# Vector Applications Unit Assignment.docx - Vector...

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Vector Applications Unit Assignment1.a.a.b=2(3)+5(6)+(7) (2).630+14.10b.u=bb.b=32+(6)2+(2)2+7.u=[37,67,27]c.cosθ=a.bab.a.b=−10.a=78.b=7.10778θ=cos1¿).θ=99.3ºd.[2,5,7].[x , y, z]=0.2x+5y7z=0.x=1,y=1,z=1.2(1)+5(1)7(1)=0.[1,1,1]2.a.
.F=(2,1,5),S=(−3,5,4).¿F.S.¿(2,1,5).(−3,5,4).¿2(3)+1(5)+5(4)=31Jb.Since,w=31=|F||S|cosθ.|F|=31|S|cosθ.|F|is minimun,whencosθ=1.|F|=31151=319+25+16.¿4.384N3.a.a=[1,3,6],b=[4,5,2].a.b=^i^i^k136452¿(6+30)^i(224)^j+(−5+12)^k¿36^i+26^j+7^kWeknow,condition of perpendicularity of twovector sayxy isx .y=0Now ,¿provethat(b)is perpendicular¿bothab,weneedto showthat:a.(a.b)=0b.(a.b)=0Now ,a.(a.b)=(^i3^j+6^k)(36^i+26^j+7^k)¿3678+42¿7878¿0
So ,ais perpendicular¿(a.b)b.(a.b)=(4^i5^j2^k)(36^i+26^j+7^k)¿14413014¿1414¿0So ,bis perpendicular¿(a.b)Also ,(a.b)is perpendicular¿bothabb.If avectorc satisfythe relationa.(b.c)=0Wecansay thata,b,c arecoplanar.Because ,ifa.(b.c)=0,then wecansay thatais perpendicular¿(b.c)Also ,bothbcis perpendicular¿(b.c)This,eacha,b ,c is perpendicular¿the vector(b.c)After allthis wecansay thata,b,care coplanar4.