Which one of the following is the unit vector perpendicular to the vectors `4hat(i) + 2hat(j) and -3hat(i) + 2hat(j)`?

Options

A. `(hat(i) + hat(j))/(2)`
B. `(hat(i)- hat(j))/(sqrt2)`
C. `hat(k)`
D. `(hat(i) + hat(j) + hat(k))/(sqrt2)`

Answer

To find the unit vector that is perpendicular to the vectors \( \mathbf{A} = 4\hat{i} + 2\hat{j} \) and \( \mathbf{B} = -3\hat{i} + 2\hat{j} \), we can follow these steps:

### Step 1: Calculate the Cross Product of Vectors A and B
The cross product of two vectors \( \mathbf{A} \) and \( \mathbf{B} \) gives a vector that is perpendicular to both. The formula for the cross product in determinant form is:
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 2 & 0 \\ -3 & 2 & 0 \end{vmatrix} \]

### Step 2: Compute the Determinant
Calculating the determinant, we expand it as follows:
\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} 2 & 0 \\ 2 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & 0 \\ -3 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & 2 \\ -3 & 2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
- For \( \hat{i} \): \( 2 \cdot 0 - 0 \cdot 2 = 0 \)
- For \( \hat{j} \): \( 4 \cdot 0 - (-3) \cdot 0 = 0 \)
- For \( \hat{k} \): \( 4 \cdot 2 - (-3) \cdot 2 = 8 + 6 = 14 \)

Thus, we have:
\[ \mathbf{A} \times \mathbf{B} = 0\hat{i} - 0\hat{j} + 14\hat{k} = 14\hat{k} \]

### Step 3: Find the Magnitude of the Cross Product
The magnitude of the vector \( \mathbf{C} = 14\hat{k} \) is:
\[ |\mathbf{C}| = \sqrt{(0)^2 + (0)^2 + (14)^2} = \sqrt{196} = 14 \]

### Step 4: Calculate the Unit Vector
The unit vector \( \mathbf{u} \) in the direction of \( \mathbf{C} \) is given by:
\[ \mathbf{u} = \frac{\mathbf{C}}{|\mathbf{C}|} = \frac{14\hat{k}}{14} = \hat{k} \]

### Final Answer
The unit vector perpendicular to the vectors \( 4\hat{i} + 2\hat{j} \) and \( -3\hat{i} + 2\hat{j} \) is:
\[ \hat{k} \]