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

Answer

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

### Step-by-step Solution:

1. **Understanding Direction Cosines**:
The angles \( \alpha, \beta, \) and \( \gamma \) are the angles that the vector \( \vec{P} \) makes with the x, y, and z axes. The cosines of these angles are known as direction cosines, typically denoted as \( l, m, \) and \( n \):
\[ l = \cos \alpha, \quad m = \cos \beta, \quad n = \cos \gamma \]

2. **Using the Property of Direction Cosines**:
The sum of the squares of the direction cosines is equal to 1:
\[ l^2 + m^2 + n^2 = 1 \]
Substituting for \( l, m, n \):
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]

3. **Relating Sine and Cosine**:
We know the trigonometric identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
From this, we can express \( \cos^2 \alpha, \cos^2 \beta, \) and \( \cos^2 \gamma \) in terms of sine:
\[ \cos^2 \alpha = 1 - \sin^2 \alpha \]
\[ \cos^2 \beta = 1 - \sin^2 \beta \]
\[ \cos^2 \gamma = 1 - \sin^2 \gamma \]

4. **Substituting into the Direction Cosine Equation**:
Substitute these expressions into the equation for direction cosines:
\[ (1 - \sin^2 \alpha) + (1 - \sin^2 \beta) + (1 - \sin^2 \gamma) = 1 \]
Simplifying this gives:
\[ 3 - (\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma) = 1 \]

5. **Isolating the Sine Squares**:
Rearranging the equation to isolate \( \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma \):
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3 - 1 = 2 \]

### Final Answer:
Thus, the value of \( \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma \) is:
\[ \boxed{2} \]