Find the angle between the line whose direction cosines are given by `l+m+n=0a n d2l^2+2m^2-n^2-0.`

Answer

To find the angle between the line whose direction cosines are given by the equations \( l + m + n = 0 \) and \( 2l^2 + 2m^2 - n^2 = 0 \), we will follow these steps:

### Step 1: Understand the equations
We have two equations involving the direction cosines \( l, m, n \):
1. \( l + m + n = 0 \)
2. \( 2l^2 + 2m^2 - n^2 = 0 \)

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

### Step 3: Substitute \( n \) into the second equation
Substituting \( n \) into the second equation:
\[ 2l^2 + 2m^2 - (- (l + m))^2 = 0 \]
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
This can be factored as:
\[ (l - m)^2 = 0 \]
Thus, we find:
\[ l = m \]

### Step 5: Substitute \( l = m \) back into the equations
Substituting \( l = m \) into the first equation:
\[ l + l + n = 0 \implies 2l + n = 0 \implies n = -2l \]

### Step 6: Use the normalization condition
The direction cosines must satisfy:
\[ l^2 + m^2 + n^2 = 1 \]
Substituting \( m = l \) and \( n = -2l \):
\[ l^2 + l^2 + (-2l)^2 = 1 \]
This simplifies to:
\[ l^2 + l^2 + 4l^2 = 1 \implies 6l^2 = 1 \implies l^2 = \frac{1}{6} \implies l = \pm \frac{1}{\sqrt{6}} \]
Thus, we have:
\[ l = m = \frac{1}{\sqrt{6}}, \quad n = -\frac{2}{\sqrt{6}} \]

### Step 7: Find the direction cosines of the second line
The direction cosines of the second line can be taken as:
\[ \left( \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}} \right) \]
And for the second line:
\[ \left( \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \right) \]

### Step 8: Calculate the angle between the two lines
Using the formula for the cosine of the angle \( \theta \) between two lines:
\[ \cos \theta = \frac{l_1 l_2 + m_1 m_2 + n_1 n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}} \]
Substituting the values:
\[ \cos \theta = \frac{\left( \frac{1}{\sqrt{6}} \right)^2 + \left( \frac{1}{\sqrt{6}} \right)^2 + \left( -\frac{2}{\sqrt{6}} \right) \left( \frac{2}{\sqrt{6}} \right)}{1} \]
Calculating:
\[ \cos \theta = \frac{\frac{1}{6} + \frac{1}{6} - \frac{4}{6}}{1} = \frac{2 - 4}{6} = -\frac{1}{3} \]

### Step 9: Find \( \theta \)
Thus, we have:
\[ \theta = \cos^{-1}\left(-\frac{1}{3}\right) \]

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