Problem 42 - Algebra and Trigonometry

Question

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 given property is proved: \((\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = |\mathbf{u}|^2 - |\mathbf{v}|^2\).

Step 1: Expand the Left Side

We start by expanding the expression \((\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v})\). Apply the distributive property \((\mathbf{a} + \mathbf{b}) \cdot \mathbf{c} = \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c}\). Thus:\[(\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = (\mathbf{u} \cdot \mathbf{u} + \mathbf{u} \cdot \mathbf{v}) - (\mathbf{v} \cdot \mathbf{u} + \mathbf{v} \cdot \mathbf{v})\].

Step 2: Simplify the Expression

Now, simplify the expanded expression. Note that the dot product is commutative, meaning \(\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}\).Thus, these terms cancel out: \(\mathbf{u} \cdot \mathbf{v} - \mathbf{v} \cdot \mathbf{u} = 0\). We are left with:\[\mathbf{u} \cdot \mathbf{u} - \mathbf{v} \cdot \mathbf{v}\].

Step 3: Express in Norm Form

Recognize that \(\mathbf{u} \cdot \mathbf{u}\) can be written as \(|\mathbf{u}|^2\) and \(\mathbf{v} \cdot \mathbf{v}\) can be written as \(|\mathbf{v}|^2\).Thus, the simplified equation becomes: \[|\mathbf{u}|^2 - |\mathbf{v}|^2\].

Step 4: Conclude the Proof

Since the simplified form matches the given property, our result \((\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = |\mathbf{u}|^2 - |\mathbf{v}|^2\) proves the property is true.

Key Concepts

Dot Product

The dot product, often simply called "dot," is a fundamental operation in vector algebra. It involves multiplying two vectors and results in a scalar. The dot product can be mathematically expressed as follows:
\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\]
for vectors \(\mathbf{a} = (a_1, a_2, a_3)\) and \(\mathbf{b} = (b_1, b_2, b_3)\).
This operation highlights several important properties:

• Commutative: The order of multiplication does not matter. Thus, \(\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}\).
• Distributive: The dot product distributes over vector addition: \((\mathbf{a} + \mathbf{b}) \cdot \mathbf{c} = \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c}\).
• Scalar Multiplication: If a vector is multiplied by a scalar, the dot product can be factored: \((k\mathbf{a}) \cdot \mathbf{b} = k(\mathbf{a} \cdot \mathbf{b})\).

The use of the dot product is central to many vector operations, including the calculation of vector magnitudes.

Distributive Property

The distributive property in vector algebra makes it possible to simplify expressions where a vector or a scalar interacts with several terms. This property is vital in proving vector identities and simplifying complex vector expressions.
For vectors \(\mathbf{a}, \mathbf{b},\) and \(\mathbf{c}\), the distributive property can be expressed as:
\[(\mathbf{a} + \mathbf{b}) \cdot \mathbf{c} = \mathbf{a} \cdot \mathbf{c} + \mathbf{b} \cdot \mathbf{c}\]
This property essentially means that you can "distribute" the vector across the sum of other vectors. This action allows you to break down and easily simplify expressions in vector calculations. In the original exercise, the distributive property was utilized to expand the term \((\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v})\) to help prove the given equation. Understanding this property is a cornerstone to mastering vector operations.

Vector Magnitude

The magnitude of a vector, often considered as its length, is a crucial concept in vector algebra. It is expressed as the square root of the sum of the squares of its components. For a vector \(\mathbf{a} = (a_1, a_2, a_3)\), the magnitude \(|\mathbf{a}|\) is given by:
\[|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}\]
The magnitude of a vector reveals its size without regard to its direction.

• A vector's dot product with itself equals the square of its magnitude: \(\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2\). This concept was used in the exercise to transform dot products into magnitudes \(|\mathbf{u}|^2\) and \(|\mathbf{v}|^2\).
• Magnitude is always non-negative and provides a way to compare the lengths of different vectors.

Learning to calculate and work with magnitudes helps in evaluating vector sizes and understanding their implications in various physical contexts.

Vector Operations

Vector operations include addition, subtraction, and multiplication involving scalars or other vectors. These operations are foundational to working with vectors in any mathematical framework. Let's delve into each:

• Vector Addition: To add vectors \(\mathbf{a} = (a_1, a_2, a_3)\) and \(\mathbf{b} = (b_1, b_2, b_3)\), add each corresponding component:
  \[\mathbf{a} + \mathbf{b} = (a_1 + b_1, a_2 + b_2, a_3 + b_3)\]
• Vector Subtraction: Similar to addition, but involves subtracting components:
  \[\mathbf{a} - \mathbf{b} = (a_1 - b_1, a_2 - b_2, a_3 - b_3)\]
• Scalar Multiplication: Multiplying a vector by a scalar changes its magnitude but not its direction (unless the scalar is negative):
  \[k\mathbf{a} = (ka_1, ka_2, ka_3)\]

These operations are not only essential in theoretical problems but also in practical applications like physics and engineering.
In solving vector equations or proving vector identities, as seen in the exercise, an understanding of these operations is vital.