Find the shortest distance between the lines $L_1$ & $L_2$ given below: $L_1$: The line passing through $(2, -1, 1)$ and parallel to $\frac{x}{1} = \frac{y}{1} = \frac{z}{3}$ $L_2: \vec{r} = \hat{i} + (3\mu + 1)\hat{j} - (\mu + 2)\hat{k}$

$\frac{3}{\sqrt{110}}$ units

The correct answer is Option (2) → $\frac{3}{\sqrt{110}}$ units ##

Given $L_1$: The line passing through $(2, -1, 1)$ and parallel to $\frac{x}{1} = \frac{y}{1} = \frac{z}{3} \dots(i)$

Direction ratios of line (i) are $(1, 1, 3)$

$∴$ Vector equation of line passing through $(2, -1, 1)$ having direction ratios is

$\vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda(\hat{i} + \hat{j} + 3\hat{k}) \dots(ii)$

Given, $L_2$: $\vec{r} = \hat{i} + (3\mu + 1)\hat{j} - (\mu + 2)\hat{k}$

or $\vec{r} = (\hat{i} + \hat{j} - 2\hat{k}) + \mu(3\hat{j} - \hat{k}) \dots(iii)$

Now, shortest distance between lines (ii) & (iii) is given by

$d = \left| \frac{(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1})}{|\vec{b_1} \times \vec{b_2}|} \right|$

Here, $\vec{a_1} = (2\hat{i} - \hat{j} + \hat{k})$, $\vec{a_2} = (\hat{i} + \hat{j} - 2\hat{k})$

$\vec{b_1} = (\hat{i} + \hat{j} + 3\hat{k})$, $\vec{b_2} = (3\hat{j} - \hat{k})$

Now, $\vec{a_2} - \vec{a_1} = (1-2)\hat{i} + (1+1)\hat{j} + (-2-1)\hat{k} = -\hat{i} + 2\hat{j} - 3\hat{k}$

$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 3 \\ 0 & 3 & -1 \end{vmatrix}$

$= \hat{i}(-1 - 9) - \hat{j}(-1 - 0) + \hat{k}(3 - 0) = -10\hat{i} + \hat{j} + 3\hat{k}$

and $|\vec{b_1} \times \vec{b_2}| = \sqrt{(-10)^2 + (1)^2 + (3)^2} = \sqrt{100 + 1 + 9} = \sqrt{110}$

$d = \left| \frac{(-\hat{i} + 2\hat{j} - 3\hat{k}) \cdot (-10\hat{i} + \hat{j} + 3\hat{k})}{\sqrt{110}} \right|$

$= \left| \frac{10 + 2 - 9}{\sqrt{110}} \right| = \frac{3}{\sqrt{110}} \text{ units}$