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Find the area of the parallelogram that has two adjacent sides \(\mathbf{u}\) and \(\mathbf{v}\) $$\mathbf{u}=8 \mathbf{i}+2 \mathbf{j}-3 \mathbf{k}, \mathbf{v}=2 \mathbf{i}+4 \mathbf{j}-4 \mathbf{k}$$

Short Answer

Expert verified
Answer: The area of the parallelogram is \(\sqrt{404}\) square units.

Step by step solution

01

Compute the cross product of \(\mathbf{u}\) and \(\mathbf{v}\)

To compute the cross product of the vectors \(\mathbf{u}\) and \(\mathbf{v}\), we use the formula: $$\mathbf{u} \times \mathbf{v} = \begin{pmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 8 & 2 & -3 \\ 2 & 4 & -4 \end{pmatrix}$$ Expand the above determinant as follows: $$\mathbf{u} \times \mathbf{v} = \mathbf{i}\begin{pmatrix}2 & -3\\4 & -4\end{pmatrix} - \mathbf{j}\begin{pmatrix}8 & -3\\2 & -4\end{pmatrix} + \mathbf{k}\begin{pmatrix}8 & 2\\2 & 4\end{pmatrix}$$ Compute the determinants: $$\mathbf{u} \times \mathbf{v} = (8\mathbf{i} + 14\mathbf{j} - 12\mathbf{k})$$
02

Compute the magnitude of the cross product

Now, find the magnitude of the cross product vector \(\mathbf{u} \times \mathbf{v}\). Use the formula: $$\|\mathbf{u} \times \mathbf{v}\| = \sqrt{(8)^2 + (14)^2 + (-12)^2}$$ Calculate the magnitude: $$\|\mathbf{u} \times \mathbf{v}\| = \sqrt{64 + 196 + 144} = \sqrt{404}$$
03

Interpret the result

We find that the magnitude of the cross product of \(\mathbf{u}\) and \(\mathbf{v}\) is \(\sqrt{404}\). This value represents the area of the parallelogram with adjacent sides \(\mathbf{u}\) and \(\mathbf{v}\). Thus, the area of the parallelogram is \(\boxed{\sqrt{404}}\) square units.

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