Chapter 11: Problem 39
Find the areas of the parallelograms whose vertices are given. $$A(0,0,0), \quad B(3,2,4), \quad C(5,1,4), \quad D(2,-1,0)$$
Short Answer
Expert verified
The area of the parallelogram is \( \sqrt{129} \).
Step by step solution
01
Identify Two Adjacent Sides
To find the area of a parallelogram in 3D, we need two vectors representing its adjacent sides. We can choose vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AD} \). These vectors are calculated as follows: \( \overrightarrow{AB} = B - A = (3, 2, 4) \) and \( \overrightarrow{AD} = D - A = (2, -1, 0) \).
02
Calculate the Cross Product
The area of a parallelogram is given by the magnitude of the cross product of the two vectors representing its sides. The cross product \( \overrightarrow{AB} \times \overrightarrow{AD} \) is \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3 & 2 & 4 \ 2 & -1 & 0 \end{vmatrix} = \mathbf{i}(2 \cdot 0 - 4 \cdot (-1)) - \mathbf{j}(3 \cdot 0 - 4 \cdot 2) + \mathbf{k}(3 \cdot (-1) - 2 \cdot 2) \]. This simplifies to \( \mathbf{i}(4) - \mathbf{j}(-8) + \mathbf{k}(-7) = (4, 8, -7) \).
03
Compute the Magnitude of the Cross Product
Find the magnitude of resultant vector from the cross product: \( \sqrt{4^2 + 8^2 + (-7)^2} = \sqrt{16 + 64 + 49} = \sqrt{129} \).
04
Conclusion
The magnitude of the cross product vector \( \sqrt{129} \) gives the area of the parallelogram formed by the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AD} \). Thus, the area of the parallelogram is \( \sqrt{129} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cross Product
The cross product is a fundamental operation in vector calculus used to find a vector that is perpendicular to two given vectors in three-dimensional space. It is particularly useful in scenarios involving physics and engineering, such as finding torque, rotational motion, or the area of a parallelogram in 3D space.
The cross product of two vectors \( \overrightarrow{a} \) and \( \overrightarrow{b} \), denoted as \( \overrightarrow{a} \times \overrightarrow{b} \), results in a third vector, \( \overrightarrow{c} \), that is orthogonal to both \( \overrightarrow{a} \) and \( \overrightarrow{b} \).
Here's a quick guide to calculating the cross product:
The cross product of two vectors \( \overrightarrow{a} \) and \( \overrightarrow{b} \), denoted as \( \overrightarrow{a} \times \overrightarrow{b} \), results in a third vector, \( \overrightarrow{c} \), that is orthogonal to both \( \overrightarrow{a} \) and \( \overrightarrow{b} \).
Here's a quick guide to calculating the cross product:
- Arrange the vectors in a matrix with the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) on the first row, and the components of the vectors on the second and third rows.
- Use the determinant of this matrix to compute the resultant vector.
- Each component (\( \mathbf{i}, \mathbf{j}, \mathbf{k} \)) is calculated by omitting the column and finding the determinant of the remaining 2x2 matrix.
- The final vector represents both direction and magnitude of the cross product.
3D Vectors
3D vectors are directed line segments in three-dimensional space with magnitude and direction. Vectors are essential in describing motion, location, and forces in real-world contexts. Each vector is represented by three components corresponding to the \( x \), \( y \), and \( z \) axes, typically written as \( \overrightarrow{v} = (x, y, z) \).
In the context of the original exercise, the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AD} \) are derived from the coordinates of points in space. Here are some concepts to help further understand 3D vectors:
In the context of the original exercise, the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AD} \) are derived from the coordinates of points in space. Here are some concepts to help further understand 3D vectors:
- Vector Addition: Vectors can be added together by summing their respective components. For example, if \( \overrightarrow{v} = (x_1, y_1, z_1) \) and \( \overrightarrow{u} = (x_2, y_2, z_2) \), then their sum is \( \overrightarrow{v} + \overrightarrow{u} = (x_1 + x_2, y_1 + y_2, z_1 + z_2) \).
- Vector Subtraction: Similarly, vectors can be subtracted by subtracting each of their components. It helps in finding the direction between two points, such as \( \overrightarrow{AB} = B - A \).
- Scalar Multiplication: Multiplying a vector by a scalar changes its magnitude without altering its direction.
Magnitude of a Vector
The magnitude of a vector, sometimes referred to as its length or norm, quantifies how long the vector is, disregarding its direction. In a three-dimensional space, a vector \( \overrightarrow{v} = (x, y, z) \) has a magnitude given by the formula:
\[ \| \overrightarrow{v} \| = \sqrt{x^2 + y^2 + z^2} \]
This calculation extends the Pythagorean theorem into three dimensions, where the magnitude is the hypotenuse of a right triangle formed by the vector’s components.
Key points to remember about vector magnitude include:
\[ \| \overrightarrow{v} \| = \sqrt{x^2 + y^2 + z^2} \]
This calculation extends the Pythagorean theorem into three dimensions, where the magnitude is the hypotenuse of a right triangle formed by the vector’s components.
Key points to remember about vector magnitude include:
- Magnitude is always a non-negative number.
- It directly represents the physical size of the vector.
- The zero vector, \( (0, 0, 0) \), has a magnitude of zero since it has no length.