Let PQ and QR be diagonals of adjacent faces of a rectangular box, with its centre at O. If `angle QOR, angle ROP and angle POQ` are `theta, phi and Psi` respectively then the value of `'costheta +cos phi+cos Psi' ` is :

Options

A. -2
B. `-sqrt(3)`
C. -1
D. 0

Answer

To solve the problem, we need to find the value of \( \cos \theta + \cos \phi + \cos \psi \) where \( \theta, \phi, \psi \) are the angles \( \angle QOR, \angle ROP, \angle POQ \) respectively in a rectangular box.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let the rectangular box have dimensions \( L \), \( B \), and \( H \).
- The center \( O \) of the box is at the midpoint of the diagonals of the faces.
- Points \( P, Q, R \) are the vertices of the box such that \( PQ \) and \( QR \) are diagonals of adjacent faces.

2. **Identifying the Angles**:
- \( \angle QOR = \theta \)
- \( \angle ROP = \phi \)
- \( \angle POQ = \psi \)

3. **Finding Lengths**:
- The length of diagonal \( PR \) (which lies in the face containing points \( P \) and \( R \)) can be calculated using the Pythagorean theorem:
\[ PR = \sqrt{L^2 + B^2} \]
- The length \( OQ \) (the distance from the center \( O \) to vertex \( Q \)) is:
\[ OQ = \frac{H}{2} \]

4. **Using Trigonometric Relationships**:
- For angle \( \phi \) (i.e., \( \angle ROP \)):
\[ \cot \phi = \frac{OQ}{PR} = \frac{\frac{H}{2}}{\sqrt{L^2 + B^2}} \]
- Hence,
\[ \cos \phi = \frac{OQ}{\sqrt{OQ^2 + PR^2}} = \frac{\frac{H}{2}}{\sqrt{\left(\frac{H}{2}\right)^2 + (L^2 + B^2)}} \]

5. **Finding \( \cos \theta \) and \( \cos \psi \)**:
- Similarly, we can find:
\[ \cos \theta = \frac{L^2 - H^2 - B^2}{L^2 + B^2 + H^2} \]
\[ \cos \psi = \frac{B^2 - L^2 - H^2}{L^2 + B^2 + H^2} \]

6. **Summing Up**:
- Now we sum \( \cos \theta + \cos \phi + \cos \psi \):
\[ \cos \theta + \cos \phi + \cos \psi = \left(\frac{L^2 - H^2 - B^2}{L^2 + B^2 + H^2}\right) + \left(\frac{B^2 - L^2 - H^2}{L^2 + B^2 + H^2}\right) + \left(\frac{H^2 - L^2 - B^2}{L^2 + B^2 + H^2}\right) \]

7. **Simplifying the Expression**:
- Combining the numerators:
\[ = \frac{(L^2 - H^2 - B^2) + (B^2 - L^2 - H^2) + (H^2 - L^2 - B^2)}{L^2 + B^2 + H^2} \]
- The numerator simplifies to:
\[ = L^2 - H^2 - B^2 + B^2 - L^2 - H^2 + H^2 - L^2 - B^2 = - (L^2 + B^2 + H^2) \]
- Thus, we have:
\[ \cos \theta + \cos \phi + \cos \psi = \frac{- (L^2 + B^2 + H^2)}{L^2 + B^2 + H^2} = -1 \]

### Final Answer:
\[ \cos \theta + \cos \phi + \cos \psi = -1 \]