Problem 43 - Algebra and Trigonometry

Question

Let \(\mathbf{v}\) and \(\mathbf{w}\) denote two nonzero vectors. Show that the vectors \(\|\mathbf{w}\| \mathbf{v}+\|\mathbf{v}\| \mathbf{w}\) and \(\|\mathbf{w}\| \mathbf{v}-\|\mathbf{v}\| \mathbf{w}\) are orthogonal.

Step-by-Step Solution

Answer

The vectors are orthogonal because their dot product equals zero.

Step 1: Define the Vectors

Define the vectors \(\textbf{a} = \|\textbf{w}\| \textbf{v} + \|\textbf{v}\| \textbf{w}\) and \(\textbf{b} = \|\textbf{w}\| \textbf{v} - \|\textbf{v}\| \textbf{w}\).

Step 2: Dot Product Definition

Recall that two vectors \(\textbf{a}\) and \(\textbf{b}\) are orthogonal if their dot product is zero: \(\textbf{a} \cdot \textbf{b} = 0\).

Step 3: Compute the Dot Product

Calculate \(\textbf{a} \cdot \textbf{b}\):\[ \textbf{a} \cdot \textbf{b} = (\|\textbf{w}\| \textbf{v} + \|\textbf{v}\| \textbf{w}) \cdot (\|\textbf{w}\| \textbf{v} - \|\textbf{v}\| \textbf{w}). \]

Step 4: Expand the Dot Product

Expand the expression: \[(\|\textbf{w}\| \textbf{v} + \|\textbf{v}\| \textbf{w}) \cdot (\|\textbf{w}\| \textbf{v} - \|\textbf{v}\| \textbf{w}) = \|\textbf{w}\|^2 ( \textbf{v} \cdot \textbf{v} ) - \|\textbf{w}\| \|\textbf{v}\|( \textbf{v} \cdot \textbf{w} ) + \|\textbf{v}\| \|\textbf{w}\|( \textbf{w} \cdot \textbf{v} ) - \|\textbf{v}\|^2 ( \textbf{w} \cdot \textbf{w} ). \]

Step 5: Simplify the Dot Product

Simplify the expanded terms: \[\textbf{a} \cdot \textbf{b} = \|\textbf{w}\|^2 ( \textbf{v} \cdot \textbf{v} ) - \|\textbf{w}\| \|\textbf{v}\|( \textbf{v} \cdot \textbf{w} ) + \|\textbf{v}\| \|\textbf{w}\|( \textbf{w} \cdot \textbf{v} ) - \|\textbf{v}\|^2 ( \textbf{w} \cdot \textbf{w} ) \]

Step 6: Recognize the Scalar Symmetry

The terms \( - \|\textbf{w}\| \|\textbf{v}\|( \textbf{v} \cdot \textbf{w} ) \) and \( + \|\textbf{v}\| \|\textbf{w}\|( \textbf{w} \cdot \textbf{v} ) \) are equal in magnitude but opposite in sign because \( \textbf{v} \cdot \textbf{w} = \textbf{w} \cdot \textbf{v} \), leading to their cancelation.

Step 7: Conclude the Orthogonality

This leads to the expression: \( \|\textbf{w}\|^2 ( \textbf{v} \cdot \textbf{v} ) - \|\textbf{v}\|^2 ( \textbf{w} \cdot \textbf{w} ) = 0, \) since \( \|\textbf{w}\|^2 ( \textbf{v} \cdot \textbf{v} ) = \|\textbf{v}\|^2 ( \textbf{w} \cdot \textbf{w} ) \). Hence \(\textbf{a} \cdot \textbf{b} = 0\), demonstrating orthogonality.

Key Concepts

vector algebra

Vector algebra helps us deal with vectors and the operations we perform on them. Vectors are quantities that have both a magnitude (length) and a direction. Some common operations in vector algebra include addition, subtraction, and scalar multiplication. In this problem, we are given two vectors, \( \textbf{v} \) and \( \textbf{w} \), and need to show that the vectors \( \|\textbf{w}\| \textbf{v} + \|\textbf{v}\| \textbf{w} \) and \( \|\textbf{w}\| \textbf{v} - \|\textbf{v}\| \textbf{w} \) are orthogonal. We'll achieve this by using vector algebra concepts to define and manipulate these vectors.

dot product

The dot product, also known as the scalar product, is a fundamental operation in vector algebra. It takes two vectors and returns a scalar (a number). For two vectors \( \textbf{a} \) and \( \textbf{b} \), their dot product is given by:
\[ \textbf{a} \cdot \textbf{b} = \| \textbf{a} \| \| \textbf{b} \| \cos \theta \]
where \( \theta \) is the angle between the two vectors. In this exercise, the dot product is crucial as it determines orthogonality. If the dot product of two vectors is zero, they are orthogonal (perpendicular). We use this property to show that the given vectors are orthogonal.

orthogonality

Orthogonality is a key concept in vector algebra. Two vectors are said to be orthogonal if their dot product is zero. This implies that the vectors are perpendicular to each other. In this problem, we need to show that the vectors \( \|\textbf{w}\| \textbf{v} + \|\textbf{v}\| \textbf{w} \) and \( \|\textbf{w}\| \textbf{v} - \|\textbf{v}\| \textbf{w} \) are orthogonal. We do this by computing and simplifying their dot product, demonstrating that it equals zero. This involves recognizing symmetry and leveraging mathematical properties of the dot product and norms.

vector norms

Vector norms measure the length or magnitude of a vector. For example, the norm of a vector \( \textbf{v} \) is denoted as \( \|\textbf{v}\| \) and is calculated as:
\[ \|\textbf{v}\| = \sqrt{\textbf{v} \cdot \textbf{v}} \]
Norms are crucial in this exercise because we are dealing with terms like \( \|\textbf{v}\| \) and \( \|\textbf{w}\| \). By understanding and using these norms, we can simplify and manipulate the vectors in question to ultimately prove their orthogonality. We will see how \( \|\textbf{v}\| \) and \( \|\textbf{w}\| \) interact when calculating the dot product.