Shortest distance between lines `(x -3)/2 = (y-15)/(-7) = (z – 9)/5` and `(x + 1)/2 = (y – 1)/1 = (z – 9)/(-3)` is equal to

Options

A. `24 sqrt 3`
B. `12 sqrt 3`
C. `6 sqrt 3`
D. `sqrt 3`

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 by:
\[ \frac{x - 3}{2} = \frac{y - 15}{-7} = \frac{z - 9}{5} \]
This can be represented in vector form as:
\[ \mathbf{r_1} = \begin{pmatrix} 3 \\ 15 \\ 9 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -7 \\ 5 \end{pmatrix} \]
where \( \mathbf{a_1} = \begin{pmatrix} 3 \\ 15 \\ 9 \end{pmatrix} \) and \( \mathbf{b_1} = \begin{pmatrix} 2 \\ -7 \\ 5 \end{pmatrix} \).

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

### Step 2: Calculate \( \mathbf{a_1} - \mathbf{a_2} \)
Now, we compute \( \mathbf{a_1} - \mathbf{a_2} \):
\[ \mathbf{a_1} - \mathbf{a_2} = \begin{pmatrix} 3 \\ 15 \\ 9 \end{pmatrix} - \begin{pmatrix} -1 \\ 1 \\ 9 \end{pmatrix} = \begin{pmatrix} 3 - (-1) \\ 15 - 1 \\ 9 - 9 \end{pmatrix} = \begin{pmatrix} 4 \\ 14 \\ 0 \end{pmatrix} \]

### Step 3: Compute the cross product \( \mathbf{b_1} \times \mathbf{b_2} \)
Next, we calculate the cross product \( \mathbf{b_1} \times \mathbf{b_2} \):
\[ \mathbf{b_1} = \begin{pmatrix} 2 \\ -7 \\ 5 \end{pmatrix}, \quad \mathbf{b_2} = \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix} \]
Using the determinant method:
\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} \]
Calculating the determinant:
\[ = \mathbf{i}((-7)(-3) - (5)(1)) - \mathbf{j}((2)(-3) - (5)(2)) + \mathbf{k}((2)(1) - (-7)(2)) \]
\[ = \mathbf{i}(21 - 5) - \mathbf{j}(-6 - 10) + \mathbf{k}(2 + 14) \]
\[ = \mathbf{i}(16) + \mathbf{j}(16) + \mathbf{k}(16) = \begin{pmatrix} 16 \\ 16 \\ 16 \end{pmatrix} \]

### Step 4: Calculate the magnitude of \( \mathbf{b_1} \times \mathbf{b_2} \)
The magnitude of the cross product is:
\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{16^2 + 16^2 + 16^2} = \sqrt{768} = 16\sqrt{3} \]

### Step 5: Calculate the shortest distance \( D \)
Using the formula for the shortest distance \( D \):
\[ D = \frac{|\mathbf{a_1} - \mathbf{a_2} \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]
First, compute the dot product:
\[ \mathbf{a_1} - \mathbf{a_2} \cdot (\mathbf{b_1} \times \mathbf{b_2}) = \begin{pmatrix} 4 \\ 14 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 16 \\ 16 \\ 16 \end{pmatrix} = 4 \cdot 16 + 14 \cdot 16 + 0 \cdot 16 = 64 + 224 + 0 = 288 \]
Thus, the distance \( D \) becomes:
\[ D = \frac{|288|}{16\sqrt{3}} = \frac{288}{16\sqrt{3}} = \frac{18}{\sqrt{3}} = 6\sqrt{3} \]

### Final Answer
The shortest distance between the two lines is:
\[ \boxed{6\sqrt{3}} \]