sections4_1and4_2

sections4_1and4_2 - Chapter 4 Vector Geometry 4.1 Vectors...

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Unformatted text preview: Chapter 4 Vector Geometry 4.1 Vectors and Lines 1. 2. 3. 10. 11. '12. (a) 3 (b) x/6 (cm (d) \/5 (e) N6 H) 3J6 (a)TV7§[7 ~15]T (b)§[e2 e12]T (a) %[ —2 4 1 ]T (b)a=i1/Hfi’l] (a) x/i (b) fl (c) \/Z_1 (d) 3 B & fizfiJrB—E:§A§+§Bé=%(AB+Bd)=§T. (b) 1375’ “m” §T+§CT§= %(E+CB) = §E. (a) Yes (b) Yes (a) N0 (d) Yes (a) 13—17 (13%? . (c) 7'5 (d) WEN?) (a) [2 2 —-3 13m (b)[——1 ——1 affix/2‘7 (c)[o 0 fl4]T;4 (d) [0 o 0]T;0 (e)[—2 o affix/Em (£)[—2 2 _2 ]T;¢1“2 (a) Q(01533) Q€i2afi1:1) (b) (i) Q(5,—1 2) (ii) Q(1,1r4) (b)f=fiw6fi+513:[—26 4 19]T (a) (a,b,c)= [10 —11 —8 ]T (b) (a,b,c):[—8 14 13 f" 47 48 Chapter 4 — Vector Geometry T:[:c; 9:2 x3]T. f1. .1 .0. 4.4 $1.”??? —1 0 1 3:2 —> —1 0 1 3:2 1 1 $3 13. (a),(b) Ifit holds then[3a+4b+c —-a+c b+c If there is to be a solution then 3121 + 3332 : 4333 must hold, and this is not satisfied in either case. 14. (a) %(9,2,—8) (b) i(5r5r2) l5. (4,0,4) and (6, e3,3). 16 “~“+T(“~u“)— s”+T” . 'U~'U1 T+Sv2 ‘Ul —-r+s'Ul T+8’Uz. 17- (a)Q(4am3:'—2) (b)Q(0,713) 18. (a)f:%[5 3 +4]T (b)_:1‘2’2115[—20 —13 14]T 19. 175:7[9 —6 3]Tomr=[—9 6 —3] 20. (a) 3(—1,—2,3) (b)S(-—1,3,2) 21. (a) True. = 0 implies 17: (l. (b) "nus. Me as on = 0 implies s — w z (1'. ((1) True. 22? = 5 implies if 2 (l. (d) False. He“ : ||—fi|f for all fibut «a: is only holds if a: 6. (e) False. H[ 2 1 2 ]T [ 2 ——1 2 F“ . (f) False. If t < 0 they have opposite directions. (g) True. If 13’ = t('5’+ 111') then if % 0 {as 13'7E 6) so ti? : (1:017. (h) False. “~51?” : 5 us?” so it fails if 6% 6. (0 True- IWEI = |1217H= 2 H1711 implieS IWH I 0- (j) False. Take 13 : ~17 where 17% 22. (a) [1 —13]T+t[2 W1 0]T (b)[3 —1 4]T+t[2 —15]T (c)[3 —14]T+t[001]T (d)[111]T+t[111]T (e)[10 s3]T+t[2 —1 3]T (f)[2 —11]T+7§[—1 0 if (g)[1 01]T+t[2 —3 3]Tand[10 1]T+t[211]T 23.' (a) P corresponds to t : 1, Q corresponds to t = —2. (b) P corresponds to t = 2, Q corresponds to t 2 5. 24. (a) P(2,3,0); t = *1, s = ~—1. (b) No intersection. Section 4.2 — Projections and Planes 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 4.2 49 (c) No intersection. If there were one, then there would exist 3 and t such that [ 3 —1 2 ]T+ t — 2s 2 —2 t[1 1 —1]T==[1 1 —2]T+s[2 0 3]T;Le t —2 Thwe Hm. .- .. ... . . . _ .1 . .-H.. ._t 1;. gsHZ: A74 . . equations have no solution. (d) P(2,21,3); t = —2, s 2 —3 The vector equation is if 2 6 + 1252 ml; t arbitrary. If it meets the XY-plane at Patna, yo, 0) then its vector equation is 172 (330,90, 0} + t(0, O, 1) som2m0,y2y0,22t. Wehavex=w0+ta,y=y0+tb,z=zg+tc. Solvefort= "fa—awn zygyn 2 [3—620 fihflfi=fl014fnmfisefi=hlgflfimmfi=fi= flt=[10M?meEB=E123=[10MLfi022mf= [iezisf I _. —} T _. —-—> _. _‘ _ erte d 2 b — 6‘: 2 [ 1 1 —1 ] , so (2’2 E + td. Then “AG” 2 2“c — a” 2 |t| ,while —> “56” Z “go all = “(1 * tlalll =11 “ tlllfllll, SO [IA—d” = 2“Boll means ltl = 2l1 * tl- Hence 19240213)?andsot2§ort22;thatis02 “T1 §]T0rC2 [ 3 1 0 ]T. -——> +AF2 ———> > > > > —> _——7 We have AE 2 BD so AB + AE 2 AD. Similarly A AD. ——> (a)0P1=—_ia’n0152=—oi3mofi3:—015EL ‘0 53k 2 m5?” is, 1 S It 3 n, if there are 2n points. + -27r (c) , (d) In general, let u: 2 e17? . Then, in the complex plane Pg 2 for)"8 for all k and the result comes from 1 + w + 102 + + w””1 2 0 (since 71:” — 1 2 0). —> A? = E+EI§ = §AC+ ————> %DB 2 %(E§+R‘+D_Cl+0—Bl) 2 Henceiffi> 2 w—r—a——)—>—)———+——+——-—) 2EF=mEZ+ZFyenMs¢AF=DA+FO=CB+FC=FB Let P, Q, and R be the vertices with position vectors 13,17 and 16. The position vector of the midpoint M of QR is fit 2 at? + 11?). Hence the point C of PM g the way from P to M has position vector 5'2 {5+ — if) 2 31-(fl'+ 17+ 11?). Similarly for the other medians. Let iii, 'u? and :7? correspond to A, B, C and D. Let E denote the centroid of triangle BOD so E has position vector 1%(17-1- 13 + f) by the previous exercise. Then the required position vector is %(17+1I5+f) + $111.2 %(’U+u7+§)} = i(11'+17+13+f). Pro ect ions and Planes (a) 0 (b) 6 (c) —4 (c1) 0 (e) or + by + cz (f) 0 (a) E or 45° (b) 1r or 180° (c) 72$ or 90° (d) g or 60° (e) 2?” or 120° (f) 2% or 120° 50 3. 4. 7. 9 10. 11. 12. 13. 14. 15. 16. W{c) s[1l '0 2 ]Tu+t[”0n 1 IOI]T Chapter 4 — Vector Geometry (aw; (a)t[5 —1 1]T (b) 1 or —17 (b)t[e1 1 2]T (d)s[1'2'0”]Ti+i[0 3 '1]? [—210]T,[0 01f" Pfi=[3 a2 4]T,13’1%:[5 5 afandcfizp 7 2]?” (a) 13—62) - 541% = 0 so angle Q is a. right angie. (b) IEFQIIZ + um? = 29 + 57 = 86 = ItP—R’EP.T E:[211 V5]T,;1—5'~_~[3 17 —15 andIB—C}':[1 6 —10]T,andnotwoare orthogonal. (a) A:%or90°,B=§0r60°,0:%0r30°. (b) A=B:C:§—or60° . . —’ T —’ T Dlrectlon vectors P0131: [ e1 0 —1 ] and P2133 : [ ——1 6 1 ] are orthogonal. (a) ge (b %e (c) %ff (d) $71 (20%[1—1 3]T+fi[16 —5 —7]T (mg-[2 #1 ~4]T+§11u[53 26 20? {c)%[31—1]T+f11"[7*165]T (d)§—§[6 —41]T+%[—3226]T (a) $v133, Q %,%—,%) (b) %\/5642, Q(%,%,%). (8') (1311—1) (0:010) (C) {12475) (42—1578) (a)10m+4y—z:21 (c) 3m—2y—-z=7 (e) 9:0 (g) 22+y—2z274 (i) x—2y+4z=% (b) —23$+32y+11z= 11 (d) 2m—y+z=5 (f) 2$+3y+22=7 (h) 2w—7y—32=—1 (j)m—y—z=3 (a) [w y z]T=[3 —14]T+t[3 e2 m1]T (b)[a: y z]T=[2 71 3]T+t[210]T (c) [x y z]T:t[111 2]T (d)[$ y z]T=[11 ~1]T+t[1 1 1]T (e) [m y z]T:[21 ~1]T+t[—110 3]T (f)[:c y z}T=[1 1 2]T+t[41—5]T (a) 4% We, 29%?) (b) 1%, Q<§=§=~%) Section 4.2 — Projections and Planes 51 17. (a) No. (if: [ 1 2 —3 ]T is not orthogonal to the normal 733 = [ 2 71 71 ]T of the plane so the line is not even parallel to the plane. Note that P(1, 2, —3) dOes lie in the plane. (b) Yes. The equation is 5:1: — 3y - 4s = 0. 18. R is on the line through P and C2216?) : (1, —2,2) and (~1,6,75) = (1,2, ~1) w 2(1, —2,2). 19. (a)[3 1 1]T+t[~1 4 7]T (b)[-2 7 0]T+t[3 e5 2]T 20. (a) P (%, 1, (b) None (0) Every point on the line. (d) P (%, —'{—3, %) 21. (a) 2m + y + 32 = d; d arbitrary. (1)) 3m + 2z = d; d arbitrary. (c) as: + by + cz m G; a, b, c not all zero. (d) a(x fl 3) + My — 2) + 6(2 + 4) = 0; a, b, and c not all zero. (9) 2am + by + az 2 o + b; (1,?) not both zero. (f) as: + by + (b — a)z = a; a and b not both zero. (g) or + cry + cz :— 3o; 0., c not both zero. (11) a2: + by + (a — 25);: 2 5a — 45; a and b not both zero. 22. The plane has vector equation 71' ~ 15': d where 71’ is the normal. If 131 and 162 are the position vectors of P1 and P2, then every point P on the line has position vector 15': 131 + t(35'2 — 331). Thus it v 13’: d + t(d — d) = at, so P lies on the plane. 23. (a) :3; 5 (b) (if) 24. Mfifi‘i’ 3— %),B(%,1,%) v: a -%)a Be, as (d) a, A(1—3?,2, a, as, 13,0) 25. Direction vectors are all = [ 1 m1 ]T and d; m [ 1 m2 ]T, and J1 - dz = 1 + mlmg. 26. (a) Assume each edge is of length a, one vertex is at the origin, and three edges lie along the positive axes. Then the diagonals are :l:[ a a a ]T, i[ —a a a ]T, j: [ a —a a. ] and :l: [ a a —a ]T. The dot products of these equal iag gé 0. T (b) Consider the diagonal of: [ a a a F. The six face diagonals in question are :E [ a 0 so, ]T, :l: [ 0 o. —a ]T and i [ a 7e 0 1T. All these are orthogonal to The result works for the other diagonals by symmetry. 27. Position the solid as in 26(a) with the x/i-side along the Y—axis. The diagonal is it = .t' _. 2 [1 fl 1]Tandthelongsideisfi=[0 «2' 0 ]T.Thuscos 9: u v 1 so 9: 1 or 45°. IEfl'll [an 2 m 2 vs 4 52 . Chapter 4 — Vector Geometry 28. Position the solid as in 26(a) and let the edges be of lengths a, b, c. Then'the diagonals are :l:[a b c]T, i[ —o. b c]T,:i:[a —b C]Tandi[ a b —c lT.Thedotproducts . . squat eke? t b2..+.c2)a is? -. 52. to??? .i(a2.+ .52 T 62.): 29. Letzfizfl 1 e1 1]Tandg—6’:s[2 0 —1]T.ThenB—Cl=zT5'iE§:[2s—t t —s—t ]T is orthogonal to A—C)’ (because the angle at C is a right angle), and hence to [ 2 0 —1 ]T. This implies t = '53, and so R)” z [ —33 55 —63 ]T. Hence a” = [3 —5 6 ]T is T = a direction vector for the line through B and C, and so the equation is [ x y z [2 e1 1]T+T[3 —5 6 fur" a parameter. 30. Let if and 11} be vectors along adjacent sides of the parallelogram, with a common tail. Then the diagonals are 1?+ 711’ and if“ a? so insisting that “17+ “(Eng 2 H1? — ‘LUHZ implies 41'}. 1E? : 0. Hence 6’ - 217 z 0 and the figure is a rectangle. '9" a " . . -:' . v 2 = ’U -'i = (m,y,z) - (1,0,0) 2 LL‘. Hence the prOJeCthfl along t is m. The others are 31. 4 2' similar. 7: 32. No. If 9 is the angle between 11’ and if, then cos 9 = fi %7. This is impossible. 33. (fi+fi)-(s—s =s—eus-sw-sss-s: Hank "an? 34. (a) Compute “13+ 17H2 2 (175+ 17) - (11+ 1?) and “11’— fi||2 as in Exercise 33, and add. (1)) The squares of the lengths of. the diagonals add to the sum of the squares of the lengths of the four sides. 35. If “:3 and 27 are the sides then the diagonals are 11‘ + if and a? — 43. Hence 0 = (13+ 6‘) - (21’ — ") = “175'”:2 i Since and are both non—negative, this implies = 36. To 2 s— s and as : 175+17(sinee 3—6 = a). Hence R5 . E : “an? —|;s1|2 : 0. 37. Clear from Theorem 3 since it - 17 = O. 38. (a) {5- {17+ 13) = “J- ’D' + "It" 13 : 0 so 11' and fi+ 13 are orthogonal. Hence using Exercise 37 twice H134“ 17+ wilz = "all2 +ll17+ 13H2 2 Hle + ll’r’ll2 +ll13llz- a. 71+ *+ r r 2 1 = = w since Me! = “at = Hell. The same ilulillwv +wll nun llullg + livllg + lel2 is true of if and 13 so the cosines of the angles agree. (b) 39. (a) If P1{;t1,y1) and P2(a:2,y2) are On the line, then or; +by1 + c 2 0 and (13:2 + byg + c = Ofi —v Hence d = P1132 : (332 .— Iia’yz * yi), SO 71‘ ' d = a($2 - x1.) + We — 2/1) = (6132 + 5:92) “ (owl + byl) : —e+ c rm 0. (b) Let 150, 351 be the position vectors of P0, P1, so 175' = 170 W171. Then “Li'- fi' = 70 -fi' e #1 47?: = (13-73 fl lfi- 771'! Hfill2 llfill ’ (are + byo) — (excl + byl) = emu + byg + 0. Hence the distance is as required. Section 4.3 7 The Cross Product 53 Elfillllifill (since they are not parallel) and the cosine of the angle between 71’ and u? is l ifl'll llfill _, . Similarly for 17 and 13. ' 41. (a) "E - 5: a. so cos 0: = a/H'U“ follows. The others are similar. as) “all? z a? + b2 + 42. Have 15': if] for some t. Now (f - I?" = 0 so 175- f)’ = 17 = tllfi’flz. Hence t : (fl‘fiflnfi'fl? as required. . . '_. _,. “Pi-(0.17) _, 2(fi-fi)_, (ii-“5)., .. - . .7 43. ThepmJectlon ofuon av 15 _, 2 cm = a. 2 _, 21) = _, 2 1;, and the IS the prOJectmn 4 A Havli lal M nvu ~ of u on 1). 44- (a) lfi'fil = |||”f1'|li|17|100S 9| = Flfl'llllfilllCOS 9E S llfillflfill- (b) Ifflgélgvrthenlcos 9! :lmeans cos 9=i1;ie. 6:00r9=7r;ie. flandfiare ' parallel. (c) Takefl': [ m1 yl 21]Tand17:[3:2 yg 22 ]T T T . (d) Take[:c1 yl 21] =[a3 y z]Tand[m2 yg z2]T=[y z m]111[c). (e) Take[:1:1 yl zl ]T:[:c y z]Tand[a:2 y; z2]T=[1 11]Tin(c). u+v v 45. u Compare lengths of sides: Hfi-i— 17” g + ...
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This note was uploaded on 09/17/2011 for the course MATH 115 taught by Professor Dunbar during the Fall '07 term at Waterloo.

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sections4_1and4_2 - Chapter 4 Vector Geometry 4.1 Vectors...

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