Given two equations describing the direction cosines of two lines: \( l + m + n = 0 \) and \( m^2 + n^2 - l^2 = 0 \). The direction cosines are denoted by \( l \), \( m \), and \( n \), representing the cosines of the angles with the \( x \), \( y \), and \( z \)-axes respectively. The objective is to determine the angle between these lines.
Step 1: Express \( n \) using the first equation.From \( l + m + n = 0 \), we get \( n = -l - m \).
Step 2: Substitute into the second equation.Substituting \( n \) into \( m^2 + n^2 - l^2 = 0 \) yields \( m^2 + (-l - m)^2 - l^2 = 0 \). Expanding and simplifying results in \( 2m^2 + 2lm = 0 \).
Step 3: Factor the equation.Factoring \( 2m^2 + 2lm = 0 \) gives \( 2m(m + l) = 0 \), implying either \( m = 0 \) or \( m = -l \).
Step 4: Solve for the angle.Consider the case \( m = -l \). Substituting this into \( n = -l - m \) gives \( n = -l - (-l) = 0 \). The direction cosines are thus \( l, -l, 0 \). The formula for the angle \( \theta \) between two lines with direction cosines \( l_1, m_1, n_1 \) and \( l_2, m_2, n_2 \) is \( \cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2 \). Using \( l_1 = l, m_1 = -l, n_1 = 0 \) and \( l_2 = l, m_2 = -l, n_2 = 0 \), we get \( \cos \theta = l^2 + (-l)^2 + 0 = 2l^2 \). If \( \cos \theta = \frac{1}{2} \), then \( l^2 = \frac{1}{4} \), so \( l = \frac{1}{2} \). The angle \( \theta \) is \( \cos^{-1} \left( \frac{1}{2} \right) = 60^\circ \).Thus, the angle between the lines is \( \boxed{60^\circ} \).