Problem 39 - University Calculus: Early Transcendentals

Question

Find the areas of the parallelograms whose vertices are given. $$A(0,0,0), \quad B(3,2,4), \quad C(5,1,4), \quad D(2,-1,0)$$

Step-by-Step Solution

Answer

The area of the parallelogram is \( \sqrt{129} \).

Step 1: 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 \( \overrightarrow{AD} = D - A = (2, -1, 0) \).

Step 2: 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 \( \mathbf{i}(4) - \mathbf{j}(-8) + \mathbf{k}(-7) = (4, 8, -7) \).

Step 3: Compute the Magnitude of the Cross Product

Find the magnitude of resultant vector from the cross product: \( \sqrt{4^2 + 8^2 + (-7)^2} = \sqrt{16 + 64 + 49} = \sqrt{129} \).

Step 4: 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 \( \sqrt{129} \).

Key Concepts

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:

• 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.

Understanding the cross product and its computation is crucial for analyzing and solving problems involving parallelograms and vector properties.

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:

• 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.

These operations form the basis of manipulating and understanding vectors in three-dimensional space and are crucial when working with problems that consider points and lines in space.

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:

• 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.

In the given exercise, calculating the magnitude of the cross product vector \( (4, 8, -7) \) provides the area of the parallelogram formed by the original vectors. This directly links the concept of vector magnitude to real-world physical quantities, like areas and distances.