The area of a parallelogram formed by two vectors $\mathbf{u}$ and $\mathbf{v}$ can be found using the cross product of these vectors. The magnitude of the cross product gives the area of the parallelogram.

Given vectors: $\mathbf{u} = \mathbf{i} - 3\mathbf{j} + 3\mathbf{k}$ $\mathbf{v} = 3\mathbf{j} - \mathbf{k}$

First, find the cross product $\mathbf{u} \times \mathbf{v}$.

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -3 & 3 \\ 0 & 3 & -1 \end{vmatrix}$$ The determinant of this matrix is: $$\mathbf{u} \times \mathbf{v} = \mathbf{i} \left((-3)(-1) - (3)(3)\right) - \mathbf{j} \left(1(-1) - 3(0)\right) + \mathbf{k} \left(1(3) - (-3)(0)\right)$$ Simplifying: $$\mathbf{u} \times \mathbf{v} = \mathbf{i} (3 - 9) - \mathbf{j} (-1 - 0) + \mathbf{k} (3 - 0)$$ $$\mathbf{u} \times \mathbf{v} = -6\mathbf{i} + \mathbf{j} + 3\mathbf{k}$$ The cross product vector is: $$\mathbf{u} \times \mathbf{v} = -6\mathbf{i} + \mathbf{j} + 3\mathbf{k}$$ Next, find the magnitude of this vector to get the area of the parallelogram: $$\text{Area} = \sqrt{(-6)^2 + 1^2 + 3^2}$$ $$\text{Area} = \sqrt{36 + 1 + 9} = \sqrt{46}$$ So, the area of the parallelogram is $$\sqrt{46}$$ square units. Would you like more details or have any questions? Here are some related questions: 1. How do you compute the cross product of two vectors? 2. What is the geometric interpretation of the cross product? 3. How does the cross product help in determining the orientation of the plane formed by two vectors? 4. What is the difference between a parallelogram and a rectangle in terms of their area calculation? 5. Can the cross product be used to find areas in dimensions other than 3D? **Tip:** The magnitude of the cross product of two vectors gives the area of the parallelogram spanned by those vectors in 3D space.