Let `vec(A)=hat(i)A cos theta+hat(j)A sin theta`, be any vector. Another vector `vec(B)` which is normal to `vec(A)` is :-

Options

A. `hat(i)B cos theta + hat(j)B sin theta`
B. `hat(i)B sin theta + hat(j)B cos theta`
C. `hat(i)B sin theta - hat(j)B cos theta`
D. `hat(i)A cos theta + hat(j)A sin theta`

Explanation

To find the vector \(\vec{B}\) which is normal to the vector \(\vec{A}\), we can follow these steps:

### Step 1: Define the given vector \(\vec{A}\)
The vector \(\vec{A}\) is given as:
\[ \vec{A} = \hat{i} A \cos \theta + \hat{j} A \sin \theta \]

### Step 2: Understand the condition for normal vectors
Two vectors \(\vec{A}\) and \(\vec{B}\) are normal (orthogonal) to each other if their dot product is zero:
\[ \vec{A} \cdot \vec{B} = 0 \]

### Step 3: Express the vector \(\vec{B}\)
Let’s assume the vector \(\vec{B}\) can be expressed in the form:
\[ \vec{B} = \hat{i} B_x + \hat{j} B_y \]
where \(B_x\) and \(B_y\) are the components of vector \(\vec{B}\).

### Step 4: Calculate the dot product \(\vec{A} \cdot \vec{B}\)
Now, we compute the dot product:
\[ \vec{A} \cdot \vec{B} = (A \cos \theta)(B_x) + (A \sin \theta)(B_y) \]
Setting this equal to zero for orthogonality:
\[ A \cos \theta \cdot B_x + A \sin \theta \cdot B_y = 0 \]

### Step 5: Simplify the equation
Dividing the entire equation by \(A\) (assuming \(A \neq 0\)):
\[ \cos \theta \cdot B_x + \sin \theta \cdot B_y = 0 \]

### Step 6: Solve for \(B_y\) in terms of \(B_x\)
Rearranging the equation gives:
\[ \sin \theta \cdot B_y = -\cos \theta \cdot B_x \]
Thus, we can express \(B_y\) as:
\[ B_y = -\frac{\cos \theta}{\sin \theta} B_x = -B_x \cot \theta \]

### Step 7: Choose a specific form for \(\vec{B}\)
To find a specific vector \(\vec{B}\), we can choose \(B_x = B\) (a constant) and substitute:
\[ B_y = -B \cot \theta \]
Thus, we can express \(\vec{B}\) as:
\[ \vec{B} = \hat{i} B + \hat{j} \left(-B \cot \theta\right) \]

### Conclusion
The vector \(\vec{B}\) which is normal to \(\vec{A}\) can be expressed as:
\[ \vec{B} = \hat{i} B - \hat{j} B \cot \theta \]
where \(B\) is any non-zero scalar.