Three vector `vec(A)`,`vec(B)`, `vec(C )` satisfy the relation `vec(A)*vec(B)=0`and `vec(A).vec(C )=0`. The vector `vec(A)` is parallel to

Options

A. `vec(B)`
B. `vec(C )`
C. `vec(B).vec(C )`
D. `vec(B)xxvec(C )`

Explanation

To solve the problem, we need to analyze the given conditions involving the vectors \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\).

### Step-by-Step Solution:

1. **Understanding the Given Conditions**:
- We are given that \(\vec{A} \cdot \vec{B} = 0\).
- We are also given that \(\vec{A} \cdot \vec{C} = 0\).

2. **Interpreting the Dot Product**:
- The dot product of two vectors is given by:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \]
where \(\theta\) is the angle between the vectors \(\vec{A}\) and \(\vec{B}\).
- Since \(\vec{A} \cdot \vec{B} = 0\), it implies that \(\cos \theta = 0\). Therefore, \(\theta = 90^\circ\).
- This means that \(\vec{A}\) is perpendicular to \(\vec{B}\).

3. **Applying the Same Logic to \(\vec{C}\)**:
- Similarly, for the second condition \(\vec{A} \cdot \vec{C} = 0\):
\[ \vec{A} \cdot \vec{C} = |\vec{A}| |\vec{C}| \cos \phi \]
where \(\phi\) is the angle between \(\vec{A}\) and \(\vec{C}\).
- Since \(\vec{A} \cdot \vec{C} = 0\), it follows that \(\cos \phi = 0\) and thus \(\phi = 90^\circ\).
- This means that \(\vec{A}\) is also perpendicular to \(\vec{C}\).

4. **Conclusion About the Orientation of Vectors**:
- Since \(\vec{A}\) is perpendicular to both \(\vec{B}\) and \(\vec{C}\), it means that \(\vec{A}\) is orthogonal to the plane formed by \(\vec{B}\) and \(\vec{C}\).
- Therefore, \(\vec{A}\) is parallel to the normal vector of the plane formed by \(\vec{B}\) and \(\vec{C}\).

5. **Final Answer**:
- The vector \(\vec{A}\) is parallel to the direction that is orthogonal to the plane formed by vectors \(\vec{B}\) and \(\vec{C}\).