If `vec(a)= 3hat(i) + hat(j) + 2hat(k) and vec(b)= 2hat(i)-2hat(j) + 4hat(k)`, then the magnitude of `vec(b) xx vec(a)` is

Options

A. 8
B. `-8`
C. `8sqrt3`
D. None of these

Answer

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

### Step 1: Write down the vectors
Given:
\[ \vec{a} = 3\hat{i} + 1\hat{j} + 2\hat{k} \]
\[ \vec{b} = 2\hat{i} - 2\hat{j} + 4\hat{k} \]

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

### Step 3: Calculate the determinant
To compute the determinant, we can expand it as follows:
\[ \vec{b} \times \vec{a} = \hat{i} \begin{vmatrix} -2 & 4 \\ 1 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 4 \\ 3 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -2 \\ 3 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. For \(\hat{i}\):
\[ \begin{vmatrix} -2 & 4 \\ 1 & 2 \end{vmatrix} = (-2)(2) - (4)(1) = -4 - 4 = -8 \]
So, we have \(-8\hat{i}\).

2. For \(-\hat{j}\):
\[ \begin{vmatrix} 2 & 4 \\ 3 & 2 \end{vmatrix} = (2)(2) - (4)(3) = 4 - 12 = -8 \]
So, we have \(+8\hat{j}\) (since it is subtracted).

3. For \(\hat{k}\):
\[ \begin{vmatrix} 2 & -2 \\ 3 & 1 \end{vmatrix} = (2)(1) - (-2)(3) = 2 + 6 = 8 \]
So, we have \(+8\hat{k}\).

Putting it all together:
\[ \vec{b} \times \vec{a} = -8\hat{i} + 8\hat{j} + 8\hat{k} \]

### Step 4: Find the magnitude of the cross product
The magnitude of \(\vec{b} \times \vec{a}\) is given by:
\[ |\vec{b} \times \vec{a}| = \sqrt{(-8)^2 + (8)^2 + (8)^2} \]
Calculating:
\[ = \sqrt{64 + 64 + 64} = \sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3} \]

### Final Answer
The magnitude of \(\vec{b} \times \vec{a}\) is:
\[ \boxed{8\sqrt{3}} \]