A vector perpendicular to both the vector `2hati -3hatj and 3hati - 2hatj ` is

Options

A. `hatj + 5hatk`
B. `hatj -5hatk`
C. `6hatk`
D. `hati +hatj+ hatk`

Explanation

To find a vector that is perpendicular to both vectors \( \mathbf{A} = 2\hat{i} - 3\hat{j} \) and \( \mathbf{B} = 3\hat{i} - 2\hat{j} \), we can use the cross product. The cross product of two vectors results in a third vector that is perpendicular to both of the original vectors.

### Step-by-step Solution:

1. **Identify the vectors**:
- Let \( \mathbf{A} = 2\hat{i} - 3\hat{j} \)
- Let \( \mathbf{B} = 3\hat{i} - 2\hat{j} \)

2. **Set up the cross product**:
The cross product \( \mathbf{C} = \mathbf{A} \times \mathbf{B} \) can be calculated using the determinant of a matrix formed by the unit vectors and the components of \( \mathbf{A} \) and \( \mathbf{B} \):

\[ \mathbf{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 0 \\ 3 & -2 & 0 \end{vmatrix} \]

3. **Calculate the determinant**:
Expanding the determinant, we have:

\[ \mathbf{C} = \hat{i} \begin{vmatrix} -3 & 0 \\ -2 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 0 \\ 3 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -3 \\ 3 & -2 \end{vmatrix} \]

- The first determinant \( \begin{vmatrix} -3 & 0 \\ -2 & 0 \end{vmatrix} = 0 \)
- The second determinant \( \begin{vmatrix} 2 & 0 \\ 3 & 0 \end{vmatrix} = 0 \)
- The third determinant \( \begin{vmatrix} 2 & -3 \\ 3 & -2 \end{vmatrix} = (2)(-2) - (3)(-3) = -4 + 9 = 5 \)

Therefore, we have:

\[ \mathbf{C} = 0\hat{i} - 0\hat{j} + 5\hat{k} = 5\hat{k} \]

4. **Conclusion**:
The vector that is perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \) is \( 5\hat{k} \).

### Final Answer:
A vector perpendicular to both \( 2\hat{i} - 3\hat{j} \) and \( 3\hat{i} - 2\hat{j} \) is \( 5\hat{k} \).