If `vecA=3hati+hatj+2hatk` and `vecB=2hati-2hatj+4hatk`, then value of `|vecA X vecB|` will be

Options

A. `8sqrt2`
B. `8sqrt3`
C. `8sqrt5`
D. `5sqrt8`

Explanation

To solve the problem of finding the magnitude of the cross product \(|\vec{A} \times \vec{B}|\) where \(\vec{A} = 3\hat{i} + \hat{j} + 2\hat{k}\) and \(\vec{B} = 2\hat{i} - 2\hat{j} + 4\hat{k}\), we can follow these steps:

### Step 1: Write down the vectors
We have:
\[ \vec{A} = 3\hat{i} + 1\hat{j} + 2\hat{k} \]
\[ \vec{B} = 2\hat{i} - 2\hat{j} + 4\hat{k} \]

### Step 2: Set up the determinant for the cross product
The cross product \(\vec{A} \times \vec{B}\) can be calculated using the determinant of a matrix formed by the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) and the components of \(\vec{A}\) and \(\vec{B}\):
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 2 & -2 & 4 \end{vmatrix} \]

### Step 3: Calculate the determinant
Expanding the determinant, we get:
\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} 1 & 2 \\ -2 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 2 \\ 2 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 1 \\ 2 & -2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \(\begin{vmatrix} 1 & 2 \\ -2 & 4 \end{vmatrix} = (1)(4) - (2)(-2) = 4 + 4 = 8\)
2. \(\begin{vmatrix} 3 & 2 \\ 2 & 4 \end{vmatrix} = (3)(4) - (2)(2) = 12 - 4 = 8\)
3. \(\begin{vmatrix} 3 & 1 \\ 2 & -2 \end{vmatrix} = (3)(-2) - (1)(2) = -6 - 2 = -8\)

Substituting these back into the equation gives:
\[ \vec{A} \times \vec{B} = 8\hat{i} - 8\hat{j} - 8\hat{k} \]

### Step 4: Simplify the result
Thus, we have:
\[ \vec{A} \times \vec{B} = 8\hat{i} - 8\hat{j} - 8\hat{k} \]

### Step 5: Find the magnitude of the cross product
The magnitude of \(\vec{A} \times \vec{B}\) is given by:
\[ |\vec{A} \times \vec{B}| = \sqrt{(8)^2 + (-8)^2 + (-8)^2} \]
Calculating this:
\[ |\vec{A} \times \vec{B}| = \sqrt{64 + 64 + 64} = \sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3} \]

### Final Answer
Thus, the value of \(|\vec{A} \times \vec{B}|\) is:
\[ \boxed{8\sqrt{3}} \]