To find the unit vector perpendicular to both lines \( L_1 \) and \( L_2 \), we can follow these steps:
### Step 1: Write the equations of the lines in vector form
The given lines are:
- Line \( L_1: \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2} \)
- Line \( L_2: \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3} \)
For line \( L_1 \):
- The direction ratios are \( (3, 1, 2) \) and a point on the line is \( (-1, -2, -1) \).
- In vector form, it can be written as:
\[
\vec{r_1} = (-1, -2, -1) + \lambda(3, 1, 2)
\]
For line \( L_2 \):
- The direction ratios are \( (1, 2, 3) \) and a point on the line is \( (2, -2, 3) \).
- In vector form, it can be written as:
\[
\vec{r_2} = (2, -2, 3) + \mu(1, 2, 3)
\]
### Step 2: Identify the direction vectors of the lines
From the vector forms:
- The direction vector of \( L_1 \) is \( \vec{d_1} = (3, 1, 2) \)
- The direction vector of \( L_2 \) is \( \vec{d_2} = (1, 2, 3) \)
### Step 3: Find the cross product of the direction vectors
The cross product \( \vec{d_1} \times \vec{d_2} \) will give us a vector perpendicular to both lines.
\[
\vec{d_1} \times \vec{d_2} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & 2 \\
1 & 2 & 3
\end{vmatrix}
\]
Calculating the determinant:
\[
= \hat{i} \begin{vmatrix}
1 & 2 \\
2 & 3
\end{vmatrix} - \hat{j} \begin{vmatrix}
3 & 2 \\
1 & 3
\end{vmatrix} + \hat{k} \begin{vmatrix}
3 & 1 \\
1 & 2
\end{vmatrix}
\]
Calculating each of the 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:
\[
\vec{d_1} \times \vec{d_2} = -\hat{i} - 7\hat{j} + 5\hat{k}
\]
### Step 4: Find the magnitude of the cross product
\[
|\vec{d_1} \times \vec{d_2}| = \sqrt{(-1)^2 + (-7)^2 + (5)^2} = \sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3}
\]
### Step 5: Find the unit vector
The unit vector \( \hat{n} \) perpendicular to both lines is given by:
\[
\hat{n} = \frac{\vec{d_1} \times \vec{d_2}}{|\vec{d_1} \times \vec{d_2}|} = \frac{-\hat{i} - 7\hat{j} + 5\hat{k}}{5\sqrt{3}} = -\frac{1}{5\sqrt{3}} \hat{i} - \frac{7}{5\sqrt{3}} \hat{j} + \frac{5}{5\sqrt{3}} \hat{k}
\]
### Final Answer
Thus, the unit vector perpendicular to both lines \( L_1 \) and \( L_2 \) is:
\[
\hat{n} = -\frac{1}{5\sqrt{3}} \hat{i} - \frac{7}{5\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k}
\]
