Find the angle between the lines whose direction cosines are given by the equations `3l + m + 5n = 0` and `6mn - 2nl + 5lm = 0`

Options

A. parallel
B. perpendicular
C. inclined at `cos(-1)((1)/(6))`
D. none of these

Explanation

To find the angle between the lines whose direction cosines are given by the equations \(3l + m + 5n = 0\) and \(6mn - 2nl + 5lm = 0\), we will proceed step by step.

### Step 1: Rewrite the equations
We have two equations:
1. \(3l + m + 5n = 0\) (Equation 1)
2. \(6mn - 2nl + 5lm = 0\) (Equation 2)

### Step 2: Express \(m\) in terms of \(l\) and \(n\)
From Equation 1, we can express \(m\) as:
\[ m = -3l - 5n \]

### Step 3: Substitute \(m\) into Equation 2
Now, substitute \(m\) into Equation 2:
\[ 6(-3l - 5n)n - 2nl + 5l(-3l - 5n) = 0 \]
Expanding this gives:
\[ -18ln - 30n^2 - 2nl - 15l^2 - 25ln = 0 \]
Combining like terms:
\[ -15l^2 - 48ln - 30n^2 = 0 \]

### Step 4: Factor the equation
We can factor this equation:
\[ l^2 + 3ln + 2n^2 = 0 \]
This can be factored as:
\[ (l + 2n)(l + n) = 0 \]

### Step 5: Find the relationships between \(l\), \(m\), and \(n\)
From the factored form, we have two cases:
1. \(l + 2n = 0\) (Equation 3)
2. \(l + n = 0\) (Equation 4)

### Step 6: Solve for direction ratios
From Equation 3:
\[ l = -2n \implies \frac{l}{1} = \frac{n}{-\frac{1}{2}} \implies \frac{l}{1} = \frac{m}{2} = \frac{n}{-1} \]
Let’s denote this as Equation 5.

From Equation 4:
\[ l = -n \implies \frac{l}{1} = \frac{m}{-1} = \frac{n}{-1} \]
Let’s denote this as Equation 6.

### Step 7: Find the angle between the two lines
The direction ratios from Equation 5 are \( (1, 2, -1) \) and from Equation 6 are \( (1, -1, -1) \).

Using the formula for the cosine of the angle \( \theta \) between two lines:
\[ \cos \theta = \frac{l_1l_2 + m_1m_2 + n_1n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}} \]
Substituting the values:
\[ \cos \theta = \frac{(1)(1) + (2)(-1) + (-1)(-1)}{\sqrt{1^2 + 2^2 + (-1)^2} \sqrt{1^2 + (-1)^2 + (-1)^2}} \]
Calculating the numerator:
\[ 1 - 2 + 1 = 0 \]
Calculating the denominator:
\[ \sqrt{1 + 4 + 1} \sqrt{1 + 1 + 1} = \sqrt{6} \sqrt{3} \]
Thus,
\[ \cos \theta = \frac{0}{\sqrt{6} \sqrt{3}} = 0 \]
This implies that \( \theta = 90^\circ \).

### Final Answer
The angle between the lines is \(90^\circ\).