To find the shortest distance between the two given lines, we can follow these steps:
### Step 1: Write the lines in vector form
The lines are given in symmetric form. We can convert them into vector form.
For the first line:
\[
\frac{x-2}{2} = \frac{y-3}{2} = \frac{z-0}{1}
\]
This can be represented as:
\[
\mathbf{r_1} = (2, 3, 0) + \lambda (2, 2, 1)
\]
where \( \mathbf{A_1} = (2, 3, 0) \) and \( \mathbf{B_1} = (2, 2, 1) \).
For the second line:
\[
\frac{x+4}{-1} = \frac{y-7}{8} = \frac{z-5}{4}
\]
This can be represented as:
\[
\mathbf{r_2} = (-4, 7, 5) + \mu (-1, 8, 4)
\]
where \( \mathbf{A_2} = (-4, 7, 5) \) and \( \mathbf{B_2} = (-1, 8, 4) \).
### Step 2: Calculate \( \mathbf{A_2} - \mathbf{A_1} \)
Now, we need to find \( \mathbf{A_2} - \mathbf{A_1} \):
\[
\mathbf{A_2} - \mathbf{A_1} = (-4, 7, 5) - (2, 3, 0) = (-6, 4, 5)
\]
### Step 3: Calculate the cross product \( \mathbf{B_1} \times \mathbf{B_2} \)
Next, we need to calculate the cross product \( \mathbf{B_1} \times \mathbf{B_2} \):
\[
\mathbf{B_1} = (2, 2, 1), \quad \mathbf{B_2} = (-1, 8, 4)
\]
Using the determinant method for the cross product:
\[
\mathbf{B_1} \times \mathbf{B_2} = \begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 2 & 1 \\
-1 & 8 & 4
\end{vmatrix}
\]
Calculating the determinant:
\[
= \mathbf{i}(2 \cdot 4 - 1 \cdot 8) - \mathbf{j}(2 \cdot 4 - 1 \cdot (-1)) + \mathbf{k}(2 \cdot 8 - 2 \cdot (-1))
\]
\[
= \mathbf{i}(8 - 8) - \mathbf{j}(8 + 1) + \mathbf{k}(16 + 2)
\]
\[
= 0\mathbf{i} - 9\mathbf{j} + 18\mathbf{k} = (0, -9, 18)
\]
### Step 4: Calculate the dot product \( (\mathbf{A_2} - \mathbf{A_1}) \cdot (\mathbf{B_1} \times \mathbf{B_2}) \)
Now, we calculate the dot product:
\[
(\mathbf{A_2} - \mathbf{A_1}) \cdot (\mathbf{B_1} \times \mathbf{B_2}) = (-6, 4, 5) \cdot (0, -9, 18)
\]
\[
= (-6)(0) + (4)(-9) + (5)(18) = 0 - 36 + 90 = 54
\]
### Step 5: Calculate the magnitude of \( \mathbf{B_1} \times \mathbf{B_2} \)
Next, we find the magnitude of \( \mathbf{B_1} \times \mathbf{B_2} \):
\[
\|\mathbf{B_1} \times \mathbf{B_2}\| = \sqrt{0^2 + (-9)^2 + 18^2} = \sqrt{0 + 81 + 324} = \sqrt{405} = 9\sqrt{5}
\]
### Step 6: Calculate the shortest distance
Finally, we can use the formula for the shortest distance \( d \):
\[
d = \frac{|(\mathbf{A_2} - \mathbf{A_1}) \cdot (\mathbf{B_1} \times \mathbf{B_2})|}{\|\mathbf{B_1} \times \mathbf{B_2}\|}
\]
\[
d = \frac{54}{9\sqrt{5}} = \frac{6}{\sqrt{5}} = \frac{6\sqrt{5}}{5}
\]
### Step 7: Determine the interval
To find the interval in which the shortest distance lies, we approximate \( \sqrt{5} \) (which is approximately between 2 and 3):
\[
\frac{6\sqrt{5}}{5} \approx \frac{6 \times 2.236}{5} \approx \frac{13.416}{5} \approx 2.6832
\]
Thus, the shortest distance lies between 2 and 3.
### Final Answer
The shortest distance between the lines lies in the interval \( (2, 3) \).
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