A) 7 unit
B) 13 unit
C) 8 unit
D) 9 unit
E) 11 unit
Correct Answer: D
Solution :
Here, \[\overrightarrow{{{a}_{1}}}=-3\hat{i}+6\hat{j},\overrightarrow{{{a}_{2}}}=-2\hat{i}+7\hat{k},\] \[\overrightarrow{{{b}_{1}}}=-4\hat{i}+3\hat{j}+2\hat{k},\]and\[\overrightarrow{{{b}_{2}}}=-4\hat{i}+\hat{j}+\hat{k}\] Now, \[\overrightarrow{{{a}_{2}}}-\overrightarrow{{{a}_{1}}}=-2\hat{i}+7\hat{k}+3\hat{i}-6\hat{j}\] \[=\hat{i}-6\hat{j}+7\hat{k}\] and \[\overrightarrow{{{b}_{1}}}\times \overrightarrow{{{b}_{2}}}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ -4 & 3 & 2 \\ -4 & 1 & 1 \\ \end{matrix} \right|\] \[=\hat{i}(1)-\hat{j}(4)+\hat{k}(8)\] \[=\hat{i}-4\hat{j}+8\hat{k}\] \[\Rightarrow \]\[|\overrightarrow{{{b}_{1}}}\times \overrightarrow{{{b}_{2}}}|=\sqrt{1+16+64}=\sqrt{81}=9\] \[\therefore \] \[(\overrightarrow{{{a}_{2}}}-\overrightarrow{{{a}_{1}}}).(\overrightarrow{{{b}_{1}}}\times \overrightarrow{{{b}_{2}}})\] \[=(\hat{i}-6\hat{j}+7\hat{k}).(\hat{i}-4\hat{j}+8\hat{k})\] \[=1+24+56=81\] \[\therefore \]Shortest distance, \[d=\left| \frac{(\overrightarrow{{{a}_{2}}}-\overrightarrow{{{a}_{1}}}).(\overrightarrow{{{b}_{1}}}\times \overrightarrow{{{b}_{2}}})}{|\overrightarrow{{{b}_{1}}}\times \overrightarrow{{{b}_{2}}}|} \right|\] \[=\frac{81}{9}=9\,unit\]
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