Let `overset harr A = overset ^ i Acos theta + overset ^ j A sin theta,` be any vector. Another vector `overset rarr B` which is normal to `overset rarr A` is :

Options

A. `overset ^ i B cos theta + overset ^ j B sin theta `
B. `overset ^ i B sin theta + overset ^ j B cos theta `
C. `overset ^ i B sin theta - overset ^ j B cos theta `
D. `overset ^ i A sin theta - overset ^ j A cos theta `

Explanation

To find a vector \( \vec{B} \) that is normal (perpendicular) to the vector \( \vec{A} = \hat{i} A \cos \theta + \hat{j} A \sin \theta \), we can follow these steps:

### Step 1: Understand the components of vector \( \vec{A} \)
The vector \( \vec{A} \) can be expressed in terms of its components along the x-axis and y-axis:
- The x-component is \( A_x = A \cos \theta \)
- The y-component is \( A_y = A \sin \theta \)

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

### Step 3: Express vector \( \vec{B} \)
Let’s assume vector \( \vec{B} \) has components \( B_x \) and \( B_y \):
\[ \vec{B} = \hat{i} B_x + \hat{j} B_y \]

### Step 4: Set up the dot product equation
The dot product of \( \vec{A} \) and \( \vec{B} \) is given by:
\[ \vec{A} \cdot \vec{B} = (A \cos \theta)(B_x) + (A \sin \theta)(B_y) = 0 \]

### Step 5: Solve for \( B_y \) in terms of \( B_x \)
Rearranging the equation gives:
\[ A \cos \theta \cdot B_x + A \sin \theta \cdot B_y = 0 \]
Dividing through by \( A \) (assuming \( A \neq 0 \)):
\[ \cos \theta \cdot B_x + \sin \theta \cdot B_y = 0 \]
From this, we can express \( B_y \) in terms of \( B_x \):
\[ B_y = -\frac{\cos \theta}{\sin \theta} B_x = -B_x \cot \theta \]

### Step 6: Choose a specific value for \( B_x \)
To simplify, we can choose \( B_x = 1 \):
\[ B_y = -\cot \theta \]

### Step 7: Write the final expression for vector \( \vec{B} \)
Thus, the vector \( \vec{B} \) that is perpendicular to \( \vec{A} \) can be expressed as:
\[ \vec{B} = \hat{i} (1) + \hat{j} \left(-\cot \theta\right) = \hat{i} - \hat{j} \cot \theta \]

### Final Answer
The vector \( \vec{B} \) that is normal to \( \vec{A} \) is:
\[ \vec{B} = \hat{i} - \hat{j} \cot \theta \]

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