Find the distance between the lines: $\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$; $\vec{r} = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(4\hat{i} + 6\hat{j} + 12\hat{k})$

$\frac{\sqrt{293}}{7}$ units

The correct answer is Option (2) → $\frac{\sqrt{293}}{7}$ units ##

Given:

$\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$

$\vec{r} = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(4\hat{i} + 6\hat{j} + 12\hat{k})$

These lines are parallel

$∴$ Distance between two parallel lines

$= \frac{|\vec{b} \times (\vec{a_2} - \vec{a_1})|}{|\vec{b}|}$

$\vec{a_1} = \hat{i} + 2\hat{j} - 4\hat{k}$

$\vec{a_2} = 3\hat{i} + 3\hat{j} - 5\hat{k}$

$\vec{a_2} - \vec{a_1} = 2\hat{i} + \hat{j} - \hat{k}$

and $|\vec{b}| = \sqrt{4+9+36} = 7$

$\vec{b} \times (\vec{a_2} - \vec{a_1}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 2 & 1 & -1 \end{vmatrix}$

$= \hat{i}(-3 - 6) - \hat{j}(-2 - 12) + \hat{k}(2 - 6)$

$= -9\hat{i} + 14\hat{j} - 4\hat{k}$

$|\vec{b} \times (\vec{a_2} - \vec{a_1})| = |\sqrt{9^2 + 14^2 + 4^2}|$

$= |\sqrt{81 + 196 + 16}|$

$= \sqrt{293} \text{ units}$

$\text{Shortest distance} = \frac{\sqrt{293}}{7} \text{ units}$