Find the shortest distance between the lines l_1 and l_2 whose vector…

The correct answer is Option (1) → $\frac{10}{\sqrt{59}}$ ##

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

and $\vec{r} = (2\hat{i} + \hat{j} - \hat{k}) + \mu(3\hat{i} - 5\hat{j} + 2\hat{k}) \quad \dots(ii)$

On comparing the eqs. (i) & (ii) with $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \lambda\vec{b_2}$, we get:

$\vec{a_1} = \hat{i} + \hat{j}, \quad \vec{b_1} = 2\hat{i} - \hat{j} + \hat{k}$

$\vec{a_2} = 2\hat{i} + \hat{j} - \hat{k} \quad \text{and} \quad \vec{b_2} = 3\hat{i} - 5\hat{j} + 2\hat{k}$

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

We know that,

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

Now, $(\vec{b_1} \times \vec{b_2}) = (2\hat{i} - \hat{j} + \hat{k}) \times (3\hat{i} - 5\hat{j} + 2\hat{k})$

$= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix}$

$= \hat{i}(-2 + 5) - \hat{j}(4 - 3) + \hat{k}(-10 + 3)$

$= 3\hat{i} - \hat{j} - 7\hat{k}$

Also, $|\vec{b_1} \times \vec{b_2}| = \sqrt{3^2 + (-1)^2 + (-7)^2} = \sqrt{59}$

Hence, $\vec{d} = \left| \frac{(3\hat{i} - \hat{j} - 7\hat{k}) \cdot (\hat{i} - \hat{k})}{\sqrt{59}} \right|$

$= \left| \frac{3 - 0 + 7}{\sqrt{59}} \right| = \frac{10}{\sqrt{59}}$