The shortest distance between the lines $\vec r = (\hat i+2\hat j+3\hat k) + λ(2\hat i +3\hat j + 4\hat k)$ and $\vec r = (2\hat i+4\hat j+5\hat k) + μ(4\hat i + 6\hat j + 8\hat k)$ is equal to

$\sqrt{\frac{5}{29}}$

The correct answer is Option (4) → $\sqrt{\frac{5}{29}}$

Given two lines:

Line 1: $\vec{r}_1 = \vec{a}_1 + \lambda \vec{b}_1$, where $\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$

Line 2: $\vec{r}_2 = \vec{a}_2 + \mu \vec{b}_2$, where $\vec{a}_2 = 2\hat{i} + 4\hat{j} + 5\hat{k}$ and $\vec{b}_2 = 4\hat{i} + 6\hat{j} + 8\hat{k}$

Note: $\vec{b}_2 = 2\vec{b}_1$, so the lines are parallel.

Shortest distance between two parallel lines is:

$D = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}_1|}{|\vec{b}_1|}$

$\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$

$\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$

Compute cross product:

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 3 & 4 \\ \end{vmatrix} = \hat{i}(2 \cdot 4 - 2 \cdot 3) - \hat{j}(1 \cdot 4 - 2 \cdot 2) + \hat{k}(1 \cdot 3 - 2 \cdot 2)$

$= \hat{i}(8 - 6) - \hat{j}(4 - 4) + \hat{k}(3 - 4) = 2\hat{i} + 0\hat{j} -1\hat{k}$

Magnitude: $|\vec{a} \times \vec{b}| = \sqrt{2^2 + 0^2 + (-1)^2} = \sqrt{5}$

$|\vec{b}_1| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{29}$

Shortest distance = $\frac{\sqrt{5}}{\sqrt{29}} = \sqrt{\frac{5}{29}}$ units