the acute angle between two lines such that the direction cosines l, m, n of each of them satisfy the equations `l + m + n = 0` and `l^2 + m^2 - n^2 = 0` is

Options

A. `phi`
B. `(phi)/(3)`
C. `(phi)/(4)`
D. `(phi)/(6)`

Answer

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

### Step 1: Analyze the given equations
We have two equations:
1. \( l + m + n = 0 \) (Equation 1)
2. \( l^2 + m^2 - n^2 = 0 \) (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 \( n \) in Equation 2
Substituting \( n \) into Equation 2:
\[ l^2 + m^2 - (- (l + m))^2 = 0 \]
This simplifies to:
\[ l^2 + m^2 - (l^2 + 2lm + m^2) = 0 \]
\[ l^2 + m^2 - l^2 - 2lm - m^2 = 0 \]
\[ -2lm = 0 \]

### Step 4: Solve for \( l \) and \( m \)
From \( -2lm = 0 \), we can conclude:
\[ lm = 0 \]
This implies either \( l = 0 \) or \( m = 0 \).

### Step 5: Case analysis
1. **Case 1**: If \( l = 0 \):
- From \( l + m + n = 0 \), we have \( m + n = 0 \) which gives \( m = -n \).
- The direction cosines are \( (0, -n, n) \).

2. **Case 2**: If \( m = 0 \):
- From \( l + m + n = 0 \), we have \( l + n = 0 \) which gives \( l = -n \).
- The direction cosines are \( (-n, 0, n) \).

### Step 6: Determine the direction vectors
From Case 1:
- Direction vector \( \mathbf{h_1} = (0, -1, 1) \) (assuming \( n = 1 \)).

From Case 2:
- Direction vector \( \mathbf{h_2} = (-1, 0, 1) \) (assuming \( n = 1 \)).

### Step 7: Calculate the angle between the two vectors
The cosine of the angle \( \theta \) between two vectors \( \mathbf{h_1} \) and \( \mathbf{h_2} \) is given by:
\[ \cos \theta = \frac{\mathbf{h_1} \cdot \mathbf{h_2}}{|\mathbf{h_1}| |\mathbf{h_2}|} \]

### Step 8: Compute the dot product and magnitudes
1. **Dot product**:
\[ \mathbf{h_1} \cdot \mathbf{h_2} = (0)(-1) + (-1)(0) + (1)(1) = 1 \]

2. **Magnitude of \( \mathbf{h_1} \)**:
\[ |\mathbf{h_1}| = \sqrt{0^2 + (-1)^2 + 1^2} = \sqrt{2} \]

3. **Magnitude of \( \mathbf{h_2} \)**:
\[ |\mathbf{h_2}| = \sqrt{(-1)^2 + 0^2 + 1^2} = \sqrt{2} \]

### Step 9: Substitute into the cosine formula
\[ \cos \theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2} \]

### Step 10: Find the angle \( \theta \)
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \]

### Final Answer
The acute angle between the two lines is \( \frac{\pi}{3} \).

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