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 between the diagonals of adjacent faces of a rectangular box with center at \( O \).

### Step 1: Define the coordinates of the points
Let the rectangular box have dimensions \( a, b, c \). We can define the coordinates of the vertices of the box as follows:
- \( P(0, 0, c) \)
- \( Q(0, b, 0) \)
- \( R(a, b, 0) \)

The center \( O \) of the rectangular box is at:
\[ O\left(\frac{a}{2}, \frac{b}{2}, \frac{c}{2}\right) \]

### Step 2: Define the vectors
We can define the vectors from the center \( O \) to the points \( P, Q, R \):
- \( \vec{OP} = \left(0 - \frac{a}{2}, 0 - \frac{b}{2}, c - \frac{c}{2}\right) = \left(-\frac{a}{2}, -\frac{b}{2}, \frac{c}{2}\right) \)
- \( \vec{OQ} = \left(0 - \frac{a}{2}, b - \frac{b}{2}, 0 - \frac{c}{2}\right) = \left(-\frac{a}{2}, \frac{b}{2}, -\frac{c}{2}\right) \)
- \( \vec{OR} = \left(a - \frac{a}{2}, b - \frac{b}{2}, 0 - \frac{c}{2}\right) = \left(\frac{a}{2}, \frac{b}{2}, -\frac{c}{2}\right) \)

### Step 3: Calculate the cosines of the angles
To find \( \cos \theta \), \( \cos \phi \), and \( \cos \psi \), we will use the dot product formula:
\[ \cos \theta = \frac{\vec{OQ} \cdot \vec{OR}}{|\vec{OQ}| |\vec{OR}|} \]
\[ \cos \phi = \frac{\vec{OR} \cdot \vec{OP}}{|\vec{OR}| |\vec{OP}|} \]
\[ \cos \psi = \frac{\vec{OP} \cdot \vec{OQ}}{|\vec{OP}| |\vec{OQ}|} \]

### Step 4: Calculate the dot products
1. **For \( \cos \theta \)**:
\[ \vec{OQ} \cdot \vec{OR} = \left(-\frac{a}{2}\right)\left(\frac{a}{2}\right) + \left(\frac{b}{2}\right)\left(\frac{b}{2}\right) + \left(-\frac{c}{2}\right)\left(-\frac{c}{2}\right) = -\frac{a^2}{4} + \frac{b^2}{4} + \frac{c^2}{4} \]
\[ |\vec{OQ}| = \sqrt{\left(-\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2 + \left(-\frac{c}{2}\right)^2} = \frac{1}{2}\sqrt{a^2 + b^2 + c^2} \]
\[ |\vec{OR}| = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2 + \left(-\frac{c}{2}\right)^2} = \frac{1}{2}\sqrt{a^2 + b^2 + c^2} \]
Thus,
\[ \cos \theta = \frac{-\frac{a^2}{4} + \frac{b^2}{4} + \frac{c^2}{4}}{\left(\frac{1}{2}\sqrt{a^2 + b^2 + c^2}\right)^2} = \frac{-\frac{a^2}{4} + \frac{b^2}{4} + \frac{c^2}{4}}{\frac{1}{4}(a^2 + b^2 + c^2)} = \frac{-a^2 + b^2 + c^2}{a^2 + b^2 + c^2} \]

2. **For \( \cos \phi \)** and **\( \cos \psi \)**, similar calculations yield:
\[ \cos \phi = \frac{-b^2 + a^2 + c^2}{a^2 + b^2 + c^2} \]
\[ \cos \psi = \frac{-c^2 + a^2 + b^2}{a^2 + b^2 + c^2} \]

### Step 5: Sum the cosines
Now we can sum the cosines:
\[ \cos \theta + \cos \phi + \cos \psi = \frac{-a^2 + b^2 + c^2}{a^2 + b^2 + c^2} + \frac{-b^2 + a^2 + c^2}{a^2 + b^2 + c^2} + \frac{-c^2 + a^2 + b^2}{a^2 + b^2 + c^2} \]
Combining these, we find:
\[ = \frac{-a^2 - b^2 - c^2 + 3(a^2 + b^2 + c^2)}{a^2 + b^2 + c^2} = \frac{2(a^2 + b^2 + c^2)}{a^2 + b^2 + c^2} = 2 \]

### Final Answer
Thus, the value of \( \cos \theta + \cos \phi + \cos \psi \) is:
\[ \boxed{2} \]