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

Options

A. `60^@`
B. `cos^(-1)((1)/(3))`
C. `cos^(-1)(-(1)/(3))`
D. none of these

Explanation

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

### Step 2: Substitute \( n \) into the second equation
Now, substitute \( n \) into the second equation:
\[ 2l^2 + 2m^2 - n^2 = 0 \]
Substituting \( n = -(l + m) \):
\[ 2l^2 + 2m^2 - (-(l + m))^2 = 0 \]
This simplifies to:
\[ 2l^2 + 2m^2 - (l^2 + 2lm + m^2) = 0 \]
\[ 2l^2 + 2m^2 - l^2 - 2lm - m^2 = 0 \]
\[ l^2 + m^2 - 2lm = 0 \]

### Step 3: Factor the equation
This can be factored as:
\[ (l - m)^2 = 0 \]
Thus, we have:
\[ l = m \]

### Step 4: Substitute \( l \) back to find \( n \)
Substituting \( l = m \) into the equation \( n = -(l + m) \):
\[ n = -2l \]

### Step 5: Use the identity \( l^2 + m^2 + n^2 = 1 \)
Now, using the identity \( l^2 + m^2 + n^2 = 1 \):
\[ l^2 + l^2 + (-2l)^2 = 1 \]
\[ 2l^2 + 4l^2 = 1 \]
\[ 6l^2 = 1 \implies l^2 = \frac{1}{6} \implies l = \pm \frac{1}{\sqrt{6}} \]
Thus, we have:
\[ m = \pm \frac{1}{\sqrt{6}}, \quad n = \mp \frac{2}{\sqrt{6}} \]

### Step 6: Calculate the direction cosines
The direction cosines can be:
1. \( \left( \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, -\frac{2}{\sqrt{6}} \right) \)
2. \( \left( -\frac{1}{\sqrt{6}}, -\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \right) \)

### Step 7: Find the angle between the lines
The cosine of the angle \( \theta \) between two lines with direction cosines \( (l_1, m_1, n_1) \) and \( (l_2, m_2, n_2) \) is given by:
\[ \cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2 \]
Calculating for the first case:
\[ \cos \theta = \left( \frac{1}{\sqrt{6}} \right) \left( -\frac{1}{\sqrt{6}} \right) + \left( \frac{1}{\sqrt{6}} \right) \left( -\frac{1}{\sqrt{6}} \right) + \left( -\frac{2}{\sqrt{6}} \right) \left( \frac{2}{\sqrt{6}} \right) \]
\[ = -\frac{1}{6} - \frac{1}{6} - \frac{4}{6} = -\frac{6}{6} = -1 \]

### Step 8: Conclusion
Thus, the angle \( \theta \) is:
\[ \theta = \cos^{-1}(-1) = 180^\circ \]