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Chapter 5: Problem 28

Find the area of the parallelogram that has the vectors as adjacent sides. $$\mathbf{u}=(2,-1,0), \quad \mathbf{v}=(-1,2,0)$$

Short Answer

Expert verified
The area of the parallelogram formed by vectors \(\mathbf{u} = (2,-1,0)\) and \(\mathbf{v} = (-1,2,0)\) is 3 units.

Step by step solution

01

Calculate the Cross Product

To begin with, one must compute the cross product of the given vectors \(\mathbf{u}\) and \(\mathbf{v}\). The cross product of \(\mathbf{u}\) and \(\mathbf{v}\) is defined as \(\mathbf{u} \times \mathbf{v} = (u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1)\)Substitute values, \(\mathbf{u} = (2,-1,0)\) and \(\mathbf{v} = (-1,2,0)\) into the formula, we get \(\mathbf{u} \times \mathbf{v} = ((-1)(0) - (0)(2), (0)(-1) - (2)(0), (2)(2) - (-1)(-1)) = (0, 0, 3)\)
02

Compute the magnitude of the Cross Product

The magnitude of the cross product vector \(\mathbf{u} \times \mathbf{v} = (0, 0, 3)\) gives the area of the parallelogram. It's calculated as\(|\mathbf{u} \times \mathbf{v}| = \sqrt{(0)^2 + (0)^2 + (3)^2} = 3\)
03

Final Answer

The area of the parallelogram formed by vectors \(\mathbf{u} = (2,-1,0)\) and \(\mathbf{v} = (-1,2,0)\) is equal to the magnitude of the cross product of these vectors, which is calculated as 3 units. So, the final answer is 3 units.

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Most popular questions from this chapter

Prove that \(\mathbf{u} \times \mathbf{v}\) is orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\).

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