Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the area of the parallelogram with \(\mathbf{a}=-\mathbf{i}+\mathbf{j}-3 \mathbf{k}\) and \(\mathbf{b}=4 \mathbf{i}+2 \mathbf{j}-4 \mathbf{k}\) as the adjacent sides.

Step-by-Step Solution

Answer

The area of the parallelogram is \(2\sqrt{74}\).

Step 1: Recall the Area Formula for a Parallelogram

The area of a parallelogram with adjacent sides \(\mathbf{a}\) and \(\mathbf{b}\) can be found by calculating the magnitude of their cross product, \(\|\mathbf{a} \times \mathbf{b}\|\). We will use this formula to find the area.

Step 2: Compute the Cross Product \(\mathbf{a} \times \mathbf{b}\)

To find \(\mathbf{a} \times \mathbf{b}\), compute the determinant of the following matrix:\[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-1 & 1 & -3 \4 & 2 & -4\end{vmatrix}\]Expanding this determinant along the first row will give us the cross product.

Step 3: Calculate the Determinant

Calculate the determinant by expanding along the first row:\[\mathbf{a} \times \mathbf{b} = \mathbf{i}(1\cdot(-4) - (-3)\cdot2) - \mathbf{j}((-1)\cdot(-4) - (-3)\cdot4) + \mathbf{k}((-1)\cdot2 - 1\cdot4)\]Simplifying each component, we get:\[\mathbf{a} \times \mathbf{b} = \mathbf{i}(-4 + 6) - \mathbf{j}(4 + 12) + \mathbf{k}(-2 - 4)\]\[\mathbf{a} \times \mathbf{b} = 2\mathbf{i} - 16\mathbf{j} - 6\mathbf{k}\]

Step 4: Find the Magnitude of the Cross Product

To find the area, calculate the magnitude of \(\mathbf{a} \times \mathbf{b}\):\[\|\mathbf{a} \times \mathbf{b}\| = \sqrt{(2)^2 + (-16)^2 + (-6)^2}\]\[\|\mathbf{a} \times \mathbf{b}\| = \sqrt{4 + 256 + 36}\]\[\|\mathbf{a} \times \mathbf{b}\| = \sqrt{296}\]

Step 5: Simplify the Magnitude

Review the expression \(\sqrt{296}\) and simplify it if possible. Since \(\sqrt{296} = \sqrt{4 \times 74} = 2\sqrt{74}\), the area of the parallelogram is \(2\sqrt{74}\).

Key Concepts

Cross Product

The cross product is a crucial operation in vector calculus, especially when dealing with three-dimensional space. It involves two vectors, say \(\mathbf{a}\) and \(\mathbf{b}\), and produces another vector that is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\). This perpendicular vector can be beneficial for finding areas and volumes in spatial geometry.

To find the cross product \(\mathbf{a} \times \mathbf{b}\), use the determinant of a matrix composed of unit vectors \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\), and the components of \(\mathbf{a}\) and \(\mathbf{b}\). The produced vector represents a direction perpendicular to the plane formed by the two original vectors.

• Aids in determining orthogonality.
• Provides a way to compute areas such as parallelograms.

It's important to remember that the cross product does not exist for spaces with less than three dimensions, as a minimum of three dimensions is required for a perpendicular vector.

Determinants

Determinants are a fundamental concept when computing the cross product. For a 3x3 matrix formed by vector components and unit vectors, the determinant helps find the new vector's components.

The determinant is calculated by expanding across the components of the row that contains the unit vectors:\[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-1 & 1 & -3 \4 & 2 & -4\end{vmatrix}\]For vector \(\mathbf{i}\), multiply the minor determinant of the submatrix excluding its row and column; do similarly for \(\mathbf{j}\) and \(\mathbf{k}\). Then, sum these calculations according to the positive, negative signs, resulting in a new vector:

• Used in computing cross products.
• Provides the scalar multiplication for finding components of the new vector.

Determinants give a systematic approach to handle vectors algebraically, ensuring clarity in deriving perpendicular vectors.

Vector Magnitude

Vector magnitude is an important concept, particularly when assessing the length or size of vectors. In the context of the cross product, the magnitude reflects the length of the resulting vector.

To find the magnitude of a vector \(\mathbf{v} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}\), use the formula:
\[\|\mathbf{v}\| = \sqrt{(a)^2 + (b)^2 + (c)^2}\]
For optimized understanding, always compute the squares of individual components, sum them, and take the square root of the result. This process provides the Euclidean length of the vector:

• Helps quantify the vector's size.
• Used to compute distances and areas.

Magnitude offers critical insight into spatial dimensions, allowing us to convert vector operations into measurable distances and areas.

Parallelogram Area

The area of a parallelogram formed by two vectors, \(\mathbf{a}\) and \(\mathbf{b}\), is found using the cross product. It's computed as the magnitude of the cross product, \(\|\mathbf{a} \times \mathbf{b}\|\).

A parallelogram's area, in geometric terms, represents how much space it covers on a plane. If the vectors \(\mathbf{a}\) and \(\mathbf{b}\) are adjacent sides of the parallelogram, then the cross product captures the spatial spread between these sides.

• Calculated via cross product magnitude.
• Useful in physics and engineering for solving surface problems.

The cross product’s magnitude gives an immediate measure of area without requiring traditional geometric formulas, making vector calculus a powerful tool in modern mathematical applications.