Problem 32 - 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 \(\mathbf{w}=\mathbf{i}+5 \mathbf{j}-3 \mathbf{k}\). $$ \operatorname{proj}_{\mathbf{u}}(\mathbf{w}+\mathbf{v}) $$

Step-by-Step Solution

Answer

The projection is \( \frac{45}{14} \mathbf{i} + \frac{15}{7} \mathbf{j} + \frac{15}{14} \mathbf{k} \).

Step 1: Calculate Sum of Vectors w and v

First, find the vector sum \( \mathbf{w} + \mathbf{v} \). Given \( \mathbf{w} = \mathbf{i} + 5\mathbf{j} - 3\mathbf{k} \) and \( \mathbf{v} = 2\mathbf{i} - \mathbf{k} \), the sum is calculated as follows: \[ \mathbf{w} + \mathbf{v} = (\mathbf{i} + 5\mathbf{j} - 3\mathbf{k}) + (2\mathbf{i} - \mathbf{k}) = 3\mathbf{i} + 5\mathbf{j} - 4\mathbf{k}. \]

Step 2: Find Dot Product of u and (w+v)

Next, calculate the dot product of \( \mathbf{u} \) and \( \mathbf{w} + \mathbf{v} \). \( \mathbf{u} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} \) and \( \mathbf{w} + \mathbf{v} = 3\mathbf{i} + 5\mathbf{j} - 4\mathbf{k} \). Use the formula for the dot product: \[ \mathbf{u} \cdot (\mathbf{w} + \mathbf{v}) = (3)(3) + (2)(5) + (1)(-4) = 9 + 10 - 4 = 15. \]

Step 3: Calculate Magnitude Squared of u

Compute the magnitude squared of \( \mathbf{u} \). The magnitude is calculated by: \[ \| \mathbf{u} \|^2 = (3^2) + (2^2) + (1^2) = 9 + 4 + 1 = 14. \]

Step 4: Compute Projection of (w+v) onto u

Using the formula for projection, \( \text{proj}_{\mathbf{u}}(\mathbf{w} + \mathbf{v}) = \frac{\mathbf{u} \cdot (\mathbf{w} + \mathbf{v})}{\|\mathbf{u}\|^2} \mathbf{u} \), calculate: \[ \text{proj}_{\mathbf{u}}(\mathbf{w} + \mathbf{v}) = \frac{15}{14} \times (3\mathbf{i} + 2\mathbf{j} + \mathbf{k}). \] This results in: \[ \frac{45}{14} \mathbf{i} + \frac{30}{14} \mathbf{j} + \frac{15}{14} \mathbf{k}. \]

Step 5: Simplify the Projection

Simplify the terms in the projection vector: \( \frac{45}{14} \mathbf{i} + \frac{30}{14} \mathbf{j} + \frac{15}{14} \mathbf{k} \) becomes \( \frac{45}{14} \mathbf{i} + \frac{15}{7} \mathbf{j} + \frac{15}{14} \mathbf{k} \).

Key Concepts

Dot Product

The dot product, also called the scalar product, is a fundamental operation in vector mathematics. It combines two vectors to yield a single number, known as a scalar. Here's how it works:

To find the dot product of two vectors \( \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \) and \( \mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k} \), use the formula:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)

For our exercise, we computed the dot product of \( \mathbf{u} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} \) and \( \mathbf{w}+\mathbf{v} = 3\mathbf{i} + 5\mathbf{j} - 4\mathbf{k} \). This results in the calculation:

• \( (3)(3) + (2)(5) + (1)(-4) = 9 + 10 - 4 = 15 \)

The dot product provides significant insight, especially when determining angles between vectors or projections.

Vector Addition

Vector addition involves summing the corresponding components of vectors to form a new vector. This follows the principle of combining like terms, making it simple and straightforward.

To add two 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 result is:

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

In our exercise, the vectors \( \mathbf{w} \) and \( \mathbf{v} \) add to form \( \mathbf{w} + \mathbf{v} = 3\mathbf{i} + 5\mathbf{j} - 4\mathbf{k} \).

• \( (\mathbf{i} + 5\mathbf{j} - 3\mathbf{k}) + (2\mathbf{i} - \mathbf{k}) = 3\mathbf{i} + 5\mathbf{j} - 4\mathbf{k} \)

This process of adding vectors is very visual. It can quickly show displacement or change in direction.

Magnitude of a Vector

The magnitude of a vector represents its length. This is essential when determining vector norms, which can be thought of as the vector's natural 'size'.

For a vector \( \mathbf{u} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \), the magnitude is:

• \( \| \mathbf{u} \| = \sqrt{a_1^2 + a_2^2 + a_3^2} \)

In our calculation, we find the magnitude of \( \mathbf{u} = 3\mathbf{i} + 2\mathbf{j} + \mathbf{k} \):

• \( \| \mathbf{u} \|^2 = 3^2 + 2^2 + 1^2 = 9 + 4 + 1 = 14 \)

This squared result is used for projections. Understanding magnitude helps in defining distances and analyzing vector relations.

Calculus Problem Solving

Calculus problem solving often involves vector calculus, which combines algebra with limits and rates of change. For vector operations, calculus helps in understanding motion over time as well as field behavior.

In this exercise, the projection is found by using the dot product and magnitude:

• Projection formula: \( \text{proj}_{\mathbf{u}}(\mathbf{w} + \mathbf{v}) = \frac{\mathbf{u} \cdot (\mathbf{w} + \mathbf{v})}{\|\mathbf{u}\|^2} \mathbf{u} \)

This requires calculating constants like dot products and magnitudes. Calculus translates these insights to real-world contexts, explaining direction and size changes dynamically.
In essence, calculus helps in creating a framework to predict and understand complex vector systems.