Question:
If the direction ratios of two lines are given by $ l + m + n = 0 $ and $ mn - 2ln + lm = 0 $, then the angle between the lines is:
If the direction ratios of two lines are given by $ l + m + n = 0 $ and $ mn - 2ln + lm = 0 $, then the angle between the lines is:
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In problems involving homogeneous equations in $ l, m, n $, always aim to reduce the system to a quadratic in terms of a ratio like $ l/m $ or $ m/n $.
Updated On: May 13, 2026
- $ \frac{\pi}{4} $
- $ \frac{\pi}{3} $
- $ \frac{\pi}{2} $
- $ 0 $
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The Correct Option is C
Solution and Explanation
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 dot product $ l_1 l_2 + m_1 m_2 + n_1 n_2 = 0 $, the lines are perpendicular ($ \theta = \pi/2 $).
Step 1: Substituting one variable.
From $ l + m + n = 0 $, we have $ n = -(l + m) $. Substitute this into the second equation $ mn - 2ln + lm = 0 $: $$ m(-(l + m)) - 2l(-(l + m)) + lm = 0 $$
$$ -ml - m^2 + 2l^2 + 2lm + lm = 0 $$
$$ 2l^2 + 2lm - m^2 = 0 $$
Step 2: Forming a quadratic in $ (l/m) $.
Divide the equation by $ m^2 $:
$$ 2\left(\frac{l}{m}\right)^2 + 2\left(\frac{l}{m}\right) - 1 = 0 $$
Let the two roots (direction ratios of the two lines) be $ \frac{l_1}{m_1} $ and $ \frac{l_2}{m_2} $.
From the properties of quadratic equations:
Product of roots: $ \frac{l_1 l_2}{m_1 m_2} = \frac{c}{a} = \frac{-1}{2} \Rightarrow 2l_1 l_2 = -m_1 m_2 \Rightarrow 2l_1 l_2 + m_1 m_2 = 0 $.
Step 3: Finding the final condition.
We also have $ n_1 = -(l_1 + m_1) $ and $ n_2 = -(l_2 + m_2) $.
Product $ n_1 n_2 = (l_1 + m_1)(l_2 + m_2) = l_1 l_2 + l_1 m_2 + l_2 m_1 + m_1 m_2 $.
Using the sum of roots $ \frac{l_1}{m_1} + \frac{l_2}{m_2} = -\frac{b}{a} = -1 $, we get $ l_1 m_2 + l_2 m_1 = -m_1 m_2 $.
Substitute this back: $ n_1 n_2 = l_1 l_2 - m_1 m_2 + m_1 m_2 = l_1 l_2 $.
Now calculate $ l_1 l_2 + m_1 m_2 + n_1 n_2 $:
Using $ m_1 m_2 = -2l_1 l_2 $:
$$ l_1 l_2 + (-2l_1 l_2) + l_1 l_2 = 0 $$
Since the sum of the products of the direction ratios is zero, the lines are perpendicular.
$$ \cos \theta = \frac{|l_1 l_2 + m_1 m_2 + n_1 n_2|}{\sqrt{l_1^2+m_1^2+n_1^2}\sqrt{l_2^2+m_2^2+n_2^2}} $$
If the dot product $ l_1 l_2 + m_1 m_2 + n_1 n_2 = 0 $, the lines are perpendicular ($ \theta = \pi/2 $).
Step 1: Substituting one variable.
From $ l + m + n = 0 $, we have $ n = -(l + m) $. Substitute this into the second equation $ mn - 2ln + lm = 0 $: $$ m(-(l + m)) - 2l(-(l + m)) + lm = 0 $$
$$ -ml - m^2 + 2l^2 + 2lm + lm = 0 $$
$$ 2l^2 + 2lm - m^2 = 0 $$
Step 2: Forming a quadratic in $ (l/m) $.
Divide the equation by $ m^2 $:
$$ 2\left(\frac{l}{m}\right)^2 + 2\left(\frac{l}{m}\right) - 1 = 0 $$
Let the two roots (direction ratios of the two lines) be $ \frac{l_1}{m_1} $ and $ \frac{l_2}{m_2} $.
From the properties of quadratic equations:
Product of roots: $ \frac{l_1 l_2}{m_1 m_2} = \frac{c}{a} = \frac{-1}{2} \Rightarrow 2l_1 l_2 = -m_1 m_2 \Rightarrow 2l_1 l_2 + m_1 m_2 = 0 $.
Step 3: Finding the final condition.
We also have $ n_1 = -(l_1 + m_1) $ and $ n_2 = -(l_2 + m_2) $.
Product $ n_1 n_2 = (l_1 + m_1)(l_2 + m_2) = l_1 l_2 + l_1 m_2 + l_2 m_1 + m_1 m_2 $.
Using the sum of roots $ \frac{l_1}{m_1} + \frac{l_2}{m_2} = -\frac{b}{a} = -1 $, we get $ l_1 m_2 + l_2 m_1 = -m_1 m_2 $.
Substitute this back: $ n_1 n_2 = l_1 l_2 - m_1 m_2 + m_1 m_2 = l_1 l_2 $.
Now calculate $ l_1 l_2 + m_1 m_2 + n_1 n_2 $:
Using $ m_1 m_2 = -2l_1 l_2 $:
$$ l_1 l_2 + (-2l_1 l_2) + l_1 l_2 = 0 $$
Since the sum of the products of the direction ratios is zero, the lines are perpendicular.
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