/*! 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 21 Show that \((a) \hat{\mathbf{i}}... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show that \((a) \hat{\mathbf{i}} \times \hat{\mathbf{i}}=\hat{\mathbf{j}} \times \hat{\mathbf{j}}=\hat{\mathbf{k}} \times \hat{\mathbf{k}}=0\) (b) \(\hat{\mathbf{i}} \times \hat{\mathbf{j}}=\hat{\mathbf{k}}\) \(\hat{\mathbf{i}} \times \hat{\mathbf{k}}=-\hat{\mathbf{j}},\) and \(\hat{\mathbf{j}} \times\hat{\mathbf{k}}=\hat{\mathbf{i}}\)

Short Answer

Expert verified
(a) Zero since vectors are parallel. (b) Use the right-hand rule for direction.

Step by step solution

01

Understand Cross Product Property

The cross product of two vectors \(\mathbf{A}\) and \(\mathbf{B}\), denoted \(\mathbf{A} \times \mathbf{B}\), is a vector that is perpendicular to both \(\mathbf{A}\) and \(\mathbf{B}\). The magnitude of \(\mathbf{A} \times \mathbf{B}\) is given by \(|\mathbf{A}||\mathbf{B}|\sin\theta\), where \(\theta\) is the angle between \(\mathbf{A}\) and \(\mathbf{B}\). If \(\mathbf{A}\) and \(\mathbf{B}\) are parallel (or identical), the angle \(\theta = 0\), making \(\sin\theta = 0\) and thus the cross product is zero.
02

Part (a) Solution

Given that the unit vectors \(\hat{\mathbf{i}}\), \(\hat{\mathbf{j}}\), and \(\hat{\mathbf{k}}\) are orthogonal unit vectors directed along the axes of a right-handed coordinate system. If we take the cross product of any unit vector with itself, such as \(\hat{\mathbf{i}} \times \hat{\mathbf{i}}\), the vectors are parallel, hence \(\hat{\mathbf{i}} \times \hat{\mathbf{i}} = 0\). Similarly, \(\hat{\mathbf{j}} \times \hat{\mathbf{j}} = 0\) and \(\hat{\mathbf{k}} \times \hat{\mathbf{k}} = 0\). Therefore, \[\hat{\mathbf{i}} \times \hat{\mathbf{i}} = \hat{\mathbf{j}} \times \hat{\mathbf{j}} = \hat{\mathbf{k}} \times \hat{\mathbf{k}} = 0.\]
03

Part (b.i) Cross Product of \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\)

The cross product \(\hat{\mathbf{i}} \times \hat{\mathbf{j}}\) follows the right-hand rule: if the fingers of the right hand curve from \(\hat{\mathbf{i}}\) to \(\hat{\mathbf{j}}\), the thumb points in the direction of the cross product, which is \(\hat{\mathbf{k}}\). Thus, \(\hat{\mathbf{i}} \times \hat{\mathbf{j}} = \hat{\mathbf{k}}\).
04

Part (b.ii) Cross Product of \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{k}}\)

Using the right-hand rule, if the fingers curl from \(\hat{\mathbf{i}}\) to \(\hat{\mathbf{k}}\), the thumb points in the reverse direction of \(\hat{\mathbf{j}}\), hence \(\hat{\mathbf{i}} \times \hat{\mathbf{k}} = -\hat{\mathbf{j}}\).
05

Part (b.iii) Cross Product of \(\hat{\mathbf{j}}\) and \(\hat{\mathbf{k}}\)

Following the right-hand rule again, if the fingers curve from \(\hat{\mathbf{j}}\) to \(\hat{\mathbf{k}}\), the thumb points in the direction of \(\hat{\mathbf{i}}\). Therefore, \(\hat{\mathbf{j}} \times \hat{\mathbf{k}} = \hat{\mathbf{i}}\).

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 Multiplication
Vector multiplication, often referred to as the cross product, is a mathematical operation performed on two vectors in three-dimensional space. The result of this operation is another vector that is orthogonal, or perpendicular, to both of the original vectors.
  • The cross product of vectors \(\mathbf{A}\) and \(\mathbf{B}\) is denoted as \(\mathbf{A} \times \mathbf{B}\).
  • The direction of the resulting vector is determined by the orientation of the original vectors in space.
  • The magnitude of the cross product is calculated using the formula \(|\mathbf{A}||\mathbf{B}|\sin\theta\), where \(\theta\) is the angle between \(\mathbf{A}\) and \(\mathbf{B}\).
When vectors are parallel or identical, meaning the angle between them is zero, their cross product is zero because the sine of zero degrees is zero. Understanding this helps explain why the cross product of any unit vector with itself, such as \(\hat{\mathbf{i}} \times \hat{\mathbf{i}}\), results in zero.
Right-Hand Rule
The right-hand rule is a simple and effective tool to determine the direction of the cross product vector. To apply it, imagine using your right hand:
  • Align your fingers in the direction of the first vector, \(\mathbf{A}\).
  • Then, curl your fingers towards the second vector, \(\mathbf{B}\).
  • Your thumb will point in the direction of the resultant vector of the cross product, \(\mathbf{A} \times \mathbf{B}\).
This rule is particularly useful when determining the cross product of the standard unit vectors:
  • \(\hat{\mathbf{i}} \times \hat{\mathbf{j}} = \hat{\mathbf{k}}\) - Curl from \(\hat{\mathbf{i}}\) to \(\hat{\mathbf{j}}\), thumb points in direction of \(\hat{\mathbf{k}}\).
  • \(\hat{\mathbf{i}} \times \hat{\mathbf{k}} = -\hat{\mathbf{j}}\) - Curl from \(\hat{\mathbf{i}}\) to \(\hat{\mathbf{k}}\), thumb opposes \(\hat{\mathbf{j}}\).
  • \(\hat{\mathbf{j}} \times \hat{\mathbf{k}} = \hat{\mathbf{i}}\) - Curl from \(\hat{\mathbf{j}}\) to \(\hat{\mathbf{k}}\), thumb aligns with \(\hat{\mathbf{i}}\).
Remember, the right-hand rule only works for cross products in a right-handed coordinate system.
Unit Vectors
Unit vectors are vectors that have a magnitude of one and are typically used to indicate direction. In three-dimensional space, the coordinate axes have standard unit vectors:
  • \(\hat{\mathbf{i}}\) for the x-axis,
  • \(\hat{\mathbf{j}}\) for the y-axis,
  • \(\hat{\mathbf{k}}\) for the z-axis.
These unit vectors are fundamental since they define the direction along each axis but without contributing to the magnitude. In the context of the cross product:
  • Any unit vector crossed with itself results in a vector of zero magnitude -> \(\hat{\mathbf{i}} \times \hat{\mathbf{i}} = 0\).
  • The cross product of different unit vectors follows the rules determined by the coordinate system.
Unit vectors are crucial for building and understanding more complex vector operations.
Orthogonal Vectors
Orthogonal vectors are vectors that are at right angles to each other, meaning they have a dot product of zero. In the context of the unit vectors \(\hat{\mathbf{i}}\), \(\hat{\mathbf{j}}\), and \(\hat{\mathbf{k}}\), each pair is orthogonal:
  • \(\hat{\mathbf{i}} \cdot \hat{\mathbf{j}} = 0\),
  • \(\hat{\mathbf{i}} \cdot \hat{\mathbf{k}} = 0\),
  • \(\hat{\mathbf{j}} \cdot \hat{\mathbf{k}} = 0\).
They form the basis of the right-handed coordinate system, making them indispensable for vector mathematics and physics. In vector multiplication contexts, orthogonality is essential because it often ensures the resultant vector from a cross product is meaningful and aligned perfectly perpendicular to the original vectors:
  • If two vectors are orthogonal, the cross product produces a non-zero vector.
  • When using unit vectors, this guides the expected direction of the resultant vector according to the right-hand rule.
Understanding orthogonality helps in visualizing and solving vector-related problems.

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

A uniform horizontal rod of mass \(M\) and length \(\ell\) rotates with angular velocity \(\omega\) about a vertical axis through its center. Attached to each end of the rod is a small mass \(m\). Determine the angular momentum of the system about the axis.

$$ \begin{array}{l}{\text { (1) Show that }(a) \hat{\mathbf{i}} \times \hat{\mathbf{i}}=\hat{\mathbf{j}} \times \hat{\mathbf{j}}=\hat{\mathbf{k}} \times \hat{\mathbf{k}}=0,(b) \hat{\mathbf{i}} \times \hat{\mathbf{j}}=\hat{\mathbf{k}},} \\ {\hat{\mathbf{i}} \times \hat{\mathbf{k}}=-\hat{\mathbf{j}}, \text { and } \hat{\mathbf{j}} \times \hat{\mathbf{k}}=\hat{\mathbf{i}}}\end{array} $$

(II) A figure skater can increase her spin rotation rate from an initial rate of 1.0 rev every 1.5 \(\mathrm{s}\) a final rate of 2.5 \(\mathrm{rev} / \mathrm{s} .\) If her initial moment of inertia was 4.6 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) , what is her final moment of inertia? How does she physi- cally accomplish this change?

A uniform stick \(1.0 \mathrm{~m}\) long with a total mass of \(270 \mathrm{~g}\) is pivoted at its center. A 3.0 -g bullet is shot through the stick midway between the pivot and one end (Fig. \(11-36\) ). The bullet approaches at \(250 \mathrm{~m} / \mathrm{s}\) and leaves at \(140 \mathrm{~m} / \mathrm{s}\). With what angular speed is the stick spinning after the collision?

(1I) A 4.2 -m-diameter merry-go-round is rotating frecly with an angular velocity of 0.80 rad/s. Its total moment of inertia is 1760 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) Four people standing on the ground, cach of mass 65 \(\mathrm{kg}\) suddenly step onto the edge of the merry-go-round.What is the angular velocity of the merry-go-round now? What if the people were on it initially and then jumped off in a radial direction (relative to the merry-go-round)?

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.