The angle between the lines whose direction cosines satisfy the equations `l+m+n=""0` and `l^2=m^2+n^2` is (1) `pi/3` (2) `pi/4` (3) `pi/6` (4) `pi/2`

Options

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

Answer

To find the angle between the lines whose direction cosines satisfy the equations \( l + m + n = 0 \) and \( l^2 = m^2 + n^2 \), we can follow these steps:

### Step 1: Express \( l \) in terms of \( m \) and \( n \)
From the equation \( l + m + n = 0 \), we can express \( l \) as:
\[ l = -m - n \]

### Step 2: Substitute \( l \) into the second equation
Now substitute \( l \) into the second equation \( l^2 = m^2 + n^2 \):
\[ (-m - n)^2 = m^2 + n^2 \]
Expanding the left side:
\[ m^2 + 2mn + n^2 = m^2 + n^2 \]

### Step 3: Simplify the equation
Subtract \( m^2 + n^2 \) from both sides:
\[ 2mn = 0 \]
This implies:
\[ mn = 0 \]

### Step 4: Consider the cases for \( m \) and \( n \)
Since \( mn = 0 \), we have two cases:
1. \( m = 0 \)
2. \( n = 0 \)

### Step 5: Case 1: \( m = 0 \)
If \( m = 0 \), then from \( l + m + n = 0 \):
\[ l + n = 0 \implies n = -l \]
Substituting \( n = -l \) into \( l^2 = m^2 + n^2 \):
\[ l^2 = 0 + (-l)^2 \implies l^2 = l^2 \]
This is always true. We can choose \( l = \frac{1}{\sqrt{2}}, n = -\frac{1}{\sqrt{2}}, m = 0 \).

### Step 6: Case 2: \( n = 0 \)
If \( n = 0 \), then from \( l + m + n = 0 \):
\[ l + m = 0 \implies m = -l \]
Substituting \( m = -l \) into \( l^2 = m^2 + n^2 \):
\[ l^2 = (-l)^2 + 0 \implies l^2 = l^2 \]
This is also always true. We can choose \( l = \frac{1}{\sqrt{2}}, m = -\frac{1}{\sqrt{2}}, n = 0 \).

### Step 7: Find the angle between the lines
The direction cosines for the two cases are:
1. Case 1: \( (l_1, m_1, n_1) = \left(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}}\right) \)
2. Case 2: \( (l_2, m_2, n_2) = \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0\right) \)

Using the formula for the cosine of the angle \( \theta \) between two lines:
\[ \cos \theta = \frac{l_1 l_2 + m_1 m_2 + n_1 n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}} \]

### Step 8: Calculate the dot product
Calculating the dot product:
\[ l_1 l_2 = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} \]
\[ m_1 m_2 = 0 \cdot -\frac{1}{\sqrt{2}} = 0 \]
\[ n_1 n_2 = -\frac{1}{\sqrt{2}} \cdot 0 = 0 \]
Thus,
\[ l_1 l_2 + m_1 m_2 + n_1 n_2 = \frac{1}{2} + 0 + 0 = \frac{1}{2} \]

### Step 9: Calculate the magnitudes
Calculating the magnitudes:
\[ \sqrt{l_1^2 + m_1^2 + n_1^2} = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + 0^2 + \left(-\frac{1}{\sqrt{2}}\right)^2} = \sqrt{\frac{1}{2} + 0 + \frac{1}{2}} = \sqrt{1} = 1 \]
\[ \sqrt{l_2^2 + m_2^2 + n_2^2} = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 + 0^2} = \sqrt{\frac{1}{2} + \frac{1}{2} + 0} = \sqrt{1} = 1 \]

### Step 10: Final calculation
Thus,
\[ \cos \theta = \frac{\frac{1}{2}}{1 \cdot 1} = \frac{1}{2} \]
This implies:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \]

### Conclusion
The angle between the lines is \( \frac{\pi}{3} \).