The shortest distance between lines
`L_(1):(x+1)/(3)=(y+2)/(1)=(z+1)/(2)`,
`L_(2):(x-2)/(1)=(y+2)/(2)=(z-3)/(3)` is

Options

A. 0
B. `(17)/(sqrt(3))`
C. `(41)/(5sqrt(3))`
D. `(17)/(5sqrt(3))`

Answer

To find the shortest distance between the two lines \( L_1 \) and \( L_2 \), we can use the formula for the distance between two skew lines in space. The formula is given by:

\[ d = \frac{|(\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]

where:
- \( \mathbf{a_1} \) and \( \mathbf{a_2} \) are points on lines \( L_1 \) and \( L_2 \) respectively,
- \( \mathbf{b_1} \) and \( \mathbf{b_2} \) are direction vectors of lines \( L_1 \) and \( L_2 \) respectively.

### Step 1: Identify points and direction vectors

From the equations of the lines, we can extract the points and direction vectors.

For line \( L_1: \frac{x+1}{3} = \frac{y+2}{1} = \frac{z+1}{2} \):
- A point on \( L_1 \) can be taken as \( \mathbf{a_1} = (-1, -2, -1) \)
- The direction vector \( \mathbf{b_1} = (3, 1, 2) \)

For line \( L_2: \frac{x-2}{1} = \frac{y+2}{2} = \frac{z-3}{3} \):
- A point on \( L_2 \) can be taken as \( \mathbf{a_2} = (2, -2, 3) \)
- The direction vector \( \mathbf{b_2} = (1, 2, 3) \)

### Step 2: Calculate \( \mathbf{a_2} - \mathbf{a_1} \)

Now, we calculate the vector \( \mathbf{a_2} - \mathbf{a_1} \):

\[ \mathbf{a_2} - \mathbf{a_1} = (2, -2, 3) - (-1, -2, -1) = (2 + 1, -2 + 2, 3 + 1) = (3, 0, 4) \]

### Step 3: Calculate the cross product \( \mathbf{b_1} \times \mathbf{b_2} \)

Next, we calculate the cross product of the direction vectors \( \mathbf{b_1} \) and \( \mathbf{b_2} \):

\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 1 & 2 \\ 1 & 2 & 3 \end{vmatrix} \]

Calculating the determinant:

\[ = \mathbf{i} \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 3 & 2 \\ 1 & 3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = (1)(3) - (2)(2) = 3 - 4 = -1 \)
2. \( \begin{vmatrix} 3 & 2 \\ 1 & 3 \end{vmatrix} = (3)(3) - (2)(1) = 9 - 2 = 7 \)
3. \( \begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix} = (3)(2) - (1)(1) = 6 - 1 = 5 \)

Putting it all together:

\[ \mathbf{b_1} \times \mathbf{b_2} = -1 \mathbf{i} - 7 \mathbf{j} + 5 \mathbf{k} = (-1, -7, 5) \]

### Step 4: Calculate the magnitude of \( \mathbf{b_1} \times \mathbf{b_2} \)

Now we calculate the magnitude of the cross product:

\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{(-1)^2 + (-7)^2 + (5)^2} = \sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3} \]

### Step 5: Calculate the dot product \( (\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2}) \)

Now we calculate the dot product:

\[ (\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2}) = (3, 0, 4) \cdot (-1, -7, 5) = (3)(-1) + (0)(-7) + (4)(5) = -3 + 0 + 20 = 17 \]

### Step 6: Calculate the distance \( d \)

Finally, we can substitute these values into the distance formula:

\[ d = \frac{|17|}{5\sqrt{3}} = \frac{17}{5\sqrt{3}} = \frac{17\sqrt{3}}{15} \]

Thus, the shortest distance between the lines \( L_1 \) and \( L_2 \) is:

\[ \boxed{\frac{17\sqrt{3}}{15}} \]