Let u = (1, 1, 8). Find ||||. Use that to find a unit vector that points in the direction of u. Write your answers in exact form, using sqrt() to indicate square roots. Enter the vector using < and > as enclosing brackets. ||ū|| = Unit vector that points in the direction of u:
Let u = (1, 1, 8). Find ||||. Use that to find a unit vector that points in the direction of u. Write your answers in exact form, using sqrt() to indicate square roots. Enter the vector using < and > as enclosing brackets. ||ū|| = Unit vector that points in the direction of u:
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![To solve for the magnitude of vector \( \vec{u} \) and a corresponding unit vector, follow these steps:
Given:
\[ \vec{u} = \langle -1, -1, 8 \rangle \]
**Step 1: Find the magnitude \( \|\vec{u}\| \)**
The magnitude of a vector \( \vec{u} = \langle u_1, u_2, u_3 \rangle \) is calculated using the formula:
\[ \|\vec{u}\| = \sqrt{u_1^2 + u_2^2 + u_3^2} \]
Here, \( u_1 = -1 \), \( u_2 = -1 \), \( u_3 = 8 \).
So,
\[ \|\vec{u}\| = \sqrt{(-1)^2 + (-1)^2 + 8^2} \]
\[ \|\vec{u}\| = \sqrt{1 + 1 + 64} \]
\[ \|\vec{u}\| = \sqrt{66} \]
**Step 2: Find a unit vector in the direction of \( \vec{u} \)**
A unit vector in the direction of \( \vec{u} \) is given by:
\[ \hat{u} = \frac{\vec{u}}{\|\vec{u}\|} \]
\[ \hat{u} = \frac{\langle -1, -1, 8 \rangle}{\sqrt{66}} \]
Thus, the unit vector is:
\[ \hat{u} = \left\langle \frac{-1}{\sqrt{66}}, \frac{-1}{\sqrt{66}}, \frac{8}{\sqrt{66}} \right\rangle \]
**Final Answers:**
Magnitude \( \|\vec{u}\| \):
\[ \|\vec{u}\| = \sqrt{66} \]
Unit vector in the direction of \( \vec{u} \):
\[ \left\langle \frac{-1}{\sqrt{66}}, \frac{-1}{\sqrt{66}}, \frac{8}{\sqrt{66}} \right\rangle \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0fdef185-de8b-42f5-933e-8018dcf36541%2Ff7e5e801-3a94-4a81-bca3-01ecec3315c5%2Fbaq1jnv_processed.png&w=3840&q=75)
Transcribed Image Text:To solve for the magnitude of vector \( \vec{u} \) and a corresponding unit vector, follow these steps:
Given:
\[ \vec{u} = \langle -1, -1, 8 \rangle \]
**Step 1: Find the magnitude \( \|\vec{u}\| \)**
The magnitude of a vector \( \vec{u} = \langle u_1, u_2, u_3 \rangle \) is calculated using the formula:
\[ \|\vec{u}\| = \sqrt{u_1^2 + u_2^2 + u_3^2} \]
Here, \( u_1 = -1 \), \( u_2 = -1 \), \( u_3 = 8 \).
So,
\[ \|\vec{u}\| = \sqrt{(-1)^2 + (-1)^2 + 8^2} \]
\[ \|\vec{u}\| = \sqrt{1 + 1 + 64} \]
\[ \|\vec{u}\| = \sqrt{66} \]
**Step 2: Find a unit vector in the direction of \( \vec{u} \)**
A unit vector in the direction of \( \vec{u} \) is given by:
\[ \hat{u} = \frac{\vec{u}}{\|\vec{u}\|} \]
\[ \hat{u} = \frac{\langle -1, -1, 8 \rangle}{\sqrt{66}} \]
Thus, the unit vector is:
\[ \hat{u} = \left\langle \frac{-1}{\sqrt{66}}, \frac{-1}{\sqrt{66}}, \frac{8}{\sqrt{66}} \right\rangle \]
**Final Answers:**
Magnitude \( \|\vec{u}\| \):
\[ \|\vec{u}\| = \sqrt{66} \]
Unit vector in the direction of \( \vec{u} \):
\[ \left\langle \frac{-1}{\sqrt{66}}, \frac{-1}{\sqrt{66}}, \frac{8}{\sqrt{66}} \right\rangle \]
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