Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the volume of the parallelepiped with edges \(3 \mathbf{i}-4 \mathbf{j}+2 \mathbf{k},-\mathbf{i}+2 \mathbf{j}+\mathbf{k}\), and \(3 \mathbf{i}-2 \mathbf{j}+5 \mathbf{k}\).

Step-by-Step Solution

Answer

The volume of the parallelepiped is 12.

Step 1: Understand the Vectors

The problem provides three vectors which are the edges of the parallelepiped: \ \( \mathbf{a} = 3 \mathbf{i} - 4 \mathbf{j} + 2 \mathbf{k} \), \ \( \mathbf{b} = -\mathbf{i} + 2 \mathbf{j} + \mathbf{k} \), \ \( \mathbf{c} = 3 \mathbf{i} - 2 \mathbf{j} + 5 \mathbf{k} \).

Step 2: Set Up the Triple Scalar Product

The volume of the parallelepiped can be found using the triple scalar product of the vectors \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \), which is given by \( |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| \).

Step 3: Compute the Cross Product \( \mathbf{b} \times \mathbf{c} \)

Use the determinant formula for the cross product: \ \[ \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & 2 & 1 \ 3 & -2 & 5 \end{vmatrix} \] \ This equals \( \mathbf{i}(2 \times 5 - 1 \times -2) - \mathbf{j}(-1 \times 5 - 1 \times 3) + \mathbf{k}(-1 \times -2 - 2 \times 3) \), which simplifies to \( 12 \mathbf{i} + 8 \mathbf{j} - 8 \mathbf{k} \).

Step 4: Compute the Dot Product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \)

Calculate the dot product: \ \( \mathbf{a} \cdot (12 \mathbf{i} + 8 \mathbf{j} - 8 \mathbf{k}) = 3 \cdot 12 + (-4) \cdot 8 + 2 \cdot (-8) \) \ which simplifies to \( 36 - 32 - 16 \), resulting in \( -12 \).

Step 5: Determine the Volume

The volume of the parallelepiped is the absolute value of the dot product computed: \ \( | -12 | = 12 \).

Key Concepts

Vector Cross Product

The vector cross product is a fundamental operation in linear algebra, particularly when dealing with three-dimensional vectors. It is used to find a third vector that is perpendicular to two given vectors. The cross product of vectors \( \mathbf{b} \) and \( \mathbf{c} \) is denoted as \( \mathbf{b} \times \mathbf{c} \). To compute this, we use a formula involving the determinant of a matrix:

• The first row of the matrix contains the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \).
• The second and third rows are formed by the components of the vectors \( \mathbf{b} \) and \( \mathbf{c} \), respectively.

By solving this determinant, we can find the components of the resulting vector, which are computed sequentially. In our example, the result of \( \mathbf{b} \times \mathbf{c} \) was \( 12 \mathbf{i} + 8 \mathbf{j} - 8 \mathbf{k} \). This vector is crucial for finding the volume of a parallelepiped as it later forms part of the triple scalar product.

Triple Scalar Product

The triple scalar product is an important tool in linear algebra used to calculate the volume of a parallelepiped in three-dimensional space. This product combines the cross product and the dot product in a specific order. Given three vectors \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \), the triple scalar product is formulated as \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \). Here's how it works:

• The cross product \( \mathbf{b} \times \mathbf{c} \) gives a vector that is orthogonal to both \( \mathbf{b} \) and \( \mathbf{c} \).
• The subsequent dot product with \( \mathbf{a} \) calculates a scalar value, which physically represents the volume of the parallelepiped formed by these three vectors.

The result's magnitude gives the parallelepiped's volume, meaning we take the absolute value to ensure a positive measurement. In our solution, the triple scalar product calculated to \( -12 \), and its absolute value, \( | -12 | \), yields a volume of \( 12 \).

Dot Product

The dot product is an operation that takes two equal-length sequences of numbers (usually vectors in linear algebra) and returns a single number. It's crucial for various calculations, including determining projections and angles between vectors. For vectors \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \), the dot product is computed as:\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\]This straightforward operation summarizes how much one vector extends in the direction of another.In the context of calculating volumes, after obtaining \( \mathbf{b} \times \mathbf{c} \), the result is a vector, and taking its dot product with \( \mathbf{a} \) gives a scalar. This scalar can represent the volume when combined appropriately with cross product operations. For our specific exercise, we arrived at the dot product value of \(-12\).

Linear Algebra

Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. At its core are operations such as vector addition, scalar multiplication, and products like dot and cross products, which are essential in higher-dimensional analysis. Applications of linear algebra extend across various fields:

• Computer Science: Algorithms, graphics, and data structures are underpinned by linear algebra.
• Physics: Physics uses vector calculations for forces, moments, and fields.
• Engineering: System designs and optimizations rely heavily on matrix operations.

Understanding vector operations, as seen in our parallelepiped problem, showcases linear algebra’s ability to solve complex spatial queries with ease. It combines theoretical mathematics with practical applications, giving robust tools for accurate and efficient problem solving.