A) 5
B) 2
C) 2
D) 1
Correct Answer: A
Solution :
Comparing the given equations of lines \[\overrightarrow{r}=(4\hat{i}-\hat{j})+\lambda (\hat{i}+2\hat{j}-3\hat{k})\] and\[\overrightarrow{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu (2\hat{i}+4\hat{j}-5\hat{k})\] with the general equations of lines \[\overrightarrow{r}={{\overrightarrow{a}}_{1}}+\lambda {{\overrightarrow{b}}_{1}}\] and \[\overrightarrow{r}={{\overrightarrow{a}}_{2}}+\mu {{\overrightarrow{b}}_{2}}\] we get \[{{\overrightarrow{a}}_{1}}=4\hat{i}-\hat{j},\text{ }{{\overrightarrow{a}}_{2}}=\hat{i}-\hat{j}+2\hat{k}\] \[{{\overrightarrow{b}}_{1}}=\hat{i}+2\hat{j}-3\hat{k},{{\overrightarrow{b}}_{2}}=2\hat{i}+4\hat{j}-5\hat{k},\] We know that the shortest distance between the lines \[\overrightarrow{r}={{\overrightarrow{a}}_{1}}+\lambda {{\overrightarrow{b}}_{1}}\] and\[\overrightarrow{r}={{\overrightarrow{a}}_{2}}+\mu {{\overrightarrow{b}}_{2}}\] is given by \[d=\left| \frac{({{\overrightarrow{a}}_{2}}-{{\overrightarrow{a}}_{1}}).({{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}})}{|{{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}}|} \right|\] where\[{{\overrightarrow{a}}_{2}}-{{\overrightarrow{a}}_{1}}=(\hat{i}-\hat{j}-2\hat{k})-(4\hat{i}-\hat{j})\] \[=-3\hat{i}+2\hat{k}\] \[=-3\hat{i}+0\hat{j}+2\hat{k}\] and\[{{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \\ \end{matrix} \right|\] \[=2\hat{i}-\hat{j}+0\hat{k},\] \[\Rightarrow \] \[|{{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}}|=\sqrt{4+1+0}=\sqrt{5}\] \[\therefore \] \[d=\left| \frac{(-3\hat{i}+0\hat{j}+2\hat{k}).(2\hat{i}-\hat{j}+0k)}{\sqrt{5}} \right|\] \[=\left| -\frac{6}{\sqrt{5}} \right|\] \[\Rightarrow \] \[d=\frac{6}{\sqrt{5}}\]
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