ALL 10 CHAPTERS COVERED
m m m
Answer key
,2 SolutionsmtomExercise
s
ProblemmSetm1.1,mpagem8
1 Themcombinationsmgivem(a)m amlineminmR3 (b)m amplaneminmR3 (c)m allmofmR3.
2 vm+mwm =m(2,m3)mandmvm−mwm =m(6,m−1)mwillmbemthemdiagonalsmofmthemparallelogrammwith
vmandmwmasmtwomsidesmgoingmoutmfromm(0,m0).
3 Thismproblemm givesmthemdiagonalsmvm +mwm andmvm −m wm ofmthemparallelogrammandmasksm
formthemsides:mThemoppositemofmProblemm2.mInmthismexamplemvm=m(3,m3)mandmwm=m(
2,m−2).
4 3vm+mwm =m(7,m5)mandmcvm+mdwm=m(2cm+md,mcm+m2d).
5 u+vm=m(−2,m3,m1)mandmu+v+w m =m(0,m0,m0)mandm2u+2v+wm=m(maddmfirstmanswe
rs)m=m(−2,m3,m1).mThemvectorsmu,mv,mwmareminmthemsamemplanembecausemamcombina
tionmgivesm(0,m0,m0).mStatedmanothermway:mum=m−vm−mwmisminmthemplanemofmvmand
m w.
6 Them componentsmofm everym cvm +mdwm addm tom zerom becausem them componentsmofm vm andm ofm w
addmtomzero.m cm=m3mandmdm=m9mgivem(3,m3,m−6).m Theremismnomsolutionmtomcv+dwm =m(3,m3,m6)
becausem3m+m3m+m6mismnotmzero.
7 Themninemcombinationsmc(2,m1)m+md(0,m1)mwithmcm=m0,m1,m2mandmdm=m(0,m1,m2)mwillmliemon
a lattice.m Ifmwemtookmallmwholemnumbersmcmandmd,mthemlatticemwouldmliemovermthemwhole
m m
mplane.
8 Themothermdiagonalmismvm−mwm(ormelsemwm−mv).m Addingmdiagonalsmgivesm2vm(orm2w).
9 Themfourthmcornermcanmbem(4,m4)morm(4,m0)morm(−2,m2).m Threempossiblemparallelograms!
10 im−mjm=m(1,m1,m0)misminmthembasem(x-
ymplane).mim+mjm+mkm=m(1,m1,m1)mismthemoppositemcornermfromm(0,m0,m0).mPointsminmth
emcubemhavem0m≤mxm≤m1,m0m≤mym≤m1,m0m≤mzm≤m1.
11 Fourm morem cornersm (1,m1,m0),m(1,m0,m1),m(0,m1,m1),m(1,m1,m1).m Them centerm pointm ism (m1m,m1m,m 1m).
2m 2m 2
Centersmofmfacesm arem (m1m,mm1m,m0),m(m1m,mm 1m,m1)mandm (0,m1mm,m1m),m(1,mm1m,m 1m)m andm(m1m,m0,m1m),m(m1m,m1,m1m).
2m 2 2m 2 2m 2 2 2 2 2
2m
2
12 Them combinationsm ofm im =m (1,m0,m0)m andm im +m jm =m (1,m1,m0)m fillm them xym planem inm xyzm space.
13 Summ=mzeromvector.mSumm=m−2:00mvectorm=m8:00mvector.m2:00mism30◦m frommhorizontal
√
=m(cosmπm, msinmπm)m=m( 3/2, m1/2).
6 6
14 Movingmthemoriginmtom6:00maddsmjm=m(0,m1)mtomeverymvector.mSomthemsummofmtwelvemv
ectorsmchangesmfromm0mtom12jm=m(0,m12).
,SolutionsmtomExercise 3
s
3m 1m
15 Them pointm vm +m wm ism three-fourthsm ofm them waym tom vm startingm fromm w.m m Them vector
4 4
1 1 1 1
vm+m wmismhalfwaymtomum=m mvm+m w.mThemvectormvm+mwmism2um(themfarmcornermofmthe
4 4 2 2
parallelogram).
16 Allm combinations m withm cm+mdm =m 1m arem onm them linem thatm passesm throughm vm and
m w.mThempointmVm =m−vm+m2wmismonmthatmlinembutmitmismbeyondmw.
17 Allmvectorsmcvm+mcwmaremonmthemlinempassingmthroughm(0,m0)mandmum =m 1mvm+m 1mw.m That
2 2
linemcontinuesmoutmbeyondmvm+mwmandmbackmbeyondm(0,m0).m Withmcm≥m0,mhalfmofmthis
mlinemismremoved,mleavingmamraymthatmstartsmatm(0,m0).
18 Themcombinationsmcvm+mdwmwithm0m≤mcm≤m1mandm0m≤mdm≤m1mfillmthemparallelogramm
withmsidesmvmandmw.m Formexample,mifmvm=m(1, m0)mandmwm=m(0,m1)mthenmcvm+mdwmfills
m themunitmsquare.mButmwhenmvm=m(a,m0)mandmwm=m(b,m0)mthesemcombinationsmonlymfillm
amsegmentmofmamline.
19 Withmcm≥m0mandmdm≥m0mwemgetmtheminfinitem“cone”morm“wedge”mbetweenmvmandmw.
m Formexample,mifmvm=m(1,m0)mandmwm=m(0,m1),mthenmthemconemismthemwholemquadrant
x ≥m0,mym≥
m m
0.m Question:m Whatm ifm wm =m −v?m Them conem opensmtom am half-
space.m Butm them combinationsmofmvm=m(1,m0)mandmwm=m(−1,m0)monlymfillmamline.
20 (a)m 1mum+m 1mvm+m 1mwm ismthemcentermofmthemtrianglembetweenmu,mvm andmw;m 1mum+m 1mwm lies
3 3 3 2 2
betweenmumandmw (b)m Tomfillmthemtrianglemkeepmcm≥m0,mdm≥m0,mem≥m0,mandmcm+mdm+mem=m1.
21 Themsummism(vm−mu)m+(wm−mv)m+(um−mw)m=mzeromvector.m Thosemthreemsidesmofmamtria
nglemareminmthemsamemplane!
22 Themvectorm1m(um+mvm+mw)mismoutsidemthempyramidmbecausemcm+mdm+mem=m1m+m1m+m1m>m1.
2 2 2 2
23 Allm vectorsm arem combinationsm ofm u,mv,mwm asm drawnm (notm inm them samem plane).m Startm
bymseeingmthatmcum+mdvm fillsmamplane,mthenmaddingmewm fillsmallmofmR3.
24 Themcombinationsmofmumandmvm fillmonemplane.m Themcombinationsmofmvm andmwm fillmanother
mplane.m Thosemplanesmmeetminmamline:m onlymthemvectorsmcvm areminmbothmplanes.
25 (a)m Formamline,mchoosemum =mvm =mwm =manymnonzeromvector (b)m Formamplane,mchoose
um andm vm inm differentm directions.m Am combinationm likem wm =m um +m vm ism inm them samem plane.
, 4 SolutionsmtomExercise
s
26 Twomequationsmcomemfrommthemtwomcomponents:m cm+m3dm=m14mandm2cm+mdm=m8.m T
hemsolutionmismcm=m2mandmdm=m4.mThenm2(1,m2)m+m4(3,m1)m=m(14,m8).
27 Amfour-dimensionalmcubemhasm24m=m 16mcornersmandm2m·m4m =m 8mthree-
dimensionalmfacesmandm24mtwo-
dimensionalmfacesmandm32medgesminmWorkedmExamplem2.4mA.
28 Theremarem6munknownmnumbersmv1,mv2,mv3,mw1,mw2,mw3.m Themsixmequationsmcomemfrommthem
componentsmofmvm+mwm =m(4,m5,m6)mandmvm−mwm =m(2,m5,m8).m Addmtomfindm2vm =m(6,m
10,m14)
somvm=m(3,m5,m7)mandmwm =m(1,m0,m−1).
29 Factm:m Form anymthreem vectorsm u,mv,mwm inmthem plane,m somem combinationmcum+mdvm+mewm i
smthem zerom vectorm (beyondm them obviousmcm=m dm =m em =m 0).m Som ifm therem ism onem combi
nationmCum+mDvm+mEwmthatmproducesmb,m theremwillmbemmanymmore—
justmaddmc,md,memorm2c,m2d,m2emtomthemparticularmsolutionmC,mD,mE.
Themexamplemhasm3um−m2vm+mwm =m 3(1, m3)m−m2(2, m7)m+m1(1, m5)m =m (0, m0).m Itmalsomhas
−2um+m1vm+m0wm =mbm =m(0,m1).m Addingmgivesmum−mvm+mwm =m(0,m1).m Inmthismcasemc,md,me
equalm3,m−2,m1m andmC,mD,mEm =m −2,m1,m0.
Couldmanothermexamplemhavemu,mv,mwmthatmcouldmNOTmcombinemtomproducembm?m Yes.m T
hemvectorsm(1,m1),m(2,m2),m(3,m3)maremonmamlinemandmnomcombinationmproducesmb.m Wemca
nmeasilymsolvemcum+mdvm+mewm =m 0mbutmnotmCum+mDvm +mEwm =m b.
30 Themcombinationsmofmvmandmwmfillmthemplanemunlessmvmandmwmliemonmthemsamemlinemthrough
m(0,m0).m Fourmvectorsmwhosemcombinationsmfillm4-
dimensionalmspace:m onemexamplemismthem“standardmbasis”m(1,m0,m0,m0),m(0,m1,m0,m0),m(0,m
0,m1,m0),mandm(0,m0,m0,m1).
31 Themequationsmcum+mdvm+mewm=mbmare
2cm −d =m 1 Somdm=m2e cm=m3/4
−cm+2dm −em=m0 m thenmcm=m dm=m2/4
−dm+2em=m 0 3emthenm4e em=m1/4
m=m1
m m m
Answer key
,2 SolutionsmtomExercise
s
ProblemmSetm1.1,mpagem8
1 Themcombinationsmgivem(a)m amlineminmR3 (b)m amplaneminmR3 (c)m allmofmR3.
2 vm+mwm =m(2,m3)mandmvm−mwm =m(6,m−1)mwillmbemthemdiagonalsmofmthemparallelogrammwith
vmandmwmasmtwomsidesmgoingmoutmfromm(0,m0).
3 Thismproblemm givesmthemdiagonalsmvm +mwm andmvm −m wm ofmthemparallelogrammandmasksm
formthemsides:mThemoppositemofmProblemm2.mInmthismexamplemvm=m(3,m3)mandmwm=m(
2,m−2).
4 3vm+mwm =m(7,m5)mandmcvm+mdwm=m(2cm+md,mcm+m2d).
5 u+vm=m(−2,m3,m1)mandmu+v+w m =m(0,m0,m0)mandm2u+2v+wm=m(maddmfirstmanswe
rs)m=m(−2,m3,m1).mThemvectorsmu,mv,mwmareminmthemsamemplanembecausemamcombina
tionmgivesm(0,m0,m0).mStatedmanothermway:mum=m−vm−mwmisminmthemplanemofmvmand
m w.
6 Them componentsmofm everym cvm +mdwm addm tom zerom becausem them componentsmofm vm andm ofm w
addmtomzero.m cm=m3mandmdm=m9mgivem(3,m3,m−6).m Theremismnomsolutionmtomcv+dwm =m(3,m3,m6)
becausem3m+m3m+m6mismnotmzero.
7 Themninemcombinationsmc(2,m1)m+md(0,m1)mwithmcm=m0,m1,m2mandmdm=m(0,m1,m2)mwillmliemon
a lattice.m Ifmwemtookmallmwholemnumbersmcmandmd,mthemlatticemwouldmliemovermthemwhole
m m
mplane.
8 Themothermdiagonalmismvm−mwm(ormelsemwm−mv).m Addingmdiagonalsmgivesm2vm(orm2w).
9 Themfourthmcornermcanmbem(4,m4)morm(4,m0)morm(−2,m2).m Threempossiblemparallelograms!
10 im−mjm=m(1,m1,m0)misminmthembasem(x-
ymplane).mim+mjm+mkm=m(1,m1,m1)mismthemoppositemcornermfromm(0,m0,m0).mPointsminmth
emcubemhavem0m≤mxm≤m1,m0m≤mym≤m1,m0m≤mzm≤m1.
11 Fourm morem cornersm (1,m1,m0),m(1,m0,m1),m(0,m1,m1),m(1,m1,m1).m Them centerm pointm ism (m1m,m1m,m 1m).
2m 2m 2
Centersmofmfacesm arem (m1m,mm1m,m0),m(m1m,mm 1m,m1)mandm (0,m1mm,m1m),m(1,mm1m,m 1m)m andm(m1m,m0,m1m),m(m1m,m1,m1m).
2m 2 2m 2 2m 2 2 2 2 2
2m
2
12 Them combinationsm ofm im =m (1,m0,m0)m andm im +m jm =m (1,m1,m0)m fillm them xym planem inm xyzm space.
13 Summ=mzeromvector.mSumm=m−2:00mvectorm=m8:00mvector.m2:00mism30◦m frommhorizontal
√
=m(cosmπm, msinmπm)m=m( 3/2, m1/2).
6 6
14 Movingmthemoriginmtom6:00maddsmjm=m(0,m1)mtomeverymvector.mSomthemsummofmtwelvemv
ectorsmchangesmfromm0mtom12jm=m(0,m12).
,SolutionsmtomExercise 3
s
3m 1m
15 Them pointm vm +m wm ism three-fourthsm ofm them waym tom vm startingm fromm w.m m Them vector
4 4
1 1 1 1
vm+m wmismhalfwaymtomum=m mvm+m w.mThemvectormvm+mwmism2um(themfarmcornermofmthe
4 4 2 2
parallelogram).
16 Allm combinations m withm cm+mdm =m 1m arem onm them linem thatm passesm throughm vm and
m w.mThempointmVm =m−vm+m2wmismonmthatmlinembutmitmismbeyondmw.
17 Allmvectorsmcvm+mcwmaremonmthemlinempassingmthroughm(0,m0)mandmum =m 1mvm+m 1mw.m That
2 2
linemcontinuesmoutmbeyondmvm+mwmandmbackmbeyondm(0,m0).m Withmcm≥m0,mhalfmofmthis
mlinemismremoved,mleavingmamraymthatmstartsmatm(0,m0).
18 Themcombinationsmcvm+mdwmwithm0m≤mcm≤m1mandm0m≤mdm≤m1mfillmthemparallelogramm
withmsidesmvmandmw.m Formexample,mifmvm=m(1, m0)mandmwm=m(0,m1)mthenmcvm+mdwmfills
m themunitmsquare.mButmwhenmvm=m(a,m0)mandmwm=m(b,m0)mthesemcombinationsmonlymfillm
amsegmentmofmamline.
19 Withmcm≥m0mandmdm≥m0mwemgetmtheminfinitem“cone”morm“wedge”mbetweenmvmandmw.
m Formexample,mifmvm=m(1,m0)mandmwm=m(0,m1),mthenmthemconemismthemwholemquadrant
x ≥m0,mym≥
m m
0.m Question:m Whatm ifm wm =m −v?m Them conem opensmtom am half-
space.m Butm them combinationsmofmvm=m(1,m0)mandmwm=m(−1,m0)monlymfillmamline.
20 (a)m 1mum+m 1mvm+m 1mwm ismthemcentermofmthemtrianglembetweenmu,mvm andmw;m 1mum+m 1mwm lies
3 3 3 2 2
betweenmumandmw (b)m Tomfillmthemtrianglemkeepmcm≥m0,mdm≥m0,mem≥m0,mandmcm+mdm+mem=m1.
21 Themsummism(vm−mu)m+(wm−mv)m+(um−mw)m=mzeromvector.m Thosemthreemsidesmofmamtria
nglemareminmthemsamemplane!
22 Themvectorm1m(um+mvm+mw)mismoutsidemthempyramidmbecausemcm+mdm+mem=m1m+m1m+m1m>m1.
2 2 2 2
23 Allm vectorsm arem combinationsm ofm u,mv,mwm asm drawnm (notm inm them samem plane).m Startm
bymseeingmthatmcum+mdvm fillsmamplane,mthenmaddingmewm fillsmallmofmR3.
24 Themcombinationsmofmumandmvm fillmonemplane.m Themcombinationsmofmvm andmwm fillmanother
mplane.m Thosemplanesmmeetminmamline:m onlymthemvectorsmcvm areminmbothmplanes.
25 (a)m Formamline,mchoosemum =mvm =mwm =manymnonzeromvector (b)m Formamplane,mchoose
um andm vm inm differentm directions.m Am combinationm likem wm =m um +m vm ism inm them samem plane.
, 4 SolutionsmtomExercise
s
26 Twomequationsmcomemfrommthemtwomcomponents:m cm+m3dm=m14mandm2cm+mdm=m8.m T
hemsolutionmismcm=m2mandmdm=m4.mThenm2(1,m2)m+m4(3,m1)m=m(14,m8).
27 Amfour-dimensionalmcubemhasm24m=m 16mcornersmandm2m·m4m =m 8mthree-
dimensionalmfacesmandm24mtwo-
dimensionalmfacesmandm32medgesminmWorkedmExamplem2.4mA.
28 Theremarem6munknownmnumbersmv1,mv2,mv3,mw1,mw2,mw3.m Themsixmequationsmcomemfrommthem
componentsmofmvm+mwm =m(4,m5,m6)mandmvm−mwm =m(2,m5,m8).m Addmtomfindm2vm =m(6,m
10,m14)
somvm=m(3,m5,m7)mandmwm =m(1,m0,m−1).
29 Factm:m Form anymthreem vectorsm u,mv,mwm inmthem plane,m somem combinationmcum+mdvm+mewm i
smthem zerom vectorm (beyondm them obviousmcm=m dm =m em =m 0).m Som ifm therem ism onem combi
nationmCum+mDvm+mEwmthatmproducesmb,m theremwillmbemmanymmore—
justmaddmc,md,memorm2c,m2d,m2emtomthemparticularmsolutionmC,mD,mE.
Themexamplemhasm3um−m2vm+mwm =m 3(1, m3)m−m2(2, m7)m+m1(1, m5)m =m (0, m0).m Itmalsomhas
−2um+m1vm+m0wm =mbm =m(0,m1).m Addingmgivesmum−mvm+mwm =m(0,m1).m Inmthismcasemc,md,me
equalm3,m−2,m1m andmC,mD,mEm =m −2,m1,m0.
Couldmanothermexamplemhavemu,mv,mwmthatmcouldmNOTmcombinemtomproducembm?m Yes.m T
hemvectorsm(1,m1),m(2,m2),m(3,m3)maremonmamlinemandmnomcombinationmproducesmb.m Wemca
nmeasilymsolvemcum+mdvm+mewm =m 0mbutmnotmCum+mDvm +mEwm =m b.
30 Themcombinationsmofmvmandmwmfillmthemplanemunlessmvmandmwmliemonmthemsamemlinemthrough
m(0,m0).m Fourmvectorsmwhosemcombinationsmfillm4-
dimensionalmspace:m onemexamplemismthem“standardmbasis”m(1,m0,m0,m0),m(0,m1,m0,m0),m(0,m
0,m1,m0),mandm(0,m0,m0,m1).
31 Themequationsmcum+mdvm+mewm=mbmare
2cm −d =m 1 Somdm=m2e cm=m3/4
−cm+2dm −em=m0 m thenmcm=m dm=m2/4
−dm+2em=m 0 3emthenm4e em=m1/4
m=m1
