Find the angle between the lines whose direction cosines have the relations `l+m+n=0 and 2l^(2)+2m^(2)-n^(2)=0`.

Answer

To find the angle between the lines whose direction cosines satisfy the equations \( l + m + n = 0 \) and \( 2l^2 + 2m^2 - n^2 = 0 \), we can follow these steps:

### Step 1: Write down the equations
We have two equations:
1. \( l + m + n = 0 \) (Equation 1)
2. \( 2l^2 + 2m^2 - n^2 = 0 \) (Equation 2)

### Step 2: Express \( n \) in terms of \( l \) and \( m \)
From Equation 1, we can express \( n \):
\[ n = - (l + m) \]

### Step 3: Substitute \( n \) into Equation 2
Substituting \( n \) into Equation 2:
\[ 2l^2 + 2m^2 - (- (l + m))^2 = 0 \]
This simplifies to:
\[ 2l^2 + 2m^2 - (l^2 + 2lm + m^2) = 0 \]
Combining like terms gives:
\[ 2l^2 + 2m^2 - l^2 - 2lm - m^2 = 0 \]
\[ l^2 + m^2 - 2lm = 0 \]

### Step 4: Factor the equation
We can factor this as:
\[ (l - m)^2 = 0 \]
This implies:
\[ l = m \]

### Step 5: Substitute \( l = m \) back into Equation 1
Substituting \( l = m \) into Equation 1:
\[ l + l + n = 0 \implies 2l + n = 0 \implies n = -2l \]

### Step 6: Use the condition \( l^2 + m^2 + n^2 = 1 \)
Since \( l = m \) and \( n = -2l \), we can substitute these into the normalization condition:
\[ l^2 + l^2 + (-2l)^2 = 1 \]
This simplifies to:
\[ 2l^2 + 4l^2 = 1 \implies 6l^2 = 1 \implies l^2 = \frac{1}{6} \implies l = \pm \frac{1}{\sqrt{6}} \]

### Step 7: Find \( m \) and \( n \)
Since \( l = m \):
\[ m = \pm \frac{1}{\sqrt{6}} \]
And for \( n \):
\[ n = -2l = \mp \frac{2}{\sqrt{6}} = \mp \frac{\sqrt{6}}{3} \]

### Step 8: Determine the direction cosines
Thus, the direction cosines are:
1. \( \left( \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{\sqrt{6}}{3} \right) \)
2. \( \left( \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{\sqrt{6}}{3} \right) \)
3. \( \left( -\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, \frac{\sqrt{6}}{3} \right) \)
4. \( \left( -\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, -\frac{\sqrt{6}}{3} \right) \)

### Step 9: Calculate the angle between the lines
Using the formula for the angle \( \theta \) between two lines given by their direction cosines:
\[ \cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2 \]
Taking two sets of direction cosines:
\[ \cos \theta = \left( \frac{1}{\sqrt{6}} \cdot \frac{1}{\sqrt{6}} \right) + \left( \frac{1}{\sqrt{6}} \cdot \frac{1}{\sqrt{6}} \right) + \left( \frac{\sqrt{6}}{3} \cdot -\frac{\sqrt{6}}{3} \right) \]
Calculating this gives:
\[ \cos \theta = \frac{1}{6} + \frac{1}{6} - \frac{2}{3} = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3} \]

### Step 10: Find the angle
Thus, the angle \( \theta \) is:
\[ \theta = \cos^{-1}\left(-\frac{1}{3}\right) \]

### Final Answer
The angle between the lines is \( \theta = \cos^{-1}\left(-\frac{1}{3}\right) \).

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