Problem 42 - Precalculus Mathematics for Calculus

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 property is proven by expanding the dot product and simplifying.

Step 1: Expand the Dot Product Expression

The expression \((\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v})\) is a dot product between two vectors. Using the distributive property of dot products, expand it as follows:\[(\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = \mathbf{u} \cdot \mathbf{u} + \mathbf{u} \cdot \mathbf{v} - \mathbf{v} \cdot \mathbf{u} - \mathbf{v} \cdot \mathbf{v}\]

Step 2: Simplify Using Dot Product Properties

Apply properties of the dot product: \(\mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u}\). This simplifies the expanded expression:\[\mathbf{u} \cdot \mathbf{u} + \mathbf{u} \cdot \mathbf{v} - \mathbf{v} \cdot \mathbf{u} - \mathbf{v} \cdot \mathbf{v} = \mathbf{u} \cdot \mathbf{u} - \mathbf{v} \cdot \mathbf{v}\]

Step 3: Express in Terms of Magnitudes

The terms \(\mathbf{u} \cdot \mathbf{u}\) and \(\mathbf{v} \cdot \mathbf{v}\) represent the squares of the magnitudes of \(\mathbf{u}\) and \(\mathbf{v}\) respectively. Thus, we have:\[(\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = |\mathbf{u}|^2 - |\mathbf{v}|^2\]

Step 4: Conclusion

By utilizing the distributive property and the symmetry of the dot product, we have successfully shown that the original equation simplifies to the desired form:\[(\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = |\mathbf{u}|^2 - |\mathbf{v}|^2\]

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector algebra. It combines two vectors and results in a single scalar value. The dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is calculated using their components and the cosine of the angle between them:\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos(\theta) \]Where \( |\mathbf{a}| \) and \( |\mathbf{b}| \) are the magnitudes of the vectors, and \( \theta \) is the angle between them. Instead of angles, you can break it down into components as well.
If \( \mathbf{a} = [a_1, a_2, a_3] \) and \( \mathbf{b} = [b_1, b_2, b_3] \), then the dot product can be expressed by:

• \( \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \)

Understanding this is crucial since it simplifies calculations involving angles and magnitudes simultaneously.

Vector Magnitudes

Vector magnitude, often referred to as the length of a vector, is a measure of how long the vector is. It's the distance from the origin to the point represented by the vector in space.
For a vector \( \mathbf{v} = [v_1, v_2, v_3] \), the magnitude is calculated as:
\[ |\mathbf{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]
This formula arises from the Pythagorean theorem applied in three dimensions, representing the hypotenuse of the triangle formed by the vector components.

• Magnitude is always a non-negative number.
• If the magnitude is zero, it indicates the vector is a zero vector.

Understanding vector magnitudes allows for clearer calculations in operations such as normalization and dot products, as seen by the terms \( \mathbf{u} \cdot \mathbf{u} = |\mathbf{u}|^2 \) in our solution, showing how these relate to squared magnitudes.

Distributive Property of Dot Product

The distributive property is one of the key properties of the dot product, making it highly useful in simplifying calculations involving vectors. It states that for any vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \), the following is true:
\[ \mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} \]
In the original exercise, the distributive property is employed to expand the expression \((\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v})\).

• This allows breaking down complex vector expressions into simpler components.
• The property simplistically rearranges terms for straightforward solutions.

The symmetry of the dot product, where \( \mathbf{u} \cdot \mathbf{v} = \mathbf{v} \cdot \mathbf{u} \), complements the distributive property, which was instrumental in reaching the solution by resolving intermediate terms to arrive at the squared magnitudes difference, \( |\mathbf{u}|^2 - |\mathbf{v}|^2 \).