Problem 8 - Calculus: Early Transcendentals

Question

Find and explain the value of \((\boldsymbol{i} \times \boldsymbol{j}) \times \boldsymbol{k}\) and \((\boldsymbol{i}+\boldsymbol{j}) \times(\boldsymbol{i}-\boldsymbol{j}) .\)

Step-by-Step Solution

Answer

\((\boldsymbol{i} \times \boldsymbol{j}) \times \boldsymbol{k} = \boldsymbol{0}\); \((\boldsymbol{i}+\boldsymbol{j}) \times (\boldsymbol{i}-\boldsymbol{j}) = -2\boldsymbol{k}\).

Step 1: Understand the Cross Product

The cross product of two vectors \( \boldsymbol{a} \) and \( \boldsymbol{b} \) is denoted by \( \boldsymbol{a} \times \boldsymbol{b} \), resulting in a vector perpendicular to both \( \boldsymbol{a} \) and \( \boldsymbol{b} \). For unit vectors \( \boldsymbol{i} \), \( \boldsymbol{j} \), and \( \boldsymbol{k} \), we have: \( \boldsymbol{i} \times \boldsymbol{j} = \boldsymbol{k} \), \( \boldsymbol{j} \times \boldsymbol{k} = \boldsymbol{i} \), and \( \boldsymbol{k} \times \boldsymbol{i} = \boldsymbol{j} \). Additionally, \( \boldsymbol{i} \times \boldsymbol{i} = \boldsymbol{j} \times \boldsymbol{j} = \boldsymbol{k} \times \boldsymbol{k} = \boldsymbol{0} \).

Step 2: Calculate \((\boldsymbol{i} \times \boldsymbol{j}) \times \boldsymbol{k}\)

First, calculate \( \boldsymbol{i} \times \boldsymbol{j} \), which is \( \boldsymbol{k} \). Now compute \( \boldsymbol{k} \times \boldsymbol{k} \). Since any vector cross product with itself is the zero vector, \( \boldsymbol{k} \times \boldsymbol{k} = \boldsymbol{0} \). Therefore, the expression \((\boldsymbol{i} \times \boldsymbol{j}) \times \boldsymbol{k}\) simplifies to \( \boldsymbol{0} \).

Step 3: Expand Expression \((\boldsymbol{i}+\boldsymbol{j}) \times (\boldsymbol{i}-\boldsymbol{j})\)

Use the distributive property of the cross product, \((\boldsymbol{a} + \boldsymbol{b}) \times \boldsymbol{c} = \boldsymbol{a} \times \boldsymbol{c} + \boldsymbol{b} \times \boldsymbol{c}\), to expand:\( (\boldsymbol{i} + \boldsymbol{j}) \times (\boldsymbol{i} - \boldsymbol{j}) = \boldsymbol{i} \times \boldsymbol{i} - \boldsymbol{i} \times \boldsymbol{j} + \boldsymbol{j} \times \boldsymbol{i} - \boldsymbol{j} \times \boldsymbol{j} \).

Step 4: Simplify Each Term

Since \( \boldsymbol{i} \times \boldsymbol{i} = \boldsymbol{0} \) and \( \boldsymbol{j} \times \boldsymbol{j} = \boldsymbol{0} \), these terms vanish. For \( \boldsymbol{i} \times \boldsymbol{j} = \boldsymbol{k} \) and \( \boldsymbol{j} \times \boldsymbol{i} = -\boldsymbol{k} \). Thus, the expression becomes:\( \boldsymbol{0} - \boldsymbol{k} + (-\boldsymbol{k}) + \boldsymbol{0} = -2\boldsymbol{k} \).

Key Concepts

Vector Operations

Vector operations are fundamental in mathematics and physics. They involve mathematical procedures applied to vectors, which are quantities having both magnitude and direction. Two primary vector operations are the dot product (scalar product) and the cross product (vector product). Let's focus on the cross product here, as it's vital for understanding the given exercise.

The cross product of two vectors, \( \boldsymbol{a} \) and \( \boldsymbol{b} \), denoted as \( \boldsymbol{a} \times \boldsymbol{b} \), results in a third vector that is perpendicular to both \( \boldsymbol{a} \) and \( \boldsymbol{b} \). This operation is unique to three-dimensional space. When dealing with standard unit vectors \( \boldsymbol{i}, \boldsymbol{j}, \boldsymbol{k} \) (the basic unit vectors along x, y, z axes respectively), the cross products can be remembered as:

• \( \boldsymbol{i} \times \boldsymbol{j} = \boldsymbol{k} \)
• \( \boldsymbol{j} \times \boldsymbol{k} = \boldsymbol{i} \)
• \( \boldsymbol{k} \times \boldsymbol{i} = \boldsymbol{j} \)
• \( \boldsymbol{i} \times \boldsymbol{i} = \boldsymbol{j} \times \boldsymbol{j} = \boldsymbol{k} \times \boldsymbol{k} = \boldsymbol{0} \) (a vector crossing with itself gives the zero vector)

Understanding these properties helps simplify more complex vector expressions.

Unit Vectors

Unit vectors play a crucial role in simplifying vector calculations. They are vectors with a magnitude of one and serve as the building blocks for any vector space.

In three-dimensional Cartesian coordinates, the standard unit vectors \( \boldsymbol{i}, \boldsymbol{j}, \boldsymbol{k} \) define the directions along the x, y, and z axes. Any vector in this space can be expressed as a combination of these unit vectors with appropriate coefficients.

These unit vectors are especially handy because their cross products are straightforward and follow specific rules. For example, \( \boldsymbol{i} \times \boldsymbol{j} \) results in \( \boldsymbol{k} \), reflecting a 90-degree rotation around the coordinate system. Such operations allow us to navigate more complex vector interactions easily.

Unit vectors hold the property of being orthogonal, meaning any two different unit vectors such as \( \boldsymbol{i} \) and \( \boldsymbol{j} \) are perpendicular to each other. This assists in understanding and simplifying vector operations, making unit vectors a key concept in vector mathematics.

Distributive Property of Vector Products

The distributive property is a critical tool when working with vector operations, particularly the cross product. This property allows us to break down and simplify expressions involving multiple vectors.

Mathematically, for any vectors \( \boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c} \), it holds that:

• \((\boldsymbol{a} + \boldsymbol{b}) \times \boldsymbol{c} = \boldsymbol{a} \times \boldsymbol{c} + \boldsymbol{b} \times \boldsymbol{c}\)

This principle applies to the exercise where we have the product \((\boldsymbol{i} + \boldsymbol{j}) \times (\boldsymbol{i} - \boldsymbol{j})\). By using the distributive property, the expression expands to multiple individual cross products, which can then be evaluated using known unit vector properties.

Each of these cross products involves simple calculations, such as \( \boldsymbol{i} \times \boldsymbol{i} = \boldsymbol{0} \). Applying the distributive property simplifies complex vector operations, facilitating the computation of cross products in a more streamlined manner.