0534552080_98853

Advanced Engineering Mathematics

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Unformatted text preview: 106 Section 6.1 Chapterlfiix w Vectors and. Vector Spaces Section 6.1 The Algebra and Geometry of Vectors 1‘ F+G = (2-i— y/g)i+3j;F—G : (2 m flfi— 9j+10k;|EFi| : = fl; 2F = 4'1— 63' + 101:;38 : sfli + 18j w 151: 23F+G=i+gw3mF~Gzimg—3mww:«Emmpng22L4mfiG=1m 3PVJE:3b4quchw1m+kJFW=VfifiGfl23fliflfi2M—1m; 38=3HJfim3k 4- F—G: (fi+8)i+j—4k;F*G= (fl—8)i+j—8k;llFli =x/E;||G|| = v”6_8; 2F = 2x/§i+ 23' — 12k;3G = 24: +6k 5. F+G=3i—j+3k;F—G=-~i+3j—k;f|F||:fi;]}GIE=2\/§;2F=2i+2j+2k; M3=6L—fi+6k 6F+G=i+®F—G:iwfi TF+G=$—%F—G:i Section 6‘] 107 &F+Gmw$+flFwG=—fi—fi 9. F+Gn2i—5j;F—-G=j m.F+G:wfi+$F*G:i+fi G: —2i—3j 108 Section 6.1 11. 051%“?ng 12. aF=12iw4j 13. 05F = 12j 14. 0:3“ = 3i — 33 l arF peeps; Section 6.1 109 16. A vector from (1,0,4) to (2,1,1) is M u i+j — 3k. A vector from (1,0,4) on L to ($,y,z) on L is — 1)i + (y # 0)j + (z — 4jk, which must be parallei to M. Thus :1: m 1 = igy — {l : t,z — 4 2 —3t, for some scalar t. Parametric equations of the line are a: z 1 + t,y : t, z z 4 —~ 3t:L for —00 < t < 00. Eiiminating the parameter t gives normal 3: —~ 1 y z — ec nations 2 — — . i 1 1 ~3 1.7. Parametric equations are a: z 3 + fit, y 2 wt, 2 = 0; M00 < t < 00; Normal equations are a: — 3 y 6 ' Z “fin—1., Z z 18. Parametric equations are a: x 2,3,! 2 1,2: x 1 — 3t; ~00 < t < 00; Normal equations are a: z 2, y 2 1, z = 1 H 3t. rI‘his line is parallel to the z axis through (2,1,0). 19. Parametric equations are :5 : [Ly = 1 + t,z = 3 + 2t; —00 < t < 00; Normal equations :5 w 3 ~: m 1 2 dm—. are a O,y 2 20. Parametric equations are m z 1 — 3t,y : 0 — 2t,z : —4 + 9t; ~00 < t < 00; Normal ~— 1 z + 4 equations are m I y — “3 T2 "" 9 21. Parametric equations are :1: z 2 A 3t,y : —3 + 9t,z : 6 — 2t; ~00 < t < 00; Normal uat'ons are a: m 2 m M z _6 eq 1 m3 _ 9 _ —2 ' 10 22. F : \/5 (cos 1+ sin j) : £0 +j) 23. F 2 6 (cos i + sin j) -- tore 3i 4* 110 311' 31? . 1=5 24% (cos(5 5 m w1.545i + 4.7559 )i+Sin( Section 6.2 Zflk, (3)11. — 1 27. Let the coordinates of H 2 (33,-, 3h, 23). Then F,- r {:c,.,.1 — mfli ~1— (y.,—_.,.1 — y,)j + (2,-_,__, — 71—1 and F1 -€- F2 + - -- + F114 2 Zflmrfl #— :L',.)i + {s:,.+l + (z,._,.1 —~ ,3.er m (:37, ._ 3:1)1 + r=1 + (Z71 _ zlflc Z "'"Fn: 50 F] 'i— F2 ‘i' ' ' ' “l” F11 : 1 28. Since HtFJI = z 1 implies [ti : Two choices for the scalar t are t 2 : 29. Let [1,3, 0 be the vertiees of the triangle, P and Q the feet of the altitudes from A and B to sides BC and AC‘ respectively, and S the point where these two altitudes intersect. Finally, let R be the point where the vector CS extended intersects side AB, It will be sufficient to show GR is perpendicular to AB (so the altitude from C‘ also passes through S}. We have AS ‘ BC 2 BS - AC 2 (l, and then CS-AB : CS-(AC+CB) ——— CS-AC+CS‘CB : (CB+BS)- AC+(CA+AS)-CB 2 CB‘AC+BS-AC+CA-CB+AS‘CB :— CB-AC+U+CB‘CA+0 2 CB » (AC+CA) m CB-O : U, which completes the proof. Section 6.2 The Dot Product In 1 — 6, we use the first vector as F, the second as G. 2m. Section 6.2 111 1. F - G : 2;coe[0) : : fi; not orthogonai; |F - G1 : 2 < V/l— = 2. F - G : 8;cos(9) : not orthogonal; [F‘ G! = 8 < m 3. F-G .= —23;cos(9) = not orthogonal; EF‘GI 2 23 < = 4. F-G : —63;coe(9) = not orti'logonal; 1113; = 63 2 me < 2 “melon 5. F - G = —18;cos(6) 2 Vflgfl—ljlm; not orthogonal; IF » GI 18 = < 1% = a 1'9 . G : 4330509) not orthogonal; [F - GI ’1 < 6 I 7. 3:33—y+4z=4 8.1:w2y2—1 9.11:1;w3y-i—2z225 10. —3:1:+2y :1 11. Tm+6y—5z:v~26 12. 4m+3y+z=e+fil . V - AB . 13. Let 1M = 111.1deth of BC, and V 2 AM. Then (105(8) Here-we have M = (7/2,1,—3),V : 5/21 + 33 u 9k, AB 2 21+ 2j — 5k,cos(a) : i = ..—152_... Mm 11m . . . . 101. 14. M : (—1/2,1/2,4),V 2 «7721+ 5/2; + 7k, AB = —51 + 23 + 4k,cos(9) m 3 101 m 15. M : (w3/2,1,2),V : —5/2i+3j—4k,AB = —i+6j—9k,cos(6) : 2 £53— V’125Jfi 5% 1.6. M : (4 —1,2) V : 41 —6j +3k AB :1” 73 +6k,coe(6) i : ME?” ’ ’ ‘ EM 1/5246 17. That F = 0, since F - F 21m”? = 0. 18. That F 2 0, since every vector X 2 xi + yj + 2k for some 133;? 2. Then F - X : U for every vector X and F = O by Problem 17. 19. For any unit vector U, F - U : c036, so [F - U] = ||F|Hcos€l :4 The maximum value of IF - U1 is achieved when |cos€l = 1 so 9 : O or e and hence U is parallel to F. ObServe that (F ‘ U)max = and (F - U)min : ~|§F||. 20. Let F oi—l—bj—Fck. Then F-i : (1 since (i-i) = 1, = [i‘k] = U. Si1'11ila1‘ly(F-j) b and (F - k) = c and F : (F - '1)i + (F + (F v k)k. 112 Section 6.3 Section 61;; The Cross Product i j k 1 F - G 4 1.1331132 —3 6 1 "—"81-1-2j-l—12k=~--G><F; c0519 2 2———‘;sin!9 _1 mg 1 I 1 ) élFllllGil «3—9 ( l , 15 K53“ 1 1. 15—3 1/1. —C082 2 — E 2 @flllt E||1G|gsm(9) 2 2 V21 2 “(3711+ 4+ 144 2 In Problems 2 w 10, the values are given in order for F X G;cos(6);sin(6); and the common value of 5111(0) §|F >< \f181 ~— 2 - -v’181 2. 1-1- 1.2j 4» 6k; —w_—— 1/185’ 1/185’ —12 ET; 3. —8i — 123' 2 5k; ‘/ 3 M233 vex/o5 4. 112k; 2; 55-; 112 o o {’2 W; “606 ;2\/il.606 51/5/1119 51/? 109 6. 2i+16j;0;1,2x/65 In Problems 7- 11, denote the points in order by P, Q, R and form vectors F 2 PQ, G 2 PR. The points are oolinear if and only F 2 AG. Otherwise F >< G will be a normal to the plane containing P, Q, R. If F x G 2 *rzli+112j +113k, and P = ($9,310, 20), an equation of the plane will be m (:1: M :50) + 112(3): — ya) + o3(2: — 30) 2 G. 7. F 2 31—j-~ 51c= G 2 4i—j --- 6k, so HQ, R are not collinear. N 2 F X G 2 i— 2j-+1{,a11d the plane containing P, Q, R is; given by (m + 1) 2(y — 1) + (z --— 6) 2 O or :1: ~ 2y + z 2 3. 5. 18i E 5Uj (iUk; 8. not coiinear; x + 2y + 62 2 12 9. not colinear; 233 — 113; + z 2 U 10. not colinear; 59; + 161; —— 22 2 —4 1.1. not colineaz‘; 29:1: + 37;} — 122 2 30 In Problems 12 through 16, with F and G being vectors along a pair of adjacent sides of the parallelogram, area 2 “F x 12. Form F 2 PQ 2 i—‘r4j—6k, G 2 PR 2 5i+2j—5k, and compute FXG 2 1 :1 ~1—{6 2 5 2 —El —81 — ‘25j 18k. Then area 2 F x G" 1.3. TV”? 14. 98 15. 16. 20m Section 6.3 113 In Problems 17 — 21, we take the points given as P, Q, R, 5' and form the vectors PQ, PR,PS. Then the required volume is given by IPQ - PR X PSI 17. PQ = —5‘r+j+6k,PR: 2i+4j+6k,PS : —i+0j+5k and PQ-PRXPS = —5 1 6 ' 2 4 6 2 «~92, so volume 2 i - 92l= 92. —1 0 5 ~—3 0 10 18. Volume 2 — 1 6 8 z 34- w3 —1 10 HS —2 1 19. Volume 2 — 2 —6 ——1 s 98 1 —4 —5 9 0 —4 20. Volume 2 H “2 3 w6 :1 296 3 —1 wlU 1 1 l. ‘21. Volume m — 5 I} 2 = 22 —3' 3 5 22. N:8i—j+k 23. N:i—j+2k 24. N=i—33'+2k 25. N:Ti+j—7k 26. N=4i+fij+4k 27. This proof uses a property of determinants. 1 J k 1 3 k F X (G+Hl : f1 f2 f3 3 f1 f2 f3 91+h1 92+h2 93+h3 91 92 93 :(FXG)+(FXH) 28. By a property of determinants we have 1 J k i J k i j k {Ifr 0fo Clifa ‘7 f1 f2 f3 =04 f1 f2 f3 91 92 93 0:91 0192 (193 91 92 93 i j k f1 f2 f 3 ’11 11.2 kg, As vector calculations these represent, respectively (ch) X G = F X (ch) 2 oc(F X G) 29. Since F X (G X H) is parallel to the plane of G and H and orthogonal to F we have F X (G X : aG+fiH for some scalars 05,5 and F X (G X H)-F = 0 2 a(F-G)+[3(F-H). Thus 0: : A(F - H), and {3 n —)\(F - G) for some scalar A. This gives FX(GXH):A[(F~H)G-(F-G)Hl (1) To determine A we construct a special orthogonal triple of unit vectors, e3 along H, so H : hgeg, eg in the plane of G and H, so G s 9282 + g3e3, and (—31 : e2 X e3 so F = fle1+ fgeg+f3e3. Substitute these expressions for F, G, H in (1] and calculate F X (G X H) = 114 Section 6.4 F x (92113631) : —f292h3e3 +fggghge2. Aiso AKF - I-I)G — (F - (3)11] : A[(f3h3)(ggeg ~§~ 9383) — (f292+f393)(h3e3)] Z /\[f3h3g2€2 --f2§2h383], SD A 3 1, and FX(GXH) "—" (F‘I‘I)G~(F-G)I‘I. 30. Let P = (01, b1,c1),Q = (o2,b2,cr,3) and R x ((13,133, (:3) be the three given points. Form the vectors V = = (or; — a]_)i + (112 — b1)j+(c2 w cflk and W 2 PR 2 (e3 — a1)i + (113 w ()1 + (03 ~— cflk. The points PQR are coiinear if and only if V and W are parellei (or . . . , or — a b« —— b r: - c equivalently V = A W The conthtzon that at least two of the retlos —‘;———-—-1-, 31—1, —2——} Q3 --- ft] ()3 ~--- 51 (:3 -- [:1 are different will ensure P, Q, R are not coiiuear. The area of triangie PQR is given by 1 I 1 I i k A = §|1V1E||W|1 5111(9) m §|1V >< Wli = a2 — a1 52 — bl C2 — Cr 1 ' (ls—ELI 53—51 03—81 so 2 2 2 If? 1 52 H 311 02 — CI (12 — r11 (:2 — (:1 a2 — e, 132—, area m — , ‘2 b3 — bl C3 — Cl (13 — 0.1 Cg m [:1 a3 — a1 53 — ()1 Using properties of determinants we get 52 _ bl C2 _ Cl 1 O 0 1 b1 0, b_b CNS 0 52—5102—61 = 1 53 (:2 3 1 3 1 0 133—131 C3—c1 1 53 53 and similar results for the other two terms. Thus, in terms of coordinates of P, Q, R, 1 1 151 C1 2 (L1 1 C] 2 [51 :51 I 2 “2 area ofAPszw 1 ()2 c2 ' + :12 1 02 + {L2 .52 1 2 _ 153 {:3 (1,3163 o3 531 i k 31. F-GXH = (e1i+blj+clk)- (t2 b2 (22 =o1(bQC3~63C2)—bl(o2.C3—e3cfl+ Ge IE)3 C3 0109253 " @352) 031 b} (.‘1 3 (.52 b2 Cg a3 53 C3 Section 6.4 The Vector Space R” In 1-6 we give in order, the sum of the two vectors, the dot product,-'end the cosine of the angle between them. 1. F——G=Se1 +5e2+6e3+5e4~¥e5;1*‘»G=0,cos(n9) :0 2. F —— G : 3e1+ 982 +10e3 — 7e4;F - G : 22,cos(9) : Section 6.5 115 6 3. F G:17e «~49 ——3e +6e;F-G:24;cos9 :— ‘l‘ 1 2 3 4 - 2‘? 4. F Gzlfle +49 ——6e +0e +0e+0e,;F-G=27;cos9 =—#w + 1 2 3 4 s () EVE —4’i’ 5. F + G : 6e1 + 7e? ~ 6e3 — 2e5 v as +11e7 + 4eggF - G : w94;cos(9) = i 6. F + G = “"491 ‘i‘ 382 “l‘ 393 —1584 "— 35;]? ' G "" 7. Clearly O is in S by taking :1: = y 2 z = 0. For F : (:L','y, z, a), 11:) and G : (0., b, (3, (1,0,) we have F + G 2 (a: + (1,3,: + 6,3 +c,:c + c, :1; + a) is in S and 05F = (a:r:,ay,crz,ua;,aa:) is in 5', so 5' is a subspace of R5. 8. S is a subspace of R“i since by taking a", = y : 0 we get 0 in S, and for F 2 (:s,2:1:,3:t,y) and G = (a,2e,3c,b) we have F + G = + c,2(a: + e),3(:1: + a),y + b) is in S, and (IF 2 (0:33, 2a$,3ua:,cey) is in S. 9. S is not a subspace of R6 since every vector in S has 4“ component equal to 1. Hence 0 x (0,0,0,0, 0,0) is not in S. _ 10. S is a. subspace of R3. With :5 = y 2 0 we have 0 in S; o:(0,:s,y) (0,0111, say) is in .S', and (0,:1:,y) + (the, b) x (0,3: +a,y +b) is in S. 11. S is a subspace of R4. With a: 2 y : 0 we have 0 in S; or(:i:,y,:v + y,::: — y) (aw,ay,a(m + y),oc(:i; — y)) is in S and (3:,y,a: + y,:1: — y) + (a,b,a + he — b) = (m + c,y -+- b,(:s+e) +(y+b),($+e) — (y+b)) is in S. 12. S is a subspace of R7. Clearly 0 is in 8. Also for F, G in 5,1“ '5— G and (IF will have third and fifth components equal to zero. 13. S is a subspace of R4 since 0, F-I‘G, and (IF will all have equal first and second components whenever F and G have. 14. S is not a subspace of R3 because 0 is not in 8. Note 0 2 (0,0,0) is not on the plane om+by+cz:kfork790. 15. For F,G in R“ we have + GHQ = (F + G) - [F + G) 2 (F - F) + (F - G) + [G- F) + (G - G) = [EF112 + 2(F-G)+1{Gfi2,_and ||F — (3)2 : [1F|12 m 2(F - G) + §|G||2. Thus llF + Gllg + "F — Gllz = 2(llFll2 + llGIl2l 16. For F, G in R” we have + GH2 2 F - F —i— F - G + G -F + G - G = + “GHQ, since F - G x G - F r 0 because F and G are orthogonal. 17. Yes. For any F, G in R” we have “F + G“2 : + 205‘ + G) + If in addition, F and G satisfy Pythagoras’s Theorem, then ||F + Gil2 = + It foliows that (F ’ G) 2 0, and F and G are orthogonal. Section 6.5 Linear Independence, Spanning Sets and Dimension 1. Suppose a(31+ 23') +fi(i —j) = 0. Then 30: +6 : 0 and ‘20 — B 1. But the only solution to these two equations is a = )3 2 0. Thus 3i —[— 2j and i ---- are lineariy independent 116 Section 6.5 2. Suppose 01(21) + d2(3j) + d3(5i — 12k} + 0:4(i +j + k) = 0. Then 20:; + 50:3 + (£4 = 0, 30:2 +0.21 2 U, —120:3 +054 :: 0. By inspection we see 0:1 = —17/2,o:2 :2 —4, 0:3 =1,o:.1 =12 is a non-trivial solution of these equations. It follows that 2i, 3j, 51— 12k, i +j +k are linearly dependent. 3. These vectors are linearly independent by Lemma 5.3. 4. Since 4(1, 0, 0, 0) “6(1), 1, 1, 0) + (—4, 6, 6, 0) z 0 these vectors are linearly dependent. 5. Since 2(1, 2, —v3, 1) + (4, O, 0, 2) - (6, 4, :6, 4) z 0 these vectors are linearly dependent. 6. Suppose 0:1(L,1,1,1) + a2(e3,2,4,4) + (r3(-2, 2,34, 2) + 054(1, 1, ——6,2) : 0. This gives the matrix system of equations 0 —3 —2 1 0:1 0 1 2 2 1 0:2 0 1 4 34 —6 0:3 0 1 4 2 2 a4 0 Since the determinant of the cosfiicient matrix equals 129, the system has only the trivial solution 011 = as; = a3 = 0:4. Thus the vectors are linearly independent. 7. Suppose a1(1,“~2)'l"'£1'2(4,1)+013(6,6) : 0. Then a1+4cr2+6cr3 2 U and ~~2a1+crg+fioc3 = 0. By subtracting those two equations we see 30:1 «1— 3&2 : 0. Thus choose 0:1 3 2, d2 2 —2 and 0:3 3 1 and we see the vectors are linearly dependent. 8. Suppose orfl—l,1,0,U,U)+a2(0,m1,1,0,0)+a3(0,1,1,1,U)= 0. Then ——o:1 2 {Lou —o:2+ a3 : 0, d2 + d3 3 0, 0:3 2 0 from which we see the only solution is oq : [kg = a3 = 0. Thus the vectors are linearly independent. 9. Suppose 0:1(—2,0,0,1,1)+ a2(1,U,U,U,U) + d3(0,0,0, O, 2) + 054(1, “1, 3, 3, 1) I 0. Then certainly a4 0 and —2m + 02 + 0:4 0,611 + 30:4 2 0 and a1 + 2023 + (24 z 0. The only solution of these equations is a1 = (£2 = (r3 2 d4 2 0. Thus the vectors are linearly independent. 10. Suppose a1(3,{l,0,41) + 02(2,O,0, 8) z 0. Then 30:; + 20:2 = U and 4&1 + 8&2 = 0. The only solution of these equations is 051 a (1'2 2 0. Thus the vectors are linearly independent. 11. Suppose F, G,H are linearly dependent so that QF + 6G + 7H = U with at least one of (1,117 nonzero. Specifically, suppose 7 -7£ 0 so that H 2 *EF w EG. Then [F,G,I‘I] : 7 fr F‘G XH : U by Problem 42, Section 5.3. Conversely, let F : ((11,61, c1),G : (e2, {22, c2),I~I : (e3,b3,C3) and suppose [F, G,H] m 0. Then By Problem 39, Section 5.3, the determinant 0:1 b1 Cl (11 e2 0.3 a 0 :12 b2 02 2 0. Thus the system of equations b1 132 ()3 [3 = 0 has (13 b3 Cg - C1 Cg C3 7 U a nontrivial solution, i.e. at least one of 05,117 nonzero. But then aF + 5G + 7H 2 (J and F, G, H are linearly dependent. 3 5 “+1 12. {3i+6j —k, 8i+2j —4k, i ~j +k] = 8 2 w68 36 0, so the vectors are linearly l —1 1 independent. Section 6.5 I 117 1 6 w2 13. [i + Eij w 21{, ——i + 4j w 3k,i+ lfij —7k) 3 —1 4 #3 z 0, so the vectors are iineariy 1 16 —7 dependent. 4 —3 1 14. [4i — 3j + k, 10i w 3j,2i — 6,} + 3k] 2 10 —3 0 2 0, so the vectors are linearly 2 ~6 3 dependent. 8 6 0 15. (Si "l— Eij, 2i. ~45, i+k] z 2 "4 0 2 W44 79 0, so the vectors are iineariy independent. } 0 1 12 0 —3 16. [12i m 3k,i + 2j - 14, "BE + :4 1 2 --1 z 18 7t 0, so the vectors are linearly --3 4 0 independent. 17. A basis for S is the set of vectors (1,0,0, —1) and (0,1,-1,0), so 3 has dimension two. 18. A basis for .S' is the set of vectors (1,0, 2, 0) and (0, 1, 0, 3), so 8 has dimension two. 19. A point (m,y,z) is on the plane 2.1: u- y + z : 0 if and only if y 2 22: + 2. Thus all vectors in the piane have the form (3:, 22; + z, z) and every such vector can be expressed as a linear combination 33(1, 2,0) + z(0,1,1) for some 13,3. Since (1, 2,0) and (0,1,1) are linearly independent, they form a basis for S, so 8 has dimension two as we would expect for a plane. 20. A basis for S is the set of vectors (1,0,0,1,0),(0, 1,r~1,—1,0), (0,0,0,0,1), so S has dimension three. 21. A basis for S is the set of vectors (1,0,0,0),(0,0,1,0), (0,0, 0,1), so S has dimension three. 22. A basis for Sis the set of vectors (~1, 1,0,0) and (0,0,1,2), so 8 has dimension two. 23. Every vector in R2 parallel to the line 3,; = 4:1: has the form ($,4.’IJ) for some m. A basis for S is the vector (1, 4), so .9 has dimension one, as we would expect for a line. 24. Every vector in S has the form (m,y,4a‘ + 23;) 2 $(1,0,4)+y(01112) for some 3:,y. A basis for S is the vectors (1, 0, 4) and (0,1,2), so 5' has dimension two. ...
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0534552080_98853 - 106 Section 6.1 Chapterlfiix w Vectors...

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