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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Question
![**Vector Projection Problem**
Given vectors:
\[ \mathbf{v} = \mathbf{i} + 2\mathbf{j} \]
\[ \mathbf{w} = 3\mathbf{i} + 6\mathbf{j} \]
Calculate the projection of vector \(\mathbf{v}\) onto vector \(\mathbf{w}\).
Options:
- \(\mathbf{i} - 2\mathbf{j}\)
- \(\mathbf{i} + 2\mathbf{j}\)
- \(-3\mathbf{i} - 6\mathbf{j}\)
- \(3\mathbf{i} + 6\mathbf{j}\)
Solution:
The projection of vector \(\mathbf{v}\) onto vector \(\mathbf{w}\) is calculated using the formula:
\[ \text{Proj}_{\mathbf{w}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}} \mathbf{w} \]
Let's compute it step-by-step:
1. **Dot Product \(\mathbf{v} \cdot \mathbf{w}\):**
\[ \mathbf{v} \cdot \mathbf{w} = (1 \cdot 3) + (2 \cdot 6) = 3 + 12 = 15 \]
2. **Dot Product \(\mathbf{w} \cdot \mathbf{w}\):**
\[ \mathbf{w} \cdot \mathbf{w} = (3 \cdot 3) + (6 \cdot 6) = 9 + 36 = 45 \]
3. **Calculate the projection scalar:**
\[ \frac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}} = \frac{15}{45} = \frac{1}{3} \]
4. **Find the projection vector:**
\[ \text{Proj}_{\mathbf{w}} \mathbf{v} = \frac{1}{3} \mathbf{w} = \frac{1}{3} (3\mathbf{i} + 6\mathbf{j}) = \mathbf{i} + 2\mathbf{j} \]
So, the projection of \(\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb65d250f-ce79-4155-be29-8b1588e7e3e2%2F1c271db4-d52d-43e6-aa34-99fc28d3b9ea%2Flckxbc8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Vector Projection Problem**
Given vectors:
\[ \mathbf{v} = \mathbf{i} + 2\mathbf{j} \]
\[ \mathbf{w} = 3\mathbf{i} + 6\mathbf{j} \]
Calculate the projection of vector \(\mathbf{v}\) onto vector \(\mathbf{w}\).
Options:
- \(\mathbf{i} - 2\mathbf{j}\)
- \(\mathbf{i} + 2\mathbf{j}\)
- \(-3\mathbf{i} - 6\mathbf{j}\)
- \(3\mathbf{i} + 6\mathbf{j}\)
Solution:
The projection of vector \(\mathbf{v}\) onto vector \(\mathbf{w}\) is calculated using the formula:
\[ \text{Proj}_{\mathbf{w}} \mathbf{v} = \frac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}} \mathbf{w} \]
Let's compute it step-by-step:
1. **Dot Product \(\mathbf{v} \cdot \mathbf{w}\):**
\[ \mathbf{v} \cdot \mathbf{w} = (1 \cdot 3) + (2 \cdot 6) = 3 + 12 = 15 \]
2. **Dot Product \(\mathbf{w} \cdot \mathbf{w}\):**
\[ \mathbf{w} \cdot \mathbf{w} = (3 \cdot 3) + (6 \cdot 6) = 9 + 36 = 45 \]
3. **Calculate the projection scalar:**
\[ \frac{\mathbf{v} \cdot \mathbf{w}}{\mathbf{w} \cdot \mathbf{w}} = \frac{15}{45} = \frac{1}{3} \]
4. **Find the projection vector:**
\[ \text{Proj}_{\mathbf{w}} \mathbf{v} = \frac{1}{3} \mathbf{w} = \frac{1}{3} (3\mathbf{i} + 6\mathbf{j}) = \mathbf{i} + 2\mathbf{j} \]
So, the projection of \(\
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