If 1,m ,n are the direction cosines of the line of shortest distance between the lines `(x-3)/(2) = (y+15)/(-7) = (z-9)/(5) and (x+1)/(2) = (y-1)/(1) = (z-9)/(-3) ` then :

Options

A. `31-1 5m + 9n =0`
B. `2l -7m + 5n =0`
C. `1=m =n=1 //sqrt3`
D. `21+m -3n =0`

Answer

To find the direction cosines \( L, M, N \) of the line of shortest distance between the two given lines, we can follow these steps:

### Step 1: Write the equations of the lines in parametric form
The first line is given by:
\[ \frac{x-3}{2} = \frac{y+15}{-7} = \frac{z-9}{5} \]
Let \( t \) be the parameter. Then the parametric equations are:
\[ x = 3 + 2t, \quad y = -15 - 7t, \quad z = 9 + 5t \]

The second line is given by:
\[ \frac{x+1}{2} = \frac{y-1}{1} = \frac{z-9}{-3} \]
Let \( s \) be the parameter. Then the parametric equations are:
\[ x = -1 + 2s, \quad y = 1 + s, \quad z = 9 - 3s \]

### Step 2: Identify points and direction vectors
From the parametric equations:
- For the first line, the point \( P_1(3, -15, 9) \) and direction vector \( \mathbf{d_1} = (2, -7, 5) \).
- For the second line, the point \( P_2(-1, 1, 9) \) and direction vector \( \mathbf{d_2} = (2, 1, -3) \).

### Step 3: Find the vector connecting the two points
The vector connecting points \( P_1 \) and \( P_2 \) is:
\[ \mathbf{P_1P_2} = P_2 - P_1 = (-1 - 3, 1 + 15, 9 - 9) = (-4, 16, 0) \]

### Step 4: Set up the equations for the direction cosines
The direction cosines \( L, M, N \) satisfy the equations derived from the shortest distance condition:
\[ \mathbf{d_1} \cdot \mathbf{n} = 0 \quad \text{and} \quad \mathbf{d_2} \cdot \mathbf{n} = 0 \]
where \( \mathbf{n} = (L, M, N) \) is the direction vector of the line of shortest distance.

From the first line:
\[ 2L - 7M + 5N = 0 \quad \text{(1)} \]
From the second line:
\[ 2L - M - 3N = 0 \quad \text{(2)} \]

### Step 5: Solve the system of equations
We can solve equations (1) and (2) simultaneously.

From equation (1):
\[ 2L - 7M + 5N = 0 \implies 2L = 7M - 5N \implies L = \frac{7M - 5N}{2} \quad \text{(3)} \]

Substituting (3) into equation (2):
\[ 2\left(\frac{7M - 5N}{2}\right) - M - 3N = 0 \]
This simplifies to:
\[ 7M - 5N - M - 3N = 0 \implies 6M - 8N = 0 \implies 3M = 4N \implies M = \frac{4}{3}N \quad \text{(4)} \]

### Step 6: Substitute back to find L, M, N
Substituting (4) into (3):
\[ L = \frac{7\left(\frac{4}{3}N\right) - 5N}{2} = \frac{\frac{28}{3}N - 5N}{2} = \frac{\frac{28}{3}N - \frac{15}{3}N}{2} = \frac{\frac{13}{3}N}{2} = \frac{13}{6}N \]

### Step 7: Normalize the direction cosines
The direction cosines must satisfy:
\[ L^2 + M^2 + N^2 = 1 \]
Substituting \( L \) and \( M \):
\[ \left(\frac{13}{6}N\right)^2 + \left(\frac{4}{3}N\right)^2 + N^2 = 1 \]
Calculating:
\[ \frac{169}{36}N^2 + \frac{16}{9}N^2 + N^2 = 1 \]
Converting to a common denominator:
\[ \frac{169}{36}N^2 + \frac{64}{36}N^2 + \frac{36}{36}N^2 = 1 \]
\[ \frac{269}{36}N^2 = 1 \implies N^2 = \frac{36}{269} \implies N = \pm \frac{6}{\sqrt{269}} \]

### Step 8: Find L and M
Using \( N \):
\[ M = \frac{4}{3}N = \pm \frac{8}{\sqrt{269}}, \quad L = \frac{13}{6}N = \pm \frac{13}{\sqrt{269}} \]

### Final Result
Thus, the direction cosines are:
\[ L = \frac{13}{\sqrt{269}}, \quad M = \frac{8}{\sqrt{269}}, \quad N = \frac{6}{\sqrt{269}} \]