question_answer

By computing shortest distance, determine whether the following pair of lines intersect or not.             \[\vec{r}=(4\hat{i}+5\hat{j})+\lambda (\hat{i}+2\hat{j}-3\hat{k})\] and       \[\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu (2\hat{i}+4\hat{j}-5\hat{k})\]

Answer:

Given equations of lines are \[\vec{r}=(4\hat{i}+5\hat{j})+\lambda (\hat{i}+2\hat{j}-3\hat{k})\] and       \[\vec{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu (2\hat{i}+4\hat{j}-5\hat{k})\] On comparing with \[\vec{r}={{\vec{a}}_{1}}+\lambda \,{{\vec{b}}_{1}}\] and \[\vec{r}={{\vec{a}}_{2}}+\lambda \,{{\vec{b}}_{2}},\] we get \[{{\vec{a}}_{1}}=4\hat{i}+5\hat{j},\] \[{{\vec{b}}_{1}}=\hat{i}+2\hat{j}-3\hat{k}\] and       \[{{\vec{a}}_{2}}=\hat{i}-\hat{j}+2\hat{k},\] \[{{\vec{b}}_{2}}=2\hat{i}+4\hat{j}-5\hat{k}\]    \[\therefore \] \[{{\vec{a}}_{2}}-{{\vec{a}}_{1}}=\hat{i}-\hat{j}+2\hat{k}-4\hat{i}-5\hat{j}=-\,3\hat{i}-6\hat{j}+2\hat{k}\] and       \[{{\vec{b}}_{1}}\times {{\vec{b}}_{2}}=\left| \begin{matrix}    {\hat{i}} & {\hat{j}} & {\hat{k}}  \\    1 & 2 & -\,3  \\    2 & 4 & -\,5  \\ \end{matrix} \right|\]             \[=\hat{i}(-\,10+12)-\hat{j}(-\,5+6)+\hat{k}(4-4)\]             \[=2\hat{i}-\hat{j}+0\,\,\hat{k}\] Clearly, \[|{{\vec{b}}_{1}}\times {{\vec{b}}_{2}}|=\sqrt{{{(2)}^{2}}+{{(-\,1)}^{2}}+{{(0)}^{2}}}\]             \[=\sqrt{4+1}=\sqrt{5}\]                                    Now, shortest distance between given two lines is given by             \[=\left| \frac{({{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}})\cdot ({{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}})}{({{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}})} \right|\]             \[=\left| \frac{(2\hat{i}-\hat{j}+0\hat{k})\cdot (-\,3\hat{i}-6\hat{j}+2\hat{k})}{\sqrt{5}} \right|\] \[=\left| \frac{2\times (-\,3)+(-\,1)\times (-\,6)+0\times 2}{\sqrt{5}} \right|=\left| \frac{-\,6+6+0}{\sqrt{5}} \right|=0\] Thus, the shortest distance between given two lines is zero. Hence, the two lines intersect each other.