`(l_(1),m_(1),n_(1)),(l_(2),m_(2),n_(2)),(l_(3),m_(3),n_(3))` are the direction cosines of three mutually perpendicular lines. If the line, whose direction ratios are `l_(1)+l_(2)+l_(3),m_(1)+m_(2)+m_(3),n_(1)+n_(2)+n_(3)`, makes angle `theta` with any of these three lines, then `cos theta` is equal to

Options

A. `(1)/(2)`
B. `(1)/(3)`
C. `(1)/(sqrt(3))`
D. none of these

Answer

To solve the problem, we need to find the cosine of the angle \( \theta \) between the line represented by the direction ratios \( (l_1 + l_2 + l_3, m_1 + m_2 + m_3, n_1 + n_2 + n_3) \) and any of the three mutually perpendicular lines represented by their direction cosines \( (l_1, m_1, n_1) \), \( (l_2, m_2, n_2) \), and \( (l_3, m_3, n_3) \).

### Step-by-Step Solution:

1. **Define the Direction Vectors:**
Let the direction vectors of the three mutually perpendicular lines be:
\[ \mathbf{A} = (l_1, m_1, n_1), \quad \mathbf{B} = (l_2, m_2, n_2), \quad \mathbf{C} = (l_3, m_3, n_3) \]

2. **Define the New Direction Vector:**
The direction vector of the new line is given by:
\[ \mathbf{P} = (l_1 + l_2 + l_3, m_1 + m_2 + m_3, n_1 + n_2 + n_3) \]

3. **Calculate the Cosine of the Angle:**
The cosine of the angle \( \theta \) between two vectors \( \mathbf{A} \) and \( \mathbf{P} \) can be calculated using the dot product formula:
\[ \cos \theta = \frac{\mathbf{A} \cdot \mathbf{P}}{|\mathbf{A}| |\mathbf{P}|} \]
where \( \mathbf{A} \cdot \mathbf{P} \) is the dot product of \( \mathbf{A} \) and \( \mathbf{P} \).

4. **Calculate the Dot Product:**
The dot product \( \mathbf{A} \cdot \mathbf{P} \) is:
\[ \mathbf{A} \cdot \mathbf{P} = l_1(l_1 + l_2 + l_3) + m_1(m_1 + m_2 + m_3) + n_1(n_1 + n_2 + n_3) \]
Expanding this gives:
\[ = l_1^2 + l_1l_2 + l_1l_3 + m_1^2 + m_1m_2 + m_1m_3 + n_1^2 + n_1n_2 + n_1n_3 \]

5. **Use the Properties of Direction Cosines:**
Since \( \mathbf{A}, \mathbf{B}, \mathbf{C} \) are mutually perpendicular, we know:
\[ l_1^2 + m_1^2 + n_1^2 = 1 \]
and similarly for \( \mathbf{B} \) and \( \mathbf{C} \). The cross terms \( l_1l_2, m_1m_2, n_1n_2 \) etc., will equal zero due to orthogonality.

6. **Calculate the Magnitude of the Vectors:**
The magnitude of \( \mathbf{A} \) is:
\[ |\mathbf{A}| = \sqrt{l_1^2 + m_1^2 + n_1^2} = 1 \]
The magnitude of \( \mathbf{P} \) is:
\[ |\mathbf{P}| = \sqrt{(l_1 + l_2 + l_3)^2 + (m_1 + m_2 + m_3)^2 + (n_1 + n_2 + n_3)^2} \]

7. **Final Calculation of Cosine:**
Since the direction cosines are normalized, we can simplify:
\[ \cos \theta = \frac{1}{|\mathbf{P}|} \]

8. **Conclusion:**
Therefore, the final result for \( \cos \theta \) is:
\[ \cos \theta = \frac{1}{\sqrt{(l_1 + l_2 + l_3)^2 + (m_1 + m_2 + m_3)^2 + (n_1 + n_2 + n_3)^2}} \]