The direction cosines of two lines are given by the equations `3m+n+5l=0, 6nl-2lm+5mn=0.` find the angle between them

Options

A. `1/sqrt(6),(-2)/sqrt(6), 1/sqrt(6)`
B. `(-1)/sqrt(6),(2)/sqrt(6), (-1)/sqrt(6)`
C. `(-1)/sqrt(6),1/sqrt(6), 2/sqrt(6)`
D. `(1)/sqrt(6),(-1)/sqrt(6), (-2)/sqrt(6)`

Answer

To find the angle between the two lines given their direction cosines through the equations \(3m + n + 5l = 0\) and \(6nl - 2lm + 5mn = 0\), we can follow these steps:

### Step 1: Express \(n\) in terms of \(m\) and \(l\)
From the first equation:
\[ 3m + n + 5l = 0 \implies n = -3m - 5l \]

### Step 2: Substitute \(n\) into the second equation
Substituting \(n\) into the second equation:
\[ 6(-3m - 5l)l - 2lm + 5m(-3m - 5l) = 0 \]
This simplifies to:
\[ -18ml - 30l^2 - 2lm - 15m^2 - 25ml = 0 \]
Combining like terms:
\[ -30l^2 - 43ml - 15m^2 = 0 \]

### Step 3: Multiply through by -1 and rearrange
Multiplying the entire equation by -1 gives:
\[ 30l^2 + 43ml + 15m^2 = 0 \]

### Step 4: Use the quadratic formula
This is a quadratic in \(l\). We can use the quadratic formula \(l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 30\), \(b = 43m\), and \(c = 15m^2\):
\[ l = \frac{-43m \pm \sqrt{(43m)^2 - 4 \cdot 30 \cdot 15m^2}}{2 \cdot 30} \]
Calculating the discriminant:
\[ (43m)^2 - 4 \cdot 30 \cdot 15m^2 = 1849m^2 - 1800m^2 = 49m^2 \]
Thus:
\[ l = \frac{-43m \pm 7m}{60} \]
This gives us two possible values for \(l\):
1. \(l_1 = \frac{-36m}{60} = -\frac{3m}{5}\)
2. \(l_2 = \frac{-50m}{60} = -\frac{5m}{6}\)

### Step 5: Find corresponding \(n\) values
Using \(n = -3m - 5l\):
1. For \(l_1 = -\frac{3m}{5}\):
\[ n_1 = -3m - 5\left(-\frac{3m}{5}\right) = -3m + 3m = 0 \]
2. For \(l_2 = -\frac{5m}{6}\):
\[ n_2 = -3m - 5\left(-\frac{5m}{6}\right) = -3m + \frac{25m}{6} = \frac{7m}{6} \]

### Step 6: Find direction ratios
The direction ratios for the first line are:
\[ l_1 : m : n_1 = -\frac{3m}{5} : m : 0 \implies -3 : 5 : 0 \]
The direction ratios for the second line are:
\[ l_2 : m : n_2 = -\frac{5m}{6} : m : \frac{7m}{6} \implies -5 : 6 : 7 \]

### Step 7: Calculate direction cosines
To find the direction cosines, we normalize the direction ratios:
1. For the first line:
\[ \sqrt{(-3)^2 + 5^2 + 0^2} = \sqrt{9 + 25} = \sqrt{34} \]
Thus, direction cosines are:
\[ l_1 = -\frac{3}{\sqrt{34}}, \quad m_1 = \frac{5}{\sqrt{34}}, \quad n_1 = 0 \]

2. For the second line:
\[ \sqrt{(-5)^2 + 6^2 + 7^2} = \sqrt{25 + 36 + 49} = \sqrt{110} \]
Thus, direction cosines are:
\[ l_2 = -\frac{5}{\sqrt{110}}, \quad m_2 = \frac{6}{\sqrt{110}}, \quad n_2 = \frac{7}{\sqrt{110}} \]

### Step 8: Use the formula to find the angle
The angle \(\theta\) between the two lines is given by:
\[ \cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2 \]
Substituting the values:
\[ \cos \theta = \left(-\frac{3}{\sqrt{34}}\right)\left(-\frac{5}{\sqrt{110}}\right) + \left(\frac{5}{\sqrt{34}}\right)\left(\frac{6}{\sqrt{110}}\right) + (0)\left(\frac{7}{\sqrt{110}}\right) \]
Calculating:
\[ \cos \theta = \frac{15}{\sqrt{34} \sqrt{110}} + \frac{30}{\sqrt{34} \sqrt{110}} = \frac{45}{\sqrt{34} \sqrt{110}} \]

### Step 9: Find \(\theta\)
Thus:
\[ \theta = \cos^{-1}\left(\frac{45}{\sqrt{34} \sqrt{110}}\right) \]