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Chapter 12: Problem 22

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

Short Answer

Expert verified
Answer: The area of the parallelogram is \(\sqrt{38}\).

Step by step solution

01

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

To find the cross product of the given vectors, \(\mathbf{u} \times \mathbf{v}\), use the formula: $$\mathbf{u} \times \mathbf{v} = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 0 & 2 \\ 1 & 1 & 1 \end{array} \right|$$ Evaluate the determinant: $$\mathbf{u} \times \mathbf{v} = (\mathbf{i} (0 \cdot 1 - 2 \cdot 1)) - (\mathbf{j} (-3 \cdot 1 - 2 \cdot 1)) + (\mathbf{k} (-3\cdot 1 - 0 \cdot 1))$$ $$\mathbf{u} \times \mathbf{v} = (-2 \mathbf{i}) + (5 \mathbf{j}) - (-3 \mathbf{k})$$ $$\mathbf{u} \times \mathbf{v} = -2 \mathbf{i} + 5 \mathbf{j} + 3 \mathbf{k}$$
02

Calculate the magnitude of the cross product

Use the formula for the magnitude of a vector \(||\mathbf{w}|| = \sqrt{w_{x}^2 + w_{y}^2 + w_{z}^2}\), where \(\mathbf{w}=-2 \mathbf{i} + 5 \mathbf{j} + 3 \mathbf{k}\) : $$||\mathbf{u} \times \mathbf{v}|| = \sqrt{(-2)^2 + 5^2 + 3^2}$$ $$||\mathbf{u} \times \mathbf{v}|| = \sqrt{4 + 25 + 9}$$ $$||\mathbf{u} \times \mathbf{v}|| = \sqrt{38}$$
03

Provide the area of the parallelogram

The area of the parallelogram formed by \(\mathbf{u}\) and \(\mathbf{v}\) is equal to the magnitude of their cross product: $$A = ||\mathbf{u} \times \mathbf{v}||$$ $$A = \sqrt{38}$$ The area of the parallelogram is \(\sqrt{38}\).

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Most popular questions from this chapter

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