Vector `vecP` angle `alpha,beta` and `gamma` with the x,y and z axes respectively.
Then `sin^(2)alpha+sin^(2)beta+sin^(2)gamma=`

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the problem, we need to find the value of \( \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma \) where \( \alpha, \beta, \gamma \) are the angles that a vector \( \vec{P} \) makes with the x, y, and z axes respectively.

### Step-by-Step Solution:

1. **Understanding Direction Cosines**:
The direction cosines of a vector \( \vec{P} \) making angles \( \alpha, \beta, \gamma \) with the x, y, and z axes are given by:
\[ \cos \alpha = \frac{a}{|\vec{P}|}, \quad \cos \beta = \frac{b}{|\vec{P}|}, \quad \cos \gamma = \frac{c}{|\vec{P}|} \]
where \( a, b, c \) are the components of the vector \( \vec{P} \) and \( |\vec{P}| = \sqrt{a^2 + b^2 + c^2} \).

2. **Using the Identity**:
We know from trigonometric identities that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
Therefore, we can express \( \sin^2 \alpha, \sin^2 \beta, \sin^2 \gamma \) in terms of the cosines:
\[ \sin^2 \alpha = 1 - \cos^2 \alpha, \quad \sin^2 \beta = 1 - \cos^2 \beta, \quad \sin^2 \gamma = 1 - \cos^2 \gamma \]

3. **Substituting for Cosines**:
Substituting the expressions for \( \cos^2 \alpha, \cos^2 \beta, \cos^2 \gamma \):
\[ \sin^2 \alpha = 1 - \left(\frac{a}{|\vec{P}|}\right)^2, \quad \sin^2 \beta = 1 - \left(\frac{b}{|\vec{P}|}\right)^2, \quad \sin^2 \gamma = 1 - \left(\frac{c}{|\vec{P}|}\right)^2 \]

4. **Adding the Sine Squares**:
Now, we can add these three equations:
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = \left(1 - \frac{a^2}{|\vec{P}|^2}\right) + \left(1 - \frac{b^2}{|\vec{P}|^2}\right) + \left(1 - \frac{c^2}{|\vec{P}|^2}\right) \]
Simplifying this gives:
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3 - \frac{a^2 + b^2 + c^2}{|\vec{P}|^2} \]

5. **Using the Magnitude of the Vector**:
Since \( |\vec{P}|^2 = a^2 + b^2 + c^2 \), we can substitute this into our equation:
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3 - 1 = 2 \]

### Final Result:
Thus, the final answer is:
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 2 \]