If `vec(a)=2hat(i)-3hat(j)-hat(k), vec(b)=hat(i)+4hat(j)-2hat(k)`, then what is `(vec(a)+vec(b))xx(vec(a)-vec(b))` equal to ?

Options

A. `2(vec(a)xxvec(b))`
B. `-2(vec(a)xxvec(b))`
C. `(vec(a)xxvec(b))`
D. `-(vec(a)xxvec(b))`

Explanation

To solve the problem, we need to calculate \((\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})\) where \(\vec{a} = 2\hat{i} - 3\hat{j} - \hat{k}\) and \(\vec{b} = \hat{i} + 4\hat{j} - 2\hat{k}\).

### Step 1: Calculate \(\vec{a} + \vec{b}\)
\[ \vec{a} + \vec{b} = (2\hat{i} - 3\hat{j} - \hat{k}) + (\hat{i} + 4\hat{j} - 2\hat{k}) \]
Combining the components:
\[ = (2 + 1)\hat{i} + (-3 + 4)\hat{j} + (-1 - 2)\hat{k} \]
\[ = 3\hat{i} + 1\hat{j} - 3\hat{k} \]

### Step 2: Calculate \(\vec{a} - \vec{b}\)
\[ \vec{a} - \vec{b} = (2\hat{i} - 3\hat{j} - \hat{k}) - (\hat{i} + 4\hat{j} - 2\hat{k}) \]
Combining the components:
\[ = (2 - 1)\hat{i} + (-3 - 4)\hat{j} + (-1 + 2)\hat{k} \]
\[ = 1\hat{i} - 7\hat{j} + 1\hat{k} \]

### Step 3: Calculate \((\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})\)
Now we need to calculate the cross product:
\[ (3\hat{i} + 1\hat{j} - 3\hat{k}) \times (1\hat{i} - 7\hat{j} + 1\hat{k}) \]
Using the determinant method for cross products:
\[ \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -3 \\ 1 & -7 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 1 & -3 \\ -7 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & -3 \\ 1 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 1 \\ 1 & -7 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} 1 & -3 \\ -7 & 1 \end{vmatrix} = (1)(1) - (-3)(-7) = 1 - 21 = -20\)
2. \(\begin{vmatrix} 3 & -3 \\ 1 & 1 \end{vmatrix} = (3)(1) - (-3)(1) = 3 + 3 = 6\)
3. \(\begin{vmatrix} 3 & 1 \\ 1 & -7 \end{vmatrix} = (3)(-7) - (1)(1) = -21 - 1 = -22\)

Putting it all together:
\[ = -20\hat{i} - 6\hat{j} - 22\hat{k} \]

### Final Result
Thus, the result of \((\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})\) is:
\[ -20\hat{i} - 6\hat{j} - 22\hat{k} \]