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
![**Problem: Finding the Magnitude of a Vector**
**Given:**
Let \(\mathbf{u} = \langle -5, 1, -3 \rangle\).
**Find:**
The magnitude of the vector \(\mathbf{u}\), denoted as \(|\mathbf{u}|\).
---
Explanation:
To find the magnitude (or length) of a vector \(\mathbf{u} = \langle x, y, z \rangle\), you can use the formula:
\[
|\mathbf{u}| = \sqrt{x^2 + y^2 + z^2}
\]
Here, the coordinates of vector \(\mathbf{u}\) are:
- \(x = -5\)
- \(y = 1\)
- \(z = -3\)
Substituting these values into the formula will give you the magnitude of the vector.
::- show how to substitute and solve in an educational style: -::
\[
|\mathbf{u}| = \sqrt{(-5)^2 + 1^2 + (-3)^2}
= \sqrt{25 + 1 + 9}
= \sqrt{35}
\]
So the magnitude of vector \(\mathbf{u}\) is:
\[
|\mathbf{u}| = \sqrt{35}
\]
You can leave it in this square root form or approximate it to a decimal if needed for specific applications.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F16864198-3c29-44d0-a974-0ddc23a60123%2Fe8b31066-76cb-4001-8dc1-78a5c31ef909%2F9gfkfzo_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem: Finding the Magnitude of a Vector**
**Given:**
Let \(\mathbf{u} = \langle -5, 1, -3 \rangle\).
**Find:**
The magnitude of the vector \(\mathbf{u}\), denoted as \(|\mathbf{u}|\).
---
Explanation:
To find the magnitude (or length) of a vector \(\mathbf{u} = \langle x, y, z \rangle\), you can use the formula:
\[
|\mathbf{u}| = \sqrt{x^2 + y^2 + z^2}
\]
Here, the coordinates of vector \(\mathbf{u}\) are:
- \(x = -5\)
- \(y = 1\)
- \(z = -3\)
Substituting these values into the formula will give you the magnitude of the vector.
::- show how to substitute and solve in an educational style: -::
\[
|\mathbf{u}| = \sqrt{(-5)^2 + 1^2 + (-3)^2}
= \sqrt{25 + 1 + 9}
= \sqrt{35}
\]
So the magnitude of vector \(\mathbf{u}\) is:
\[
|\mathbf{u}| = \sqrt{35}
\]
You can leave it in this square root form or approximate it to a decimal if needed for specific applications.
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