Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(\mathbf{u}+\mathbf{v}\) is orthogonal to \(\mathbf{u}-\mathbf{v}\), what can you say about the relative magnitudes of \(\mathbf{u}\) and \(\mathbf{v}\) ?

Step-by-Step Solution

Answer

The magnitudes of \( \mathbf{u} \) and \( \mathbf{v} \) are equal.

Step 1: Understand Orthogonality

Vectors \( \mathbf{u}+\mathbf{v} \) and \( \mathbf{u}-\mathbf{v} \) are orthogonal if their dot product is zero. This means we need to evaluate \( (\mathbf{u}+\mathbf{v}) \cdot (\mathbf{u}-\mathbf{v}) = 0 \).

Step 2: Distribute the Dot Product

Compute the dot product \((\mathbf{u}+\mathbf{v}) \cdot (\mathbf{u}-\mathbf{v})\) by distributing: \[ (\mathbf{u}+\mathbf{v}) \cdot (\mathbf{u}-\mathbf{v}) = \mathbf{u} \cdot \mathbf{u} - \mathbf{u} \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{u} - \mathbf{v} \cdot \mathbf{v}. \]

Step 3: Simplify the Dot Product

Using properties of dot products, we know \( \mathbf{u} \cdot \mathbf{u} = ||\mathbf{u}||^2 \) and \( \mathbf{v} \cdot \mathbf{v} = ||\mathbf{v}||^2 \), and \( \mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u} \). Therefore, the expression simplifies to: \[ ||\mathbf{u}||^2 - ||\mathbf{v}||^2 = 0. \]

Step 4: Find the Relation Between Magnitudes

Since the simplified dot product expression is \( ||\mathbf{u}||^2 - ||\mathbf{v}||^2 = 0 \), this implies that \( ||\mathbf{u}||^2 = ||\mathbf{v}||^2 \). Taking the square root of both sides, we conclude that \( ||\mathbf{u}|| = ||\mathbf{v}|| \).

Step 5: Conclusion

The magnitudes of vectors \( \mathbf{u} \) and \( \mathbf{v} \) are equal. Therefore, we can say that \( ||\mathbf{u}|| = ||\mathbf{v}|| \).

Key Concepts

Dot Product

The dot product is a fundamental operation in vector algebra that helps us understand angles and projections between two vectors. The dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is calculated as \( \mathbf{a} \cdot \mathbf{b} = ||\mathbf{a}|| ||\mathbf{b}|| \cos(\theta) \), where \( \theta \) is the angle between the vectors.
This calculation provides a scalar value. If the dot product is zero, the vectors are orthogonal, meaning they meet at a right angle. This orthogonality condition is a useful property to figure out relationships like in the original exercise.
In the given problem, we assess the orthogonality of \( \mathbf{u}+\mathbf{v} \) and \( \mathbf{u}-\mathbf{v} \) by evaluating their dot product. It leads us to insights on vector properties and their respective magnitudes.

Vector Magnitudes

Understanding vector magnitudes is crucial when comparing vectors. The magnitude of a vector \( \mathbf{a} \), denoted as \( ||\mathbf{a}|| \), represents its length or size. It is computed as \( ||\mathbf{a}|| = \sqrt{a_1^2 + a_2^2 + a_3^2} \) for a 3D vector, with similar formulas for other dimensions.
In vector algebra, vector magnitudes help establish equality or relations between vectors. In the exercise, the magnitude of \( \mathbf{u} \) and \( \mathbf{v} \) are found to be equal due to the orthogonality condition: \( ||\mathbf{u}||^2 = ||\mathbf{v}||^2 \).
This relationship is pivotal for understanding how vectors are positioned relative to each other in space, leading to the conclusion that equal magnitudes denote equally 'strong' vectors in terms of their length.

Vector Properties

Vectors have several properties that are useful in mathematical computations and understanding geometric relations. Important properties include:

• Commutativity: The order of vectors in a dot product does not affect the result, i.e., \( \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a} \).
• Distributivity: The dot product distributes over vector addition, allowing expressions like \( (\mathbf{a} + \mathbf{b}) \cdot \mathbf{c} = \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c} \).
• Orthogonality: When two vectors are orthogonal, their dot product is zero, indicating a perpendicular relationship.

Applying these properties facilitates the solving of problems, such as the question of orthogonality in the original exercise. These properties also help in resolving vectors into components, calculating angles, and determining projections, which are key operations in vector analysis.