Find the area of a parallelogram whose adjacent sides are the vectors `a=2i-2j+k" and "b=i-3j+3k`.

Answer

To find the area of the parallelogram formed by the vectors **a** and **b**, we can use the formula for the area of a parallelogram defined by two vectors, which is given by the magnitude of the cross product of the two vectors.

### Step-by-Step Solution:

1. **Identify the vectors**:
- Let **a** = \( 2\mathbf{i} - 2\mathbf{j} + \mathbf{k} \)
- Let **b** = \( \mathbf{i} - 3\mathbf{j} + 3\mathbf{k} \)

2. **Set up the cross product**:
The area of the parallelogram is given by:
\[ \text{Area} = |\mathbf{a} \times \mathbf{b}| \]
To compute the cross product \( \mathbf{a} \times \mathbf{b} \), we can use the determinant of a matrix formed by the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) and the components of vectors **a** and **b**.

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

3. **Calculate the determinant**:
Expanding the determinant, we have:
\[ \mathbf{a} \times \mathbf{b} = \mathbf{i} \begin{vmatrix} -2 & 1 \\ -3 & 3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 2 & -2 \\ 1 & -3 \end{vmatrix} \]

Now, calculating each of these 2x2 determinants:
- For \( \mathbf{i} \):
\[ \begin{vmatrix} -2 & 1 \\ -3 & 3 \end{vmatrix} = (-2)(3) - (1)(-3) = -6 + 3 = -3 \]
- For \( \mathbf{j} \):
\[ \begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} = (2)(3) - (1)(1) = 6 - 1 = 5 \]
- For \( \mathbf{k} \):
\[ \begin{vmatrix} 2 & -2 \\ 1 & -3 \end{vmatrix} = (2)(-3) - (-2)(1) = -6 + 2 = -4 \]

Putting it all together:
\[ \mathbf{a} \times \mathbf{b} = -3\mathbf{i} - 5\mathbf{j} - 4\mathbf{k} \]

4. **Find the magnitude of the cross product**:
The magnitude is calculated as follows:
\[ |\mathbf{a} \times \mathbf{b}| = \sqrt{(-3)^2 + (-5)^2 + (-4)^2} \]
\[ = \sqrt{9 + 25 + 16} = \sqrt{50} \]
\[ = 5\sqrt{2} \]

5. **Final answer**:
Therefore, the area of the parallelogram is:
\[ \text{Area} = 5\sqrt{2} \text{ square units} \]