If a vector `alpha` lie in the plane of `beta` and `gamma` , then which is correct ?

Options

A. `[alpha,beta,gamma]`=0
B. `[alpha,beta,gamma]=1`
C. `[alpha,beta,gamma]=3`
D. `[beta,gamma,alpha]=1`

Answer

To determine the relationship between the vectors \( \alpha \), \( \beta \), and \( \gamma \) given that \( \alpha \) lies in the plane formed by \( \beta \) and \( \gamma \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Plane**:
- The vectors \( \beta \) and \( \gamma \) define a plane in three-dimensional space. Any vector that lies in this plane can be expressed as a linear combination of \( \beta \) and \( \gamma \).

2. **Finding the Normal Vector**:
- The normal vector \( \mathbf{n} \) to the plane formed by \( \beta \) and \( \gamma \) can be found using the cross product:
\[ \mathbf{n} = \beta \times \gamma \]
- This vector \( \mathbf{n} \) is perpendicular to the plane defined by \( \beta \) and \( \gamma \).

3. **Condition for \( \alpha \)**:
- Since \( \alpha \) lies in the same plane as \( \beta \) and \( \gamma \), it must be perpendicular to the normal vector \( \mathbf{n} \).
- The dot product of \( \alpha \) with the normal vector \( \mathbf{n} \) must therefore equal zero:
\[ \alpha \cdot \mathbf{n} = 0 \]

4. **Expressing the Dot Product**:
- Substituting the expression for \( \mathbf{n} \):
\[ \alpha \cdot (\beta \times \gamma) = 0 \]
- This equation indicates that the scalar triple product of \( \alpha \), \( \beta \), and \( \gamma \) is zero.

5. **Conclusion**:
- The condition \( \alpha \cdot (\beta \times \gamma) = 0 \) implies that the vectors \( \alpha \), \( \beta \), and \( \gamma \) are coplanar. Therefore, if \( \alpha \) lies in the plane of \( \beta \) and \( \gamma \), the scalar triple product must be zero.

### Final Result:
The correct conclusion is that:
\[ \alpha \cdot (\beta \times \gamma) = 0 \]
This indicates that \( \alpha \), \( \beta \), and \( \gamma \) are coplanar.