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Unformatted text preview: SECTION 13.1 THREEDIMENSIONAL COORDINATE SYSTEMS ET SECTION 12.1 :1 219 ﬁnd the distances between points: 0,) Fixstwe
DE1 = (1 — m2 + [—2 — (5»2 + (4 — 5)? = «if lEFl = (3—1)2+[4—(—2)l2+(24>2=\/‘T=2\/ﬁ um: <3—0>2+[4—<—5>12+<2—s)2:ngm Since IDEI + IEFl = lDFl, the three points lie on a straight line. "‘\
® The distance from a point to the anyplane is the absolute value of the zcoordinate o is l—5l = 5. f the point. Thus, the distance e ofthe point: (3] = 3. (b) Similarly, the distance to the yzplane is the absolute value of the m—coordinat (c) The distance to the mz—plane is the absolute value of the ycoordinate of the point: (7 : 7. 7, —5) is the point (3,0, 0). (Approach the :caxis perpendicularly.) ce between these two points: .t v can (d) The point on the araxis closest to (3, The distance from (3, 7, *5) to the maxis is the distan ,/(3 — 3)2 + (7 ~ 0)2 + (—5 — 0)2 z (/77: m 8.60. (0,7,0). The distance between these points is (e) The point on the yaxis closest to (3, 7, —5) is (/(3 ~ 0)2 + (7 — 7)2 + (—5 — 0)2 = m z 5.83. (f) The point on the z—axis closest to (3, 7, —5) is (0, 0, —5). The distance between these points is (3 ﬂ 0)2 + (7 — 0)2 + [*5 — (—5)] = JBE as 7.62. 11. An equation ofthe sphere with center (1, ——4, 3) and radius 5 is (m — 1)2 + [y g (—40]2 + (z  3)2 = 52 or 3)2 = 25. The intersection of this sphere with th
(ac—1)2 +42 +(z—3)2 =25,y:00r (m — 1)2 + (y + 4)2 + (2: ~ e xzplane is the set of points on the sphere WhOse ycoordinate is 0. Putting y = 0 into the equation, we have 2 : 9, y : 0, which represents a circle in the mzplane with center (1,0,3) and radius 3. (35—1)2+(z#3) @n equation of the sphere with center (2, —6, 4) and radius 5 is (05 ~ 2
he intersection of this sphere with the any—plane is the s )2 + [y e (—6))2 + (z — 4)2 : 52 or (m—2)2+(y+6)2+(z—4)2:25.T whose z—coordinate is 0. Putting z : 0 into the equation, we have 2, —6, 0) and radius 3. To ﬁnd the intersection with the mz—plane, we set y : 0: in the :Lyplane with center ( (:1? — 2 2 + z — 4 2 = —11. Since no oints satisfy this e uation, the s here does not intersect the cm—
P q p the distance from the center of the sphere to the mzplane is greater than the radius of the sphere.) To ﬁnd the inter the yzplane, we set ac : 0'. (y + 6)2 + (z — 4)2 : 21, a: = 0, a circle in the yz— 13. The radius of the sphere is the distance between (4, 3, —1) and (3, 8, 1): r : Thus, an equation ofthe sphere is (a: — 3)2 + (y — 8)2 + (2 ~ 1)2 = 30. et of points on the sphere 2 = 9, z = 0 which represents a circle plane. (Also note that section with plane with center (0, #6, 4) and radius v21. (3—4)2 +(8—3)2 +[1 4—1)] : £6. ill‘ Sec 1%.! ET CHAPTER 12 AND THE GEOMETRY OF SPACE 3 CHAPTER 13 VECTORS 220
origin, the radius of the spher )2 z Then an equation ofthe sp + (y — 2)2 + (2 ~ 3)2 : 14. f the sphere passes through the
0)2+(2—0)2+(3~0 14. I
here is (m ~ 1)2 T: (1——
z2—6z+4y2z:11 gives :11+9+4+1 2:» (a:— s in the equation m2 + y2 +
: 25, which we (22 — 22 + 1)
ith center (3, 2, 1) and radius 5. ives(m2+8:c+16)+(y2~6y+9)+(z2+2z+1) ompleting squares in the equation g
(z + 1)2 z 9, which we recognize as an equation of a sphere with center (~4, 3, (m+4)2+(y—3)2+ 15. Completing square
3)2 + (y + 2)2 + (z e 1)2 (x2~6a:+9)+(y2+4y+4)+ =*17+16+9+1 2 ~1) and radius 3. 2 + 2z2 + 242 = 1 gives n the equation 23:2 — 8m + 2y
2? + 2y2 + 2(z + (3)2  17. Completing squaresi
2+12z+36) =1+8+72 2(m2 — 49: + 4) + 2y2 + 2(z (2: ~ 2)2 +y2 + (z +6)2 : §%,whichwer => 2(11: a
here with center (2, 0, *6) and radius ecognize as an equation of a sp '3 x/ % = 9/ ﬁ
i; “ @ompleting squares in the equation gives 4(322 ~ 233 + 1) + 4W2 + 4y + 4) + 422 : 1 + 4 + 16 :> V i 4m — n2 + My + $2 + 4Z2 : 21 :z (a: ' D2 + (y + 2)2 + Z2 = 24;, Which we recognize as an eQuation ofa sphere
1;. with center (17 ~10) and radius 2—41— : Z1) to P2(:132,y2,22) is Q : (:51 :m27y12y2121—gzz), 19. (a) If the midpoint of the line segment from P1(ZI?1, y1, then the distances \P1Q\ and \QP2\ are e
\P1P2\ = «I (:82 ~ 5802 + (112 # 111)? + (22 — 21)2 mm = qual, and each is half of \P1 P21. We verify that this is the case: “W‘VW‘WWFWQC’WWEF .t 1 . J may...“ SECTION 13.1 THREEDIMENSIONAL COORDINATE SYSTEMS ET SECTlON 12.1 221 (b)JB¥pm(a), the midpoints of sides AB, BC and CA are P1(—%, 1, 4), P2 (1, §, 5) and P3 (g, %,4). (Recall that a median ofa mangle is a line segment from a vertex to Mei:t/02+<;—2>2+<5—3>2=£3:ﬁﬁ
mm: (3+2>2+(%>2+<4—s>2=Mei—raw:E=MZ
(—g—4)2+(1—1)2+(4—5)2=i/§f+1=§\/SE the diameter (and thus the center of the sphere) is C (3, 2, 7). The radius is half the the midpoint of the opposite side.) Then the lengths of the medians are: lCP1l= 20. By Exercise 19(a), the midpoint of diameter, so r = % (4 — 2)2‘+ (3 — 1)2 + (10 — 4)2 = = Therefore an equation ofthe sphere is (z—3>2+(y—2)2+(z~7)2=11. 21. (:1) Since the sphere touches the nayplane, its radius is the distance from its center, (2, —3, 6), to the gay—plane, namely 6. Therefore 7‘ : 6 and an equation ofthe sphere is (ac — 2)2 + (y + 3)2 + (z — 6)2 = 62 : 36_ (b) The radius of this sphere is the distance from its center (2, —37 6) to the yz—plane, which is 2. Therefore, an equation is (z—2)2+(y+3)2+(z—6)2 =4.
(c) Here the radius is the distance from the center (2, ~3, 6) to the mzplane, which is 3. Therefore, an equation is
(ac2>2+(y+3)"’+(z—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, 4, 9) to any of the three coordinate planes. The shortest such distance is to the arzplane, a distance of 4. Thus an equation of the sphere is (:8 ﬂ 5)2 + (y — 4)2 + (z — 9)2 : 16.
23. The equation y = 4 represents a plane parallel to the Ltzplane and 4 units to the left of it.
24 The equation a: : 10 represents a plane parallel to the yzplane and 10 units in front of it.
25. The inequality x > 3 represents a halfspace consisting of all points in front of the plane ct = 3. 26. The inequality y 2 0 represents a halfspace consisting of all points on or to the right of the aczplane. 27 The inequality 0 g z 3 6 represents all points on or between the horizontal planes .2 = 0 (the acyplane) and z : 6. 28. The equation 22 z 1 c> z 2 i1 represents two horizontal planes; z 2 1 is parallel to the myplane, one unit above it, and z : #1 is one unit below it. QTh,.222.. .1 ..
 elnequalityq: +y +2 g 315 equivalent to a: +3; +2 3 x/g, so the region cons1sts of those pomts whose distance from the origin is at most This is the set of all points on or inside the sphere with radius \/§ and center (0, 0, 0). 30. The equation cc = z represents a plane perpendicular to the mzplane and intersecting the mzplane in the line a: = z, y : 0. lll‘ 3a 13.! ETRY OF SPACE ET CHAPTER 12
orn the y—axis is TER 13 VECTORS AND THE GEOM \/ $2 + z2 g 3 which describes the set of
onsisting of all points on or insi 222 El CHAP @ere 3:2 + z2 g 9 or equivalently
at most 3. Thus, the inequality represents the region c axis the yaxis.
1 is equivalent to t/ m2 + y2 + (z — 1)2 > 1, so the region <:> 3112+y2+(z—1)2 >
point (0,0,1) is greater than 1. This is the set of all points outside the Sphere all points in R3 whose distance fr de a circular cylinder of radius 3 with 32. The inequality 332 + y2 + z2 > 22 consists of those points whose distance from the with radius 1 and center (0, 0, 1).
33. This describes all points whose mcoordinate is between 0 and 5, that is, 0 < :1: < 5.
nter the origin and radius 2 we h 2 + y2 g 4. Also each point the disk in the qtyplane with ce
2§£0§z£8 34. For any point on or above
: 8, so the region is describe dbycr2 +y lies on or between the planes 2 : 0 and z
' gin which is greater than 7", but 5 points have a distance to the ori maller than R. So 35. This describes a region all of whose gtheregionarer < x/at2+y2+z2 < R,orr2 <m2+y2+z inequalities describin s
.
i z—coordinate to k‘ m2+y2+22§4,z_>_0.
coordinates of the point P, we project it onto L2 and  and y—axes. To ﬁnd the ect P onto either the wz—plane or the yz—plane z—coordinate, we proj
:3 or ycoordinate) and then proj (using our knowledge of its ect the r, we could draw a line parallel to Q0 resulting point onto the zaxis. (O
are (2, 1, 4). from P to the zaxis.) The coordinates of P (b) A is the intersection of L1 and L2, B is directly below the 9,.
y—intercept of L2, and (agngMnﬂPBlzﬂﬂﬁ a 4WBE=4PAF @ 38. Let P :
4((ac~6)2+(y#2)2+(z+2)2) = (m+1)2 +(y—5)2+(z~3)2 42>
i Mﬂ1%+$ﬁ—ﬁ—Qw+Mf—4yw9—f+ﬂm+Mf+4rHQ—ﬂ+6a=% a
i wP—mm+w2~®+3X+2uze5M4—w—16<¢a?~%x+f—2ywﬂ+%z=~%¥  . 25 2 2 11 2 # 332 ~ v ‘ fasphefe‘w
By completing the square three times we get ( ~ 3) + (y — 1) + (z + —3—) . T, which is an equation 0 center , 1, ~ and radius ...
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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 at Austin.
 Spring '09
 Chu

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