To solve the problem, we need to analyze the given lines \( L_1 \) and \( L_2 \) and find the necessary parameters for the line \( L \) that bisects the acute angle between them.
### Step 1: Identify the Direction Ratios of the Lines
The lines are given in symmetric form. We can extract the direction ratios from the equations:
- For line \( L_1: \frac{x-1}{1} = \frac{y}{-1} = \frac{z-1}{3} \)
- Direction ratios \( d_1 = (1, -1, 3) \)
- For line \( L_2: \frac{x-1}{-3} = \frac{y}{-1} = \frac{z-1}{1} \)
- Direction ratios \( d_2 = (-3, -1, 1) \)
### Step 2: Find the Point of Intersection of \( L_1 \) and \( L_2 \)
To find the point of intersection, we can set the equations equal to each other. We can express \( L_1 \) and \( L_2 \) in parametric form:
- \( L_1: \)
\[
x = 1 + t, \quad y = -t, \quad z = 1 + 3t
\]
- \( L_2: \)
\[
x = 1 - 3s, \quad y = -s, \quad z = 1 + s
\]
Setting these equal to find \( t \) and \( s \):
1. From \( x \): \( 1 + t = 1 - 3s \) \( \Rightarrow t + 3s = 0 \) \( \Rightarrow t = -3s \)
2. From \( y \): \( -t = -s \) \( \Rightarrow t = s \)
Substituting \( t = -3s \) into \( t = s \):
\[
-3s = s \Rightarrow 4s = 0 \Rightarrow s = 0 \Rightarrow t = 0
\]
Thus, the point of intersection is:
\[
(1, 0, 1)
\]
### Step 3: Find the Unit Direction Vectors of \( L_1 \) and \( L_2 \)
Next, we need to find the unit direction vectors of the lines:
- For \( L_1 \):
\[
\text{Magnitude of } d_1 = \sqrt{1^2 + (-1)^2 + 3^2} = \sqrt{1 + 1 + 9} = \sqrt{11}
\]
\[
\text{Unit vector } u_1 = \left( \frac{1}{\sqrt{11}}, \frac{-1}{\sqrt{11}}, \frac{3}{\sqrt{11}} \right)
\]
- For \( L_2 \):
\[
\text{Magnitude of } d_2 = \sqrt{(-3)^2 + (-1)^2 + 1^2} = \sqrt{9 + 1 + 1} = \sqrt{11}
\]
\[
\text{Unit vector } u_2 = \left( \frac{-3}{\sqrt{11}}, \frac{-1}{\sqrt{11}}, \frac{1}{\sqrt{11}} \right)
\]
### Step 4: Find the Direction Ratios of the Angle Bisector
The direction ratios of the angle bisector can be found using the formula:
\[
d = u_1 + u_2
\]
Calculating:
\[
d = \left( \frac{1}{\sqrt{11}} - \frac{3}{\sqrt{11}}, \frac{-1}{\sqrt{11}} - \frac{1}{\sqrt{11}}, \frac{3}{\sqrt{11}} + \frac{1}{\sqrt{11}} \right)
\]
\[
= \left( \frac{-2}{\sqrt{11}}, \frac{-2}{\sqrt{11}}, \frac{4}{\sqrt{11}} \right)
\]
### Step 5: Write the Equation of Line \( L \)
The line \( L \) can be expressed in symmetric form as:
\[
\frac{x - \alpha}{-2} = \frac{y - 1}{-2} = \frac{z - \gamma}{4}
\]
From this, we can identify:
- \( \alpha = 1 \)
- \( \gamma = 1 \)
- The direction ratios are \( (-2, -2, 4) \).
### Step 6: Compare with Given Form
The line \( L \) is also given as:
\[
\frac{x - \alpha}{l} = \frac{y - 1}{m} = \frac{z - \gamma}{-2}
\]
From this, we can compare:
- \( l = -2 \)
- \( m = -2 \)
### Step 7: Conclusion
Now we can summarize the values:
- \( \alpha = 1 \)
- \( \gamma = 1 \)
- \( l = -2 \)
- \( m = -2 \)
