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Find the area of the parallelogram spanned by the following vectors. \[ \begin{array}{l}\mathbf{u}=[0,0,3] \\ \mathbf{v}=[8,6,4] \\ \text { Area of parallelogram }=\text { Ex: } 1.234\end{array} \]

Ask by Smith Horton. in the United States
Feb 22,2025

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Answer

The area of the parallelogram is 30.

Solution

To find the area of the parallelogram spanned by vectors u and v, we first compute the cross product u × v. Given:  u = [0, 0, 3]  v = [8, 6, 4] The cross product is calculated as:  u × v = [ (0)(4) - (3)(6), (3)(8) - (0)(4), (0)(6) - (0)(8) ]      = [ 0 - 18, 24 - 0, 0 - 0 ]      = [ -18, 24, 0 ] The area of the parallelogram is the magnitude of u × v:  Area = |u × v| = √((-18)² + 24² + 0²)     = √(324 + 576)     = √900     = 30 Thus, the area of the parallelogram is 30.000.

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To find the area of the parallelogram spanned by the vectors \(\mathbf{u}\) and \(\mathbf{v}\), we can use the cross product of these two vectors. 1. First, compute the cross product \(\mathbf{u} \times \mathbf{v}\): \[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 0 & 3 \\ 8 & 6 & 4 \end{vmatrix} = \mathbf{i}(0 \cdot 4 - 3 \cdot 6) - \mathbf{j}(0 \cdot 4 - 3 \cdot 8) + \mathbf{k}(0 \cdot 6 - 0 \cdot 8) \] \[ = \mathbf{i}(0 - 18) - \mathbf{j}(0 - 24) + \mathbf{k}(0 - 0) = -18\mathbf{i} + 24\mathbf{j} \] 2. Next, find the magnitude of the cross product: \[ \|\mathbf{u} \times \mathbf{v}\| = \sqrt{(-18)^2 + (24)^2} = \sqrt{324 + 576} = \sqrt{900} = 30 \] Therefore, the area of the parallelogram is \(30\). Area of parallelogram = 30

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