If a vector `vec(P)` making angles `alpha, beta, gamma` respectively with the X,Y, and Z axes respectively. Then `sin^(2)theta+sin^(2)beta+sin^(2)gamma=`

Options

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

Explanation

To solve the problem, we need to derive the relationship between the angles `alpha`, `beta`, and `gamma` that a vector `vec(P)` makes with the X, Y, and Z axes respectively. The goal is to prove that:

\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 1 \]

### Step-by-Step Solution:

1. **Understanding the Angles**:
The angles `alpha`, `beta`, and `gamma` are the angles that the vector `vec(P)` makes with the X, Y, and Z axes respectively.

2. **Using the Cosine Rule**:
The cosine of each angle can be expressed in terms of the components of the vector `vec(P)`. If `P` is the vector with components \( P_x, P_y, P_z \):
- \( \cos \alpha = \frac{P_x}{|\vec{P}|} \)
- \( \cos \beta = \frac{P_y}{|\vec{P}|} \)
- \( \cos \gamma = \frac{P_z}{|\vec{P}|} \)

Here, \( |\vec{P}| = \sqrt{P_x^2 + P_y^2 + P_z^2} \) is the magnitude of the vector.

3. **Expressing Sine in Terms of Cosine**:
We can express the sine of each angle using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
- \( \sin^2 \alpha = 1 - \cos^2 \alpha = 1 - \left(\frac{P_x}{|\vec{P}|}\right)^2 \)
- \( \sin^2 \beta = 1 - \cos^2 \beta = 1 - \left(\frac{P_y}{|\vec{P}|}\right)^2 \)
- \( \sin^2 \gamma = 1 - \cos^2 \gamma = 1 - \left(\frac{P_z}{|\vec{P}|}\right)^2 \)

4. **Adding the Sine Squares**:
Now, we add these expressions together:
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = \left(1 - \frac{P_x^2}{|\vec{P}|^2}\right) + \left(1 - \frac{P_y^2}{|\vec{P}|^2}\right) + \left(1 - \frac{P_z^2}{|\vec{P}|^2}\right) \]
Simplifying this gives:
\[ = 3 - \left(\frac{P_x^2 + P_y^2 + P_z^2}{|\vec{P}|^2}\right) \]

5. **Using the Magnitude**:
Since \( |\vec{P}|^2 = P_x^2 + P_y^2 + P_z^2 \), we can substitute:
\[ = 3 - \frac{|\vec{P}|^2}{|\vec{P}|^2} = 3 - 1 = 2 \]

6. **Final Result**:
Thus, we conclude that:
\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 1 \]

### Conclusion:
The relationship holds true, and we have shown that:

\[ \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 1 \]