A) 6
B) 8
C) 12
D) 9
Correct Answer: D
Solution :
Here, \[{{\overrightarrow{a}}_{1}}=6\hat{i}+2\hat{j}+2\hat{k},\]\[{{\overrightarrow{a}}_{2}}=-\,4\hat{i}+0\hat{j}-\hat{k},\] \[{{\overrightarrow{b}}_{1}}=\hat{i}-2\hat{j}+2\hat{k}\] and \[{{\overrightarrow{b}}_{2}}=3\hat{i}-2\hat{j}-2\hat{k}\] \[\therefore \]Shortest distance\[=\left[ \frac{({{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}})\cdot ({{{\vec{b}}}_{1}}-{{{\vec{b}}}_{2}})}{|{{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}|} \right]\] \[=\left| \frac{(-10\,\hat{i}-2\hat{j}-3\hat{k})\cdot (8\hat{i}+8\hat{j}+4\hat{k})}{\sqrt{64+64+16}} \right|\] \[=\left| \frac{-108}{12} \right|=9\]
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