Find a vector of magnitude 9 in the direction of v = <- 3, 2>. O <-3√13, 2√13) O-27, 18) 9 <-3,2) 13 1 V <-3,2) 13 O None of these
Find a vector of magnitude 9 in the direction of v = <- 3, 2>. O <-3√13, 2√13) O-27, 18) 9 <-3,2) 13 1 V <-3,2) 13 O None of these
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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Question
![**Problem: Find a Vector of Specific Magnitude in a Given Direction**
**Question:**
Find a vector of magnitude 9 in the direction of \(\mathbf{v} = \langle -3, 2 \rangle\).
**Choices:**
- \(\langle -3\sqrt{13}, 2\sqrt{13} \rangle\)
- \(\langle -27, 18 \rangle\)
- \(\dfrac{9}{\sqrt{13}} \langle -3, 2 \rangle\)
- \(\dfrac{1}{\sqrt{13}} \langle -3, 2 \rangle\)
- None of these
**Explanation:**
To solve this problem, find a vector with a specified magnitude in the direction of a given vector. This involves vector normalization and scaling:
1. **Normalize the vector \(\mathbf{v}\):**
\[
\|\mathbf{v}\| = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}
\]
The unit vector in the direction of \(\mathbf{v}\) is:
\[
\mathbf{u} = \dfrac{\mathbf{v}}{\|\mathbf{v}\|} = \dfrac{\langle -3, 2 \rangle}{\sqrt{13}} = \left\langle \dfrac{-3}{\sqrt{13}}, \dfrac{2}{\sqrt{13}} \right\rangle
\]
2. **Scale the unit vector by the magnitude 9:**
\[
\mathbf{u}_{new} = 9 \mathbf{u} = 9 \left\langle \dfrac{-3}{\sqrt{13}}, \dfrac{2}{\sqrt{13}} \right\rangle = \left\langle \dfrac{9 \cdot (-3)}{\sqrt{13}}, \dfrac{9 \cdot 2}{\sqrt{13}} \right\rangle = \left\langle \dfrac{-27}{\sqrt{13}}, \dfrac{18}{\sqrt{13}} \right\rangle
\]
After analyzing the given choices, the correct simplified option is:
\[
\boxed{\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F55815b3c-2575-4201-a9dc-5a4b41a14c1d%2F3ad05bfb-887e-4cad-afcd-ce82eb6e3f46%2Forvcj7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem: Find a Vector of Specific Magnitude in a Given Direction**
**Question:**
Find a vector of magnitude 9 in the direction of \(\mathbf{v} = \langle -3, 2 \rangle\).
**Choices:**
- \(\langle -3\sqrt{13}, 2\sqrt{13} \rangle\)
- \(\langle -27, 18 \rangle\)
- \(\dfrac{9}{\sqrt{13}} \langle -3, 2 \rangle\)
- \(\dfrac{1}{\sqrt{13}} \langle -3, 2 \rangle\)
- None of these
**Explanation:**
To solve this problem, find a vector with a specified magnitude in the direction of a given vector. This involves vector normalization and scaling:
1. **Normalize the vector \(\mathbf{v}\):**
\[
\|\mathbf{v}\| = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}
\]
The unit vector in the direction of \(\mathbf{v}\) is:
\[
\mathbf{u} = \dfrac{\mathbf{v}}{\|\mathbf{v}\|} = \dfrac{\langle -3, 2 \rangle}{\sqrt{13}} = \left\langle \dfrac{-3}{\sqrt{13}}, \dfrac{2}{\sqrt{13}} \right\rangle
\]
2. **Scale the unit vector by the magnitude 9:**
\[
\mathbf{u}_{new} = 9 \mathbf{u} = 9 \left\langle \dfrac{-3}{\sqrt{13}}, \dfrac{2}{\sqrt{13}} \right\rangle = \left\langle \dfrac{9 \cdot (-3)}{\sqrt{13}}, \dfrac{9 \cdot 2}{\sqrt{13}} \right\rangle = \left\langle \dfrac{-27}{\sqrt{13}}, \dfrac{18}{\sqrt{13}} \right\rangle
\]
After analyzing the given choices, the correct simplified option is:
\[
\boxed{\
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