If `veca = 2hati -3hatj-1hatk and vecb =hati + 4hatj -2hatk " then " veca xx vecb` is

Options

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

Explanation

To find the cross product of the vectors \(\vec{a}\) and \(\vec{b}\), we can follow these steps:

Given:
\[ \vec{a} = 2\hat{i} - 3\hat{j} - 1\hat{k} \]
\[ \vec{b} = \hat{i} + 4\hat{j} - 2\hat{k} \]

### Step 1: Set up the determinant for the cross product
The cross product \(\vec{a} \times \vec{b}\) can be computed using the determinant of a matrix formed by the unit vectors and the components of the vectors \(\vec{a}\) and \(\vec{b}\).

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & -1 \\ 1 & 4 & -2 \end{vmatrix} \]

### Step 2: Calculate the determinant
To calculate the determinant, we can expand it as follows:

\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} -3 & -1 \\ 4 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -1 \\ 1 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -3 \\ 1 & 4 \end{vmatrix} \]

### Step 3: Compute the 2x2 determinants
1. For \(\hat{i}\):
\[ \begin{vmatrix} -3 & -1 \\ 4 & -2 \end{vmatrix} = (-3)(-2) - (-1)(4) = 6 + 4 = 10 \]

2. For \(\hat{j}\):
\[ \begin{vmatrix} 2 & -1 \\ 1 & -2 \end{vmatrix} = (2)(-2) - (-1)(1) = -4 + 1 = -3 \]

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

### Step 4: Substitute back into the equation
Now substituting these values back into our expression for the cross product:

\[ \vec{a} \times \vec{b} = 10\hat{i} - (-3)\hat{j} + 11\hat{k} \]
\[ = 10\hat{i} + 3\hat{j} + 11\hat{k} \]

### Final Answer
Thus, the cross product \(\vec{a} \times \vec{b}\) is:
\[ \vec{a} \times \vec{b} = 10\hat{i} + 3\hat{j} + 11\hat{k} \]

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