Chapter 1: Problem 43
Determine a vector of magnitude 5 in the direction of vector \(A B\), where \(A(2,1,5)\) and \(B(3,4,-7)\).
Short Answer
Expert verified
The vector is \( \left( \frac{5}{\sqrt{154}}, \frac{15}{\sqrt{154}}, \frac{-60}{\sqrt{154}} \right) \).
Step by step solution
01
Find the Direction Vector
Calculate the direction vector \( \mathbf{v} \) from point \( A \) to point \( B \) by subtracting the coordinates of \( A \) from \( B \). The calculation is as follows: \( \mathbf{v} = (3-2, 4-1, -7-5) = (1, 3, -12) \).
02
Calculate the Magnitude of the Direction Vector
Find the magnitude of \( \mathbf{v} = (1, 3, -12) \) using the formula \( \| \mathbf{v} \| = \sqrt{1^2 + 3^2 + (-12)^2} \). This simplifies to \( \| \mathbf{v} \| = \sqrt{1 + 9 + 144} = \sqrt{154} \).
03
Normalize the Direction Vector
Normalize the vector \( \mathbf{v} \) by dividing each component by its magnitude \( \sqrt{154} \) to find the unit vector in the direction of \( AB \). The unit vector \( \hat{\mathbf{v}} \) is \( \left( \frac{1}{\sqrt{154}}, \frac{3}{\sqrt{154}}, \frac{-12}{\sqrt{154}} \right) \).
04
Scale the Unit Vector to Desired Magnitude
Scale the unit vector \( \hat{\mathbf{v}} \) by 5 to get the vector with magnitude 5. This calculation is \( \mathbf{v'} = 5 \times \left( \frac{1}{\sqrt{154}}, \frac{3}{\sqrt{154}}, \frac{-12}{\sqrt{154}} \right) \), resulting in \( \mathbf{v'} = \left( \frac{5}{\sqrt{154}}, \frac{15}{\sqrt{154}}, \frac{-60}{\sqrt{154}} \right) \).
05
Simplify the Scaled Vector
The vector \( \mathbf{v'} \) is scaled to a magnitude of 5. Simplify if necessary to present the final vector in a clearer form.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Magnitude
The magnitude of a vector, often referred to as its "length," is a measure of how long the vector is. It’s calculated using the Pythagorean theorem in multi-dimensional space. To find the magnitude of a three-dimensional vector, such as \( \mathbf{v} = (1, 3, -12) \), we use the formula:
- \[ \| \mathbf{v} \| = \sqrt{1^2 + 3^2 + (-12)^2} \]
Direction Vector
A direction vector shows the path from one point to another in a defined space. For example, to find the direction vector from point \( A (2,1,5) \) to point \( B (3,4,-7) \), we subtract the coordinates of \( A \) from \( B \). The calculation is:
- \( \mathbf{v} = (3-2, 4-1, -7-5) = (1, 3, -12) \)
Unit Vector
A unit vector is a vector with a magnitude of 1. Its primary use is to indicate direction without concerning magnitude. To find a unit vector in the direction of a given vector, we need to divide each component of the vector by its magnitude.For the direction vector \( \mathbf{v} = (1, 3, -12) \), whose magnitude was \( \sqrt{154} \), the unit vector \( \hat{\mathbf{v}} \) becomes:
- \( \hat{\mathbf{v}} = \left( \frac{1}{\sqrt{154}}, \frac{3}{\sqrt{154}}, \frac{-12}{\sqrt{154}} \right) \)
Normalization
Normalization is an essential process of converting a vector into a unit vector. This means adjusting the vector's magnitude to 1 while maintaining its direction. In essence, the original vector's influence is kept intact, but without any scale bias.Given a vector \( \mathbf{v} \), normalize it by dividing each component by its magnitude, \( \| \mathbf{v} \| \). Consequently, our vector \( (1, 3, -12) \) with magnitude \( \sqrt{154} \) turns into:
- \( \left( \frac{1}{\sqrt{154}}, \frac{3}{\sqrt{154}}, \frac{-12}{\sqrt{154}} \right) \)