The shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-5}{8}$ is equal to

The correct answer is:
$\sqrt{\frac{5}{29}}$

Step-by-Step Explanation:

First, let us write down the Cartesian equations of the two given lines:
Line 1:
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$
Line 2:
$\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-5}{8}$

We observe the direction ratios of the two lines:
- The direction ratios of Line 1 are $(2, 3, 4)$.
- The direction ratios of Line 2 are $(4, 6, 8)$.
Since $(4, 6, 8)=2\times (2, 3, 4)$, the direction ratios are proportional. Therefore, the two lines are parallel to each other.

Let us represent the equations of the lines in vector form, $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$:
- Line 1 passes through the point $A_1(1, 2, 3)$, so its position vector is:
$\overrightarrow{a_1}=\hat{i}+2\hat{j}+3\hat{k}$
- Line 2 passes through the point $A_2(2, 4, 5)$, so its position vector is:
$\overrightarrow{a_2}=2\hat{i}+4\hat{j}+5\hat{k}$
- Both lines share the common direction vector parallel to $(2, 3, 4)$:
$\overrightarrow{b}=2\hat{i}+3\hat{j}+4\hat{k}$

The shortest distance $d$ between two parallel lines is given by the formula:
$d=\frac{\left|\right(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}|}{\left|\overrightarrow{b}\right|}$

Let us calculate the constituent terms of this formula step-by-step:
1. Find $\overrightarrow{a_2}-\overrightarrow{a_1}$:
$\overrightarrow{a_2}-\overrightarrow{a_1}=(2-1)\hat{i}+(4-2)\hat{j}+(5-3)\hat{k}$
$\overrightarrow{a_2}-\overrightarrow{a_1}=\hat{i}+2\hat{j}+2\hat{k}$

2. Find the cross product $(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}$:
We evaluate the vector cross product using the determinant of a $3\times 3$ matrix:
$(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 2\\ 2& 3& 4\end{array}\right|$

Expanding along the first row:
$=\hat{i}\left[\right(2\times 4)-(2\times 3\left)\right]-\hat{j}\left[\right(1\times 4)-(2\times 2\left)\right]+\hat{k}\left[\right(1\times 3)-(2\times 2\left)\right]$
$=\hat{i}[8-6]-\hat{j}[4-4]+\hat{k}[3-4]$
$=2\hat{i}-0\hat{j}-\hat{k}=2\hat{i}-\hat{k}$

3. Compute the magnitudes:
- The magnitude of the cross-product vector is:
$\left|\right(\overrightarrow{a_2}-\overrightarrow{a_1})\times \overrightarrow{b}|=\sqrt{2^2+{(-1)}^2}=\sqrt{4+1}=\sqrt{5}$
- The magnitude of the direction vector $\overrightarrow{b}$ is:
$\left|\overrightarrow{b}\right|=\sqrt{2^2+3^2+4^2}=\sqrt{4+9+16}=\sqrt{29}$

4. Calculate the shortest distance $d$:
Substituting these values back into the shortest distance formula:
$d=\frac{\sqrt{5}}{\sqrt{29}}=\sqrt{\frac{5}{29}}$

Thus, the shortest distance between the given lines is $\sqrt{\frac{5}{29}}$.