The cross product of the vectors `(i - 2j + 3k) and (i + 4j -2k)` is

Answer

To find the cross product of the vectors \( \mathbf{A} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k} \) and \( \mathbf{B} = \mathbf{i} + 4\mathbf{j} - 2\mathbf{k} \), we will use the determinant method.

### Step-by-Step Solution:

1. **Set Up the Determinant**:
We will create a 3x3 determinant where the first row consists of the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), the second row consists of the components of vector \( \mathbf{A} \), and the third row consists of the components of vector \( \mathbf{B} \).

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & 3 \\ 1 & 4 & -2 \end{vmatrix} \]

2. **Calculate the Determinant**:
We can expand this determinant using the rule of Sarrus or cofactor expansion.

\[ \mathbf{A} \times \mathbf{B} = \mathbf{i} \begin{vmatrix} -2 & 3 \\ 4 & -2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 3 \\ 1 & -2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & -2 \\ 1 & 4 \end{vmatrix} \]

3. **Calculate Each 2x2 Determinant**:
- For \( \mathbf{i} \):
\[ \begin{vmatrix} -2 & 3 \\ 4 & -2 \end{vmatrix} = (-2)(-2) - (3)(4) = 4 - 12 = -8 \]
- For \( \mathbf{j} \):
\[ \begin{vmatrix} 1 & 3 \\ 1 & -2 \end{vmatrix} = (1)(-2) - (3)(1) = -2 - 3 = -5 \]
- For \( \mathbf{k} \):
\[ \begin{vmatrix} 1 & -2 \\ 1 & 4 \end{vmatrix} = (1)(4) - (-2)(1) = 4 + 2 = 6 \]

4. **Combine the Results**:
Now substituting these values back into the expression for the cross product:
\[ \mathbf{A} \times \mathbf{B} = -8\mathbf{i} + 5\mathbf{j} + 6\mathbf{k} \]

5. **Final Result**:
Thus, the cross product of the vectors \( \mathbf{A} \) and \( \mathbf{B} \) is:
\[ \mathbf{A} \times \mathbf{B} = -8\mathbf{i} + 5\mathbf{j} + 6\mathbf{k} \]