Find the area of the parallelogram whose diagonals are determined by vectors `vec(a)= 3hat(i) + hat(j)-2hat(k) and vec(b)= hat(i) - 3hat(j) + 4hat(k)`

Answer

To find the area of the parallelogram whose diagonals are determined by the vectors \(\vec{a} = 3\hat{i} + \hat{j} - 2\hat{k}\) and \(\vec{b} = \hat{i} - 3\hat{j} + 4\hat{k}\), we can use the formula for the area of a parallelogram given its diagonals:

\[ \text{Area} = \frac{1}{2} \|\vec{a} \times \vec{b}\| \]

where \(\vec{a} \times \vec{b}\) is the cross product of the vectors \(\vec{a}\) and \(\vec{b}\).

### Step 1: 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 \(\vec{b}\):

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

### Step 2: Calculate the determinant

To calculate the determinant, we expand it as follows:

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

Calculating each of these 2x2 determinants:

1. For \(\hat{i}\):
\[ \begin{vmatrix} 1 & -2 \\ -3 & 4 \end{vmatrix} = (1)(4) - (-2)(-3) = 4 - 6 = -2 \]

2. For \(\hat{j}\):
\[ \begin{vmatrix} 3 & -2 \\ 1 & 4 \end{vmatrix} = (3)(4) - (-2)(1) = 12 + 2 = 14 \]

3. For \(\hat{k}\):
\[ \begin{vmatrix} 3 & 1 \\ 1 & -3 \end{vmatrix} = (3)(-3) - (1)(1) = -9 - 1 = -10 \]

Putting it all together:

\[ \vec{a} \times \vec{b} = -2\hat{i} - 14\hat{j} - 10\hat{k} \]

### Step 3: Find the magnitude of the cross product

Now, we calculate the magnitude of \(\vec{a} \times \vec{b}\):

\[ \|\vec{a} \times \vec{b}\| = \sqrt{(-2)^2 + (-14)^2 + (-10)^2} \]

Calculating each term:

\[ = \sqrt{4 + 196 + 100} = \sqrt{300} \]

### Step 4: Calculate the area of the parallelogram

Now we can find the area:

\[ \text{Area} = \frac{1}{2} \|\vec{a} \times \vec{b}\| = \frac{1}{2} \sqrt{300} = \frac{1}{2} \times 10\sqrt{3} = 5\sqrt{3} \]

Thus, the area of the parallelogram is:

\[ \boxed{5\sqrt{3}} \]