If `veca=2hati-3hatj-hatk` and `vecb=hati+4hatj-2hatk`, then `vecaxxvecb` is

Options

A. `10hati+2hatj+11hatk`
B. `10hati+3hatj+11hatk`
C. `10hati-3hatj+11hatk`
D. `10hati-3hatj-10hatk`

Answer

To find the cross product of the vectors \(\vec{a} = 2\hat{i} - 3\hat{j} - \hat{k}\) and \(\vec{b} = \hat{i} + 4\hat{j} - 2\hat{k}\), we can use the determinant method. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Write the vectors in component form:**
\[ \vec{a} = 2\hat{i} - 3\hat{j} - \hat{k} \quad \text{and} \quad \vec{b} = \hat{i} + 4\hat{j} - 2\hat{k} \]

2. **Set up the determinant for the cross product:**
The cross product \(\vec{a} \times \vec{b}\) can be calculated using the following determinant:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & -1 \\ 1 & 4 & -2 \end{vmatrix} \]

3. **Calculate the determinant:**
- For \(\hat{i}\):
\[ \hat{i} \cdot \begin{vmatrix} -3 & -1 \\ 4 & -2 \end{vmatrix} = \hat{i} \cdot ((-3)(-2) - (4)(-1)) = \hat{i} \cdot (6 + 4) = 10\hat{i} \]
- For \(\hat{j}\):
\[ -\hat{j} \cdot \begin{vmatrix} 2 & -1 \\ 1 & -2 \end{vmatrix} = -\hat{j} \cdot ((2)(-2) - (1)(-1)) = -\hat{j} \cdot (-4 + 1) = -\hat{j} \cdot (-3) = 3\hat{j} \]
- For \(\hat{k}\):
\[ \hat{k} \cdot \begin{vmatrix} 2 & -3 \\ 1 & 4 \end{vmatrix} = \hat{k} \cdot ((2)(4) - (-3)(1)) = \hat{k} \cdot (8 + 3) = 11\hat{k} \]

4. **Combine the results:**
Putting it all together, we have:
\[ \vec{a} \times \vec{b} = 10\hat{i} + 3\hat{j} + 11\hat{k} \]

### Final Answer:
\[ \vec{a} \times \vec{b} = 10\hat{i} + 3\hat{j} + 11\hat{k} \]