question_answer

Two ships in the sea were reported missing their link with ground control suddenly. They were on the lines \[\overrightarrow{r}=(2+\lambda )\hat{i}-(3+\lambda )\hat{j}\,+\,(5+\lambda )\hat{k}\]             and       \[\overrightarrow{r}=(2\mu +1)\hat{i}\,+(4\mu -1)\hat{j}\,+\,(5-3\mu )\hat{k}.\] Using   shortest   distance   formula, determine whether they met any mishappening. What value do you see in it.

Answer:

Given equation of lines can be rewritten as                         \[\vec{r}=(2\hat{i}-3\hat{j}+5\hat{k})+\lambda (\hat{i}-\hat{j}\,+\hat{k})\]             and       \[\vec{r}=(-\,\hat{i}-\hat{j}+5\hat{k})+\mu (2\hat{i}\,+4\hat{j}\,-3\hat{k})\] On comparing the above equations with standard vector form of equation of line, \[\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b},\] we get             \[{{\vec{a}}_{1}}=2\hat{i}-3\hat{j}+5\hat{k},\] \[{{\vec{b}}_{1}}=\hat{i}-\hat{j}+\hat{k},\]             \[{{\vec{a}}_{2}}=-\,\hat{i}-\hat{j}+5\hat{k}\] and \[{{\vec{b}}_{2}}=2\hat{i}+4\hat{j}-3\hat{k}\] Now, consider \[\overrightarrow{{{b}_{1}}}\times \overrightarrow{{{b}_{2}}}=\left| \begin{matrix}    {\hat{i}} & {\hat{j}} & {\hat{k}}  \\    1 & -\,1 & 1  \\    2 & 4 & -\,3  \\ \end{matrix} \right|\]             \[=\hat{i}(3-4)-\hat{j}(-\,3-2)+\hat{k}(4+2)\] \[\Rightarrow \]   \[{{\vec{b}}_{1}}\times {{\vec{b}}_{2}}=-\,\hat{i}+5\hat{j}+6\hat{k}\] \[\Rightarrow \]   \[|{{\vec{b}}_{1}}\times {{\vec{b}}_{2}}|\,\,=\sqrt{{{(-1)}^{2}}+{{(5)}^{2}}+{{(6)}^{2}}}\] \[\Rightarrow \]               \[=\sqrt{1+25+36}=\sqrt{62}\] Also,     \[{{\vec{a}}_{2}}-{{\vec{a}}_{1}}=(-\hat{i}-\hat{j}\,+5\hat{k})-(2\hat{i}-3\hat{j}+5\hat{k})\]                         \[=-\,3\hat{i}+2\hat{j}\] We know that, shortest distance between two lines is given by \[d=\left| \frac{({{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}).({{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}})}{|{{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}|} \right|\] On purring above values, we get             \[d=\left| \frac{(-\hat{i}+5\hat{j}\,+6\hat{k}).(-3\hat{i}\,+2\hat{j})}{\sqrt{62}} \right|\]             \[=\left| \frac{3+10+0}{\sqrt{62}} \right|=\frac{13}{\sqrt{62}}=\frac{13\sqrt{62}}{62}\] Hence, required shortest distance is \[\frac{13\sqrt{62}}{62}\] units. Two ships will not met any accidental collision. Value Carelessness leads to mishappening.