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Since 71 Z 2, we have k 2 1, and ~ , k :ting the equality [”ka“ : M2 m . ' s . - . e odd mteger . L 18 one of th rteger since ' an ‘1 < i < n and 1' is even, then we 0 M2’“ : T‘ M :2ke7‘M7which‘5 1” ”T l 1' 2" ‘ ~ 2 2 k thenl:l,srncel>1~l2 ‘ ws that is less than or equal to n. This sho k and V V y 'V t 2 n C en integer for e e L excel) ncludes the PYOOf' 13 El VECTORS AND THE GEOMETRY OF SPACE E] E' 13.1 Three-Dimensional Coordinate Systems E 1, We start at the origin, which has coordinates (0, 0,0). First we move 2. 4 units along the positive .1--axis, affecting only the r—coordinate, bringing us to the point ('4. 0. (J). We then move 3 units straight downward, in the negative :-direction. Thus only the z—coordinate is affected, and we arrive at (71. 0. —3). X i(4,0,~1) i—aThe distance from a point to the acz-plane is the absolute value of the y-coordinate of the point. Q(—5, —1, 4) has the y-coordinate with the smallest absolute value, so Q is the point closest to the :cz—plane. R(O, 3, 8) must lie in the yz-t since the distance from R. to the yz-plane, given by the x-coordinate of R, is 0. 4. The projection of(2, 3. 3) on the any—plane is (2. 3. 0); on the yz—plane, (0. 3. 5); on the a‘z-plane. (2. 0, 5). The length ofthe diagonal of the box is the distance between the origin and (2, 3. 5), giwn by V (2—0)2 + (3— m» + r15 40,)2 : v38 26.16 E'Dte equation J: + y 2 2 represents the set of all points in R3 whose 2:— and y-coordinates have a sum of2, or equivalently where y : 2 . .17. This is the set {(3512 — at, z) I :1: E R. r; -: R} which is a vertical plane that intersects the :ry-plane in the line y = 2 ~ .L', z : 0. 7"".(3‘) In 1R2, the equation .r : 4 represents a line parallel to y the y-axis. In R3, the equation I = 4 represents the x = 4 i set {(maya 2) I x = l}. the set ot‘all points whose D CHAPTER 13 VECTORS AND THE GEOMETRY OF SPACE ET CHAPTER 12 the sphere passes through the origin, the radius of the sphere must be the distance from the origin to the point (1, 2, 3): = x/(l — 0)2 + (2 — 0)2 + (3 — 0)2 = V14. Then an equation ofthe sphere IS (gr ~ 1)2 + (y — 2)2 + (z — 3)2 =14, ampleting squares in the equation 9:2 + y2 + 22 7 6m + 4y — 22 : 11 gives (b) By part (a), the midpoi ofa triangle is a line 5: :2-6.z’+9)+(y2+4y+4)+(z2—22—l—1):11+9+4+1 => (1» 3)2+(;1/+2)"’+(z~1)2:25,whichwe cognize as an equation of a sphere with center (37 ~2, 1) and radius 5. )mpleting squares in the equation gives (3:2 + 8:10 + 16) + (y2 — 6y + 9) + (2 3 + 2x + 1) : —17 +16 + 9 +1 :> '+ 4)2 + (y — 3)2 + (z + 1)2 = 9, which we recognize as an equation ofa sphere with center (—41. 3, —l) and radius 3. 20. By Exercise l9(a). the mid diameter, so 7" : éx /(4 — (x—3)2+(y—2)2+(z )mpleting squares in the equation 2m2 — 8m + 23/2 + 222 + 242 : 1 gives m2—4m+4)+2y2+2(22+12z+36) = 1+8+72 :> 205—2)2 +2112 +2(z+6)2 :81 2 ‘ — 2)2 + y2 + (z + 6)2 = %, which we recognize as an equation ofa sphere with center (2, (l, 43) and radius 21. (a) Since the sphere touche 8 71 : Q/fi. Therefore 1' : 6 and an (b) The radius of this spher )mpleting squares in the equation gives 4(952 — 2m + 1) + 4(3/2 + 4g + 4) + 4:2 :: 1 + 4 + 16 $ (36 2)2 (’11 ”2 cc — 1)2 + 4(3) + 2)2 + 422 = 21 :> (z — 1)2 + (y + 2)2 + 22 : %, which we recognize as an equation ofa Sphere th center (1, —2, 0) and radius % = 12:1. (93 2)2 (1/ 3V 22. The largest sphere container ‘ ' - . + a» ’1 + + z- ) Ifthe midpomt ofthe line segment from P1(a:1,y1, 21) to P2(1'2, 1/2, 22) is Q : (an 2 'll 112 Z] 2 > 2 j 2 7 2 . , ‘ . . any of the three coordinate 1 then the distances |P1Q| and |QP2| are equal, and each 18 half of 1P1P2]. We verity that this is the case: sphere is (a: — 5)2 + (:11 *4 lP1P2|= ($2-$1)2+(y2*y1)2+(z2—ZI)2 ‘ . The equation 3/ : *4 repre: IP1Q|= [%($1+;v2)—wi]2+ [%(y1+y2)—y1]2+[ m ii—i A N H N i l N H . The equation an 2 10 repres: : @332 _ %z1)2 +(%112 ~§y1)2 + 6Z2 _ %Z1)2 . The inequalityz- > 3 repres. . The inequality y 2 0 repres< : % IP1P2| . The inequality 0 g 2 g 6 re . The equation 22 : 4: IQP2l: [$2—§(w1+$2)]2+ [312‘ %(y1+y2)i2+ [22— ii“ +~2)i2 andz: —1isoneunitbelor — (i902 —%w1)2 + (5212 ~%y1)2 + ($22 #%21)2 V/(T)2 [(932 ~ 102 + (.112 - 1/1)2 + (Z2 — 202i The inequality 12 + 142 + 22 =% (m w)? +<y2 ,y1)2+ (22 ~21? : % IP1P2I 50, a \s M .4! 1’" at r, 4”: pm i .gin to the point (1, 2, 3): (9—2)? +<Z '3)? z 14‘ _ h'ch we a)2 +(z~ DZ 225”” 1 =~17+16+9+1 T Her (—4, 3, «1) and radius 3. :81 :> .6) and radius ._7 ' e ognize as an equation of a spher W £11.32 31 ”2), T‘s 2 2 at this is the case: r7131): (—g‘4)2+(1—1)2+(4—5)2= §5+1=i 85 20. By Exercise 19(a), the midpoint ofthe diameter (and thus the center ofthe sphere) is C(3, 2, 7). The radius is half diameter, so r 2 év’tzl 2)? + (3 ~ 1)2 + (10 — 4)2 2 %\/44 2 x/ll. Therefore an equation ofthe sphere is (av—3)2 +(y— 2)2 + z 7)2 : 11. i 21. |(a) Since the sphere touches the :ry-plane, its radius is the distance from its center, (2, —3, 6) , to the ey-plane, nam Therefore 7‘ 2 6 and an equation of the sphere is ( m—2)2+(y+3)2+(z—6)2 262 236. (b) The radius of this sphere is the distance from its center (2, — 3, 6) to the yz-plane, which is 2. Therefore, an equ; (£12 —2)2 + (y+ 3)2 + (z— 6)2 2 4. (c) Here the radius is the distance from the center (2, —3. 6) to the arz—plane, which is 3. Therefore, an equation is (Jr 2 2)2 + (y + a)? 4» (.3 — 6)2 : 9. 22. The largest sphere contained in the first octant must have a radius equal to the minimum distance from the center (5, any of the three coordinate planes. The shortest such distance is to the xz-plane, a distance of 4. Thus an equation 01 sphere is (1‘ — 5)2 + (i/ 7 4)‘ + (: ~ 9V 2 16. 23. The equation !/ 2 ~4 represents a plane parallel to the are—plane and 4 units to the left ofit. 24. The equation 1: 2 10 represents a plane parallel to the yz—plane and 10 units in front of it. 25. The inequality .r > 3 represents a half—space consisting of all points in front ofthe plane r 2 3. 25- The inequality y 2 0 represents a half-space consisting ofall points on or to the right ofthe ,rz-plane. 27- The inequality 0 g z :2 ti ~epresents all points on or between the horizontal planes 2 2 0 (the any-plane) and z 2 '. 8 The equation 22 2 l 2 ,: 2 il represents two horizontal planes; 2 2 1 is parallel to the qty—plane, one unit abc and Z = —1 is one unit below it. The inequality 102 + 3/2 rl- :2 5' 3 is equivalent to «:52 + y2 + Z2 3 fl, so the region consists ofthose points whose fibulthe origin is at most \ .73. 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H94 ~ a“ ‘(8—) ~ 9)! = [Cl 4! 2:1 W = MP) + z9/‘ =l @1741) : <81~ ‘6—> + <pz~ ‘01) : qg+ag‘y (81* ‘Z> : <(9—) + 31~'(g~)+ g) : q.”- (z ‘0 ‘H (J) = (L + I~ ‘9 + 3—) = (1‘9) + <1- ‘z—) w (0) (z— ‘0 ‘17) V (I ‘Z '17) 8 x (D) ‘s1u9uodu100 /‘ 'um01111 sum 1111q1001 9111 11911171111; p99ds 911101 1enb9 9pn11u3121u s1211 13101 911su91 07m 9111 11123 79 1 pue 1e1uoz11011 91111111711 001710 913% 119 s9>112u1 A 1019971 [(119019/1 911 .L = 19 91911711 ‘1399 9111 1111M 'N 8'08 m 0815 111$ 09 2 egg 1113 ’3' s1 1u9uodu103 129111911 : (011%) + (83101 : 9111mm ‘N17'68 z 088803 09 = 0885031111 513 90101 9111101u9uod1u09 191uoz11011 911119111995 9m 91nt 91111110151 11901971 S‘UBIUOM 9111 pm; 919m 9111 01 199ds91 1111M 'Lg 17 ,9 1111111121 7 a '<§/\z‘g>: (7111111): A $11111 '§/\z : §% .17 : (g/JL)u1s|A| : lAl s1 p99ds punoxfi 911 J. s11u9uod11109—fi 9111pu12 g : 1 .17 : (g/u)soo |A1 : : A s1 A 10 1u9uodu109-x 911112111 99s 9711 ‘9111811 9111 1110111 15 2 mid/x A11u9112A1nb9 9 9 9,7A , <9” ‘9/\ g‘y—>111<%i-L~> : (3 ‘17'5 )fi- 9 _ 119 S19 1118u911111m1nq u0119911p 9Lu12s 9111 H1 1019911 V ‘IOQVSODWQIPUW‘ 9 Z1 9 1911p 9111 111011 I1/1191 09 < . z 4? ‘Z‘ Q/‘Z , 1 g 1 , > I : n S1<Z 1-93 7110 u0119911p 9111 u110199A 11111112 OS Q/‘Z : 175A : 53 + 517 + 3(Z—)/ ‘ : HZ V'Z‘N 1a 9112u1p1009 9111 dn 19$ '09 <7w7> 006— : 911121 u9111 §/\ 001 31% +1‘ [‘9 — 1 9 _(>117+ [‘— 18)9 s1 u0119911p 9111125 9111 111111101391111111 911117 110119an A111181; = 18% = Z17 + 7(11) + 78% : I117 + f~18l I16119131211 >117 + f— 18 101m @1111 <1 1‘ 11> <17‘z‘17>1—“OS‘9‘J9€ ‘—zV+zZ+z(V‘) : )+ 4003-) % ldl 8“ 89 {J‘1‘I2‘:(FL+!€*)—}/\=“038_/\=ai+z —)/‘=|!‘1+18—| ;~) : 1:1 3101991101111 $\ : z9+7(9~)+az : 1119+19716|=1(>1—.fz)—(>117+.F17—1z)l:101791 9:9_/\= 717+ +:zz/‘ IEI >19 11777117:)18—r9+)18+.f8—117=(>1~fz)9+(>117+.f177fl18)z=q2+ez ES 008 + I C,09 303 OOZ 1013971 93101 u9A13 9111 '3; =0um )11:+.fz—15=(>1—fz)+()117+r17—1z):q+efi Q» : 218*1+ z1:+z1:/‘ =PIS—(127L121:|(>19+.f~1z—)—(>11:—.fz+1)l:lq—BJ WP: — +zZ+zD :19l >16+r+117- : 191+?87194197f17flz:(>19+.f71z~)£+(>19~rz+1)z=q9+ez 1+ z(00'8Z) % chI 139A 0M1 QSQLIJ J0 ums ms oz +1091 800 oz ommmjuwggm 'az >Iz+f+1— :(>19+.f—1z—)+(>18—.fz+1)=q+e‘1 ZL HSLdVHO 13 BOVdS :IO AHBWOEE) 3H]. 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