Question

Let \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle\), and \(\mathbf{w}=\) \(\left\langle w_{1}, w_{2}, w_{3}\right\rangle\). Let \(c\) be a scalar. Prove the following vector properties. \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\)

Short answer

Question: Prove the Cauchy-Schwarz inequality for three-dimensional vectors: \(|\mathbf{u} \cdot \mathbf{v}| \leq |\mathbf{u}||\mathbf{v}|\). Answer: To prove the Cauchy-Schwarz inequality, we defined a non-negative quantity \(S\), expanded and rearranged the terms, and showed that \((u_{1}^2 + u_{2}^2 + u_{3}^2)(v_{1}^2 + v_{2}^2 + v_{3}^2) \geq (u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3})^2\). Taking the square root of both sides, we obtained the inequality \(|\mathbf{u}||\mathbf{v}| \geq |\mathbf{u}\cdot\mathbf{v}|\), proving the Cauchy-Schwarz inequality.

Step-by-step solution

Define the dot product

The dot product of two vectors \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) and \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle\) is defined as \(\mathbf{u}\cdot\mathbf{v} = u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3}\).

Express the inequality in terms of dot products

We need to prove the inequality \(|\mathbf{u} \cdot \mathbf{v}| \leq |\mathbf{u}||\mathbf{v}|\). Using the definition of dot product and the property that the dot product of a vector with itself is equal to the square of its magnitude, we have: \(|u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3}| \leq \sqrt{(u_{1}^2 + u_{2}^2 + u_{3}^2)(v_{1}^2 + v_{2}^2 + v_{3}^2)}\)

Define a non-negative quantity related to the inequality

Let's define a new non-negative quantity \(S\), such that: \(S = (u_{1}v_{2} - u_{2}v_{1})^2 + (u_{1}v_{3} - u_{3}v_{1})^2 + (u_{2}v_{3} - u_{3}v_{2})^2\)

Expand and simplify the expression for S

Expanding and simplifying the expression for \(S\), we get: \(S = u_{1}^2v_{2}^2 - 2u_{1}u_{2}v_{1}v_{2} + u_{2}^2v_{1}^2 + u_{1}^2v_{3}^2 - 2u_{1}u_{3}v_{1}v_{3} + u_{3}^2v_{1}^2 + u_{2}^2v_{3}^2 - 2u_{2}u_{3}v_{2}v_{3} + u_{3}^2v_{2}^2\)

Rearrange the terms in S

Rearranging the terms in \(S\), we have: \(S = u_{1}^2(v_{2}^2 + v_{3}^2) + u_{2}^2(v_{1}^2 + v_{3}^2) + u_{3}^2(v_{1}^2 + v_{2}^2) - 2(u_{1}u_{2}v_{1}v_{2} + u_{1}u_{3}v_{1}v_{3} + u_{2}u_{3}v_{2}v_{3})\)

Note that S is non-negative

Since \(S\) is a sum of squares, it must be non-negative. Therefore, we have: \(S \geq 0\)

Simplify the inequality

Now, we can simplify the inequality in Step 5: \(u_{1}^2(v_{2}^2 + v_{3}^2) + u_{2}^2(v_{1}^2 + v_{3}^2) + u_{3}^2(v_{1}^2 + v_{2}^2) \geq 2(u_{1}u_{2}v_{1}v_{2} + u_{1}u_{3}v_{1}v_{3} + u_{2}u_{3}v_{2}v_{3})\) Divide both sides of the inequality by 2: \((u_{1}^2 + u_{2}^2 + u_{3}^2)(v_{1}^2 + v_{2}^2 + v_{3}^2) \geq (u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3})^2\) Taking the square root of both sides of the inequality: \(|\mathbf{u}||\mathbf{v}| \geq |\mathbf{u}\cdot\mathbf{v}|\) And this completes the proof of the Cauchy-Schwarz inequality.

Key concepts

Dot Product

The dot product, also known as the scalar product, is a crucial concept in vector algebra. It measures the extent to which two vectors align with each other. Given two vectors \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \) and \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \), their dot product is defined as:

• \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \).

In simpler terms, it is the sum of the products of their corresponding components. This operation results in a scalar and not a vector.
A valuable property of the dot product is that if the product is zero, it indicates the vectors are perpendicular.
The dot product also reflects how closely the vectors point in the same direction. The larger the dot product, the more aligned the vectors are.

Vector Magnitude

The magnitude of a vector, sometimes referred to as its length, quantifies the size of the vector in space. For any vector \( \mathbf{u} = \langle u_1, u_2, u_3 \rangle \), its magnitude is calculated using:

• \( |\mathbf{u}| = \sqrt{u_1^2 + u_2^2 + u_3^2} \).

This formula represents the distance from the origin to the point \( (u_1, u_2, u_3) \) in a three-dimensional coordinate system.
It’s similar to finding the length of the hypotenuse of a right triangle using the Pythagorean theorem.
Vector magnitude is a non-negative value and is pivotal in determining the direction of a vector when normalized.

Inequality Proof

The Cauchy-Schwarz inequality is a foundational element in mathematics and vector algebra. It provides an upper bound for the magnitude of the dot product of two vectors in this form:

• \(|\mathbf{u} \cdot \mathbf{v}| \leq |\mathbf{u}||\mathbf{v}|\)

To prove this inequality, one defines a non-negative quantity which, upon simplification, relates directly to the inequality. Introducing non-negative \(S\), we have:

• \(S = (u_{1}v_{2} - u_{2}v_{1})^2 + (u_{1}v_{3} - u_{3}v_{1})^2 + (u_{2}v_{3} - u_{3}v_{2})^2\)

Upon expanding and rearranging \(S\), the terms align to simplify the inequality as a relationship between sums of squares.
The conclusion comes from the property of non-negative sums ensuring that \(S \geq 0\). Thus, proving the inequality involves showing this relationship holds under all possible vector scenarios.

Vector Algebra

Vector algebra involves operations and rules that allow manipulation of vectors in a mathematical setting. Key operations include addition, subtraction, dot product, and cross product.

Understanding vector magnitude and direction plays a critical role in applications ranging from physics to computer graphics.
The manipulation of vectors helps in solving problems related to force, velocity, and displacement. In proving inequalities or exploring vector properties, algebraic tools allow expressing complex relationships, like the Cauchy-Schwarz inequality:

• This combines both the arithmetic of vector components and spatial understanding of their magnitudes.

Vectors simplify expressions and outcomes that are difficult or impossible to express with scalars alone. Their algebra enriches mathematical analysis by providing a structured approach to multi-dimensional problems.