2. VECTORS AND MATRICES IN 3 DIMENSIONS


 Agatha Merritt
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1 2 VECTORS AND MATRICES IN 3 DIMENSIONS 21 Extending the Theory of 2dimensionl Vectors x A point in 3dimensionl spce cn e represented y column vector of the form y z zxis yxis z x y xxis Most of the theory of 2dimensionl vectors cn e extended in strightforwrd wy to 3 dimensionl vectors In most cses the only chnge tht needs to e mde is to chnge 2 into 3 x 1 x 2 nd to put in n extr component For exmple the distnce etween two points y 1 nd y 2 is z 1 z 2 (x 1 x 2 ) 2 + (y 1 y 2 ) 2 + (z 1 z 2 ) 2 Some of the less ovious differences re listed here The mtrix corresponding to rottion through n ritrry ngle out n ritrry xis is too complicted to discuss here However where the cos sin xis is the zxis, nd the ngle of rottion is the mtrix of the rottions is sin cos 1 The determinnt of the 3 3 mtrix (,, c), tht is whose columns re the vectors, nd c, is defined to e the signed volume of the prllelepiped (3dimensionl version of the prllelogrm solid figure whose six fces re prllelogrms) whose vertices re,,, c, +, + c, + c, + + c c + c + c c The signed volume is positive if the vectors form wht is clled positive triple nd negtive if they form negtive triple These re defined s follows Three noncoplnr vectors,, c, in tht order, form positive triple if, when viewed from c, it tkes positive rottion (less thn 18) to rotte from to nd negtive triple otherwise A hndy wy to rememer this is to imgine tht you re inside room, looking up to corner The origin is tht corner of room nd the vectors,, c re the three edges coming out 31
2 from tht corner If moving round the vectors in order you re moving in positive (nticlockwise) direction, the triple is positive one Otherwise it is negtive one c c (,, c) is positive triple (,, c) is negtive triple Note tht if the three vectors lie in plne they re neither positive nor negtive triple So wht sign do we give to the determinnt? Fortuntely if the prllelepiped lies in plne the volume is zero, nd is the sme s + Adpting the rgument for 2 2 mtrices to the 3dimensionl cse we cn see tht when closed region, hving certin volume, is trnsformed y the liner trnsformtion v Av, where A is 3 3 mtrix, the resulting region will hve volume A times the originl one, if we ignore the sign This mkes the fct tht AB A B plusile Multiplying volumes y A then multiplying y B is equivlent to multiplying y A B The only difficulty is in reconciling the signs When we come to generl n n mtrices we shll give proof tht is independent of geometry The formul for 2 2 determinnt does not immeditely generlise to the 3 3 cse The pproprite formul for 3 3 determinnt is: 1 1 c c 2 c 1 ( ) c 2 ( ) + c 3 ( ) 3 3 c 3 We could prove this long the lines of the 2 2 cse, dissecting certin volume into smll pieces, ut this would e incredily complicted Wht is needed is some new concept tht mkes this formul seem nturl, geometriclly 22 Vector Products We hve defined 3 3 determinnt in terms of the volume of prllelepiped Tht is the re of the se times the perpendiculr height To find the perpendiculr height we must e le to find vector tht is perpendiculr to ny two given vectors Suppose n is orthogonl to oth nd Then n nd n Tht is, if the components of re 1, 2, 3 nd similrly for nd n: 1 n n n 3 nd 1 n n n 3 We eliminte n 3 from these equtions y multiplying the first y 3 nd the second y 3 nd sutrcting Hence ( )n 1 + ( )n 2 Writing these coefficients s 2 2 determinnts we get n n 3 One of the mny solutions to this eqution is 32
3 n 1 3 3, n Eliminting n 2 in similr wy we otin n n 3 nd tking the vlue of n 1 chosen ove, this gives 1 1 n T Hence the column vector 3 3, 3 3, 2 2 is perpendiculr to oth nd We define the vector product of nd to e T 3 3, 3 3, 2 2 Note tht the 2 2 determinnts tht occur in this definition cn e otined y forming the 3 2 mtrix (, ), tht is y writing the two vectors s the two columns, nd then deleting respectively the first, the second nd the third rows Exmple 1: If 5 nd, find 2 7 Solution: 5 2 7, , T (35, 23, 5) T 23 Geometricl Interprettion of Vector Products Theorem 1: 2 2 () 2 Proof: ( ) 2 + ( ) 2 + ( ) 2 ( 1 2 ) 2 + ( 1 3 ) 2 + ( 2 1 ) 2 + ( ( 3 1 ) 2 + ( 3 2 ) 2 2( ) ( )( ) ( ) 2 2 ( ) () 2 Theorem 2: sin Proof: () ( cos ) (1 cos 2 ) ( sin ) 2 We tke the squre roots of oth sides Since nd sin (s <), the result follows from tking the nonnegtive squre roots Theorem 3: is the re of the prllelogrm with vertices,, nd + Proof: Tking the length of the se of this prllelogrm to e, the perpendiculr height is sin The re of prllelogrm is the re of the se times the perpendiculr height nd so the result follows + sin 33
4 The vector product of two nonprllel vectors is perpendiculr to the plne determined y nd nd its length is equl to the re of the prllelogrm formed from nd There re two such vectors, one on either side of the plne Which of the two is??? Without loss of generlity we my rotte the system so tht lies long the positive hlf of the xxis nd lies in the xy plne, nd hence we my tke s r cos nd r sin where is the ngle from to (viewed from the positive hlf of the zxis) nd r, s re the distnces of the points from the origin r Then If <, s in the digrm, comes out of the pge nd so,, rs sin form positive triple If >, goes into the pge, nd once gin,, form positive triple s s r We hve thus proved the following theorem Theorem 4: If, re not prllel vectors, the vector is the unique vector c such tht: (1) c is perpendiculr to oth nd ; (2),, c is positive triple; (3) c is the re of the prllelogrm formed from nd 34
5 Theorem 5: if nd only if, re prllel Proof: This follows from (3) of Theorem 5 The re of prllelogrm is zero if nd only if the djcent sides re prllel Theorem 6: ( ) Proof: since they re oth equl to the re of the sme prllelogrm Both nd re perpendiculr to the plne determined y nd However since,, is positive triple,,, is negtive triple nd so the direction of is opposite tht of 24 Vector Products nd Determinnts We re now in position to find the determinnt of 3 3 mtrix Theorem 7: If A (,, c) is 3 3 mtrix then A ( )c Proof: Cse I:,, c is zero triple Then A (for the prllelepiped hs zero volume) The vectors,, c re coplnr Since is orthogonl to this plne it is orthogonl to c Thus ( )c Cse II:,, c is positive triple The perpendiculr height is the length of the projection of c onto c c cos c ( )c ( )c If is the ngle etween nd c then this length is c cos c c Then A volume of the prllelepiped formed from,, c re of se perpendiculr height c ( ) c Cse III:,, c is negtive triple This is s for cse II except tht the projection of c onto ( )c is now in the opposite direction to nd so is nd A is minus the volume of the prllelepiped 1 1 c 1 Theorem 8: 2 2 c c c 3 3 c c c 1 1 Proof: By Theorem 8 the determinnt is equl to 3 3 c 2 c expression which gives the required 35
6 25 The Adjoint of Mtrix The djoint of 3 3 mtrix A (,, c) is defined to e dja ( c, c, ) T Theorem 9: (dj A)A A I for ll 3 3 mtrices A Proof: (dja)a ( c, c, ) T (,, c) ( c) ( c) ( c)c (c ) (c ) (c )c ( ) ( ) ( )c A A since ( c), (c ) nd ( )c re ll equl to the signed volume of the sme A prllelepiped, the one formed from,, c nd they ll hve the sme orienttion Corollry: A 3 3 mtrix is invertile if nd only if A For such mtrices A 1 1 A dj A Theorem 1: If A there exists nonzero vector v such tht Av Proof: Suppose A (,, c) nd A Then,, c re coplnr If, re not prllel then c + for some, Hence (,, c) 1 c If nd then (,, c) 1 1 If then (,, c) Theorem 11: ( + c) + c Proof: There does not pper to e simple geometric proof of this ut it is esily proved lgericlly A convenient wy to compute vector products is to express them in terms of the three stndrd sis vectors i (1,, ), j (, 1, ) nd k (,, 1) A generl vector product cn e reduced, y the distriutive lw, to the vector products etween i, j nd k For reference we note these elow: i j k, j k i, k i j i i j j k k 36
7 Exmple 2: If (3, 5, ) nd (2,, 1) find Solution: (3i + 5j) (2i + k) 3j +1k + 5i (5, 3, 1) Another useful wy of clculting vector products is to set up 3 3 mtrix with the vectors i, j, k s the first row, with the components of the two vectors occupying the second nd third rows This will ecome more useful when we discuss the evlution of determinnts in generl, in lter chpter Exmple 3: Find (2, 6, 1) (1, 3, 5) i j k Solution: The vector product is i j k i 11j + k (33, 11, ) Since the cross product is liner in ech vrile the mp v v is liner trnsformtion nd corresponds to multipliction y 3 3 mtrix X() Theorem 12: X()X() X()X() X( ) 1 v 1 2 v 3 3 v v 1 Proof: Let 2 nd v v 2 Then v 3 v 1 1 v v 3 1 v 2 2 v v 2 v So X() 3 1 nd X() 3 1 The conclusion of the theorem is now esily checked Corollry: For ll vectors,, c ( c) + (c ) + c ( ) Proof: ( c) + (c ) + c ( ) X()X()c X()X()c X( )c 26 Appliction of Vector Products to Electromgnetic Theory The vector product is vlule tool in physics, especilly in connection with electricity nd mgnetism Typicl of its use in this re is the formul for the voltge generted in stright piece of wire tht is moving in uniform mgnetic field: E (v B)dL where E is the voltge generted, in volts, in certin direction (if E is negtive this indictes tht the flow is in the opposite direction); v is the velocity of the piece of wire (the mgnitude is the speed in metres per second nd the direction is the direction of movement of the wire); B is vector representing the mgnetic field (in weers per squre metre); nd L is vector representing the length (in metres) nd orienttion of the piece of wire B B L v One conclusion from this formul is tht if the wire lies in certin plne, nd moves only in tht plne, nd if the mgnetic field is prllel to tht plne, the voltge generted is zero This is 37
8 ecuse (v B)dL is the signed volume of the prllelogrm formed from v, B nd dl, which is zero if these vectors re coplnr Exmple 4: Find, s function of time, the voltge generted in circulr loop of wire of rdius r tht is rotting with constnt ngulr velocity of rdins per second out the zxis in uniform mgnetic field of M weers per squre metre in the direction of () the positive hlf of the zxis; () the negtive hlf of the xxis Solution: R z dl z v t r Let r e the vector representing x the point under considertion sin cos t Then R r sin sin t cos Since v is orthogonl to oth R nd k (k eing the unit vector long the positive xxis) nd k, R, v form positive triple, v hs the sme direction s k R Adjusting mgnitudes we see tht v v (k R) k R r sin r sin sin t k R r sin cos t sin sin t r sin r sin r sin cos t sin t r sin cos t Since dl is orthogonl to oth R nd v nd R, v, dl form positive triple dl dl (R v) R v r d R v r 2 cos sin cos t r 2 cos sin sin t r 2 sin 2 cos cos t r d r 2 sin r2 sin cos sin t sin cos cos t r d cos sin t sin cos t () B Then v B Mr sin sin t whence M y r sin cos t x t r sin sin t r sin y 38
9 E 2 Mr 2 cos sin d Mr 2 sin 2 d M () B Then v B Mr sin whence cos t E 2 Mr 2 sin 2 cos t d 2Mr 2 cos t sin 2 d r 2 M cos t NOTE: Cse () is gretly simplified ccount of how lternting current is generted EXERCISES FOR CHAPTER 2 Exercise 1: If (2 1, 1) (1, 2, 1) nd c (1, 2, 1) find () ; () ( + c); (c) ( c); (d) ( c) Exercise 2: If (1, 1, 3), (2, 1, 2) nd c (, 1, 1) find () ; () ( + c); (c) ( c); (d) ( c) 5 2 Exercise 3: Evlute (i) ; (ii) Exercise 4: Find 3 vlues of such tht I A where A is the mtrix of 3(i) For ech vlue of find corresponding nonzero vector v such tht (I A)v Exercise 5: Find nonzero solution to the system of equtions 3x + 2y + 7z x y z 2x + y Exercise 6: () Find nonzero vector tht is perpendiculr to (3, 2, 7) nd (1, 1, 1) () Find nonzero vector tht is perpendiculr to 5i 2k nd 2i + j Exercise 7: If (, 4, 2) nd (3, 2, 1) find nd
10 Exercise 8: If (1, 1, 2), (3, 1, 1) nd c (, 2, 3) find (i) ; (ii) c; (iii) c If A (,, c) find A 1 Exercise 9: Find 3 3 rel mtrix A such tht the liner trnsformtion X x Y A y Z z trnsforms the unit sphere x 2 + y 2 + z 2 1 into the ellipsoid X2 2 + Y2 2 + Z2 c 2 1 Find A nd hence find the volume of the ove ellipsoid Exercise 1: The six fces of prllelepiped lie on the following plnes: 2x y 2x y 1 x + y x + y 1 2x + y z 2x + y z 1 Find 3 3 mtrix A which trnsforms this prllelepiped into cue Hence find the volume of the prllelepiped Exercise 11: Write the following system of equtions 3x 2y + 5z 4 x + 7y 9 4x + 8y 5z 21 x in the form Av where v y nd A is 3 3 mtrix z Exercise 12: Find the point of intersection of the plnes 2x + 5y z 17 3x + 7y + 4z 13 12x + 5z 1 Exercise 13: () Find the re of the prllelogrm whose vertices re (,, ), (2, 3, 3), (4, 1, ) nd (6, 4, 3) () Find the re of the prllelogrm whose vertices re (1, 1, 2), (3, 2, 1), (5,, 2) nd (7, 3, 1) (c) Find the re of the tringle whose vertices re (1, 1, 2), (3, 2, 1) nd (5,, 2) Exercise 14: Find the re of the prllelogrm, three of whose vertices re (1,, 2), (5, 3, 4) nd (2, 1, 1) Exercise 15: Find the ngle etween the vectors (2, 2, 1) nd (1,, 1) Exercise 16: Find the ngles of the tringle whose vertices re A(1 2, 1), B(3,, 2) nd C(4, 4, 1) 4
11 Exercise 17: Prove tht the re of the tringle with vertices, c is ½ + c + c Exercise 18: Find the eqution of the plne through tht is perpendiculr to the vector (1, 2, 3) Exercise 19: Prove tht the eqution of the plne tht psses through the points, nd c is ( + c + c)v (c) Exercise 2: Prove tht if, re 3dimensionl column vectors nd if A is the 3 2 mtrix (, ) then A T A 2 Exercise 21: The 3dimensionl column vectors, re noncolliner nd d is the foot of the perpendiculr from c to the plne through, nd A is the 3 2 mtrix (, ) Prove tht: (i) (c d) (c d) ; (ii) A T c A T d; (iii) d A for some rel, ; (iv) d A(A T A) 1 A T c; (v) Explin why one cnnot use the identity (MN) 1 N 1 M 1 to conclude tht d c (vi) By considering the geometric interprettion of the liner trnsformtion v A(A T A) 1 A T v, or otherwise, show tht nd 1 re eigenvlues of A(A T A) 1 A T nd tht, nd re its eigenvectors Exercise 22: A 3dimensionl vector function is function tht ssigns to every vlue of rel vrile t 3dimensionl rel vector r(t) If the components of r(t) re differentile the derivtive of r(t) with respect to t is defined y differentiting ech component Prove tht if r(t) nd s(t) re differentile 3dimensionl vector functions then: d dt (r(t)s(t)) dr(t) dt s(t) + r(t) ds(t) dt Prove tht if the length of r(t) is constnt s t vries then dr(t) dt is perpendiculr to r(t) for every vlue of t Exercise 23: Prove tht if (t) nd (t) re differentile 3dimensionl vector functions of t then d dt ((t) (t)) d(t) dt (t) + (t) d(t) dt Exercise 24: If,, c re 3dimensionl unit vectors tht re not coplnr nd if nd c re perpendiculr prove tht the ngle etween nd the plne through, nd c is where sin ( c) Exercise 25: Exmine the following theorem nd proof nd comment on its vlidity Theorem: If,, c, d re nonzero vectors with ( ) (c d) then,,, c, d lie in plne Proof: Suppose ( ) (c d) Then is prllel to c d Since these re perpendiculr to the plnes,, nd, c, d respectively, these plnes must coincide nd hence ll five points must lie in this plne 41
12 SOLUTIONS FOR CHAPTER 2 Exercise 1: () (3, 3, 3); () (4,, 8); (c) 12; (d) (4, 4, 4) Exercise 2: () (1, 8, 3); () (5, 7, 4); (c) 11; (d) (4, 7, 1) Exercise 3: (i) 35; (ii) 5 2 Exercise 4: ( 1)( 5)( 7) so 2 5 1, 5, 7 re the three vlues When 1, I A so 2 is suitle vector When 5, I A so is suitle vector When 7, I A so 1 is suitle vector 1 Exercise 5: so z 1, y 2, z 1 is nonzero solution Exercise 6: () (3, 2, 7) (1, 1, 1) (5, 1, 5) is suitle vector () (5i 2k) (2i + j) 1(i i) +5(i j) + 4(k i) 2(k j) + 5k + 4j + 2i 2i + 4j + 5k Exercise 7: (8, 6, 12) If c (3, 1, 2) the determinnt is ( )c 54 Exercise 8: (i) (1, 7, 4); (ii) (5, 9, 6); (iii) (7, 3, 2) dja ( c, c, ) T A ( )c 26 so A
13 x x Exercise 9: The liner trnsformtion tkes y to y z cz 1 The required mtrix is A c A c The volume of the unit sphere is 4 3 so the volume of the ellipsoid is 4 3 c Exercise 1: Let X 2x y, Y x + y, Z 2x + y z Then the six fces re X, X 1, Y, Y 1, Z, Z 1 which ound unit cue x X 2 1 The mtrix of the liner trnsformtion tht tkes y to Y is A 1 1 z Z Since A 3, volumes re multiplied y 3 y the liner trnsformtion (The minus sign merely reltes to the orienttion of the vertices) Hence the volume of the prllelepiped is x Exercise 11: The system of equtions cn e written s y z Exercise 12: We must solve the eqution x y 13 z Let 3, 7, c Then the djoint of the mtrix is ( c, c, ) T The determinnt is 469 so the inverse is The solution is so there is unique solution x, y 3, z 2 Exercise 13: () (2, 3, 13) (4, 1, ) (13, 52, 1) whose length is () We must trnslte (1, 1, 2) to the origin The points ecome (,, ), (2, 3, 3), (4, 1, ) nd (6, 4, 3) This is the sme prllelogrm s efore, so the re is 2973 (c) The re of this tringle is hlf tht of the previous prllelogrm, tht is it is ½ 2973 Exercise 14: Trnslting (1,, 2) to the origin we get (,, ), (4, 3, 2) nd (1, 1, 1) 43
14 (4, 3, 2) (1, 1, 1) (5, 6, 1) whose length is Exercise 15: cos Exercise 16: Trnslting (1, 2, 1) to the origin we get the vectors (2, 2, 1) nd (3, 6, ) 18 If is the ngle t A then cos so Trnslting (3,, 2) to the origin we get the vectors (2, 2, 1) nd (1, 4, 1) 18 If is the ngle t A then cos so The remining ngle is 964 Exercise 17: The re is ½ ( ) (c ) ½ ( c) ( ) ( c) + ( ) ½ ( c) + ( ) + (c ) Exercise 18: The plne is x + 2y + 3z Exercise 19: Let v lie on the plne Trnslte to the origin The resulting vectors re,, c nd v A vector perpendiculr to this plne is ( ) (c ) + c + c Since v lies on this plne ( + c + c)(v ) so ( + c + c)v ( + c + c) (c) since the other terms re zero Exercise 2: Suppose A (, ) Then A T A 2 2 so A T A 2 2 () ( cos ) 2 where is the ngle etween, 2 2 (1 cos 2 ) 2 2 sin 2 2 Exercise 21: (i) The line from c to d is perpendiculr to oth, so (c d) (c d) (ii) A T T c T c c c d d AT d (iii) Since d lies in the plne determined y,, we must hve d + for some rel, 1 If 2 etc then d , d nd nd d Hence d A (iv) A T c A T d A T A so (AT A) 1 A T c nd hence d A A(AT A) 1 A T c 44
15 (v) A nd A T re 3 2 mtrices nd hence cnnot hve inverses (vi) The liner trnsformtion v A(A T A) 1 A T v projects vector in R 3 onto the plne through,, Vectors lying on this plne re fixed nd so, re eigenvectors corresponding to the eigenvlue 1 Vectors perpendiculr to this plne, such s, re mpped to nd so re eigenvectors corresponding to the eigenvlue r 1 (t) Exercise 22: Let r(t) r 2 (t) etc r 3 (t) Then r(t)s(t) r 1 (t)s 1 (t) + r 2 (t)s 2 (t) + r 2 (t)s 2 (t) Hence d dt (r(t)s(t)) d dt (r 1(t)s 1 (t) + r 2 (t)s 2 (t) + r 3 (t)s 3 (t)) d dt r 1(t) s 1 (t) + d dt r 2(t) s 2 (t) + d dt r 3(t) s 3 (t)) + r 1 (t) d dt s 1(t) s 1 (t) + r 2 (t) d dt s 2(t) s 2 (t) + r 3 (t) d dt s 3(t) s 3 (t)) dr 1 (t) dt, dr 2(t) dt, dr 3(t) dt (s 1 (t), s 2 (t), s 3 (t)) + (r 1 (t), r 2 (t), r 3 (t)) ds 1 (t) dt, ds 2(t) dt, ds 3(t) dt dr(t) dt s(t) + r(t) ds(t) dt r 1 (t) Exercise 23: Let r(t) r 2 (t) etc nd let denote derivtives with respect to t r 3 (t) 2 (t) 3 (t) 3 (t) 2 (t) Then (t) (t) 3 (t) 1 (t) 1 (t) 3 (t) 1 (t) 2 (t) 2 (t) 1 (t) Hence d 2 (t) 3 (t) 3 (t) 2 (t) + 2 (t) 3 (t) 3 (t) 2 (t) dt ((t) (t)) 3 (t) 1 (t) 1 (t) 3 (t) + 3 (t) 1 (t) 1 (t) 3 (t) 1 (t) 2 (t) 2 (t) 1 (t) + 1 (t) 2 (t) 2 (t) 1 (t) 2 (t) 3 (t) 3 (t) 2 (t) 2 (t) 3 (t) 3 (t) 2 (t) 3 (t) 1 (t) 1 (t) 3 (t) + 3 (t) 1 (t) 1 (t) 3 (t) 1 (t) 2 (t) 2 (t) 1 (t) 1 (t) 2 (t) 2 (t) 1 (t) d(t) dt (t) + (t) d(t) dt Exercise 24: The volume of the prllelepiped formed from,,, c is ( c) However the re of the se is 1 nd the perpendiculr height is sin so the volume is sin Exercise 25: The given identity holds if, in which cse, re prllel However the line through these vectors need not lie in the plne, c, d Counterexmple (i i) (j k), yet the points, i, j, k do not lie in plne 45
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