Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Simplification of Vector Expression
#### Problem Statement:
Simplify the following expression given that the magnitudes of vectors \( \mathbf{v} \) and \( \mathbf{w} \) are both 12:
\[ \|\mathbf{v}\| = 12 \quad \text{and} \quad \|\mathbf{w}\| = 12 \]
Given Expression:
\[ (\mathbf{v} + \mathbf{w}) \cdot (\mathbf{v} + \mathbf{w}) - 2 \mathbf{v} \cdot \mathbf{w} \]
#### Explanation:
Here is the step-by-step process to simplify the given expression:
1. **Expand the dot product:**
\[ (\mathbf{v} + \mathbf{w}) \cdot (\mathbf{v} + \mathbf{w}) \]
Using the distributive property of the dot product, we get:
\[ = \mathbf{v} \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{w} + \mathbf{w} \cdot \mathbf{v} + \mathbf{w} \cdot \mathbf{w} \]
Since the dot product is commutative (\( \mathbf{v} \cdot \mathbf{w} = \mathbf{w} \cdot \mathbf{v} \)):
\[ = \mathbf{v} \cdot \mathbf{v} + 2(\mathbf{v} \cdot \mathbf{w}) + \mathbf{w} \cdot \mathbf{w} \]
2. **Include the second part of the original expression:**
We subtract \( 2 \mathbf{v} \cdot \mathbf{w} \) from the expression obtained above:
\[ \mathbf{v} \cdot \mathbf{v} + 2(\mathbf{v} \cdot \mathbf{w}) + \mathbf{w} \cdot \mathbf{w} - 2 \mathbf{v} \cdot \mathbf{w} \]
Simplifying this further:
\[ = \mathbf{v} \cdot \mathbf{v} + \mathbf{w} \cdot \mathbf{w} \]
3. **Substitute the magnitudes:**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc3d3bee5-0c39-4b7b-a553-642953afa184%2Fc45d95ce-5cc6-420f-b0df-ad9c0c23f261%2F9jb9f18.png&w=3840&q=75)
Transcribed Image Text:### Simplification of Vector Expression
#### Problem Statement:
Simplify the following expression given that the magnitudes of vectors \( \mathbf{v} \) and \( \mathbf{w} \) are both 12:
\[ \|\mathbf{v}\| = 12 \quad \text{and} \quad \|\mathbf{w}\| = 12 \]
Given Expression:
\[ (\mathbf{v} + \mathbf{w}) \cdot (\mathbf{v} + \mathbf{w}) - 2 \mathbf{v} \cdot \mathbf{w} \]
#### Explanation:
Here is the step-by-step process to simplify the given expression:
1. **Expand the dot product:**
\[ (\mathbf{v} + \mathbf{w}) \cdot (\mathbf{v} + \mathbf{w}) \]
Using the distributive property of the dot product, we get:
\[ = \mathbf{v} \cdot \mathbf{v} + \mathbf{v} \cdot \mathbf{w} + \mathbf{w} \cdot \mathbf{v} + \mathbf{w} \cdot \mathbf{w} \]
Since the dot product is commutative (\( \mathbf{v} \cdot \mathbf{w} = \mathbf{w} \cdot \mathbf{v} \)):
\[ = \mathbf{v} \cdot \mathbf{v} + 2(\mathbf{v} \cdot \mathbf{w}) + \mathbf{w} \cdot \mathbf{w} \]
2. **Include the second part of the original expression:**
We subtract \( 2 \mathbf{v} \cdot \mathbf{w} \) from the expression obtained above:
\[ \mathbf{v} \cdot \mathbf{v} + 2(\mathbf{v} \cdot \mathbf{w}) + \mathbf{w} \cdot \mathbf{w} - 2 \mathbf{v} \cdot \mathbf{w} \]
Simplifying this further:
\[ = \mathbf{v} \cdot \mathbf{v} + \mathbf{w} \cdot \mathbf{w} \]
3. **Substitute the magnitudes:**
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