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Chapter 3: Problem 12

The vectors \(\mathbf{A}=a \mathbf{i}+b \mathbf{j}\) and \(\mathbf{B}=c \mathbf{i}+d \mathbf{j}\) form two sides of a parallelogram. Show that the area of the parallelogram is given by the absolute value of the following determinant. (Also see Chapter 6, Section 3.) $$\left|\begin{array}{ll} a & b \\ c & d \end{array}\right|$$

Short Answer

Expert verified
The area of the parallelogram is \( |ad - bc| \).

Step by step solution

01

- Understand vector notation

The vectors \(\mathbf{A}=a \mathbf{i}+b \mathbf{j}\) and \(\mathbf{B}=c \mathbf{i}+d \mathbf{j}\) represent two sides of a parallelogram in vector form. Here, \(a, b, c,\) and \(d\) are constants while \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors along the x and y-axis, respectively.
02

- Recall the formula for the area of a parallelogram

The area of a parallelogram formed by two vectors \(\mathbf{A}\) and \(\mathbf{B}\) can be found using the cross product of the vectors: \(\text{Area} = ||\mathbf{A} \times \mathbf{B}||\). For vectors in two dimensions, this can be represented as the determinant of a 2x2 matrix.
03

- Set up the determinant

The determinant of the matrix formed by vectors \(\mathbf{A}\) and \(\mathbf{B}\) is given by: \[\text{det} = \begin{vmatrix} a & b \ c & d \end{vmatrix}\]
04

- Calculate the determinant

The determinant of the matrix \[\begin{vmatrix} a & b \ c & d \end{vmatrix}\] is calculated as: \[\text{det} = ad - bc\]
05

- Find the absolute value

The area of the parallelogram is given by the absolute value of the determinant: \[\text{Area} = |ad - bc|\]

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Key Concepts

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

Vector Cross Product
The vector cross product is a fundamental operation in vector calculus. It is used to determine a vector perpendicular to two given vectors in a three-dimensional space. However, in the context of two-dimensional vectors, like \(\mathbf{A}=a \mathbf{i}+b \mathbf{j} \) and \(\mathbf{B}=c \mathbf{i}+d \mathbf{j}\), the cross product helps us find the area of the parallelogram they form.
  • Basic Definition: If we have two vectors \(\mathbf{u} \) and \(\mathbf{v} \), their cross product \(\mathbf{u} \times \mathbf{v} \) is a vector perpendicular to both.
  • Formula: For two vectors \(\mathbf{A} \) and \(\mathbf{B} \) in 2D, the magnitude of the cross product is given by:
    \[ ||\mathbf{A} \times \mathbf{B}|| = |a \cdot d - b \cdot c| \]
We can use the above result to compute the area of the parallelogram formed by the vectors. Even though the cross product traditionally applies to 3D vectors, the magnitude gives us a scalar value suitable for our purposes in 2D contexts.
Determinants
Determinants are mathematical objects derived from square matrices. They provide important properties like volume scaling factors, and in our case, they help compute the area of a parallelogram formed by two vectors.
  • Matrix Representation: The determinant of a 2x2 matrix \[ \begin{vmatrix} a & b \ c & d \end{vmatrix} \] is given by the formula:
    \[ \text{det} = ad - bc \]
  • Geometric Interpretation: The absolute value of the determinant of matrix \ \begin{vmatrix} a & b \ c & d \end{vmatrix} \ gives us the area of the parallelogram whose sides are represented by the rows or columns of the matrix.
By setting up a matrix using the coefficients of vector \[\mathbf{A} = a \mathbf{i} + b \mathbf{j}\] and vector \[\mathbf{B} = c \mathbf{i} + d \mathbf{j},\] computing its determinant leads to the area formula for the parallelogram. Just take the absolute value of that result.
Area of Parallelogram
When dealing with vectors forming a parallelogram, determining the area can be simplified using determinants and cross products.
  • Parallelogram Sides: Vectors \(\mathbf{A}=a \mathbf{i}+b \mathbf{j} \) and \(\mathbf{B}=c \mathbf{i}+d \mathbf{j}\) are two adjacent sides of the parallelogram.
  • Area Formula: The area of the parallelogram formed by these vectors in 2D is given by the absolute value of the determinant of the matrix they form:
    \[ \text{Area} = \left| \begin{array}{ll} a & b \ c & d \end{array} \right|\]
  • Calculation: Following these steps:
    1. Write the coefficient matrix:
      \[ \begin{vmatrix} a & b \ c & d \end{vmatrix} \]
    2. Compute the determinant:
      \[ \text{det} = ad - bc \]
    3. Take its absolute value to get the area:
      \[ \text{Area} = |ad - bc| \]
These steps give a quick and effective way to find the area using vector properties and matrix operations.

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Most popular questions from this chapter

Show that each of the following matrices is orthogonal and find the rotation and/or reflection it produces as an operator acting on vectors. If a rotation, find the axis and angle; if a reflection, find the reflecting plane and the rotation, if any, about the normal to that plane. $$\frac{1}{2}\left(\begin{array}{ccc} 1 & \sqrt{2} & -1 \\ \sqrt{2} & 0 & \sqrt{2} \\ 1 & -\sqrt{2} & -1 \end{array}\right)$$

Find a vector perpendicular to both \(\mathbf{i}+\mathbf{j}\) and \(\mathbf{i}-2 \mathbf{k}\).

Verify that each of the following matrices is Hermitian. Find its eigenvalues and eigenvectors, write a unitary matrix U which diagonalizes \(\mathrm{H}\) by a similarity transformation, and show that \(U^{-1} H U\) is the diagonal matrix of eigenvalues. $$\left(\begin{array}{cc} 2 & i \\ -i & 2 \end{array}\right)$$

Find the eigenvalues and eigenvectors of the following matrices. Do some problems by hand to be sure you understand what the process means. Then check your results by computer. $$\left(\begin{array}{rr} 3 & -2 \\ -2 & 0 \end{array}\right)$$

The following matrix product is used in discussing two thin lenses in air: $$\mathrm{M}=\left(\begin{array}{cc} 1 & -1 / f_{2} \\ 0 & 1 \end{array}\right)\left(\begin{array}{cc} 1 & 0 \\ d & 1 \end{array}\right)\left(\begin{array}{cc} 1 & -1 / f_{1} \\ 0 & 1 \end{array}\right)$$ where \(f_{1}\) and \(f_{2}\) are the focal lengths of the lenses and \(d\) is the distance between them. As in Problem 9, element \(M_{12}\) is \(-1 / f\) where \(f\) is the focal length of the combination. Find \(\mathrm{M},\) det \(\mathrm{M},\) and \(1 / f\).
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