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# homework_1_-_solutions - ECE 280—Homework#1 Solution 1...

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Unformatted text preview: ECE 280—Homework #1 Solution 1. GivenvecroreA =i—3i and B=5i+2§'—éi.deremﬁne: a] lei—Bil b} 53—3 0] The component of .4 along d) A unit vector parallel Ioﬂd + B . 1. e}|.4_1’+§|= |(1+5)£+(D+2jﬁ+(3—6]2"I=|6f+2§3—32|=V'W=7.D b] 5£—§=5(e+3e)—(5e+2§—523=(55a+152)—(5£+2§—52)= [5—5)f+(0—2)j3+(15+6)§=[if—253+21E:—2§+212 c} The component of; along}? = d] 3£+§=3re+3ej+rse+2ﬁ—52j=si+253+3§ -> —) A unit vector parallel to 3A + B: 33+§ 8£+2j3+32 8f+2ﬁ+3§ 3£+2j=+3§ 8£+2§+3£ ﬂ: #2 q—2 f 2—: lamI +B| I83: + 23: + Ezl ﬁlm)- + (2)2 + (3)2 ‘0'? 8.??5 (1.912.? + 0. 228? + II}. 3422 2. A11 ailplane has a glouncﬂ speed of 350 knm'hr in I11:- direction (in: worn. If there is a wind blowing northwest at 40 kin-"hr. calculate the nu: air speed and direction of the airplane. 2. Airplane ground speed: 350kmfhrw : Lt} = —3502 IWind Speed : 40kmfhr NW : W = 40ccs£l35°j 55 + 40 sin(135")j§ = —28.2342 + 23.2843? Trueairspeed :? : E=§+ﬁ NW _, “I W 45 ‘ a g a = g + E = (—3509) + [—23.28% + 28.284?) = —3?e.234£ + 23.2343? {— True airspeed = |3| = #1473234? + (28.234)2 2 areaqokmym- _ 28.284 Direction: 5' = tan L( ) .3732“ = —4-.2?I5° + 180” = 1?5.?24—° = 4.276a'north of West True air speed and direction: 3?Q.340kmfhr @ £1.35" north of West a. a] 11' .4=i+3i and gator—5113“ 653 L1) DCEDIIIPOEC .4 into co111ponc11ts parallel and perpendicular To B . .3 3. a] .4 - E = AXE); +£43.33. + .438; = |j||§| cosIZSAB} = {2131(5) + (03(2) + our—s) = (la-{11:12 +(31l2)(l-r532+{12)= +(—61}2)ws<oo = 5 + o — 18 = [y'ﬁjﬁ-‘Ej cosreﬂ) —13 —t: #IUﬁES = cosCSAS} —> —EI.51|3 = cosEEAB} —> Bag = cos—L{:—0.510:l= 120.664" —: a,” = 120. 664° lo] Parallel Con‘po’re’rz: A I BE = : _DI200§ : _0IZQU{5’2J_5} : 32 (V-{:5)2+(2)2+r—6122 55 =—1§—o.45:+ 1.22=ﬁ m: Peroewdicula' Componen: Fl 2 .df— H = {1,0,3} — {1,—0.4.1.2} = {2J0.4.1.8} —3 A A A n: 2x+D.4y+ 1.82 [Note: sub—.ractir‘g the parallel component from tl‘e vector will result ir‘ tl‘e [znerpendioular component.) 4. If A- : 2.". — 37(- —4_' and 3 = —IS.'i —4_f —f . fund The scalar and vector cornpmlems of f = .4 - along [11: dil'ccTLDn of = ‘._'—I1_'+_:. a _: _, 3? j? :2 56 j: 3fﬁ+24§3—82A 4 C=AxB= 2 3 —4 2 3 = +-16-\'-2}’+182 —6 —4— 1 —6 —4 —133’E + 22ft3 + 1023 = C _3 n m n A A n r— r— r— _. D 1.1'—1}-‘+12 1x—1}'+1z V A r.-' A V n Du=——r=+=+=—X——jﬁ+—Z |D| w-'|I1_i2+('l)3+(1_i2 v3 3 3 3 J. —o —3 —r Scalar component of C ir‘ the direction of D: C - D“ 5-3:.- = (—13) + {22) + (10) = is"? 3 44.434 Vector Corrpor‘er‘ts: F. Givcu t11: \‘cctors . = .3 —2i' + E and B 2 li— Ei— 3i . ﬁnd a 1111ii "in:ch1::111c11tliculm'to both :1 and E. (Hint: L's: the cross product.) 5. iiiectr:r perper‘dicular to both and E": E = 2:1" 1' g szlezz _2 22 _2= 1 —2 31 —2 c=—23—4§r—23 —. —3 —3 Unit vector aeramdicula' to both A 3rd 3: C“ _ 5 {—2.—4,—2) (—2,—4.4) (—2,—4J—2) {—‘l.—2,—1} CHZTZW+=—r_ :—l__ :—r_ : ICI tat—2} +(—-1.i2 + r_—2)2 m m we wig“ xign Ugh _x__y__z d. The verTice-s ofa n'iangLe ale given by the points. P(1 .1.0j. Qﬁiﬁﬂ}. and R[:2.5.C-). Find the three interior angle-5 of The triangle. (Hinl: ensign team-s in each aide Of the triangle and use the dot pl'oducr}. _, 5. PQ=(4—1)i+(2—1)_ﬁ=3£+1j¥ y— _3 R PR={2—1).{'+[5—1)j?=1£+4j3 — :27? = (2—4).w?+{j5—2)j‘r = —2:E+3§r —— — —) —3 — PQ -PR 2 IPQHPRICGSIS‘p _ Q (3;){1} + {1:}(4) = cos 6;, —> _— P 7" | | 3? = cas_1( _ _) = 57 529° l I I I I I x 1, 101517 E; . 9—1: = lﬁg’llﬁl cos|:180° — aq) (SM—2) + (121(3) 2 [vﬁﬂu’ﬁj cos[180° — 55?} —; 180° — HQ = cos—L = 105.255” —> 3". = 180" — 105.255cl = 74.745" 15101513 Fran" the properties of a :ria ngle, tI'e SU'T‘I oftI'e twee interimr ar‘gles must equal 180“. Therefore; 6,, = 180” — 6p — [in = 180“ — 5?.529" — 74.745” = 41726” 7. The clcchstatic potential in space is given by the smlar field . _ . 3 _ 3.13"] =VI_.\'.I1'.: | = e' — {2x} — ice-st Ii| Volts Find the potential difference between the points PILLI'J and Q [-3. 0. 0] . [Hint Don‘t forget to put your calculator into radian made! And don‘t forget units!) 7. VG?) = V{x,y,z) = a" + (2203 — 3cosﬁy) Volta P(1J'l.1:l. QﬁOJDﬁ) VIIP) = Val-1.1) = em + (2(1))3 — 3cos{1) = 9.09? V W?) = WED-0] = 8m + (2(0))3 — 3035(0) = —2.000V Potential Difference 2 VII?) — W0) 2 (9. 097 V) — (—2. 000 V) = 11. 097 V 3. The electric field in a certain region of space is given by the vector field Er}::=E[I1J'5:-}={I—:]i+ |'I+:‘I|f-' V.-'m Filidally two points P[_1]\"L'].:1J| and QI:_‘i'3.I'LE.I:} such that I11: electric ﬁeld. at P is IJEIIJEI‘JdiCHlFIl' to [he EIECiIiC ﬁeld at Q. Evaluate the electric field at each ofIllc-s-e two points. III-lint: Use the dotproduct._] —.= —3 8. ELF) = Efxbyllzlj = (x1 — 21H + (x1 +21ij me Em) = Erx2,y2, 22) = (x2 — 22:.2 + [x2 + 22);? me If the electric ﬁelds are perpendicular, then - = 0 Thus; E‘CPJxEﬁQh +£(P)}'E(Q)y = 0 {x1 —21)l:x2 — 22) + [x1 +lefx2 + 22) = CI (x1362 — x122 — x221 + 2122}+[x1x3 + 2cle + X221 + 2122) = 0 23513:; + 22122 = D —> 2x1x2 = —22122 —> xlxz = —2122 Any two points P and chat satisfy the equation xlxz = —zlzz can be used and will have the electric ﬁeld at P be perpendicularto the electric ﬁeld at (1. One possibility.r is Pi1,1,1} and 011,151}. §(P)=E(1.1,1] =(1 —1]£ + (1 + 1)? Wm = Of + 2? Vim EEQ)=E(1;1.—1)=(1 + 1):? + [1 —'1}j§ me = 2? + of Wm ...
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## This note was uploaded on 11/04/2009 for the course ECE 280 taught by Professor Mukkamala/udpa during the Spring '08 term at Michigan State University.

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homework_1_-_solutions - ECE 280—Homework#1 Solution 1...

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