Chapter 11: Problem 8
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.
Short Answer
Expert verified
The area of the parallelogram is \(\sqrt{146}\).
Step by step solution
01
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.
02
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 \(\mathbf{b} = [-1, 1, -4]\).
03
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: \[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 2 & -1 \ -1 & 1 & -4 \end{vmatrix} = \mathbf{i}(2(-4) + 1) - \mathbf{j}(2(-4) - 1) + \mathbf{k}(2(1) + 2)\].
04
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, \(\mathbf{a} \times \mathbf{b} = [-7, 9, 4]\).
05
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: \[\|\mathbf{a} \times \mathbf{b}\| = \sqrt{(-7)^2 + 9^2 + 4^2}\].
06
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 \(\sqrt{146}\).
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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 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.
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.
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.