Chapter 14: Problem 16
Find the unit vector that has the same direction as a $$ \mathbf{a}=2\langle-2,5,-1\rangle $$
Short Answer
Expert verified
The unit vector is \( \mathbf{u} = \langle \frac{-2\sqrt{30}}{15}, \frac{\sqrt{30}}{3}, \frac{-\sqrt{30}}{15} \rangle \).
Step by step solution
01
Understand the Unit Vector
A unit vector is a vector with a magnitude (length) of 1, pointing in the same direction as the given vector. To find the unit vector of any vector \( \mathbf{a} \), you need to divide the vector by its magnitude.
02
Find the Magnitude of the Vector \( \mathbf{a} \)
Calculate the magnitude of the vector \( \mathbf{a} = 2\langle -2, 5, -1 \rangle \) using the formula for the magnitude of a vector: \[ ||\mathbf{a}|| = \sqrt{(-2)^2 + 5^2 + (-1)^2} \] Compute the squares: \[ (-2)^2 = 4, \quad 5^2 = 25, \quad (-1)^2 = 1 \] Add them up and take the square root:\[ ||\mathbf{a}|| = \sqrt{4 + 25 + 1} = \sqrt{30} \]
03
Compute the Unit Vector
Divide each component of vector \( \mathbf{a} = 2\langle -2, 5, -1 \rangle \) by its magnitude \( \sqrt{30} \):\[ \mathbf{u} = \frac{1}{\sqrt{30}} \cdot 2\langle -2, 5, -1 \rangle = \langle \frac{-4}{\sqrt{30}}, \frac{10}{\sqrt{30}}, \frac{-2}{\sqrt{30}} \rangle \]
04
Simplify the Unit Vector Components
Simplify each component of the unit vector to make it more readable:\[ \mathbf{u} = \langle \frac{-4}{\sqrt{30}}, \frac{10}{\sqrt{30}}, \frac{-2}{\sqrt{30}} \rangle \] If desired, you can rationalize the denominator by multiplying numerators and denominators by \( \sqrt{30} \), resulting in:\[ \mathbf{u} = \langle \frac{-4\sqrt{30}}{30}, \frac{10\sqrt{30}}{30}, \frac{-2\sqrt{30}}{30} \rangle \] Finally, simplify the fractions:\[ \mathbf{u} = \langle \frac{-2\sqrt{30}}{15}, \frac{\sqrt{30}}{3}, \frac{-\sqrt{30}}{15} \rangle \]
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.
Magnitude of a Vector
When working with vectors, one fundamental concept is understanding the magnitude of a vector. The magnitude gives us the "length" or "size" of the vector in the space it inhabits. It is always a positive value and provides insights into how far the vector extends from its origin to its endpoint.
The formula for calculating the magnitude of a vector \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \) in three-dimensional space includes squaring each component, summing those squares, and then taking the square root of the result. Mathematically, it is expressed as follows:
The formula for calculating the magnitude of a vector \( \mathbf{v} = \langle v_1, v_2, v_3 \rangle \) in three-dimensional space includes squaring each component, summing those squares, and then taking the square root of the result. Mathematically, it is expressed as follows:
- \[ ||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]
Vector Components
Vector components are essentially the building blocks of a vector in any given coordinate system. They tell us how much the vector extends in each direction, typically represented as \( x \), \( y \), and \( z \) in a Cartesian coordinate system.
To find a vector's components, imagine breaking it down into parts for each axis. For example, a vector \( \mathbf{a} = 2\langle -2, 5, -1 \rangle \) can be decomposed into individual parts showing its influence in each direction:
To find a vector's components, imagine breaking it down into parts for each axis. For example, a vector \( \mathbf{a} = 2\langle -2, 5, -1 \rangle \) can be decomposed into individual parts showing its influence in each direction:
- -4 in the x-direction
- 10 in the y-direction
- -2 in the z-direction
Rationalizing the Denominator
Rationalizing the denominator involves the process of eliminating any irrational numbers, like square roots, from a denominator of a fraction. This technique helps simplify expressions, making them neater and often easier to interpret or further manipulate.
To rationalize a denominator, we multiply both the numerator and denominator by a number that will remove the square root from the denominator, thus "rationalizing" it. In the case of the unit vector components like \( \frac{-4}{\sqrt{30}} \), multiplying top and bottom by \( \sqrt{30} \) transforms it to:
To rationalize a denominator, we multiply both the numerator and denominator by a number that will remove the square root from the denominator, thus "rationalizing" it. In the case of the unit vector components like \( \frac{-4}{\sqrt{30}} \), multiplying top and bottom by \( \sqrt{30} \) transforms it to:
- \[ \frac{-4\sqrt{30}}{30} \]