\(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) are nonzero vectors. Which from the followings is true ? (A) \(\left|\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right|^{2}-\left|\mathrm{A}^{\rightarrow}-\mathrm{B}^{\rightarrow}\right|^{2}=2\left(\mathrm{~A}^{2}+\mathrm{B}^{2}\right)\) (B) \(\left|\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right|^{2}-\left|\mathrm{A}^{\rightarrow}-\mathrm{B}^{\rightarrow}\right|^{2}=2\left(\mathrm{~A}^{2}+\mathrm{B}^{2}\right)\) (C) \(\left|\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right|^{2}-\left|\mathrm{A}^{\rightarrow}-\mathrm{B}^{\rightarrow}\right|^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}\) (D) \(\left|\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right|^{2}-\left|\mathrm{A}^{\rightarrow}-\mathrm{B}^{\rightarrow}\right|^{2}=\mathrm{A}^{2}-\mathrm{B}^{2}\)

Short Answer

Expert verified
None of the given options are correct. The actual simplification for the expression \(\left|\mathrm{A}^{\rightarrow}+\mathrm{B}^{\rightarrow}\right|^{2}-\left|\mathrm{A}^{\rightarrow}-\mathrm{B}^{\rightarrow}\right|^{2}\) is \(4\mathrm{A}\cdot\mathrm{B}\).

Step by step solution

01

Option A

Option A is: \(\left|\mathrm{A}^\rightarrow + \mathrm{B}^\rightarrow \right|^2 - \left|\mathrm{A}^\rightarrow - \mathrm{B}^\rightarrow \right|^2 = 2\left(\mathrm{A}^2 + \mathrm{B}^2 \right)\) First, let's find the square of the magnitudes of \(\mathrm{A}^\rightarrow + \mathrm{B}^\rightarrow\) and \(\mathrm{A}^\rightarrow - \mathrm{B}^\rightarrow\): 1. \(\left|\mathrm{A}^\rightarrow + \mathrm{B}^\rightarrow \right|^2 = (\mathrm{A} + \mathrm{B})\cdot(\mathrm{A} + \mathrm{B}) = \mathrm{A}^2 + 2\mathrm{A}\cdot\mathrm{B} + \mathrm{B}^2\) 2. \(\left|\mathrm{A}^\rightarrow - \mathrm{B}^\rightarrow \right|^2 = (\mathrm{A} - \mathrm{B})\cdot(\mathrm{A} - \mathrm{B}) = \mathrm{A}^2 - 2\mathrm{A}\cdot\mathrm{B} + \mathrm{B}^2\) Now let's compute the difference between the two squared magnitudes: \(\left|\mathrm{A}^\rightarrow + \mathrm{B}^\rightarrow \right|^2 - \left|\mathrm{A}^\rightarrow - \mathrm{B}^\rightarrow \right|^2 = (\mathrm{A}^2 + 2\mathrm{A}\cdot\mathrm{B} + \mathrm{B}^2) - (\mathrm{A}^2 - 2\mathrm{A}\cdot\mathrm{B} + \mathrm{B}^2) = 4\mathrm{A}\cdot\mathrm{B}\) Option A simplifies to \(4\mathrm{A}\cdot\mathrm{B}\), and it is different from the given expression. So option A is incorrect.
02

Option B

Option B is the same as Option A: \(\left|\mathrm{A}^\rightarrow + \mathrm{B}^\rightarrow \right|^2 - \left|\mathrm{A}^\rightarrow - \mathrm{B}^\rightarrow \right|^2 = 2\left(\mathrm{A}^2 + \mathrm{B}^2 \right)\) Option B also simplifies to \(4\mathrm{A}\cdot\mathrm{B}\), which is different from the given expression. So option B is also incorrect.
03

Option C

Option C is: \(\left|\mathrm{A}^\rightarrow + \mathrm{B}^\rightarrow \right|^2 - \left|\mathrm{A}^\rightarrow - \mathrm{B}^\rightarrow \right|^2 = \mathrm{A}^2 + \mathrm{B}^2\) Using the calculation of the difference between the squared magnitudes that we found in Option A, the actual simplification is: \(\left|\mathrm{A}^\rightarrow + \mathrm{B}^\rightarrow \right|^2 - \left|\mathrm{A}^\rightarrow - \mathrm{B}^\rightarrow \right|^2 = 4\mathrm{A}\cdot\mathrm{B}\) Option C is also incorrect because it does not simplify to the correct expression.
04

Option D

Option D is: \(\left|\mathrm{A}^\rightarrow + \mathrm{B}^\rightarrow \right|^2 - \left|\mathrm{A}^\rightarrow - \mathrm{B}^\rightarrow \right|^2 = \mathrm{A}^2 - \mathrm{B}^2\) Using the calculation of the difference between the squared magnitudes that we found in Option A, the actual simplification is: \(\left|\mathrm{A}^\rightarrow + \mathrm{B}^\rightarrow \right|^2 - \left|\mathrm{A}^\rightarrow - \mathrm{B}^\rightarrow \right|^2 = 4\mathrm{A}\cdot\mathrm{B}\) Option D is also incorrect because it does not simplify to the correct expression. None of the options given are correct. The actual simplification of the given difference between the squared magnitudes is \(4\mathrm{A}\cdot\mathrm{B}\).

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 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) \(4 \mathrm{~ms}^{-1}\)

Stone \(\mathrm{A}\) is thrown in horizontal direction with velocity of $10 \mathrm{~ms}^{-1}\( at the same time stone \)\mathrm{B}$ freely falls vertically in downward direction. Calculate the velocity of \(\mathrm{B}\) with respect to \(\mathrm{A}\) after 10 second. (A) \(10 \mathrm{~ms}^{-1}\) (B) \(\sqrt{(101)} \mathrm{ms}^{-1}\) (C) \(10 \sqrt{(101)} \mathrm{ms}^{-1}\) (D) 0

$\mathrm{A}^{\boldsymbol{\longrightarrow}}=2 \mathrm{i} \wedge+2 \mathrm{j} \wedge-\mathrm{k} \wedge\( and \)\mathrm{B}^{\rightarrow}=2 \mathrm{i} \wedge-\mathrm{j} \wedge-2 \mathrm{k} \wedge\( Find \)3 \mathrm{~A}^{\rightarrow}-2 \mathrm{~B}^{\rightarrow}$ (A) \(2 \mathrm{i} \wedge+7 \mathrm{j} \wedge+\mathrm{k} \wedge\) (B) \(2 \mathrm{i} \wedge+8 \mathrm{j} \wedge-\mathrm{k} \wedge\) (C) \(2 \mathrm{i} \wedge+8 \mathrm{j} \wedge+\mathrm{k} \wedge\) (D) \(\mathrm{i} \wedge+7 \mathrm{j} \wedge+\mathrm{k} \wedge\)

A particle is projected with initial speed of \(\mathrm{V}_{0}\) and angle of \(\theta\). Find the horizontal displacement when its velocity is perpendicular to initial velocity. (A) $\left[\left(\mathrm{V}_{0}^{2}\right) /(\operatorname{gtan} \theta)\right]$ (B) \(\left[\left(\mathrm{V}_{0}^{2}\right) /(\mathrm{g} \sin \theta)\right]\) (C) \(\left[\left(\mathrm{V}_{0} \sin \theta\right) / \mathrm{g}\right]\) (D) \(\left[\left(\mathrm{V}_{0}^{2}\right) /(\tan \theta)\right]\)

A particle moves on a plane along the path \(\mathrm{y}=\mathrm{Ax}^{3}+\mathrm{B}\) in such a way that $(\mathrm{dx} / \mathrm{dt})=\mathrm{c} . \mathrm{c}, \mathrm{A}, \mathrm{B}$ are constant. Calculate the acceleration of the particle. (A) \(3 \mathrm{Ax} \mathrm{cj} \mathrm{ms}^{-2}\) (B) \(6 \mathrm{Axc}^{2} \hat{\mathrm{j}} \mathrm{ms}^{-2}\) (C) \(3 \mathrm{Axc}^{2} \hat{\mathrm{j}} \mathrm{ms}^{-2}\) (D) \(\left[c \hat{i}+3 A x c^{2} \hat{j}\right] m s^{-2}\)

See all solutions

Recommended explanations on English 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.

Sign-up for free