Vectors which is perpendicular to `( a cos theta hat (i) + b sin theta hat(j))` is

Options

A. `(1) b sin theta hat(i) - a cos theta hat(j)`
B. `(1)/(a) sin theta hat(i) - (1)/(b) cos theta hat (j)`
C. `5 hat (k)`
D. All of these

Explanation

To find a vector that is perpendicular to the vector \( \mathbf{V} = a \cos \theta \hat{i} + b \sin \theta \hat{j} \), we can use the property of the dot product. Two vectors are perpendicular if their dot product is zero.

### Step-by-Step Solution:

1. **Identify the Given Vector**:
The given vector is:
\[ \mathbf{V} = a \cos \theta \hat{i} + b \sin \theta \hat{j} \]

2. **Define a General Perpendicular Vector**:
Let’s denote a general vector that we want to check for perpendicularity as:
\[ \mathbf{W} = x \hat{i} + y \hat{j} \]
We need to find \( x \) and \( y \) such that \( \mathbf{V} \cdot \mathbf{W} = 0 \).

3. **Calculate the Dot Product**:
The dot product of \( \mathbf{V} \) and \( \mathbf{W} \) is given by:
\[ \mathbf{V} \cdot \mathbf{W} = (a \cos \theta)(x) + (b \sin \theta)(y) \]
Setting this equal to zero for perpendicularity:
\[ a \cos \theta \cdot x + b \sin \theta \cdot y = 0 \]

4. **Choose Specific Values for \( x \) and \( y \)**:
To find a specific vector, we can choose:
\[ x = b \sin \theta \quad \text{and} \quad y = -a \cos \theta \]
This gives us:
\[ \mathbf{W} = b \sin \theta \hat{i} - a \cos \theta \hat{j} \]

5. **Verify the Perpendicularity**:
Now, substituting \( x \) and \( y \) back into the dot product:
\[ \mathbf{V} \cdot \mathbf{W} = a \cos \theta (b \sin \theta) + b \sin \theta (-a \cos \theta) = ab \sin \theta \cos \theta - ab \sin \theta \cos \theta = 0 \]
Thus, the vectors are indeed perpendicular.

6. **Conclusion**:
The vector that is perpendicular to \( a \cos \theta \hat{i} + b \sin \theta \hat{j} \) is:
\[ \mathbf{W} = b \sin \theta \hat{i} - a \cos \theta \hat{j} \]