/*! 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} Q. 67 Let u and v be vectors in 鈩3 a... [FREE SOLUTION] | 91影视

91影视

Let u and v be vectors in 3 and let c be a scalar. Prove that c(uv)=(cu)v=u(cv). (This is Theorem 10.28).

Short Answer

Expert verified

Hence, we prove thatc(uv)=(cu)v=u(cv).

Step by step solution

01

Step 1. Given Information

Let u and v be vectors in 3 and let c be a scalar. Prove that c(uv)=(cu)v=u(cv).

02

Step 2. We have to prove c(u×v) =(cu)×v =u×(cv)

Let u=(u1,u2,u3)andv=(v1,v2,v3)

Firstly finding the value of c(uv)

role="math" localid="1649918741570" c(uv)=c{(u1,u2,u3)(v1,v2,v3)}c(uv)=cijku1u2u3v1v2v3c(uv)=ciu2u3v2v3-ju1u3v1v3+ku1u2v1v2c(uv)=ci(u2v3u3v2)-j(u1v3-u3v1)+k(u1v2u2v1)c(uv)=c(u2v3u3v2,u1v3+u3v1,u1v2u2v1)c(uv)=(cu2v3cu3v2,cu1v3+cu3v1,cu1v2cu2v1)

03

Step 3. Now finding the value of (cu)×v

(cu)v=(cu1,cu2,cu3)(v1,v2,v3)(cu)v=ijkcu1cu2cu3v1v2v3(cu)v=icu2cu3v2v3-jcu1cu3v1v3+kcu1cu2v1v2(cu)v=i(cu2v3cu3v2)-j(cu1v3-cu3v1)+k(cu1v2cu2v1)(cu)v=(cu2v3cu3v2,cu1v3+cu3v1,cu1v2cu2v1)

04

Step 4. Now finding the value of u×(cv)

u(cv)=(u1,u2,u3)(cv1,cv2,cv3)u(cv)=ijku1u2u3cv1cv2cv3u(cv)=iu2u3cv2cv3-ju1u3cv1cv3+ku1u2cv1cv2u(cv)=i(cu2v3cu3v2)-j(cu1v3-cu3v1)+k(cu1v2cu2v1)u(cv)=(cu2v3cu3v2,cu1v3+cu3v1,cu1v2cu2v1)

Hence, prove thatc(uv)=(cu)v=u(cv)

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 91影视!

One App. One Place for Learning.

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

Get started for free

Study anywhere. Anytime. Across all devices.