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
1.2
![The problem states:
"Let \(\vec{a} = -\vec{i} + 4\vec{j} + 2\vec{k}\). Find \(\|\vec{a}\|\)."
This task involves finding the magnitude of the vector \(\vec{a}\).
### Explanation:
The vector \(\vec{a}\) is given in component form as \(-\vec{i} + 4\vec{j} + 2\vec{k}\), where:
- \(-\vec{i}\) indicates a component of \(-1\) in the x-direction,
- \(4\vec{j}\) indicates a component of \(4\) in the y-direction,
- \(2\vec{k}\) indicates a component of \(2\) in the z-direction.
To find the magnitude \(\|\vec{a}\|\), you use the formula for the magnitude of a vector \(\vec{a} = a_1\vec{i} + a_2\vec{j} + a_3\vec{k}\):
\[
\|\vec{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}
\]
Substitute the components of \(\vec{a}\):
\[
\|\vec{a}\| = \sqrt{(-1)^2 + 4^2 + 2^2} = \sqrt{1 + 16 + 4} = \sqrt{21}
\]
Thus, the magnitude of the vector is \(\sqrt{21}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F21e8a523-579f-40b9-8f0a-805524283a8d%2F93e11dc5-ec1e-44c8-832e-d0a115fb01f0%2F9v5i63h_processed.png&w=3840&q=75)
Transcribed Image Text:The problem states:
"Let \(\vec{a} = -\vec{i} + 4\vec{j} + 2\vec{k}\). Find \(\|\vec{a}\|\)."
This task involves finding the magnitude of the vector \(\vec{a}\).
### Explanation:
The vector \(\vec{a}\) is given in component form as \(-\vec{i} + 4\vec{j} + 2\vec{k}\), where:
- \(-\vec{i}\) indicates a component of \(-1\) in the x-direction,
- \(4\vec{j}\) indicates a component of \(4\) in the y-direction,
- \(2\vec{k}\) indicates a component of \(2\) in the z-direction.
To find the magnitude \(\|\vec{a}\|\), you use the formula for the magnitude of a vector \(\vec{a} = a_1\vec{i} + a_2\vec{j} + a_3\vec{k}\):
\[
\|\vec{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}
\]
Substitute the components of \(\vec{a}\):
\[
\|\vec{a}\| = \sqrt{(-1)^2 + 4^2 + 2^2} = \sqrt{1 + 16 + 4} = \sqrt{21}
\]
Thus, the magnitude of the vector is \(\sqrt{21}\).
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