+6)+2(12+6)+1(72-36)}{\sqrt{1764}}\] \[=\frac{84}{\sqrt{1764}}=\frac{84}{42}=2\]
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