Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 29-34, find each of the given projections if \(\mathbf{u}=3 \mathbf{i}+2 \mathbf{j}+\mathbf{k}, \mathbf{v}=2 \mathbf{i}-\mathbf{k}\), and \(\mathbf{w}=\mathbf{i}+5 \mathbf{j}-3 \mathbf{k}\). $$ \operatorname{proj}_{\mathbf{k}} \mathbf{u} $$

Step-by-Step Solution

Answer

The projection of \( \mathbf{u} \) onto \( \mathbf{k} \) is \( \mathbf{k} \).

Step 1: Understand Projection

The projection of a vector \( \mathbf{a} \) onto another vector \( \mathbf{b} \) is a vector in the direction of \( \mathbf{b} \). It is calculated with the formula \( \operatorname{proj}_{\mathbf{b}} \mathbf{a} = \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \right) \mathbf{b} \).

Step 2: Identify Vectors

For this problem, \( \mathbf{a} = \mathbf{u} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} \) and \( \mathbf{b} = \mathbf{k} \).

Step 3: Compute Dot Product \( \mathbf{u} \cdot \mathbf{k} \)

The dot product \( \mathbf{u} \cdot \mathbf{k} \) is calculated as follows: \( 3\times0 + 2\times0 + 1\times1 = 1 \).

Step 4: Compute Dot Product \( \mathbf{k} \cdot \mathbf{k} \)

The dot product \( \mathbf{k} \cdot \mathbf{k} \) is calculated as: \( 0\times0 + 0\times0 + 1\times1 = 1 \).

Step 5: Apply Projection Formula

Use the projection formula: \( \operatorname{proj}_{\mathbf{k}} \mathbf{u} = \left( \frac{1}{1} \right) \mathbf{k} = \mathbf{k} \).

Key Concepts

Dot Product

The dot product is a fundamental operation in vector mathematics that helps compare and analyze vectors. It is also known as the scalar product.
When you take the dot product of two vectors, the result is a scalar (a single number), rather than another vector.

• For vectors \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is defined as \( a_1b_1 + a_2b_2 + a_3b_3 \).
• It effectively combines the magnitudes of two vectors and makes use of the cosine of the angle between them.

In general, if the dot product is zero, the vectors are orthogonal, meaning they form a 90-degree angle to each other.
In this exercise, you calculate the dot product of \( \mathbf{u} \) and \( \mathbf{k} \), which results in \( 1 \), indicating some alignment along the \( \mathbf{k} \) direction.

Vector Operations

Working with vectors involves several key operations, including addition, subtraction, and multiplication. Here, we focus on understanding their properties relative to projections.
Understanding these operations is essential for manipulating vector equations and solving complex problems in physics and engineering.

• **Addition and Subtraction**: Vectors are added or subtracted component-wise. For example, when adding \( \mathbf{u} \) and \( \mathbf{v} \), you would add the corresponding \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) components.
• **Scalar Multiplication**: This involves multiplying each component of a vector by a scalar (a real number). It's used in scaling vectors to change their magnitude while keeping the direction the same.

By understanding these operations, you can effectively simplify complex vector tasks and handle problems such as scaling vectors for projections, like in the problem to project \( \mathbf{u} \) onto \( \mathbf{k} \).

Projection Formula

The projection formula allows you to find the component of one vector in the direction of another. This is helpful when you want to determine how much of one vector lies along another vector.
The projection of vector \( \mathbf{a} \) onto vector \( \mathbf{b} \) is given by the formula:\[\operatorname{proj}_{\mathbf{b}} \mathbf{a} = \left( \frac{\mathbf{a} \cdot \mathbf{b}}{\mathbf{b} \cdot \mathbf{b}} \right) \mathbf{b}\]

• **Numerator**: The dot product \( \mathbf{a} \cdot \mathbf{b} \) finds how much \( \mathbf{a} \) aligns with \( \mathbf{b} \).
• **Denominator**: The dot product \( \mathbf{b} \cdot \mathbf{b} \) finds the magnitude squared of \( \mathbf{b} \), ensuring the scaling is correct.

By using the projection formula, you extract the component of \( \mathbf{a} \) along \( \mathbf{b} \).
In this specific exercise, applying it determines how much of \( \mathbf{u} \) projects along the unit vector \( \mathbf{k} \), yielding \( \mathbf{k} \) itself.