Find the shortest distance between the lines whose vector equations are $\overrightarrow{r}=(3s+2)\hat{i}-3\hat{j}+(4s+4)\hat{k}$ and $\overrightarrow{r}=(3t+2)\hat{i}-3\hat{j}+(4t)\hat{k}$

Concept:

The shortest distance between parallel lines $\overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b}$ and $\overrightarrow{r}=\overrightarrow{a_2}+\mu \overrightarrow{b}$ is given by: $d=\left|\frac{\overrightarrow{b}\times (\overrightarrow{a_2}-\overrightarrow{a_1})}{|b|}\right|$

Calculation:

L₁: $\overrightarrow{r}=(3s+2)\hat{i}-3\hat{j}+(4s+4)\hat{k}$ can be written as $\overrightarrow{r}=2\hat{i}-3\hat{j}+4\hat{k}+s(3\hat{i}+4\hat{k})$.

L₂: $\overrightarrow{r}=(3t+2)\hat{i}-3\hat{j}+(4t)\hat{k}$ can be written as $\overrightarrow{r}=2\hat{i}-3\hat{j}+t(3\hat{i}+4\hat{k})$.

Here, we see both lines are parallel and $\overrightarrow{a_1}=2\hat{i}-3\hat{j}+4\hat{k}$ , $\overrightarrow{a_2}=2\hat{i}-3\hat{j}$ and $\overrightarrow{b}=3\hat{i}+4\hat{k}$.

$\therefore$ The shortest distance between parallel lines L₁ and L₂: 

$d=\left|\frac{(3\hat{i}+4\hat{k})\times [(2\hat{i}-3\hat{j})-(2\hat{i}-3\hat{j}+4\hat{k})]}{|3\hat{i}+4\hat{k}|}\right|$

⇒ $d=\left|\frac{(3\hat{i}+4\hat{k})\times (-4\hat{k})}{\sqrt{9+16}}\right|$

⇒ $d=\left|\frac{12\hat{j}}{5}\right|$ ⇒ $d=\frac{12}{5}=2.4$ unit.

Hence, option 1 is correct.