To find the shortest distance between the two lines given by the equations:
1. \(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}\)
2. \(\frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}\)
we will follow these steps:
### Step 1: Identify the direction ratios and points on each line
From the first line, we can express it in the parametric form:
- \(x = 1 + 2t\)
- \(y = 2 + 3t\)
- \(z = 3 + 4t\)
The direction ratios of the first line, \(\mathbf{n_1}\), are \((2, 3, 4)\), and a point on the line, \(\mathbf{a}\), can be taken as \((1, 2, 3)\).
From the second line, we can express it in the parametric form:
- \(x = 2 + 3s\)
- \(y = 4 + 4s\)
- \(z = 5 + 5s\)
The direction ratios of the second line, \(\mathbf{n_2}\), are \((3, 4, 5)\), and a point on the line, \(\mathbf{b}\), can be taken as \((2, 4, 5)\).
### Step 2: Calculate the vector \(\mathbf{a} - \mathbf{b}\)
Now, we calculate the vector from point \(\mathbf{b}\) to point \(\mathbf{a}\):
\[
\mathbf{a} - \mathbf{b} = (1 - 2, 2 - 4, 3 - 5) = (-1, -2, -2)
\]
### Step 3: Calculate the cross product \(\mathbf{n_1} \times \mathbf{n_2}\)
Next, we calculate the cross product of the direction ratios:
\[
\mathbf{n_1} = (2, 3, 4), \quad \mathbf{n_2} = (3, 4, 5)
\]
Using the determinant form:
\[
\mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 3 & 4 \\
3 & 4 & 5
\end{vmatrix}
\]
Calculating the determinant:
\[
= \mathbf{i}(3 \cdot 5 - 4 \cdot 4) - \mathbf{j}(2 \cdot 5 - 4 \cdot 3) + \mathbf{k}(2 \cdot 4 - 3 \cdot 3)
\]
\[
= \mathbf{i}(15 - 16) - \mathbf{j}(10 - 12) + \mathbf{k}(8 - 9)
\]
\[
= -\mathbf{i} + 2\mathbf{j} - \mathbf{k}
\]
So, \(\mathbf{n_1} \times \mathbf{n_2} = (-1, 2, -1)\).
### Step 4: Calculate the magnitude of the cross product
Now we find the magnitude of the cross product:
\[
|\mathbf{n_1} \times \mathbf{n_2}| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}
\]
### Step 5: Calculate the shortest distance \(d\)
The formula for the shortest distance \(d\) between two skew lines is given by:
\[
d = \frac{|\mathbf{a} - \mathbf{b} \cdot (\mathbf{n_1} \times \mathbf{n_2})|}{|\mathbf{n_1} \times \mathbf{n_2}|}
\]
Calculating the dot product:
\[
\mathbf{a} - \mathbf{b} = (-1, -2, -2)
\]
\[
\mathbf{n_1} \times \mathbf{n_2} = (-1, 2, -1)
\]
\[
(\mathbf{a} - \mathbf{b}) \cdot (\mathbf{n_1} \times \mathbf{n_2}) = (-1)(-1) + (-2)(2) + (-2)(-1) = 1 - 4 + 2 = -1
\]
So, the absolute value is:
\[
|(\mathbf{a} - \mathbf{b}) \cdot (\mathbf{n_1} \times \mathbf{n_2})| = |-1| = 1
\]
Finally, substituting into the distance formula:
\[
d = \frac{1}{\sqrt{6}}
\]
### Final Answer
The shortest distance between the two lines is:
\[
d = \frac{1}{\sqrt{6}} \text{ units}
\]
