Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 2: Problem 275

\(\mathrm{y}\) component of \(\mathrm{A} \rightarrow \times \mathrm{B}^{-}\) is. $\mathrm{A}^{\rightarrow}=\mathrm{Ax} \hat{\imath}+\mathrm{Ay} \hat{\mathrm{j}}+\mathrm{A} z \mathrm{k}^{\wedge}$ $\mathrm{B}^{\rightarrow}=\mathrm{Bx} \hat{1}+\mathrm{By} \hat{\jmath}+\mathrm{Bzk}^{\wedge}$ (A) \(A x B y-A y B x\) (B) \(\mathrm{Az} \mathrm{Bx}-\mathrm{AxBz}\) (C) \(\mathrm{AxBz}-\mathrm{AzBx}\) (D) \(\mathrm{AzBy}-\mathrm{AyBz}\)

Short Answer

Expert verified
The short answer is: \(A \times B\) y-component = \(-(\mathrm{AxBz} - \mathrm{AzBx})\), which corresponds to option (B).

Step by step solution

01

Recall the cross product formula

To find the cross product of two vectors, we use the formula: \[\begin{bmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{bmatrix}\]
02

Compute the cross product

Compute the determinant of the matrix in the formula. We have: \[\mathrm{A} \times \mathrm{B} = \left(\mathrm{AyBz} - \mathrm{AzBy}\right) \hat{\imath} - \left(\mathrm{AxBz} - \mathrm{AzBx}\right) \hat{\jmath} + \left(\mathrm{AxBy} - \mathrm{AyBx}\right) \hat{k}\]
03

Identify the y component of the cross product

Based on the cross product computed in step 2, we see that the y component of A x B is the term with the \(\hat{\jmath}\) symbol: \[- \left(\mathrm{AxBz} - \mathrm{AzBx}\right) \]
04

Match the result with the given options

Compare the y component that we found in step 3 with the given options. This corresponds to option (B): \[\mathrm{Az} \mathrm{Bx}-\mathrm{AxBz}\]

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 cartasian equation of a projectile is given by $\mathrm{y}=2 \mathrm{x}-5 \mathrm{x}^{2}$ Calculate its initial velocity. (A) \(\sqrt{10 \mathrm{~ms}^{-1}}\) (B) \(\sqrt{5 \mathrm{~ms}^{-1}}\) (C) \(\sqrt{2} \mathrm{~ms}^{-1}\) (D) \(4 \mathrm{~ms}^{-1}\)

The relation between velocity and position of a particle is \(\mathrm{V}=\mathrm{Ax}+\mathrm{B}\) where \(\mathrm{A}\) and \(\mathrm{B}\) are constants. Acceleration of the particle is \(10 \mathrm{~ms}^{-2}\) when its velocity is \(\mathrm{V}\), How much is the acceleration when its velocity is $2 \mathrm{~V}$ (A) \(20 \mathrm{~ms}^{-2}\) (B) \(10 \mathrm{~ms}^{-1}\) (C) \(5 \mathrm{~ms}^{-2}\) (D) 0

If $\mathrm{A}^{\rightarrow}=2 \mathrm{i} \wedge+5 \mathrm{j} \wedge-\mathrm{k} \wedge\( and \)\mathrm{B}^{\rightarrow}=3 \mathrm{i} \wedge-2 \mathrm{j} \wedge-4 \mathrm{k} \wedge$ the angle between \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) is (A) 0 (B) \((\pi / 2)\) (C) \((\pi / 4)\) (D) \((\pi / 6)\)

A stone is projected with an angle \(\theta\) and velocity \(\mathrm{V}_{0}\) from point \(P\). It strikes the ground at point \(Q\). If the both \(P\) and \(Q\) are on same horizontal line, then find average velocity. (A) \(V_{0} \cos \theta\) (B) \(\mathrm{V}_{0} \sin \theta\) (C) \(\mathrm{V}_{0} \cos (\theta / 2)\) (D) \(\mathrm{V}_{0} \sin (\theta / 2)\)

A particle moves in one direction with acceleration \(2 \mathrm{~ms}^{-2}\) and initial velocity \(3 \mathrm{~ms}^{-1}\).After what time its displacement will be \(10 \mathrm{~m}\) ? (A) \(1 \mathrm{~s}\) (B) \(2 \mathrm{~s}\) (C) \(3 \mathrm{~s}\) (D) \(4 \mathrm{~s}\)
See all solutions

Recommended explanations on English Textbooks

Summary Text

Read Explanation

5 Paragraph Essay

Read Explanation

Phonology

Read Explanation

International English

Read Explanation

Prosody

Read Explanation

English Grammar

Read Explanation
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.

Sign-up for free