The shortest distance between the lines `frac{x – 3}{4} = frac{y + 7}{-11} = frac{z – 1}{5}` and `frac{x – 5}{3} = frac{y – 9}{-6} = frac{z + 2}{1}` is:

Options

A. `frac{179}{sqrt 563}`
B. `frac{178}{sqrt 563}`
C. `frac{187}{sqrt 563}`
D. `frac{185}{sqrt 563}`

Answer

To find the shortest distance between the two given lines, we can follow these steps:

### Step 1: Write the equations of the lines in vector form
The first line is given as:
\[ \frac{x - 3}{4} = \frac{y + 7}{-11} = \frac{z - 1}{5} \]
This can be written in vector form as:
\[ \mathbf{r_1} = \begin{pmatrix} 3 \\ -7 \\ 1 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ -11 \\ 5 \end{pmatrix} \]
where \(\mathbf{a_1} = \begin{pmatrix} 3 \\ -7 \\ 1 \end{pmatrix}\) and \(\mathbf{b_1} = \begin{pmatrix} 4 \\ -11 \\ 5 \end{pmatrix}\).

The second line is given as:
\[ \frac{x - 5}{3} = \frac{y - 9}{-6} = \frac{z + 2}{1} \]
This can be written in vector form as:
\[ \mathbf{r_2} = \begin{pmatrix} 5 \\ 9 \\ -2 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -6 \\ 1 \end{pmatrix} \]
where \(\mathbf{a_2} = \begin{pmatrix} 5 \\ 9 \\ -2 \end{pmatrix}\) and \(\mathbf{b_2} = \begin{pmatrix} 3 \\ -6 \\ 1 \end{pmatrix}\).

### Step 2: Find the vector connecting points on the lines
Now we find the vector \(\mathbf{a_2} - \mathbf{a_1}\):
\[ \mathbf{a_2} - \mathbf{a_1} = \begin{pmatrix} 5 \\ 9 \\ -2 \end{pmatrix} - \begin{pmatrix} 3 \\ -7 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 16 \\ -3 \end{pmatrix} \]

### Step 3: Calculate the cross product of direction vectors
Next, we calculate the cross product \(\mathbf{b_1} \times \mathbf{b_2}\):
\[ \mathbf{b_1} = \begin{pmatrix} 4 \\ -11 \\ 5 \end{pmatrix}, \quad \mathbf{b_2} = \begin{pmatrix} 3 \\ -6 \\ 1 \end{pmatrix} \]
Using the determinant formula for the cross product:
\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -11 & 5 \\ 3 & -6 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ = \mathbf{i} \left((-11)(1) - (5)(-6)\right) - \mathbf{j} \left((4)(1) - (5)(3)\right) + \mathbf{k} \left((4)(-6) - (-11)(3)\right) \]
\[ = \mathbf{i} (-11 + 30) - \mathbf{j} (4 - 15) + \mathbf{k} (-24 + 33) \]
\[ = \mathbf{i} (19) + \mathbf{j} (11) + \mathbf{k} (9) = \begin{pmatrix} 19 \\ 11 \\ 9 \end{pmatrix} \]

### Step 4: Calculate the shortest distance
The formula for the shortest distance \(d\) between the two lines is given by:
\[ d = \frac{|(\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]

First, we calculate the dot product:
\[ (\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2}) = \begin{pmatrix} 2 \\ 16 \\ -3 \end{pmatrix} \cdot \begin{pmatrix} 19 \\ 11 \\ 9 \end{pmatrix} \]
\[ = 2 \cdot 19 + 16 \cdot 11 + (-3) \cdot 9 = 38 + 176 - 27 = 187 \]

Next, we calculate the magnitude of \(\mathbf{b_1} \times \mathbf{b_2}\):
\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{19^2 + 11^2 + 9^2} = \sqrt{361 + 121 + 81} = \sqrt{563} \]

Finally, substituting these values into the distance formula:
\[ d = \frac{|187|}{\sqrt{563}} = \frac{187}{\sqrt{563}} \]

### Final Answer
Thus, the shortest distance between the two lines is:
\[ \frac{187}{\sqrt{563}} \]