Problem 36 - Algebra and Trigonometry

Question

33-36 Let \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) be vectors and let \(a\) be a scalar. Prove the given property. $$(\mathbf{u}-\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}-|\mathbf{v}|^{2}$$

Step-by-Step Solution

Answer

The expression simplifies to \(|\mathbf{u}|^2 - |\mathbf{v}|^2\), proving the property.

Step 1: Expand the Dot Product

First, expand the left-hand side \((\mathbf{u}-\mathbf{v}) \cdot (\mathbf{u}+\mathbf{v})\). The distributive property of the dot product allows us to write it as \\( (\mathbf{u} \cdot \mathbf{u}) + (\mathbf{u} \cdot \mathbf{v}) - (\mathbf{v} \cdot \mathbf{u}) - (\mathbf{v} \cdot \mathbf{v})\).

Step 2: Simplify Using Properties of Dot Product

Recall the commutative property of the dot product which states that \( \mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u} \). Use this to show that \( (\mathbf{u} \cdot \mathbf{v}) - (\mathbf{v} \cdot \mathbf{u}) = 0\), which simplifies the expression to \\(\mathbf{u} \cdot \mathbf{u} - \mathbf{v} \cdot \mathbf{v}\).

Step 3: Interpret the Dot Product as Magnitude Squared

Recognize that \( \mathbf{u} \cdot \mathbf{u} \) is \(|\mathbf{u}|^2\), which is the square of the magnitude of \(\mathbf{u}\), and \\( \mathbf{v} \cdot \mathbf{v} \) is \(|\mathbf{v}|^2\), the square of the magnitude of \(\mathbf{v}\). Substitute these into the expression.

Step 4: Conclude the Proof

Given our simplifications, we have \\(\mathbf{u} \cdot \mathbf{u} - \mathbf{v} \cdot \mathbf{v} = |\mathbf{u}|^2 - |\mathbf{v}|^2\). Thus, the original equation \\((\mathbf{u}-\mathbf{v}) \cdot (\mathbf{u}+\mathbf{v}) = |\mathbf{u}|^2 - |\mathbf{v}|^2\) is proved.

Key Concepts

Dot Product

The dot product is a fundamental operation in vector algebra. It takes two vectors and returns a scalar. This operation is computed by multiplying corresponding components of the vectors and summing those products. For vectors \(\mathbf{a} = (a_1, a_2, a_3)\) and \(\mathbf{b} = (b_1, b_2, b_3)\), the dot product \(\mathbf{a} \cdot \mathbf{b}\) is given by:

• \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\)

This operation is also expressed using angle \(\theta\) between the vectors, as \(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta\).
This highlights how the dot product relates to the magnitude and direction of the vectors.

Understanding dot product is crucial for identifying the orthogonality of vectors. If the dot product is zero, the vectors are orthogonal, meaning they are perpendicular to each other.

In our exercise, we utilized the distributive property to expand the expression involving dot products, which shows how the operation interacts in multi-term expressions.

Vector Properties

Vectors have several important properties that assist in calculations and proofs. These properties include:

• Commutative Property: The dot product of vectors is commutative, meaning \(\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}\).
• Distributive Property: Dot product distributes over vector addition, so \((\mathbf{a} + \mathbf{b}) \cdot \mathbf{c} = \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c}\).
• Scalar Multiplication: Multiplying a vector by a scalar scales its magnitude: \(c(\mathbf{a} \cdot \mathbf{b}) = (c\mathbf{a}) \cdot \mathbf{b} = \mathbf{a} \cdot (c\mathbf{b})\).

In our step-by-step solution, we utilized these properties, especially the commutative property, to simplify the expression.
This facilitated showing that certain terms cancel out, thus leading to a simpler result. By understanding these properties, the manipulation of vector expressions becomes more straightforward.

Vectors aren't just about magnitude and direction but also how they interact in a mathematical framework.

Magnitude of a Vector

The magnitude of a vector is a measure of its length. For a vector \(\mathbf{a} = (a_1, a_2, a_3)\), its magnitude \(|\mathbf{a}|\) is calculated as:

• \(|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}\)

This metric is essential in understanding the size of vectors without regard to direction.
The magnitude can be squared, in which case it eliminates the square root, providing \(|\mathbf{a}|^2 = a_1^2 + a_2^2 + a_3^2\).

In our solution, we noticed that the dot product of a vector with itself results in the squared magnitude, i.e., \(\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2\).
This identification is key in simplifying and proving expressions related to vectors, by translating the algebraic expression into more intuitive geometric terms.
The ability to express the length of vectors through magnitude improves your fluency in understanding vector-related problems and their solutions.