Question

Find the shortest distance between the lines $\overrightarrow{r}=(4\hat{i}-\hat{j})+\lambda (\hat{i}+2\hat{j}-3\hat{k})$ and $\overrightarrow{r}=(\hat{i}-\hat{j}+2\hat{k})+\mu (2\hat{i}+4\hat{j}-5\hat{k})$.

Answer 1

Shortest distance between lines with vector equations

$\overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$ and $\overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$

$\to$$\left|\frac{(\overrightarrow{b_1}\times \overrightarrow{b_2}).(\overrightarrow{a_2}-\overrightarrow{a_1})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|$

$\overrightarrow{r}=4\hat{i}-\hat{j}+\lambda (\hat{i}+2\hat{j}-3\hat{k})$

Comparing with $\overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}$

$\to \overrightarrow{a_1}=4\hat{i}-\hat{j}$

$\overrightarrow{b_1}=\hat{i}+2\hat{j}-3\hat{k}$

$\overrightarrow{r}=\hat{i}-\hat{j}+2\hat{k}+\mu (2\hat{i}+4\hat{j}-5\hat{k})$

Comparing with $\overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$

$\to \overrightarrow{a_2}=\hat{i}-\hat{j}+2\hat{k}$

$\overrightarrow{b_2}=2\hat{i}+4\hat{j}-5\hat{k}$

$\overrightarrow{a_2}-\overrightarrow{a_1}=$$(\hat{i}-\hat{j}+2\hat{k})-$$(4\hat{i}-\hat{j})$$=-3\hat{i}+2\hat{k}$

$\overrightarrow{b_1}\times \overrightarrow{b_2}$$=2\hat{i}-\hat{j}$

$|\overrightarrow{b_1}\times \overrightarrow{b_2}|$$=\sqrt{5}$

$\to$$(\overrightarrow{b_1}\times \overrightarrow{b_2}).(\overrightarrow{a_2}-\overrightarrow{a_1})=(-3\times 2)=-6$

Shortest distance $\to$$\left|\frac{(\overrightarrow{b_1}\times \overrightarrow{b_2}).(\overrightarrow{a_2}-\overrightarrow{a_1})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|=\left|\frac{-6}{\sqrt{5}}\right|$

Or, Shortest Distance $=\frac{6}{\sqrt{5}}$