Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find a simple expression for each of the following for an arbitrary vector \(\mathbf{u}\). (a) \(\operatorname{proj}_{\mathbf{u}}(-\mathbf{u})\) (b) \(\operatorname{proj}_{-\mathbf{u}}(-\mathbf{u})\)

Step-by-Step Solution

Answer

Both (a) and (b) result in \(-\mathbf{u}\).

Step 1: Understanding Projection Formula

The formula for projecting a vector \( \mathbf{a} \) onto a vector \( \mathbf{b} \) is given by \( \operatorname{proj}_{\mathbf{b}}(\mathbf{a}) = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{b} \). This will form the basis for solving both parts of this exercise.

Step 2: Calculating Projection for Part (a)

To find \( \operatorname{proj}_{\mathbf{u}}(-\mathbf{u}) \), substitute \( \mathbf{a} = -\mathbf{u} \) and \( \mathbf{b} = \mathbf{u} \) into the formula. This gives: \( \operatorname{proj}_{\mathbf{u}}(-\mathbf{u}) = \frac{(-\mathbf{u}) \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} \). Note that \( (-\mathbf{u}) \cdot \mathbf{u} = - (\mathbf{u} \cdot \mathbf{u}) \). Thus, \( \operatorname{proj}_{\mathbf{u}}(-\mathbf{u}) = -\frac{\mathbf{u} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u} = -\mathbf{u} \).

Step 3: Calculating Projection for Part (b)

For \( \operatorname{proj}_{-\mathbf{u}}(-\mathbf{u}) \), substitute \( \mathbf{a} = -\mathbf{u} \) and \( \mathbf{b} = -\mathbf{u} \). The formula becomes: \( \operatorname{proj}_{-\mathbf{u}}(-\mathbf{u}) = \frac{(-\mathbf{u}) \cdot (-\mathbf{u})}{(-\mathbf{u}) \cdot (-\mathbf{u})} (-\mathbf{u}) \). Note that \( (-\mathbf{u}) \cdot (-\mathbf{u}) = \mathbf{u} \cdot \mathbf{u} \). Hence, \( \operatorname{proj}_{-\mathbf{u}}(-\mathbf{u}) = \frac{\mathbf{u} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} (-\mathbf{u}) = -\mathbf{u} \).

Step 4: Final Answer Verification

For both parts (a) and (b), the projection results in \( -\mathbf{u} \), confirming that the solution is consistent regardless of the direction of the projection vector.

Key Concepts

Dot Product

The dot product, also sometimes referred to as the scalar product, is a fundamental operation in the realm of vector mathematics. It essentially combines two vectors to return a scalar (a single real number). To calculate the dot product of two vectors, say \( \mathbf{a} = [a_1, a_2, ..., a_n] \) and \( \mathbf{b} = [b_1, b_2, ..., b_n] \), you perform the following steps:

• Multiply each corresponding pair of components from the vectors: \( a_1 \cdot b_1, a_2 \cdot b_2, ..., a_n \cdot b_n \).
• Add all these products together: \( a_1b_1 + a_2b_2 + ... + a_nb_n \).

In mathematical notation, the dot product is expressed as \( \mathbf{a} \cdot \mathbf{b} = \sum_{i=1}^{n} a_i b_i \).
A critical property of the dot product is that it is commutative, meaning \( \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a} \). It also indicates orthogonality: if the dot product is zero, this means the vectors are perpendicular.

Unit Vectors

Unit vectors play an essential role in simplifying vector operations. A unit vector is a vector that has a magnitude of 1. The purpose of a unit vector is to indicate direction.
They are often represented using a "hat", like \( \mathbf{u} \text{̂} \), which depicts that they have been normalized or converted.

• To convert any vector \( \mathbf{v} \) into a unit vector, divide it by its magnitude. Mathematically, this is \( \mathbf{u} = \frac{\mathbf{v}}{||\mathbf{v}||} \).
• The magnitude of a vector \( \mathbf{v} = [v_1, v_2, ..., v_n] \) is calculated using \( ||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2 + ... + v_n^2} \).

By using unit vectors, calculations become clearer since the focus shifts from both direction and magnitude to only direction. This is particularly useful in settings like physics and engineering, where direction matters but length does not.

Orthogonal Projection

Orthogonal projection is a concept that allows you to project one vector onto another. This is analogous to casting the shadow of one vector onto another, capturing its component in one direction.
To find the projection of vector \( \mathbf{a} \) onto vector \( \mathbf{b} \), use the formula:

• \( \operatorname{proj}_{\mathbf{b}}(\mathbf{a}) = \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \mathbf{b} \)

This expression describes how much of \( \mathbf{a} \) is in the direction of \( \mathbf{b} \).
Think of orthogonal projections as the 2D shadow of a 3D object under a light ray parallel to the plane of \( \mathbf{b} \). Key characteristics of projections include:

• The direction of the resulting vector is the same as \( \mathbf{b} \).
• The magnitude is less than or equal to that of \( \mathbf{a} \).

This method is frequently used in computer graphics and physics to resolve forces and motions into components along various axes.

Vector Operations

Vector operations encompass the various ways vectors can be manipulated or interacted within math and physics. Key operations include:

• Addition and Subtraction: This involves either joining vectors end-to-end or performing an inverse operation, respectively. Two vectors \( \mathbf{a} = [a_1, a_2, ..., a_n] \) and \( \mathbf{b} = [b_1, b_2, ..., b_n] \) can be added to produce \( \mathbf{c} = [a_1 + b_1, a_2 + b_2, ..., a_n + b_n] \).
• Scalar Multiplication: Multiplying a vector by a scalar stretches or shrinks its magnitude without altering its direction. If scalar is negative, it also inverses direction.
• Magnitude: The length of a vector, calculated as \( ||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2 + ... + v_n^2} \).
• Normalization: Converting a vector to a unit vector, hence reducing its magnitude to 1, keeping direction.

Understanding these operations enables the effective resolution of many practical and theoretical problems, ranging from physics to computer science. Each operation provides tools to analyze and manipulate directional data.