Problem 12 - Essential Calculus Early Transcendentals

Question

Find the vector, not with determinants, but by using properties of cross products $$(\mathbf{i}+\mathbf{j}) \times(\mathbf{i}-\mathbf{j})$$

Step-by-Step Solution

Answer

The vector is \(-2 \mathbf{k}\).

Step 1: Identify Components of Vectors

The given vectors are \(\mathbf{A} = \mathbf{i} + \mathbf{j}\) and \(\mathbf{B} = \mathbf{i} - \mathbf{j}\). Each vector is written in terms of unit vectors \(\mathbf{i}\) and \(\mathbf{j}\).

Step 2: Apply the Distributive Property

Cross products adhere to the distributive property over addition. Therefore, the expression \((\mathbf{A} \times \mathbf{B})\) becomes \[(\mathbf{i} + \mathbf{j}) \times (\mathbf{i} - \mathbf{j}) = \mathbf{i} \times \mathbf{i} - \mathbf{i} \times \mathbf{j} + \mathbf{j} \times \mathbf{i} - \mathbf{j} \times \mathbf{j}.\]

Step 3: Use Properties of Cross Products

There are key properties of unit vectors to consider: \(\mathbf{i} \times \mathbf{i} = \mathbf{j} \times \mathbf{j} = \mathbf{0}\) and \(\mathbf{i} \times \mathbf{j} = \mathbf{k}\), \(\mathbf{j} \times \mathbf{i} = -\mathbf{k}\). Apply these properties: 1. \(\mathbf{i} \times \mathbf{i} = \mathbf{0}\).2. \(- \mathbf{i} \times \mathbf{j} = -\mathbf{k}\).3. \(\mathbf{j} \times \mathbf{i} = -\mathbf{k}\).4. \(\mathbf{j} \times \mathbf{j} = \mathbf{0}\).

Step 4: Simplify the Expression

Continue simplifying by substituting the results from Step 3:- \(\mathbf{0} - \mathbf{k} + (-\mathbf{k}) + \mathbf{0} = -\mathbf{k} - \mathbf{k} = -2 \mathbf{k}\). Therefore, the cross product \((\mathbf{i} + \mathbf{j}) \times (\mathbf{i} - \mathbf{j})\) simplifies to \(-2 \mathbf{k}\).

Key Concepts

Vector Algebra

Vector algebra is a branch of mathematics that deals with quantities that have both magnitude and direction. We often represent vectors in terms of their components along the axes of a coordinate system. For example, vectors in a 3D space are typically expressed using unit vectors such as \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\), which denote the unit vectors along the x, y, and z axes, respectively.
To perform operations like addition, subtraction, and cross product within vector algebra, we treat vectors as entities that can be manipulated according to geometric and algebraic rules. When dealing with cross products, it's important to note that unlike scalar multiplication, the cross product results in another vector.
Thus, the cross product encapsulates not only a magnitude but also a direction, perpendicular to both vectors involved. This is why in the exercise we aimed to find the cross product of \((\mathbf{i} + \mathbf{j})\) and \((\mathbf{i} - \mathbf{j})\). In this context, understanding how to decompose and reassemble vectors using algebraic methods is crucial.

Distributive Property

The distributive property is a fundamental principle in mathematics that also applies to vector operations, including cross products. It states that distributing a mathematical operation over a sum or difference is permissible. This means \((\mathbf{a} + \mathbf{b}) \times \mathbf{c}\) can be expanded to \(\mathbf{a} \times \mathbf{c} + \mathbf{b} \times \mathbf{c}\).
This property helps simplify complex vector expressions, as seen in the given exercise. By applying the distributive property, the expression \((\mathbf{i} + \mathbf{j}) \times (\mathbf{i} - \mathbf{j})\) can be broken apart into simpler, manageable pieces:

• \(\mathbf{i} \times \mathbf{i}\)
• - \(\mathbf{i} \times \mathbf{j}\)
• + \(\mathbf{j} \times \mathbf{i}\)
• - \(\mathbf{j} \times \mathbf{j}\)

Distributing cross products over addition and subtraction allows us to systematically tackle the problem, simplifying each component based on known properties of unit vectors.

Unit Vectors

Unit vectors are vectors with a magnitude of one, primarily used to specify directions. In three-dimensional space, the main unit vectors are \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\). Each represents directionality and serves as the foundational components for constructing more complex vectors.
One of the important properties of unit vectors is how they interact in cross product operations. Specifically:

• The cross product \(\mathbf{i} \times \mathbf{j} = \mathbf{k}\)
• The reverse order, \(\mathbf{j} \times \mathbf{i} = -\mathbf{k}\)
• \(\mathbf{i} \times \mathbf{i} = \mathbf{j} \times \mathbf{j} = \mathbf{k} \times \mathbf{k} = \mathbf{0}\)

These properties facilitate solving cross product problems efficiently, as they allow simplification by zeroing out the cross products where unit vectors align. It's this interaction that led us to easily simplify the exercise's equation to \(-2 \mathbf{k}\), demonstrating the elegant symmetry and utility of unit vectors in vector algebra.