hw-8-sec-13.2-solns

hw-8-sec-13.2-solns - ng‘ §au[5.1 i ETRY OF SPACE ET...

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Unformatted text preview: ng‘ §au [5.1 i ETRY OF SPACE ET CHAPTER 12 R 13 VECTORS AND THE GEOM 224 CHAPTE '4 ——«> .——-> ——» 'a (0) First we consider QS - PS as QS + (~PS Then smce —PS has the same length as PS but points in the opposite ———» ’9 ———-> ’4 ’a ’y ,4) direction, we have ~PS : SP and so QS — PS : QS + SP : QP. i d4) ———) ’——> —4> ———> A ’4 (a —4—> (d) We use the Triangle Law twice: RS + SP + PQ I <RS + SP) + PQ : RP + PQ : RQ 5. (a) u (b) 11 ~ v (c) v + w (d) u + v -V u flw V (b) (c) —b a 2a a) a " b g a It. (d) (f) “4] b M 1‘ {I ‘L —3a b - 33 3- a = (5 fl (—2) ,3 fl (—2)) = (7,5) CHAPTER 13 VECTORS AND THE GEOMETRY OF SPACE ET CHAPTER 12 3‘ Z I' :3 1 ,’ 226 3k)+(—2i—j+5k):—i+j+2k )+3(—2i—j+5k) 19.a+b:(i+2j— 2a+3b:2(i+2j~3k \al 2 x/12+22+(—3)2 \3i+3318k\: N/32+32+(—8)2 :Jéfi j-3k)—(-2i—j+5k)\: :2i+4j~6k~6i~3j+15k=~4i+j+9k la—bl=\(i+2 @a+b:(2i—4j+4k)+(2j—k):2i—2j+3k 2a+3b:2(2i—4j+4k)+3(2j~k):4i—8j+8k+6j—3k:4i#2j+5k \al : M22 +(—4)2 +42 = ¢§G=6 = J22 +(—6)2 +52 : \/6'5' \a~b\:l(2i—4j+4k)—(2j—k)l:l2i—6j+5kl 1 3 7 21. ~3i+7‘: ——32+72: 58,sou:——— ~3i+7' :—————i+——-'. ‘ 3‘ Vi ) V m( 3) J58 «a? 22. \(~4,2,4)\ : t/(-4)2 +22 +42 = J3EE : 6, sou = ét—4,2,4> = <—§, §, 2’5). 8 + (—1)2 + 42 : \/—8—1 : 9, so by Equation 4 the unit vector with l " 23. ThevectorSi—j+4khaslengthl8i~j+4k| = '1 'i_ thesamedirectionis%(8i—j+4k)=%i—%j+%k. lit. 24. \(—2, 4, 2)\ : «(—2? + 42 + 22 = «[2- : 2 \/6, so aunit vector in the direction 0f'<—2, 4, 2) is u : $176 <—2,4,2). , 1 6 12 6 6is6u:6- 576 <—2,4,2) = <~7€7€QE> or (—Jé,2¢6,¢'6>. A vector in the same direction but with length e figure, we see that the m—component of v is = 2 and the y-component is 25. From th v1 : lvlcos(7r/3) : 4% g : 2J3. Thus '02 = \vlsin(7r/3) : 4- <U1,1}2>=<2,2\/§>. V: nt of the we see that the horizontal compone 26. From the figure, : 50 cos 38° m 39.4 N, and the force F is cos 38° vertical component is sin 38° 2 50 sin 38° % 30.8 N. 27. The velocity vector v makes an angle of 40° with the horizontal and ch the football was thrown, i has magnitude equal to the speed at whi i From the figure, we see that the horizontal component of v is nt is \vl cos 40° : 60 cos 400 m 45.96 ft/s and the vertical compone \vl sin40° : 60 sin 40° m 38.57 ft/s. SECTION 13.2 VECTORS ET SECTION 12.2 force vectors can be expressed in terms of their horizontal and vertical components as given ' oi+20sin45°jz10fii+10¢§jand16cos30° (10fi+8\/§) i + (10\/2— 8)j e 28.00i+6.14j. Then we have positive m—axis, - 2000645 i # 16 sin 30°j = 8 fii — 8j. The resultant force F is the sum of these two vectors: F = L g (28.00)2 + (6.14:)2 m 28.7 lb and, letting 9 be the angle F makes with the ' iofi—s #1(10\/2—8> Q .—/—- :> 6=tan W124. tau"Hot/flaw? 10¢§+8¢§ ressed in terms of their horizontal and vertical components as —300i and , The given force vectors can be exp i 200 COS 600 i + 200 Sin 600j : 200(2) H’ 200 j = 100i + 100 The resultant force F is the sum of these two vectors: F = (—300 + 100) i + (0 + 100 J5 ) j : 400i + 1oo¢§ j. Then we have m z t/ (~200)2 + (100 fi )2 = 3/70' ,0" 00 : 100 f! s 264.6 N. Let 0 be the angle F makes with the positive m—axis. Then tan 0 = 1026? = fl}? and the terminal point of F lies in the second quadrant, so 6 = tan'1 + 180° N —40.9° + 1800 : 139.10. north is the positive y—direction, and east is the positive ar-direction. The wind is blowing at ocity vector is 50 km / h S45°E, which can be written as ity vector of the plane is 250 km / h N 60° E, or 30. ° ' up the coordinate axes so that bu m/ h from the direction N45°W, so that its vel d = 50(cos 45°i — sin 45° j). With respect to the still air, the veloc vwin equivalently Vplane : 250(cos 30°i + sin 30° j = (50 cos 45° + 250 cos 30) i + (—50 sin 45° + 250 sin 30°) j ). The velocity of the plane relative to the ground is V : vwind + Vplane = (25\/§+125\/§)i+ (125—25fi)j z251.9i+89.6j (251.9)2 + (89.6)2 m 267 km/h. The angle the velocity vector makes with the ac—axis is The ground speed is lvl m 0 % tan'1(28591'§9) % 20°. Therefore, the true course of the plane is about N(90 # 20)°E : N 700E. ctor sum of the velocity of the ship with respect to the water, elocity with respect to the ship. If we let north be the positive y-direction, then t/9 + 484 m 22.2 mi/h. The vector v makes an angle 6 Q1) With respect to the water’s surface, the woman’s velocrty is the ve and the woman’s v V 2 (0, 22) + (—3,0) : <—3,22). The woman’s speed is |v| : with the east, where 0 = tan'1 :6 98°. Therefore, the woman’s direction is about N(98 ‘ 90)“W : N8°W. g to the ropes of length 3 m and 5 m. In terms of vertical and horizontal 32. Call the two tensile forces T3 and T5, correspondin components, T3 : 7 [Tgl cos 520i + |T3| sin 52°j (1) and T5 = [T5l cos 40°i + [T51 sin40°j (2) rec of gravity acting on the decoration [which is The resultant of these forces, T3 + T5, counterbalances the to igj m —5(9.8)j = -49j]. So T3 + T5 : 49j. Hence ...
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This note was uploaded on 05/02/2011 for the course MATH 408D taught by Professor Chu during the Spring '09 term at University of Texas.

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hw-8-sec-13.2-solns - ng‘ §au[5.1 i ETRY OF SPACE ET...

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