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) B cos theta - hat(j) B sin theta`

Answer

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

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

### Step 2: Condition for Perpendicular Vectors
Two vectors \(\vec{A}\) and \(\vec{B}\) are perpendicular if their dot product is zero:
\[ \vec{A} \cdot \vec{B} = 0 \]

### Step 3: Assume a General Form for \(\vec{B}\)
Let's assume \(\vec{B}\) can be expressed in the form:
\[ \vec{B} = \hat{i} B_x + \hat{j} B_y \]
where \(B_x\) and \(B_y\) are components of \(\vec{B}\).

### Step 4: Calculate the Dot Product
Now, we compute the dot product \(\vec{A} \cdot \vec{B}\):
\[ \vec{A} \cdot \vec{B} = (A \cos \theta)(B_x) + (A \sin \theta)(B_y) \]

### Step 5: Set the Dot Product to Zero
To satisfy the condition for perpendicularity, we set the dot product to zero:
\[ A \cos \theta \cdot B_x + A \sin \theta \cdot B_y = 0 \]

### Step 6: Solve for \(B_y\) in Terms of \(B_x\)
Rearranging the equation gives:
\[ A \sin \theta \cdot B_y = -A \cos \theta \cdot B_x \]
Dividing both sides by \(A\) (assuming \(A \neq 0\)):
\[ \sin \theta \cdot B_y = -\cos \theta \cdot B_x \]
Thus, we can express \(B_y\) as:
\[ B_y = -\frac{\cos \theta}{\sin \theta} B_x = -B_x \cot \theta \]

### Step 7: Choose Specific Values for \(B_x\)
For simplicity, we can choose \(B_x = B\) (a constant) and then:
\[ B_y = -B \cot \theta \]

### Step 8: Write the Final Form of \(\vec{B}\)
Thus, the vector \(\vec{B}\) can be expressed as:
\[ \vec{B} = \hat{i} B + \hat{j} \left(-B \cot \theta\right) \]
This can also be simplified to:
\[ \vec{B} = B \left(\hat{i} - \hat{j} \cot \theta\right) \]

### Conclusion
The vector \(\vec{B}\) that is normal to \(\vec{A}\) can be expressed in terms of a constant \(B\):
\[ \vec{B} = B \left(\hat{i} - \hat{j} \cot \theta\right) \]