To find the acute angle between the two straight lines whose direction cosines are given by the equations \( l + m + n = 0 \) and \( l^2 + m^2 - n^2 = 0 \), we will follow these steps:
### Step 1: Set up the equations
We have the following equations:
1. \( l + m + n = 0 \) (Equation 1)
2. \( l^2 + m^2 - n^2 = 0 \) (Equation 2)
Additionally, we know that the sum of the squares of the direction cosines is equal to 1:
3. \( l^2 + m^2 + n^2 = 1 \) (Equation 3)
### Step 2: Express \( n \) in terms of \( l \) and \( m \)
From Equation 1, we can express \( n \) as:
\[ n = - (l + m) \]
### Step 3: Substitute \( n \) into Equation 2
Substituting \( n \) into Equation 2 gives:
\[ l^2 + m^2 - (- (l + m))^2 = 0 \]
Expanding this, we have:
\[ l^2 + m^2 - (l^2 + 2lm + m^2) = 0 \]
This simplifies to:
\[ l^2 + m^2 - l^2 - 2lm - m^2 = 0 \]
Thus:
\[ -2lm = 0 \]
This implies:
\[ lm = 0 \]
### Step 4: Determine values for \( l \) and \( m \)
Since \( lm = 0 \), either \( l = 0 \) or \( m = 0 \). We will consider both cases.
#### Case 1: \( l = 0 \)
If \( l = 0 \), then from Equation 1:
\[ 0 + m + n = 0 \Rightarrow n = -m \]
Substituting into Equation 3:
\[ 0^2 + m^2 + (-m)^2 = 1 \]
This simplifies to:
\[ 2m^2 = 1 \Rightarrow m^2 = \frac{1}{2} \Rightarrow m = \pm \frac{1}{\sqrt{2}} \]
Thus, \( n = \mp \frac{1}{\sqrt{2}} \).
#### Case 2: \( m = 0 \)
If \( m = 0 \), then from Equation 1:
\[ l + 0 + n = 0 \Rightarrow n = -l \]
Substituting into Equation 3:
\[ l^2 + 0^2 + (-l)^2 = 1 \]
This simplifies to:
\[ 2l^2 = 1 \Rightarrow l^2 = \frac{1}{2} \Rightarrow l = \pm \frac{1}{\sqrt{2}} \]
Thus, \( n = \mp \frac{1}{\sqrt{2}} \).
### Step 5: Identify direction cosines
From both cases, we find the direction cosines:
1. \( (0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}) \)
2. \( (0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \)
### Step 6: Calculate 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:
\[
\cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2
\]
Substituting the values:
\[
\cos \theta = 0 \cdot 0 + \frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}} + -\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = -\frac{1}{2} - \frac{1}{2} = -1
\]
### Step 7: Find the acute angle
Since \( \cos \theta = -\frac{1}{2} \), we have:
\[
\theta = \cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3}
\]
However, since we need the acute angle, we take:
\[
\text{Acute angle} = \pi - \frac{2\pi}{3} = \frac{\pi}{3}
\]
Thus, the acute angle between the two straight lines is:
\[
\boxed{\frac{\pi}{3}}
\]
