Step 1: Understanding the Concept:
The angle \(\theta\) between two lines with direction ratios \((l_1, m_1, n_1)\) and \((l_2, m_2, n_2)\) is found using:
\[ \cos \theta = \frac{l_1 l_2 + m_1 m_2 + n_1 n_2}{\sqrt{l_1^2+m_1^2+n_1^2} \sqrt{l_2^2+m_2^2+n_2^2}} \]
If the sum of products of direction ratios is zero (\(l_1l_2 + m_1m_2 + n_1n_2 = 0\)), then \(\cos \theta = 0\), which means \(\theta = \frac{\pi}{2}\) (the lines are perpendicular).
Step 2: Detailed Explanation:
We are given two equations for the direction ratios \(l, m, n\):
1) \(l + m + n = 0 \implies n = -(l+m)\)
2) \(mn - 2ln + lm = 0\)
Substitute equation (1) into equation (2):
\[ m(-(l+m)) - 2l(-(l+m)) + lm = 0 \]
\[ -ml - m^2 + 2l^2 + 2lm + lm = 0 \]
Combine like terms:
\[ 2l^2 + 2lm - m^2 = 0 \]
Divide the whole equation by \(m^2\) (assuming \(m \neq 0\)):
\[ 2 \left( \frac{l}{m} \right)^2 + 2 \left( \frac{l}{m} \right) - 1 = 0 \]
This is a quadratic in the ratio \(\frac{l}{m}\). Let the two roots (representing the two lines) be \(\frac{l_1}{m_1}\) and \(\frac{l_2}{m_2}\).
From quadratic properties:
Product of roots: \(\frac{l_1 l_2}{m_1 m_2} = \frac{\text{constant term}}{\text{coeff of } x^2} = \frac{-1}{2}\)
So, \(2 l_1 l_2 = -m_1 m_2 \implies 2 l_1 l_2 + m_1 m_2 = 0\). (Eq. A)
Now we need to relate this to \(n\). From \(n = -(l+m)\):
\(n_1 n_2 = (-(l_1 + m_1)) \cdot (-(l_2 + m_2)) = l_1 l_2 + l_1 m_2 + l_2 m_1 + m_1 m_2\)
Sum of roots: \(\frac{l_1}{m_1} + \frac{l_2}{m_2} = \frac{-b}{a} = \frac{-2}{2} = -1\)
Multiply by \(m_1 m_2\): \(l_1 m_2 + l_2 m_1 = -m_1 m_2\).
Substitute this back into the expression for \(n_1 n_2\):
\(n_1 n_2 = l_1 l_2 + (-m_1 m_2) + m_1 m_2 = l_1 l_2\).
Check the perpendicularity condition \(l_1 l_2 + m_1 m_2 + n_1 n_2\):
Substitute \(n_1 n_2 = l_1 l_2\):
\[ l_1 l_2 + m_1 m_2 + l_1 l_2 = 2 l_1 l_2 + m_1 m_2 \]
From Equation A, we already found that \(2 l_1 l_2 + m_1 m_2 = 0\).
Thus, the sum of products of direction ratios is zero.
Step 3: Final Answer:
Since the condition for perpendicularity is met, the angle is \(\frac{\pi}{2}\).