If `vec(A)` makes an angle `alpha, beta` and `gamma` from x,y and z axis 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 the vector \( \vec{A} \) makes with the x, y, and z axes, respectively.

### Step-by-step Solution:

1. **Understanding Direction Cosines:**
The direction cosines of a vector \( \vec{A} \) are defined as:
\[ \cos \alpha = \frac{A_x}{|\vec{A}|}, \quad \cos \beta = \frac{A_y}{|\vec{A}|}, \quad \cos \gamma = \frac{A_z}{|\vec{A}|} \]
where \( A_x, A_y, A_z \) are the components of the vector along the x, y, and z axes, respectively, 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 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. **Summing the Sine Squares:**
Now, we can sum these expressions:
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = (1 - \cos^2 \alpha) + (1 - \cos^2 \beta) + (1 - \cos^2 \gamma) \]
Simplifying this gives:
\[ = 3 - (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) \]

4. **Using the Property of Direction Cosines:**
The sum of the squares of the direction cosines is given by:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]
This is derived from the fact that the components of the vector must satisfy the equation of a sphere in three-dimensional space.

5. **Final Calculation:**
Substituting this result back into our expression:
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3 - 1 = 2 \]

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