Three Dimensional Geometry – Shortest Distance – JEE Main 2025 Shift 2

Solution:

Let the two lines be:
$L_1: \vec{r} = \vec{a} + \lambda \vec{p}$, where $\vec{a} = (2, 1, -3)$ and $\vec{p} = (1, 2, -3)$.
$L_2: \vec{r} = \vec{b} + \mu \vec{q}$, where $\vec{b} = (-1, -3, -5)$ and $\vec{q} = (2, 4, -5)$.

The shortest distance (SD) between two skew lines is given by:
$$SD = \frac{|(\vec{b} – \vec{a}) \cdot (\vec{p} \times \vec{q})|}{|\vec{p} \times \vec{q}|}$$

First, find $\vec{b} – \vec{a}$:
$$\vec{b} – \vec{a} = (-1-2) \hat{i} + (-3-1) \hat{j} + (-5-(-3)) \hat{k}$$
$$\vec{b} – \vec{a} = -3\hat{i} – 4\hat{j} – 2\hat{k}$$

Next, find $\vec{p} \times \vec{q}$:
$$\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix}$$
$$= \hat{i}(-10 – (-12)) – \hat{j}(-5 – (-6)) + \hat{k}(4 – 4)$$
$$= \hat{i}(2) – \hat{j}(1) + \hat{k}(0)$$
$$= 2\hat{i} – \hat{j}$$

Now, calculate the dot product and magnitude:
Numerator: $|(\vec{b} – \vec{a}) \cdot (\vec{p} \times \vec{q})| = |(-3)(2) + (-4)(-1) + (-2)(0)| = |-6 + 4| = |-2| = 2$.
Denominator: $|\vec{p} \times \vec{q}| = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}$.

Therefore,
$$SD = \frac{2}{\sqrt{5}}$$
$$SD^2 = \frac{4}{5}$$

Given $SD^2 = \frac{m}{n} = \frac{4}{5}$.
Since 4 and 5 are coprime, $m=4$ and $n=5$.
We need $m+n$:
$$m+n = 4 + 5 = 9$$

Ans. (2)