Find the distance between the lines $l_1$ and $l_2$ given by $\vec{r} = \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$ and $\vec{r} = 3 \hat{i} + 3 \hat{j} - 5 \hat{k} + \mu (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$.

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

The correct answer is Option (1) → $\frac{\sqrt{293}}{7}$ units ##

The two lines are parallel. We have

$\vec{a}_1 = \hat{i} + 2 \hat{j} - 4 \hat{k}, \quad \vec{a}_2 = 3 \hat{i} + 3 \hat{j} - 5 \hat{k} \quad \text{and} \quad \vec{b} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$

Therefore, the distance between the lines is given by

$d = \left| \frac{\vec{b} \times (\vec{a}_2 - \vec{a}_1)}{|\vec{b}|} \right| = \left|\frac{ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 2 & 1 & -1 \end{vmatrix} }{\sqrt{4 + 9 + 36}}\right|$

or $= \frac{|-9\hat{i} + 14\hat{j} - 4\hat{k}|}{\sqrt{49}} = \frac{\sqrt{293}}{\sqrt{49}} = \frac{\sqrt{293}}{7}$