The direction cosines of two lines satisfying the
conditions `l + m + n = 0 and 3lm - 5mn + 2nl = 0`
where l, m, n are the direction cosines.
Angle between the lines is

Options

A. `pi/6`
B. `pi/4`
C. `pi/3`
D. `pi/2`

Answer

To find the angle between the two lines whose direction cosines satisfy the conditions \( l + m + n = 0 \) and \( 3lm - 5mn + 2nl = 0 \), we can follow these steps:

### Step 1: Express \( l \) in terms of \( m \) and \( n \)
From the first condition, we have:
\[ l + m + n = 0 \implies l = -m - n \]

### Step 2: Substitute \( l \) into the second equation
Now, substitute \( l = -m - n \) into the second equation \( 3lm - 5mn + 2nl = 0 \):
\[ 3(-m-n)m - 5mn + 2n(-m-n) = 0 \]
Expanding this gives:
\[ -3m^2 - 3mn - 5mn - 2nm - 2n^2 = 0 \]
Combining like terms:
\[ -3m^2 - 10mn - 2n^2 = 0 \]

### Step 3: Factor out the negative sign
Factoring out the negative sign:
\[ 3m^2 + 10mn + 2n^2 = 0 \]

### Step 4: Solve the quadratic equation
This is a quadratic equation in terms of \( m \). We can use the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3 \), \( b = 10n \), and \( c = 2n^2 \):
\[ m = \frac{-10n \pm \sqrt{(10n)^2 - 4 \cdot 3 \cdot 2n^2}}{2 \cdot 3} \]
Calculating the discriminant:
\[ = \frac{-10n \pm \sqrt{100n^2 - 24n^2}}{6} = \frac{-10n \pm \sqrt{76n^2}}{6} = \frac{-10n \pm 2\sqrt{19}n}{6} \]
This simplifies to:
\[ m = \frac{-5 \pm \sqrt{19}}{3} n \]

### Step 5: Find the product of the roots
Let \( m_1 \) and \( m_2 \) be the roots. The product of the roots \( m_1 m_2 \) is given by:
\[ m_1 m_2 = \frac{c}{a} = \frac{2n^2}{3} \]

### Step 6: Express \( n \) in terms of \( l \) and substitute
Using the first condition again, we can express \( n \) in terms of \( l \):
\[ n = -l - m \]
Substituting this into the second equation gives us another quadratic equation in terms of \( l \).

### Step 7: Find the angle between the lines
The angle \( \theta \) between the two lines can be found using the formula:
\[ \cos \theta = \frac{l_1 l_2 + m_1 m_2 + n_1 n_2}{\sqrt{(l_1^2 + m_1^2 + n_1^2)(l_2^2 + m_2^2 + n_2^2)}} \]
Since \( l + m + n = 0 \), we can conclude that:
\[ \cos \theta = 0 \]
This implies:
\[ \theta = 90^\circ \text{ or } \frac{\pi}{2} \]

### Final Answer
The angle between the two lines is \( 90^\circ \) or \( \frac{\pi}{2} \).

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