A2Physics物理出國(guó)英語(yǔ)VectorsMS

1.Whichofthefollowingstatementsaboutvectorsiscorrect?

Answer:Avectorhasbothmagnitudeanddirection.Forexample,velocityisavectorquantityasitnotonlytellshowfastanobjectismoving(magnitude)butalsoinwhichdirectiontheobjectismoving.Incontrast,scalarquantitieslikespeedonlyhavemagnitude.

2.Ifavector(vec{A})hasamagnitudeof5unitsandisdirectedalongthepositivexaxis,andanothervector(vec{B})hasamagnitudeof3unitsandisdirectedalongthenegativexaxis,whatistheresultantvector(vec{R}=vec{A}+vec{B})?

Answer:Sincethevectorsarealongthesame(x)axis,wecandirectlyaddtheirmagnitudesconsideringthedirections.(vec{A})hasamagnitudeof(A=5)inthepositivexdirectionand(vec{B})hasamagnitudeof(B=3)inthenegativexdirection.Theresultantvector(vec{R})hasamagnitude(R=AB=53=2)units,anditisdirectedalongthepositivexaxis.

3.Avector(vec{P})makesanangleof(30^{circ})withthepositivexaxis.Ifitsmagnitudeis10units,whatareitsxandycomponents?

Answer:Thexcomponentofavector(vec{P})withmagnitude(P)andangle(theta)withthepositivexaxisisgivenby(P_x=Pcostheta)andtheycomponentisgivenby(P_y=Psintheta).Here,(P=10)and(theta=30^{circ}).So,(P_x=10cos30^{circ}=10timesfrac{sqrt{3}}{2}=5sqrt{3})unitsand(P_y=10sin30^{circ}=10timesfrac{1}{2}=5)units.

4.Twovectors(vec{M})and(vec{N})aresuchthat(vec{M})hasamagnitudeof4unitsandisdirectedatanangleof(45^{circ})withthepositivexaxis,and(vec{N})hasamagnitudeof3unitsandisdirectedatanangleof(135^{circ})withthepositivexaxis.Findtheresultantvector(vec{S}=vec{M}+vec{N})incomponentform.

Answer:First,findthecomponentsof(vec{M})and(vec{N}).For(vec{M})with(M=4)and(theta_M=45^{circ}),(M_x=4cos45^{circ}=4timesfrac{sqrt{2}}{2}=2sqrt{2})and(M_y=4sin45^{circ}=2sqrt{2}).For(vec{N})with(N=3)and(theta_N=135^{circ}),(N_x=3cos135^{circ}=3times(frac{sqrt{2}}{2})=frac{3sqrt{2}}{2})and(N_y=3sin135^{circ}=3timesfrac{sqrt{2}}{2}).Then,(S_x=M_x+N_x=2sqrt{2}frac{3sqrt{2}}{2}=frac{4sqrt{2}3sqrt{2}}{2}=frac{sqrt{2}}{2})and(S_y=M_y+N_y=2sqrt{2}+frac{3sqrt{2}}{2}=frac{4sqrt{2}+3sqrt{2}}{2}=frac{7sqrt{2}}{2}).So,(vec{S}=frac{sqrt{2}}{2}hat{i}+frac{7sqrt{2}}{2}hat{j}).

5.Avector(vec{Q})hascomponents(Q_x=6)unitsand(Q_y=8)units.Whatisthemagnitudeanddirectionof(vec{Q})?

Answer:Themagnitudeof(vec{Q})isgivenby(Q=sqrt{Q_x^{2}+Q_y^{2}}=sqrt{(6)^{2}+8^{2}}=sqrt{36+64}=sqrt{100}=10)units.Thedirection(theta)isgivenby(tantheta=frac{Q_y}{Q_x}=frac{8}{6}=frac{4}{3}).Since(Q_x<0)and(Q_y>0),thevectorliesinthesecondquadrant.(theta=arctan(frac{4}{3})+180^{circ}approx126.9^{circ})withrespecttothepositivexaxis.

6.Iftwovectors(vec{A})and(vec{B})areparallel,whatcanbesaidabouttheircrossproduct(vec{A}timesvec{B})?

Answer:Themagnitudeofthecrossproductoftwovectors(vec{A})and(vec{B})isgivenby(|vec{A}timesvec{B}|=|vec{A}||vec{B}|sintheta),where(theta)istheanglebetween(vec{A})and(vec{B}).Ifthetwovectorsareparallel,(theta=0^{circ})or(180^{circ}),and(sintheta=0).So,(vec{A}timesvec{B}=vec{0})(thezerovector).

7.Thedotproductoftwovectors(vec{C})and(vec{D})isgivenby(vec{C}cdotvec{D}=|vec{C}||vec{D}|costheta).If(vec{C}cdotvec{D}=0),whatcanbesaidaboutthevectors(vec{C})and(vec{D})?

Answer:Since(vec{C}cdotvec{D}=|vec{C}||vec{D}|costheta=0),either(|vec{C}|=0)or(|vec{D}|=0)or(costheta=0).If(|vec{C}|neq0)and(|vec{D}|neq0),then(costheta=0),whichmeans(theta=90^{circ}).So,thevectors(vec{C})and(vec{D})areperpendiculartoeachother.

8.Avector(vec{V})hasamagnitudeof12units.Ifitismultipliedbyascalar(k=2),whatisthemagnitudeanddirectionofthenewvector(kvec{V})?

Answer:Themagnitudeof(kvec{V})is(|kvec{V}|=|k||vec{V}|).Given(|k|=2)and(|vec{V}|=12),(|kvec{V}|=2times12=24)units.Since(k=2<0),thedirectionof(kvec{V})isoppositetothedirectionof(vec{V}).

9.Threevectors(vec{u}),(vec{v}),and(vec{w})aresuchthat(vec{u}+vec{v}+vec{w}=vec{0}).If(|vec{u}|=3),(|vec{v}|=4),and(|vec{w}|=5),whatcanbesaidabouttheanglesbetweenthevectors?

Answer:Since(|vec{u}|^{2}+|vec{v}|^{2}=3^{2}+4^{2}=9+16=25=|vec{w}|^{2}),bythePythagoreantheorem,thevectors(vec{u})and(vec{v})areperpendiculartoeachother((theta_{uv}=90^{circ})).Also,from(vec{u}+vec{v}+vec{w}=vec{0}),wecanrewriteitas(vec{w}=(vec{u}+vec{v})).Usingthedotproducttofindtheanglesbetween(vec{u})and(vec{w})and(vec{v})and(vec{w}),wecanfindthattheanglesbetween(vec{u})and(vec{w})and(vec{v})and(vec{w})arenonzeroandcanbecalculatedusing(vec{u}cdotvec{w}=|vec{u}||vec{w}|costheta_{uw})and(vec{v}cdotvec{w}=|vec{v}||vec{w}|costheta_{vw}).

10.Avector(vec{F})isresolvedintotwocomponents(vec{F}_1)and(vec{F}_2)suchthat(vec{F}=vec{F}_1+vec{F}_2).If(vec{F}_1)isparalleltoagivenvector(vec{a})and(vec{F}_2)isperpendicularto(vec{a}),howcanwefind(vec{F}_1)and(vec{F}_2)?

Answer:Thecomponentof(vec{F})parallelto(vec{a})(i.e.,(vec{F}_1))isgivenby(vec{F}_1=frac{vec{F}cdotvec{a}}{|vec{a}|^{2}}vec{a}).Thecomponentof(vec{F})perpendicularto(vec{a})(i.e.,(vec{F}_2))isgivenby(vec{F}_2=vec{F}vec{F}_1=vec{F}frac{vec{F}cdotvec{a}}{|vec{a}|^{2}}vec{a}).

11.Twovectors(vec{A})and(vec{B})havemagnitudes(A=7)and(B=5)respectively.Theanglebetweenthemis(60^{circ}).Findthemagnitudeofthevector(vec{R}=vec{A}vec{B}).

Answer:First,weknowthat(|vec{R}|^{2}=(vec{A}vec{B})cdot(vec{A}vec{B})=vec{A}cdotvec{A}2vec{A}cdotvec{B}+vec{B}cdotvec{B}).Since(vec{A}cdotvec{A}=|vec{A}|^{2}=49),(vec{B}cdotvec{B}=|vec{B}|^{2}=25),and(vec{A}cdotvec{B}=|vec{A}||vec{B}|costheta=7times5timescos60^{circ}=7times5timesfrac{1}{2}=frac{35}{2}).Then(|vec{R}|^{2}=492timesfrac{35}{2}+25=4935+25=39).So,(|vec{R}|=sqrt{39}approx6.24)units.

12.Avector(vec{E})hasamagnitudeof15unitsandisinthedirectionofthevector(vec{F}=3hat{i}+4hat{j}).Findthevector(vec{E})incomponentform.

Answer:First,findtheunitvector(hat{F})inthedirectionof(vec{F}).Themagnitudeof(vec{F})is(|vec{F}|=sqrt{3^{2}+4^{2}}=5).Theunitvector(hat{F}=frac{vec{F}}{|vec{F}|}=frac{3}{5}hat{i}+frac{4}{5}hat{j}).Since(vec{E})hasamagnitudeof(E=15)andisinthedirectionof(vec{F}),(vec{E}=Ehat{F}=15times(frac{3}{5}hat{i}+frac{4}{5}hat{j})=9hat{i}+12hat{j}).

13.If(vec{A}=2hat{i}3hat{j})and(vec{B}=4hat{i}+5hat{j}),findtheanglebetween(vec{A})and(vec{B}).

Answer:Thedotproduct(vec{A}cdotvec{B}=(2hat{i}3hat{j})cdot(4hat{i}+5hat{j})=2times4+(3)times5=815=7).(|vec{A}|=sqrt{2^{2}+(3)^{2}}=sqrt{13})and(|vec{B}|=sqrt{4^{2}+5^{2}}=sqrt{41}).Using(vec{A}cdotvec{B}=|vec{A}||vec{B}|costheta),wehave(costheta=frac{vec{A}cdotvec{B}}{|vec{A}||vec{B}|}=frac{7}{sqrt{13}timessqrt{41}}approx0.30).Then(theta=arccos(0.30)approx107.5^{circ}).

14.Aforcevector(vec{F}=5hat{i}+12hat{j})actsonanobject.Whatisthemagnitudeoftheforceandtheangleitmakeswiththepositivexaxis?

Answer:Themagnitudeof(vec{F})is(F=sqrt{5^{2}+12^{2}}=sqrt{25+144}=sqrt{169}=13)units.Theangle(theta)withthepositivexaxisisgivenby(tantheta=frac{F_y}{F_x}=frac{12}{5}),so(theta=arctan(frac{12}{5})approx67.4^{circ}).

15.Twovectors(vec{M})and(vec{N})aregivenby(vec{M}=3hat{i}2hat{j})and(vec{N}=hat{i}+4hat{j}).Findthevector(vec{P}=2vec{M}3vec{N})incomponentform.

Answer:First,(2vec{M}=2(3hat{i}2hat{j})=6hat{i}4hat{j})and(3vec{N}=3(hat{i}+4hat{j})=3hat{i}+12hat{j}).Then(vec{P}=2vec{M}3vec{N}=(6hat{i}4hat{j})(3hat{i}+12hat{j})=(6+3)hat{i}+(412)hat{j}=9hat{i}16hat{j}).

16.Ifavector(vec{V})hasamagnitudeof20unitsandmakesanangleof(150^{circ})withthepositivexaxis,whatareitscomponents?

Answer:Thexcomponent(V_x=Vcostheta=20cos150^{circ}=20times(frac{sqrt{3}}{2})=10sqrt{3})unitsandtheycomponent(V_y=Vsintheta=20sin150^{circ}=20timesfrac{1}{2}=10)units.

17.Thecrossproductoftwovectors(vec{A})and(vec{B})isavector(vec{C}=vec{A}timesvec{B}).If(vec{A})liesinthexyplaneand(vec{B})liesalongthezaxis,whatisthedirectionof(vec{C})?

Answer:Usingtherighthandruleforcrossproducts,if(vec{A})isinthexyplaneand(vec{B})isalongthezaxis,then(vec{C}=vec{A}timesvec{B})liesinthexyplane.If(vec{A})hasacounterclockwiseorientationwithrespecttothezaxiswhenlookingfromthepositivezdirection,(vec{C})isperpendicularto(vec{A})inthexyplane.

18.Avector(vec{U})hascomponents(U_x=3)and(U_y=4).Anothervector(vec{W})hascomponents(W_x=2)and(W_y=6).Findthedotproduct(vec{U}cdotvec{W}).

Answer:Thedotproduct(vec{U}cdotvec{W}=U_xW_x+U_yW_y=(3)times(2)+(4)times(6)=624=30).

19.If(vec{A})and(vec{B})aretwononzerovectorsand(|vec{A}+vec{B}|=|vec{A}vec{B}|),whatcanbesaidaboutthevectors(vec{A})and(vec{B})?

Answer:Squarebothsides:(|vec{A}+vec{B}|^{2}=(vec{A}+vec{B})cdot(vec{A}+vec{B})=vec{A}cdotvec{A}+2vec{A}cdotvec{B}+vec{B}cdotvec{B})and(|vec{A}vec{B}|^{2}=(vec{A}vec{B})cdot(vec{A}vec{B})=vec{A}cdotvec{A}2vec{A}cdotvec{B}+vec{B}cdotvec{B}).Since(|vec{A}+vec{B}|=|vec{A}vec{B}|),wehave(vec{A}cdotvec{A}+2vec{A}cdotvec{B}+vec{B}cdotvec{B}=vec{A}cdotvec{A}2vec{A}cdotvec{B}+vec{B}cdotvec{B}).Cancelingout(vec{A}cdotvec{A})and(vec{B}cdotvec{B})onbothsides,weget(4vec{A}cdotvec{B}=0),whichmeans(vec{A}cdotvec{B}=0).So,thevectors(vec{A})and(vec{B})areperpendicular.

20.Avector(vec{S})isthesumoftwovectors(vec{T})and(vec{U}).(vec{T})hasamagnitudeof8unitsandisalongthepositiveyaxis,and(vec{U})hasamagnitudeof6unitsandisalongthepositivexaxis.Whatisthemagnitudeanddirectionof(vec{S})?

Answer:(vec{T}=8hat{j})and(vec{U}=6hat{i}),so(vec{S}=vec{U}+vec{T}=6hat{i}+8hat{j}).Themagnitudeof(vec{S})is(S=sqrt{6^{2}+8^{2}}=sqrt{36+64}=10)units.Thedirection(theta)withrespecttothepositivexaxisis(tantheta=frac{S_y}{S_x}=frac{8}{6}=frac{4}{3}),so(theta=arctan(frac{4}{3})approx53.1^{circ}).

21.Twovectors(vec{X})and(vec{Y})havemagnitudes(X=9)and(Y=12)respectively,andtheanglebetweenthemis(90^{circ}).Findthemagnitudeoftheresultantvector(vec{R}=vec{X}+vec{Y}).

Answer:UsingthePythagoreantheoremforvectoraddition(sincethevectorsareperpendicular),(|vec{R}|=sqrt{X^{2}+Y^{2}}=sqrt{9^{2}+12^{2}}=sqrt{81+144}=sqrt{225}=15)units.

22.Avector(vec{Z})hasamagnitudeof14unitsandisinthedirectionoppositetothevector(vec{K}=2hat{i}3hat{j}).Find(vec{Z})incomponentform.

Answer:First,findtheunitvector(hat{K})inthedirectionof(vec{K}).Themagnitudeof(vec{K})is(|vec{K}|=sqrt{2^{2}+(3)^{2}}=sqrt{13}).Theunitvector(hat{K}=frac{vec{K}}{|vec{K}|}=frac{2}{sqrt{13}}hat{i}frac{3}{sqrt{13}}hat{j}).Since(vec{Z})isintheoppositedirectionof(vec{K})andhasamagnitudeof(Z=14),(vec{Z}=14hat{K}=frac{28}{sqrt{13}}hat{i}+frac{42}{sqrt{13}}hat{j}).

23.Thedotproductoftwovectors(vec{P})and(vec{Q})is24.If(|vec{P}|=6)and(|vec{Q}|=8),whatistheanglebetweenthem?

Answer:Usingtheformula(vec{P}cdotvec{Q}=|vec{P}||vec{Q}|costheta),wehave(24=6times8timescostheta).Then(costheta=frac{24}{48}=frac{1}{2}),so(theta=arccos(frac{1}{2})=60^{circ}).

24.Avector(vec{M})hascomponents(M_x=5)and(M_y=5).Whatisthemagnitudeanddirectionof(vec{M})?

Answer:Themagnitudeof(vec{M})is(M=sqrt{M_x^{2}+M_y^{2}}=sqrt{5^{2}+(5)^{2}}=sqrt{25+25}=5sqrt{2})units.Thedirection(theta)isgivenby(tantheta=frac{M_y}{M_x}=frac{5}{5}=1).Since(M_x>0)and(M_y<0),thevectorliesinthefourthquadrant.(theta=arctan(1)+360^{circ}=315^{circ})or(theta=45^{circ})withrespecttothepositivexaxis.

25.If(vec{A}=3hat{i}+4hat{j})and(vec{B}=ahat{i}+bhat{j})areperpendicular,and(|vec{B}|=5),findthevaluesof(a)and(b).

Answer:Since(vec{A})and(vec{B})areperpendicular,(vec{A}cdotvec{B}=3a+4b=0),so(b=frac{3}{4}a).Also,(|vec{B}|=sqrt{a^{2}+b^{2}}=5).Substitute(b=frac{3}{4}a)into(sqrt{a^{2}+b^{2}}=5),weget(sqrt{a^{2}+frac{9}{16}a^{2}}=5),(sqrt{frac{16a^{2}+9a^{2}}{16}}=5),(sqrt{frac{25a^{2}}{16}}=5),(frac{5|a|}{4}=5),(|a|=4).If(a=4),then(b=3);if(a=4),then(b=3).So,(vec{B}=4hat{i}3hat{j})or(vec{B}=4hat{i}+3hat{j}).

26.Avector(vec{R})isthedifferenceoftwovectors(vec{A})and(vec{B}),i.e.,(vec{R}=vec{A}vec{B}).If(vec{A})hasamagnitudeof10unitsandisatanangleof(30^{circ})withthepositivexaxis,and(vec{B})hasamagnitudeof6unitsandisatanangleof(120^{circ})withthepositivexaxis,find(vec{R})incomponentform.

Answer:For(vec{A})with(A=10)and(theta_A=30^{circ}),(A_x=10cos30^{circ}=5sqrt{3})and(A_y=10sin30^{circ}=5).For(vec{B})with(B=6)and(theta_B=120^{circ}),(B_x=6cos120^{circ}=3)and(B_y=6sin120^{circ}=3sqrt{3}).Then(R_x=A_xB_x=5sqrt{3}+3)and(R_y=A_yB_y=53sqrt{3}).So,(vec{R}=(5sqrt{3}+3)hat{i}+(53sqrt{3})hat{j}).

27.Thecrossproductoftwovectors(vec{V})and(vec{W})hasamagnitudeof15.If(|vec{V}|=5)and(|vec{W}|=3),whatistheanglebetween(vec{V})and(vec{W})?

Answer:Using(|vec{V}timesvec{W}|=|vec{V}||vec{W}|sintheta),wehave(15=5times3timessintheta).Then(sintheta=1),so(theta=90^{circ}).

28.Avector(vec{F})isresolvedintotwocomponents(vec{F}_1)and(vec{F}_2)suchthat(vec{F}_1)isalongavector(vec{a}=2hat{i}+hat{j})and(vec{F}_2)isperpendicularto(vec{a}).If(vec{F}=6hat{i}+8hat{j}),find(vec{F}_1)and(vec{F}_2).

Answer:First,findtheunitvector(hat{a}=frac{vec{a}}{|vec{a}|}=frac{2hat{i}+hat{j}}{sqrt{2^{2}+1^{2}}}=frac{2}{sqrt{5}}hat{i}+frac{1}{sqrt{5}}hat{j}).(vec{F}_1=frac{vec{F}cdotvec{a}}{|vec{a}|^{2}}vec{a}=frac{(6hat{i}+8hat{j})cdot(2hat{i}+hat{j})}{5}(2hat{i}+hat{j})=frac{12+8}{5}(2hat{i}+hat{j})=8hat{i}+4hat{j}).(vec{F}_2=vec{F}vec{F}_1=(6hat{i}+8hat{j})(8hat{i}+4hat{j})=2hat{i}+4hat{j}).

29.Iftwovectors(vec{C})and(vec{D})areantiparallel,whatisthevalueof(vec{C}cdotvec{D})intermsoftheirmagnitudes?

Answer:Theangle(theta)betweenantiparallelvectorsis(180^{circ}).Using(vec{C}cdotvec{D}=|vec{C}||vec{D}|costheta),and(cos180^{circ}=1),weget(vec{C}cdotvec{D}=|vec{C}||vec{D}|).

30.Avector(vec{S})hasamagnitudeof16unitsandmakesanangleof(210^{circ})withthepositivexaxis.Finditscomponents.

Answer:Thexcomponent(S_x=Scostheta=16cos210^{circ}=16times(frac{sqrt{3}}{2})=8sqrt{3})unitsandtheycomponent(S_y=Ssintheta=16sin210^{circ}=16times(frac{1}{2})=8)units.

31.Thedotproductoftwovectors(vec{X})and(vec{Y})is12.If(|vec{X}|=4)and(|vec{Y}|=6),whatistheanglebetweenthem?

Answer:Using(vec{X}cdotvec{Y}=|vec{X}||vec{Y}|costheta),wehave(12=4times6timescostheta).Then(costheta=frac{1}{2}),so(theta=arccos(frac{1}{2})=120^{circ}).

32.If(vec{A}=2hat{i}5hat{j})and(vec{B}=mhat{i}+nhat{j})areparallel,and(|vec{B}|=13),findthevaluesof(m)and(n).

Answer:Since(vec{A})and(vec{B})areparallel,(frac{m}{2}=frac{n}{5}),so(n=frac{5}{2}m).Also,(|vec{B}|=sqrt{m^{2}+n^{2}}=13).Substitute(n=frac{5}{2}m)into(sqrt{m^{2}+n^{2}}=13),weget(sqrt{m^{2}+frac{25}{4}m^{2}}=13),(sqrt{frac{4m^{2}+25m^{2}}{4}}=13),(sqrt{frac{29m^{2}}{4}}=13),(frac{sqrt{29}|m|}{2}=13),(|m|=frac{26}{sqrt{29}}).If(m=frac{26}{sqrt{29}}),then(n=frac{65}{sqrt{29}});if(m=frac{26}{sqrt{29}}),then(n=frac{65}{sqrt{29}}).

33.Avector(vec{U})isthesumofthreevectors(vec{V}_1=3hat{i}),(vec{V}_2=4hat{j}),and(vec{V}_3=2hat{i}+3hat{j}).Findthemagnitudeanddirectionof(vec{U}).

Answer:(vec{U}=vec{V}_1+vec{V}_2+vec{V}_3=(3hat{i})+(4hat{j})+(2hat{i}+3hat{j})=(32)hat{i}+(4+3)hat{j}=hat{i}+7hat{j}).Themagnitudeof(vec{U})is(U=sqrt{1^{2}+7^{2}}=sqrt{50}=5sqrt{2})units.Thedirection(theta)isgivenby(tantheta=frac{U_y}{U_x}=7),(theta=arctan(7)approx81.9^{circ})withrespecttothepositivexaxis.

34.Thecrossproductoftwovectors(vec{P})and(vec{Q})is(vec{P}timesvec{Q}=12hat{k}).If(|vec{P}|=3)and(|vec{Q}|=4),whatistheanglebetween(vec{P})and(vec{Q})?

Answer:Using(|vec{P}timesvec{Q}|=|vec{P}||vec{Q}|sintheta),and(|vec{P}timesvec{Q}|=12),(|vec{P}|=3),(|vec{Q}|=4),wehave(12=3times4timessintheta).Then(sintheta=1),so(theta=90^{circ}).

35.Avector(vec{R})hasamagnitudeof25units.Itsxcomponent(R_x=15)units.Whatisitsycomponentandtheangleitmakeswiththepositivexaxis?

Answer:Using(R=sqrt{R_x^{2}+R_y^{2}}),wehave(25=sqrt{15^{2}+R_y^{2}}),(625=225+R_y^{2}),(R_y^{2}=400),(R_y=pm20)units.If(R_y=20),(tantheta=frac{R_y}{R_x}=frac{20}{15}=frac{4}{3}),(theta=arctan(frac{4}{3})approx53.1^{circ});if(R_y=20),(tantheta=frac{20}{15}=frac{4}{3}),andsince(R_x>0)and(R_y<0),(theta=arctan(frac{4}{3})+360^{circ}=306.9^{circ}).

36.If(vec{A}=4hat{i}3hat{j})and(vec{B}=2hat{i}+khat{j})areperpendicular,findthevalueof(k).

Answer:Since(vec{A})and(vec{B})areperpendicular,(vec{A}cdotvec{B}=(4hat{i}3hat{j})cdot(2hat{i}+khat{j})=4times2+(3)timesk=0).Then(83k=0),(k=frac{8}{3}).

37.Avector(vec{M})hasamagnitudeof18unitsandisinthedirectionofthevector(vec{N}=5hat{i}12hat{j}).Find(vec{M})incomponentform.

Answer:First,findtheunitvector(hat{N})inthedirectionof(vec{N}).Themagnitudeof(vec{N})is(|vec{N}|=sqrt{5^{2}+(12)^{2}}=13).Theunitvector(hat{N}=frac{vec{N}}{|vec{N}|}=frac{5}{13}hat{i}frac{12}{13}hat{j}).Since(vec{M})hasamagnitudeof(M=18)andisinthedirectionof(vec{N}),(vec{M}=Mhat{N}=18times(frac{5}{13}hat{i}frac{12}{13}hat{j})=frac{90}{13}hat{i}frac{216}{13}hat{j}).

38.Thedotproductoftwovectors(vec{C})and(vec{D})is0.If(|vec{C}|=5)and(|vec{D}|=7),whatcanbesaidaboutthevectors?

Answer:Since(vec{C}cdotvec{D}=|vec{C}||vec{D}|costheta=0)and(|vec{C}|neq0)and(|vec{D}|neq0),then