Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Prove or disapprove:
![### Problem Statement
**Given**: \( \mathbf{w} = \langle w_1, w_2, w_3 \rangle \)
**Prove or disapprove**: \( \mathbf{w} \cdot \mathbf{w} = \|\mathbf{w}\|^2 \).
**Task**: Explain all details.
### Explanation
To address the problem, we need to understand the meanings of both dot product and vector norm:
1. **Dot Product** (\(\mathbf{w} \cdot \mathbf{w}\)):
- The dot product of a vector with itself is calculated by multiplying each of its components by itself and then summing the results:
\[
\mathbf{w} \cdot \mathbf{w} = w_1^2 + w_2^2 + w_3^2
\]
2. **Norm of a Vector** (\(\|\mathbf{w}\|\)):
- The norm (or magnitude) of a vector is calculated as the square root of the sum of the squares of its components:
\[
\|\mathbf{w}\| = \sqrt{w_1^2 + w_2^2 + w_3^2}
\]
3. **Square of the Norm** (\(\|\mathbf{w}\|^2\)):
- Taking the square of the vector norm cancels out the square root:
\[
\|\mathbf{w}\|^2 = (\sqrt{w_1^2 + w_2^2 + w_3^2})^2 = w_1^2 + w_2^2 + w_3^2
\]
From the above calculations, it is evident that:
\[
\mathbf{w} \cdot \mathbf{w} = \|\mathbf{w}\|^2
\]
### Conclusion
The statement \( \mathbf{w} \cdot \mathbf{w} = \|\mathbf{w}\|^2 \) is indeed true. The dot product of a vector with itself is equal to the square of its norm.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F999adab0-d3b0-43f1-b09c-cfc4d89ad581%2Fb7fa2e76-f620-486e-9cb0-d8161f236223%2Fcto9jsh_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**Given**: \( \mathbf{w} = \langle w_1, w_2, w_3 \rangle \)
**Prove or disapprove**: \( \mathbf{w} \cdot \mathbf{w} = \|\mathbf{w}\|^2 \).
**Task**: Explain all details.
### Explanation
To address the problem, we need to understand the meanings of both dot product and vector norm:
1. **Dot Product** (\(\mathbf{w} \cdot \mathbf{w}\)):
- The dot product of a vector with itself is calculated by multiplying each of its components by itself and then summing the results:
\[
\mathbf{w} \cdot \mathbf{w} = w_1^2 + w_2^2 + w_3^2
\]
2. **Norm of a Vector** (\(\|\mathbf{w}\|\)):
- The norm (or magnitude) of a vector is calculated as the square root of the sum of the squares of its components:
\[
\|\mathbf{w}\| = \sqrt{w_1^2 + w_2^2 + w_3^2}
\]
3. **Square of the Norm** (\(\|\mathbf{w}\|^2\)):
- Taking the square of the vector norm cancels out the square root:
\[
\|\mathbf{w}\|^2 = (\sqrt{w_1^2 + w_2^2 + w_3^2})^2 = w_1^2 + w_2^2 + w_3^2
\]
From the above calculations, it is evident that:
\[
\mathbf{w} \cdot \mathbf{w} = \|\mathbf{w}\|^2
\]
### Conclusion
The statement \( \mathbf{w} \cdot \mathbf{w} = \|\mathbf{w}\|^2 \) is indeed true. The dot product of a vector with itself is equal to the square of its norm.
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