Two vectors `vec(A)` and `vec(B)` lie in X-Y plane. The vector `vec(B)` is perpendicular to vector `vec(A)`. If `vec(A) = hat(i) + hat(j)`, then `vec(B)` may be :

Options

A. `hat(i) - hat(j)`
B. `-hat(i) + hat(j)`
C. `-2hat(i) + 2 hat(j)`
D. Any of the above

Answer

To solve the problem, we need to determine which vectors can be perpendicular to the given vector \(\vec{A} = \hat{i} + \hat{j}\).

### Step-by-step Solution:

1. **Understanding Perpendicular Vectors**:
- Two vectors \(\vec{A}\) and \(\vec{B}\) are perpendicular if their dot product is zero:
\[ \vec{A} \cdot \vec{B} = 0 \]

2. **Given Vector**:
- The vector \(\vec{A}\) is given as:
\[ \vec{A} = \hat{i} + \hat{j} \]

3. **Dot Product Calculation**:
- Let’s denote \(\vec{B}\) as a general vector in the form \(a\hat{i} + b\hat{j}\).
- The dot product of \(\vec{A}\) and \(\vec{B}\) is:
\[ \vec{A} \cdot \vec{B} = (\hat{i} + \hat{j}) \cdot (a\hat{i} + b\hat{j}) = a(\hat{i} \cdot \hat{i}) + b(\hat{j} \cdot \hat{j}) = a + b \]

4. **Setting the Dot Product to Zero**:
- For \(\vec{A}\) and \(\vec{B}\) to be perpendicular, we set the dot product to zero:
\[ a + b = 0 \]
- This implies:
\[ b = -a \]

5. **Finding Possible Vectors**:
- Thus, any vector \(\vec{B}\) can be expressed as:
\[ \vec{B} = a\hat{i} - a\hat{j} = a(\hat{i} - \hat{j}) \]
- This means that \(\vec{B}\) can be any scalar multiple of the vector \(\hat{i} - \hat{j}\).

6. **Examples of Vectors**:
- If we choose \(a = 1\), then \(\vec{B} = \hat{i} - \hat{j}\).
- If we choose \(a = -1\), then \(\vec{B} = -\hat{i} + \hat{j}\).
- If we choose \(a = 2\), then \(\vec{B} = 2\hat{i} - 2\hat{j}\).

7. **Conclusion**:
- Therefore, any vector of the form \(a(\hat{i} - \hat{j})\) where \(a\) is a non-zero scalar will be perpendicular to \(\vec{A}\).