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Find the area of the parallelogram with adjacent sides \(\mathbf{u}=\langle 3,2,0\rangle\) and \(\mathbf{v}=\langle 0,2,1\rangle\).

Short Answer

Expert verified
The area of the parallelogram is 7.

Step by step solution

01

Understand the problem

We need to find the area of a parallelogram defined by two vectors \( \mathbf{u} = \langle 3, 2, 0 \rangle \) and \( \mathbf{v} = \langle 0, 2, 1 \rangle \). One method involves using the cross product of these vectors.
02

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

The cross product \( \mathbf{u} \times \mathbf{v} \) is calculated as follows:\[ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3 & 2 & 0 \ 0 & 2 & 1 \end{vmatrix} \]Using the determinant, we get:\[ \mathbf{u} \times \mathbf{v} = \mathbf{i} (2 \cdot 1 - 0 \cdot 2) - \mathbf{j} (3 \cdot 1 - 0 \cdot 0) + \mathbf{k} (3 \cdot 2 - 0 \cdot 2) \]\[ = \mathbf{i} (2) - \mathbf{j} (3) + \mathbf{k} (6) \]\[ = \langle 2, -3, 6 \rangle \]
03

Calculate the magnitude of the cross product

The next step is to find the magnitude of the cross product vector \( \langle 2, -3, 6 \rangle \):\[ \| \mathbf{u} \times \mathbf{v} \| = \sqrt{2^2 + (-3)^2 + 6^2} \]\[ = \sqrt{4 + 9 + 36} \]\[ = \sqrt{49} \]\[ = 7 \]
04

Interpret the result

The magnitude of the cross product, 7, represents the area of the parallelogram spanned by vectors \( \mathbf{u} \) and \( \mathbf{v} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Area of Parallelogram
To find the area of a parallelogram formed by two vectors, you can use the cross product. Imagine a parallelogram with each pair of adjacent sides represented by vectors. The magnitude of the cross product of these vectors gives you the exact area. This method is particularly useful because it considers both the direction and the magnitude of the vectors involved.
  • For any two vectors \( \mathbf{u} \) and \( \mathbf{v} \), the cross product \( \mathbf{u} \times \mathbf{v} \) results in a vector perpendicular to the plane containing \( \mathbf{u} \) and \( \mathbf{v} \).
  • The length, or magnitude, of this resulting vector represents the area of the parallelogram.
This approach simplifies finding the area without needing to draw accurate shapes or use approximate methods. By calculating the cross product and then its magnitude, you obtain a precise measurement of the area. It’s an exact method grounded in vector mathematics.
Vector Magnitude
Vector magnitude is a measure of the "length" of a vector in space. You calculate it using the Pythagorean theorem for three-dimensions. For a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \), the magnitude is given by:\[ \| \mathbf{a} \| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]
  • Magnitude gives you a scalar quantity, representing how "long" the vector is.
  • In the context of a cross product, it helps quantify the area of a parallelogram spanned by two vectors.
The calculation is straightforward: square each component of the vector, sum them, and then take the square root of that sum. This formula works regardless of where in space (in any dimension) the vector is situated.
Determinant Calculation
Calculating a determinant is a crucial step in finding the cross product of two vectors, especially in three-dimensional space. For two 3D vectors \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), the cross product can be found using the determinants of a 3x3 matrix:\[\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix} \]
  • This determinant expands into a combination of the unit vectors \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \).
  • The calculation is performed by specific formulas resulting from the permutations of matrix components.
The resulting vector from this product is orthogonal to the original two vectors, representing the axis of the parallelogram's rotation—if you picture it twisting in 3D space. Calculating determinants bridges the abstract math of vectors with practical geometric interpretations, like finding areas and determining orientations in space.

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