/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 (a) Is the vector \((\hat{\imath... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) Is the vector \((\hat{\imath}+\hat{j}+\hat{k})\) a unit vector? Justify your answer. (b) Can a unit vector have any components with magnitude greater than unity? Can it have any negative components? In each case justify your answer. (c) If \(\overrightarrow{\boldsymbol{A}}=a(3.0 \hat{\imath}+4.0 \hat{\mathbf{y}}),\) where \(\boldsymbol{a}\) is a constant, determine the value of \(a\) that makes \(\vec{A}\) a unit vector.

Short Answer

Expert verified
(a) No, its magnitude is \(\sqrt{3}\). (b) It cannot have components > 1, but can be negative. (c) \(a = \frac{1}{5}\).

Step by step solution

01

Check if the vector is a unit vector

To determine if the vector \((\hat{\imath} + \hat{j} + \hat{k})\) is a unit vector, we need to calculate its magnitude. The magnitude of a vector \((a\hat{\imath} + b\hat{j} + c\hat{k})\) is computed as \(\sqrt{a^2 + b^2 + c^2}\). For our vector, \(a = 1, b = 1,\) and \(c = 1\). The magnitude is \(\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}\). Since \(\sqrt{3} eq 1\), the vector is not a unit vector.
02

Components of a unit vector

A unit vector is defined as a vector with a magnitude of 1. Consequently, none of its components can have a magnitude greater than 1, as this would increase the overall magnitude beyond 1. However, a unit vector can have negative components. The direction of a component or vector does not impact its magnitude; it only changes its orientation.
03

Find value of \(a\) for \(\overrightarrow{A}=a(3.0 \hat{\imath} + 4.0 \hat{\mathbf{y}})\) to be a unit vector

To find the value of \(a\) that makes \(\overrightarrow{A}\) a unit vector, we need the magnitude of \(\overrightarrow{A}\) to be 1. First calculate the magnitude: \(a\sqrt{(3.0)^2 + (4.0)^2} = a\sqrt{9 + 16} = a\sqrt{25} = 5a\). Set the magnitude equal to 1: \(5a = 1\). Solve for \(a\) by dividing both sides by 5: \(a = \frac{1}{5}\).
04

Conclusion

Summarizing: (a) The vector \((\hat{\imath} + \hat{j} + \hat{k})\) is not a unit vector because its magnitude is \(\sqrt{3}\). (b) A unit vector cannot have components with magnitude greater than 1 but can have negative components. (c) The value of \(a\) required to make \(\overrightarrow{A}\) a unit vector is \(\frac{1}{5}\).

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.

Unit Vector
A unit vector is a vector that has a magnitude of exactly 1 unit. Its primary purpose is to denote direction without affecting the magnitude. Imagine it as an arrow that points in a direction but has a fixed length of 1. This concept is very important in physics and engineering to express directions efficiently.

To determine if a given vector is a unit vector, you calculate its magnitude and check if it equals 1. Magnitude is calculated using the formula: \\( \text{magnitude} = \sqrt{a^2 + b^2 + c^2} \). For instance, for the vector \((\hat{\imath} + \hat{j} + \hat{k})\), each component is 1, so the magnitude is \(\sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}\). Because \(\sqrt{3} eq 1\), it is not a unit vector.

Unit vectors are useful because they allow you to separate direction from magnitude. When working with unit vectors, you can simply scale them to the desired magnitude for practical applications.
Magnitude of a Vector
The magnitude of a vector is its length, often thought of as the 'size' of the vector. For a vector, having components \(a\), \(b\), and \(c\), the magnitude is found using the formula: \[ \text{magnitude} = \sqrt{a^2 + b^2 + c^2} \]. This formula stems from the Pythagorean theorem applied in multiple dimensions.

The importance of vector magnitude lies in its various applications such as calculating speeds, forces, and distances in physics. Thus, understanding how to calculate the magnitude accurately is crucial in modeling real-world scenarios.

The exercise above involves confirming if a vector \((\hat{\imath} + \hat{j} + \hat{k})\) is a unit vector. We found its magnitude to be \(\sqrt{3}\), indicating that it is not a unit vector as its magnitude is not equal to 1. This step is key in determining whether a given vector, when scaled, can represent a true directionality without affecting size.
Vector Components
Every vector can be broken down into components that lie along the axes of a coordinate system. For example, a vector in three-dimensional space has three components along the x, y, and z axes. If a vector is expressed as \(a\hat{\imath} + b\hat{j} + c\hat{k}\), then \(a\), \(b\), and \(c\) are its components.

Understanding vector components allows us to analyze various attributes of the vector, such as direction and magnitude, separately. It’s crucial to remember that components can have negative values, which indicate direction. A negative component points in the opposite direction of the corresponding positive axis. However, for unit vectors, none of the components can exceed a magnitude of 1, or else the total magnitude would be greater than 1. Negative components are perfectly acceptable, as long as the overall magnitude remains 1.

Analyzing a vector based on its components helps solve problems related to direction and force distribution in physics. By altering components, you can represent any vector uniquely, making computations more convenient and interpretations more rigorous in practical scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Completed Pass. At Enormous State University (ESU), the football team records its plays using vector displacements, with the origin taken to be the position of the ball before the play starts. In a certain pass play, the receiver starts at \(+1.0 \hat{\imath}-5.0 \hat{\jmath}\) , where the units are yards, \(\hat{\mathbf{i}}\) is to the right, and \(\hat{\boldsymbol{j}}\) is downfield. Subsequent displacements of the receiver are \(+9.0 \hat{\imath}\) (in motion before the snap), \(+11.0 \hat{y}\) (breaks downfield), \(-6.0 \hat{\imath}+4.0 \hat{\jmath}\) (zigs), and \(+12.0 \hat{i}+18.0 \hat{j}\) (zags). Meanwhile, the quarterback has dropped straight back to a position \(-7.0 \hat{\jmath}\) . How far and in which direction must the quarterback throw the ball? (Like the coach, you will be well advised to diagram the situation before solving it numerically.)

A disoriented physics professor drives 3.25 \(\mathrm{km}\) north, then 4.75 \(\mathrm{km}\) west, and then 1.50 \(\mathrm{km}\) south. Find the magnitude and direction of the resultant displacement, using the method of components. In a vector addition diagram (roughly to scale), show that the resultant displacement found from your diagram is in qualitative agreement with the result you obtained using the method of components.

By making simple sketches of the appropriate vector products, show that \((a) \vec{A} \cdot \vec{B}\) can be interpreted as the product of the magnitude of \(\overrightarrow{\boldsymbol{A}}\) times the component of \(\overrightarrow{\boldsymbol{B}}\) along \(\overrightarrow{\boldsymbol{A}}\), or the magnitude of \(\vec{B}\) times the component of \(\vec{A}\) along \(\overrightarrow{\boldsymbol{B}}\) (b) \(|\overrightarrow{\boldsymbol{A}} \times \overrightarrow{\boldsymbol{B}}|\) can be interpreted as the product of the magnitude of \(\overrightarrow{\boldsymbol{A}}\) times the component of \(\overrightarrow{\boldsymbol{B}}\) perpendicular to \(\overrightarrow{\boldsymbol{A}},\) or the magnitude of \(\overrightarrow{\boldsymbol{B}}\) times the component \(\overrightarrow{\boldsymbol{A}}\) perpendicular to \(\overrightarrow{\boldsymbol{B}}\).

With a wooden ruler you measure the length of a rectangular piece of sheet metal to be 12 \(\mathrm{mm}\) . You use micrometer calipers to measure the width of the rectangle and obtain the value 5.98 \(\mathrm{mm}\). Give your answers to the following questions to the correct number of significant figures. (a) What is the area of the rectangle? (b) What is the ratio of the rectangle's width to its length? (c) What is the perimeter of the rectangle? (d) What is the difference between the length and width? (e) What is the ratio of the length to the width?

As you eat your way through a bag of chocolate chip cookies, you observe that each cookie is a circular disk with a diameter of \(8.50 \pm 0.02 \mathrm{cm}\) and a thickness of \(0.050 \pm 0.005 \mathrm{cm} .\) (a) Find the average volume of a cookie and the uncertainty in the volume. (b) Find the ratio of the diameter to the thickness and the uncertainty in this ratio.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.