The cross product of the vectors `(2i - 3j + 4k) and (I + 4j -5k)` is

Options

A. `-hati-14hatj+11hatk`
B. `hati+14hatj+11hatk`
C. `-hati+14hatj+11hatk`
D. `-hati+14hatj+5hatk`

Answer

To find the cross product of the vectors \( \mathbf{A} = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k} \) and \( \mathbf{B} = \mathbf{i} + 4\mathbf{j} - 5\mathbf{k} \), we can use the determinant method. Here’s a step-by-step solution:

### Step 1: Set up the determinant
We will set up 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 \( \mathbf{B} \).

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -3 & 4 \\ 1 & 4 & -5 \end{vmatrix} \]

### Step 2: Calculate the determinant
To calculate the determinant, we will expand it using the first row:

\[ \mathbf{A} \times \mathbf{B} = \mathbf{i} \begin{vmatrix} -3 & 4 \\ 4 & -5 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 4 \\ 1 & -5 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & -3 \\ 1 & 4 \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants
Now, we will calculate each of the 2x2 determinants:

1. For \( \mathbf{i} \):
\[ \begin{vmatrix} -3 & 4 \\ 4 & -5 \end{vmatrix} = (-3)(-5) - (4)(4) = 15 - 16 = -1 \]

2. For \( \mathbf{j} \):
\[ \begin{vmatrix} 2 & 4 \\ 1 & -5 \end{vmatrix} = (2)(-5) - (4)(1) = -10 - 4 = -14 \]

3. For \( \mathbf{k} \):
\[ \begin{vmatrix} 2 & -3 \\ 1 & 4 \end{vmatrix} = (2)(4) - (-3)(1) = 8 + 3 = 11 \]

### Step 4: Substitute back into the expression
Now substituting back into the expression for the cross product:

\[ \mathbf{A} \times \mathbf{B} = \mathbf{i}(-1) - \mathbf{j}(-14) + \mathbf{k}(11) \]

This simplifies to:

\[ \mathbf{A} \times \mathbf{B} = -\mathbf{i} + 14\mathbf{j} + 11\mathbf{k} \]

### Final Result
Thus, the cross product of the vectors \( \mathbf{A} \) and \( \mathbf{B} \) is:

\[ \mathbf{A} \times \mathbf{B} = -\mathbf{i} + 14\mathbf{j} + 11\mathbf{k} \]