Problem 37 - Thomas Calculus

Question

Find the areas of the parallelograms whose vertices are given in Exercises \(35-40 .\) $$ A(-1,2), \quad B(2,0), \quad C(7,1), \quad D(4,3) $$

Step-by-Step Solution

Answer

The area of the parallelogram is 13 square units.

Step 1: Identify matrix components

To find the area of a parallelogram with vertices \(A(-1,2)\), \(B(2,0)\), \(C(7,1)\), and \(D(4,3)\), we first select two vectors originating from a common vertex. Using \(A\) as the vertex, calculate the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AD} \). The vectors are given by: \( \overrightarrow{AB} = B - A = (2 + 1, 0 - 2) = (3, -2) \) and \( \overrightarrow{AD} = D - A = (4 + 1, 3 - 2) = (5, 1) \).

Step 2: Set up the matrix for the area

The area of the parallelogram can be obtained by calculating the magnitude of the cross product (or the determinant of a 2x2 matrix) of the two vectors. The matrix is set up with vectors as: \[\begin{bmatrix}3 & -2 \5 & 1\end{bmatrix}\]

Step 3: Calculate the determinant

Calculate the determinant of the 2x2 matrix to find the area of the parallelogram. The determinant is:\[ |3 imes 1 - (-2) imes 5| = |3 + 10| = 13\]

Step 4: Compute the absolute value

The absolute value of the determinant gives the area of the parallelogram. Therefore, the area of the parallelogram is \(13\) square units.

Key Concepts

Vectors

Vectors are mathematical entities that allow us to describe quantities that have both a magnitude and a direction. This is different from scalars, which only have magnitude. Vectors are commonly represented as arrows in a coordinate system, where the arrow's direction corresponds to the vector's direction, and its length corresponds to the vector's magnitude.

In the context of finding the area of a parallelogram, vectors help in translating the geometric shape into mathematical expressions. For example, in the given exercise, the vertices form two vectors if we choose one point as a reference. In this case, vectors \[\overrightarrow{AB}\]\ and \[\overrightarrow{AD}\]\ are found by subtracting the coordinates of the common point from the other points.

To get \[\overrightarrow{AB}\],

• Start at A and move towards B.
• Subtract \((x_A, y_A)\) from \((x_B, y_B)\) to get \((3, -2)\).

The same is done to get \[\overrightarrow{AD}\]\ using D,

• Resulting in the vector \((5, 1)\).

Working with vectors lets us calculate the properties of the parallelogram, such as area, using mathematical operations like cross products.

Cross Product

The cross product is a mathematical operation that takes two vectors and produces another vector that is perpendicular to the plane containing the initial vectors. This new vector's magnitude is proportional to the area of the parallelogram spanned by the original vectors, making it a very handy tool in geometry for finding areas in three-dimensional space.

However, when dealing with two dimensions as in our exercise, the magnitude of the cross product is equivalent to the determinant of a 2x2 matrix formed by the original vectors. For vectors \[\overrightarrow{AB} = (3, -2)\]\ and \[\overrightarrow{AD} = (5, 1)\]\, the matrix is set up as:

• Rows correspond to each vector, forming the matrix \[\begin{bmatrix}3 & -2 \5 & 1\end{bmatrix}\].

The task is to determine the area by calculating the cross product's magnitude, which directly gives you the parallelogram's area.

• This is done by evaluating the determinant of the matrix.

Determinant

The determinant is a scalar value that can be computed from the elements of a square matrix. It provides important properties about the matrix, such as the volume scaling factor in transformations represented by the matrix, and it also helps in solving systems of linear equations.

For a 2x2 matrix, the determinant is calculated using a simple formula: suppose you have a matrix \[[a & b \ c & d]\], the determinant, denoted as \|A\|, is calculated as:

• \(ad - bc\)

In the exercise, the matrix \[\begin{bmatrix}3 & -2 \5 & 1\end{bmatrix}\]\, results in a determinant calculated by:

• \(3 \times 1 - (-2) \times 5\)
• This equals \(3 + 10\)
• So, the area is \(13\) square units.

The absolute value of the determinant gives the necessary area, ensuring a non-negative result. This is crucial when determining geometrical areas, as they cannot be negative even if the vectors are defined in a direction that would initially suggest such.