A) \[3\]
B) \[2\]
C) \[1\]
D) \[0\]
Correct Answer: D
Solution :
Shortest distance \[=\frac{\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ \end{matrix} \right|}{\sqrt{\Sigma {{({{m}_{1}}{{n}_{2}}-{{m}_{2}}{{n}_{1}})}^{2}}}}\] Now, \[\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 5-2 & 1+3 & 6-1 \\ 3 & 4 & 5 \\ 1 & 2 & 3 \\ \end{matrix} \right|=\left| \begin{matrix} 3 & 4 & 5 \\ 3 & 4 & 5 \\ 1 & 2 & 3 \\ \end{matrix} \right|\] \[=0\] (\[\because \] two rows are identical) \[\therefore \] Shortest distance \[=0\]
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