If the vector product of a constant vector ` vec O A` with a variable vector ` vec O B` in a fixed plane `O A B` be a constant vector, then the locus of `B` is (a).a straight line perpendicular to ` vec O A` (b). a circle with centre `O` and radius equal to `| vec O A|` (c). a straight line parallel to ` vec O A` (d). none of these

Options

A. a straight line perpendicular to `vec(OA)`
B. a circle with centre O and radius equal to `|vec(OA)|`
C. a striaght line parallel to `vec(OA)`
D. none of these

Explanation

To solve the problem, we need to analyze the vector product of a constant vector \( \vec{OA} \) with a variable vector \( \vec{OB} \) in a fixed plane \( OAB \). We will derive the locus of point \( B \) based on the given conditions.

### Step-by-Step Solution:

1. **Define the Vectors**:
Let \( \vec{OA} \) be a constant vector. For simplicity, we can assume \( \vec{OA} = \hat{i} \) (the unit vector along the x-axis). The variable vector \( \vec{OB} \) can be expressed in terms of its components:
\[ \vec{OB} = x \hat{i} + y \hat{j} \]

2. **Calculate the Vector Product**:
The vector product \( \vec{OA} \times \vec{OB} \) can be calculated as follows:
\[ \vec{OA} \times \vec{OB} = \hat{i} \times (x \hat{i} + y \hat{j}) = \hat{i} \times x \hat{i} + \hat{i} \times y \hat{j} \]
Since the cross product of any vector with itself is zero, \( \hat{i} \times \hat{i} = 0 \). Therefore, we have:
\[ \vec{OA} \times \vec{OB} = 0 + y (\hat{i} \times \hat{j}) = y \hat{k} \]

3. **Set the Vector Product Equal to a Constant**:
We are given that the vector product \( \vec{OA} \times \vec{OB} \) is a constant vector. Let’s denote this constant vector as \( \vec{C} \):
\[ y \hat{k} = \vec{C} \]
Here, \( \vec{C} \) is a constant vector, which implies that \( y \) must also be a constant (let's denote it as \( k \)):
\[ y = k \]

4. **Determine the Locus of Point B**:
Since \( y \) is constant, the variable \( x \) can take any value. Thus, the locus of point \( B \) can be described by the equation:
\[ y = k \]
This represents a straight line in the plane where \( y \) is constant, and \( x \) can vary freely.

5. **Conclusion**:
The locus of point \( B \) is a straight line that is perpendicular to the vector \( \vec{OA} \) (which is along the x-axis). Therefore, the correct option is:
\[ \text{(a) a straight line perpendicular to } \vec{OA} \]