The shortest distance between the lines `(x-3)/(2) = (y +15)/(-7) = (z-9)/(5) and (x+1)/(2) = (y-1)/(1) = (z-9)/(-3) ` is :

Options

A. `2sqrt3`
B. `4sqrt3`
C. `3sqrt6`
D. `5sqrt6`

Answer

To find the shortest distance between the given lines, we can use the formula for the shortest distance between two skew lines in three-dimensional geometry. The lines are given in symmetric form, and we will extract the necessary parameters from them.

### Step-by-Step Solution:

1. **Identify the Lines**:
The first line is given by:
\[ \frac{x - 3}{2} = \frac{y + 15}{-7} = \frac{z - 9}{5} \]
The second line is given by:
\[ \frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{-3} \]

2. **Extract Points and Direction Ratios**:
From the first line, we can identify:
- Point \( P_1(3, -15, 9) \)
- Direction ratios \( \mathbf{a_1} = (2, -7, 5) \)

From the second line, we can identify:
- Point \( P_2(-1, 1, 9) \)
- Direction ratios \( \mathbf{a_2} = (2, 1, -3) \)

3. **Calculate the Vector Between the Points**:
The vector \( \mathbf{P_2 - P_1} \) is:
\[ \mathbf{P_2 - P_1} = (-1 - 3, 1 + 15, 9 - 9) = (-4, 16, 0) \]

4. **Set Up the Determinant**:
The shortest distance \( d \) between the two lines can be calculated using the formula:
\[ d = \frac{| \mathbf{(P_2 - P_1) \cdot (\mathbf{a_1} \times \mathbf{a_2})} |}{|\mathbf{a_1} \times \mathbf{a_2}|} \]
We first need to calculate the cross product \( \mathbf{a_1} \times \mathbf{a_2} \).

5. **Calculate the Cross Product**:
\[ \mathbf{a_1} = (2, -7, 5), \quad \mathbf{a_2} = (2, 1, -3) \]
The cross product \( \mathbf{a_1} \times \mathbf{a_2} \) is calculated as follows:
\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = \mathbf{i}((-7)(-3) - (5)(1)) - \mathbf{j}((2)(-3) - (5)(2)) + \mathbf{k}((2)(1) - (-7)(2)) \]
\[ = \mathbf{i}(21 - 5) - \mathbf{j}(-6 - 10) + \mathbf{k}(2 + 14) \]
\[ = \mathbf{i}(16) + \mathbf{j}(16) + \mathbf{k}(16) = (16, 16, 16) \]

6. **Magnitude of the Cross Product**:
\[ |\mathbf{a_1} \times \mathbf{a_2}| = \sqrt{16^2 + 16^2 + 16^2} = \sqrt{768} = 16\sqrt{3} \]

7. **Dot Product Calculation**:
Now calculate \( \mathbf{(P_2 - P_1) \cdot (\mathbf{a_1} \times \mathbf{a_2})} \):
\[ (-4, 16, 0) \cdot (16, 16, 16) = (-4)(16) + (16)(16) + (0)(16) = -64 + 256 + 0 = 192 \]

8. **Calculate the Shortest Distance**:
\[ d = \frac{|192|}{16\sqrt{3}} = \frac{192}{16\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3} \]

### Final Answer:
The shortest distance between the two lines is \( 4\sqrt{3} \) units.

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