If a vector `vec(A)` make angles `alpha , beta ` and `gamma`, respectively , with the `X , Y` and `Z` axes , 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 \( \vec{A} \) makes with the \( X, Y, \) and \( Z \) axes, respectively.

### Step-by-Step Solution:

1. **Understanding the Angles**:
The angles \( \alpha, \beta, \gamma \) are the angles between the vector \( \vec{A} \) and the coordinate axes. We know that:
\[ \cos \alpha = \frac{A_x}{|\vec{A}|}, \quad \cos \beta = \frac{A_y}{|\vec{A}|}, \quad \cos \gamma = \frac{A_z}{|\vec{A}|} \]
where \( A_x, A_y, A_z \) are the components of the vector \( \vec{A} \) along the \( X, Y, \) and \( Z \) axes, and \( |\vec{A}| \) is the magnitude of the vector.

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

3. **Substituting into the Equation**:
Now, substituting these into the original equation:
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = (1 - \cos^2 \alpha) + (1 - \cos^2 \beta) + (1 - \cos^2 \gamma) \]
Simplifying this gives:
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3 - (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) \]

4. **Finding the Sum of Cosine Squares**:
The sum of the squares of the cosines can be expressed as:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = \frac{A_x^2}{|\vec{A}|^2} + \frac{A_y^2}{|\vec{A}|^2} + \frac{A_z^2}{|\vec{A}|^2} \]
Since \( |\vec{A}|^2 = A_x^2 + A_y^2 + A_z^2 \), we can write:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = \frac{A_x^2 + A_y^2 + A_z^2}{A_x^2 + A_y^2 + A_z^2} = 1 \]

5. **Final Calculation**:
Now substituting back into the equation:
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3 - 1 = 2 \]

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