Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the area of the parallelogram with \(\mathbf{a}=2 \mathbf{i}+2 \mathbf{j}-\mathbf{k}\) and \(\mathbf{b}=-\mathbf{i}+\mathbf{j}-4 \mathbf{k}\) as the adjacent sides.

Step-by-Step Solution

Answer

The area of the parallelogram is \(\sqrt{146}\).

Step 1: Understand the Problem

We need to find the area of a parallelogram formed by two vectors \(\mathbf{a}\) and \(\mathbf{b}\). The formula to find the area involves calculating the magnitude of the cross product of these two vectors.

Step 2: Define Vectors

Identify the vector components for \(\mathbf{a}\) and \(\mathbf{b}\). Given, \(\mathbf{a} = 2\mathbf{i} + 2\mathbf{j} - \mathbf{k}\) and \(\mathbf{b} = -\mathbf{i} + \mathbf{j} - 4\mathbf{k}\). So the vectors are: \(\mathbf{a} = [2, 2, -1]\) and \(\mathbf{b} = [-1, 1, -4]\).

Step 3: Calculate the Cross Product \(\mathbf{a} \times \mathbf{b}\)

The cross product \(\mathbf{a} \times \mathbf{b}\) is given by the determinant of the matrix: \[\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 2 & -1 \ -1 & 1 & -4 \end{vmatrix}\]. This is calculated as follows: \[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 2 & -1 \ -1 & 1 & -4 \end{vmatrix} = \mathbf{i}(2(-4) + 1) - \mathbf{j}(2(-4) - 1) + \mathbf{k}(2(1) + 2)\].

Step 4: Simplify the Cross Product

Calculate each component of the cross product: \[\mathbf{i}(-8 + 1) - \mathbf{j}(-8 - 1) + \mathbf{k}(2 + 2) = -7\mathbf{i} + 9\mathbf{j} + 4\mathbf{k}\]. Therefore, \(\mathbf{a} \times \mathbf{b} = [-7, 9, 4]\).

Step 5: Find the Magnitude of the Cross Product

The magnitude of the cross product vector \(\mathbf{a} \times \mathbf{b}\) is the area of the parallelogram. The magnitude is calculated by: \[\|\mathbf{a} \times \mathbf{b}\| = \sqrt{(-7)^2 + 9^2 + 4^2}\].

Step 6: Final Calculation

Calculate the terms: \((-7)^2 = 49\), \(9^2=81\), and \(4^2=16\). Add them: \(49 + 81 + 16 = 146\). Then take the square root: \(\sqrt{146}\). Therefore, the area is \(\sqrt{146}\).

Key Concepts

Cross Product

The cross product is a crucial operation in vector calculus that is used to find a vector that is perpendicular to the plane established by two original vectors. When you compute the cross product of two vectors, the result is another vector, which can give us insights into the spatial relationship between the originating vectors.

• The cross product is denoted as \( \mathbf{a} \times \mathbf{b} \).
• The result is a vector that is orthogonal (perpendicular) to both \( \mathbf{a} \) and \( \mathbf{b} \).
• The direction of the resulting vector can be determined using the right-hand rule: if you point your index finger in the direction of \( \mathbf{a} \) and your middle finger in the direction of \( \mathbf{b} \), your thumb will point in the direction of the cross product.
• The magnitude of the cross product vector represents the area of the parallelogram formed by the two original vectors.

This means, for vectors \( \mathbf{a} = [2, 2, -1] \) and \( \mathbf{b} = [-1, 1, -4] \), their cross product is calculated using the determinant of a 3x3 matrix. This matrix has the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) in the first row, and the components of \( \mathbf{a} \) and \( \mathbf{b} \) in the second and third rows, respectively. Following the calculation, the resultant vector \( [-7, 9, 4] \) tells us not only the direction but also helps in calculating the parallelogram's area.

Parallelogram Area

To find the area of a parallelogram when two vectors are given as its adjacent sides, we can conveniently use the cross product. The cross product provides a vector whose magnitude equals the area of the parallelogram.

• The formula for the area from the cross product is \( \text{Area} = \| \mathbf{a} \times \mathbf{b} \| \).
• The magnitude of the cross product vector is calculated using the Euclidean norm, or simply finding the square root of the sum of its components squared.

For the vectors \( \mathbf{a} = [2, 2, -1] \) and \( \mathbf{b} = [-1, 1, -4] \), the cross product has resulted in the vector \( [-7, 9, 4] \). When you calculate its magnitude, you find \( \sqrt{(-7)^2 + 9^2 + 4^2} \), which simplifies to \( \sqrt{146} \). This value, \( \sqrt{146} \), is indeed the area of the parallelogram that these vectors form.

Vector Magnitude

Vector magnitude, sometimes referred to as the 'length' or 'norm' of a vector, is a measure of how long a vector is or, in the case of the cross product, a measure of the area represented by that vector.

• The magnitude of a vector \( \mathbf{v} = [x, y, z] \) is calculated using the formula \( \| \mathbf{v} \| = \sqrt{x^2 + y^2 + z^2} \).
• This formula represents the 3-dimensional space equivalent of the Pythagorean Theorem, extending it to vectors with three components.
• Magnitude is always a non-negative scalar.

In this problem, when calculating the magnitude of the cross product vector \( [-7, 9, 4] \), it's crucial to square each component, sum them, and compute the square root: \( \sqrt{(-7)^2 + 9^2 + 4^2} = \sqrt{146} \). This provides a clear measure of both the vector's size in 3D space and the parallelogram's area that the vectors \( \mathbf{a} \) and \( \mathbf{b} \) form.