Homework 4 - Vincent (jaw???) ~ HVVO4 w Gilbert ~—...

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Unformatted text preview: Vincent (jaw???) ~ HVVO4 w Gilbert ~— (57495) I 1 This print—out should have 20 questions. Multiplemchoice questions may continue on the next column or page W find ali choices before answering. 001 10.0 points Find the reiation between :13 and y when the point Q(£B, y) hasthe property that dlSt(Q, P1) "i" dlSt(Q, P2) 2 4 with respect to points The question really asks for the equation of P16), W1)» P2“), 1)' this ellipse. First notice that a 2 3:2 1- y “"3” = 1 dist(Q,P1) m x2+(y+1)2, ya while 2 $2 + W = l 3 P2) : $2 + __ 1)23 3 M £3 = I so the condition on Q requires that ' 4 3 - 2 2 i/$E+('y+1)2+ x2+(y~l)9m4. 3: y 4' "24:" + :6)" m 3‘ After squaring both sides and simplifying, this 2 becomes 5. mg—yg—mi m2+(y+1)2+$2+(y~1)2 y2 $2 +2 ($2+(y+1)2l($2+(y“1l2) m 16' 6 _ — —— m 1 r 3 4 I Thus 2 2 2 2 + "i 7. Fig—+1“:— = lcorrect x y + 2 + m4+2m2(y2+l)+(y2-1)2:8, m ‘ 2 - 3- *3“ + y E 1 which after rearranging and squaring gives Explanation: 7w— ($3+y2) m (3:2 + 92? + 2(532 _ y?) + 1. The distance condition ' , . , Hence, after squaring yet again, we obtain dlSt(Q, P1) + dlSt(Q, P2) = 4 49 W 14(332 +y2) m 2(332 __ 92) + 1‘ ensures that Q traces out an ellipse with foci Consequently) $113 coordinates 33, 9 0f Q 8313- at . isfy the equation Pl“): ml): P2“): 3:2 312 _ —--+_— 1 on the y—axis and so has graph 4 Vincent (jrnv’???) — HW04 — Gilbert — (57495) 3 as vertices at (i, 3) and (1, —1). Thus one equation for this hyperbola is em 1)? (as—1)? _, 4 1 " ’ which after expansion and simplification can be rewritten as y2—2y—4332+833 == 7 . keywords: graph, conic section, ellipse, shifted ellipse 004 10.0 points A rectanguiar box is constructed in 3—space with one corner at the origin and other ver»~ tices at ' (5,0,0), (0,1,0), (0,0,3). Find the length of the diagonai of the box. 1. length a 35 2. length m 11 x/ii 4. iength = 3\/?_) 3. length 5. length m 27 6. length = moorrect Explanation- We have to find the length of BB in the figure given that 01425, 01323. 00ml, Now by Pythagoras’ theorem, length OB 3 length A0 = M2?) . But then, again by Pythagoras, length 813 m Consequently, iength = x/g keywords: length diagonal, rectangular solid, Pythagoras’ theorem 005' 10.0 points What are the coordinates of the projection of the point Q(5, 4, ~3) onto the erg—plane? 1. (0,4, —3) 2. (~5,4, —3) 3. (5,0, W3) 4. (5, 4, 0) correct 5. (5, —4, —3) s. (5, 4, 3) Explanation: The soy—plane consists . of all points P(:c, y, 2:) whose z—coordinate is given by z m 0, and the projection of P(zt, y, 2) onto the my—plane has coordinates (as, y, 0). Con— sequently, the projection of Q(5, 4, ~3) onto the my~plane has coordinates - keywords: coordinate plane, point, coordinate, 3~space, projection, Vincent {jmv777) m HW04 w Gilbert - (57495) 5 009 10.0 points 7 Which of the following sets of inequalities describes the solid rectanguiar box in the first octant bounded by the planes 3: = 7, y = 9, and z m 1. 1. wgmgo, msgygo, “~15ng 2. a3<'7,y<9,z<1 3. 32—7, yZ—Q, zzwl 4. 0§x<7, 0$y<9, 0$z<l 5. 03x57,0§_y$9,0_<_z£1 correct Explanation: Because the box lies in the first octant, each ' point will have only nonnegative coordinates. So the inequalities describing the box are ng< 77 U<y<9a 0<z<l keywords: rectangular solid, inequalities, first octant ' 010 10.0 points Find an equation for the set of all points in 3—space equidistant from the points ' A(1, “3, m3), B(4,1,—2). 1. cc+4y~l~32~+~1 :2 0 2. m—3y—-4z—l m 0 3. 4$My+3z+1 H :3 || 0 4. 3m+4y+z+1 5. 4$+3y——z——1 w 0 6. 3$+4y+z—1 = Doorrect Explanation: We have to find the set of points P(:t, y, 2) such that _ W lAPl = EBPl- Now by the distance formula in 3—space, m2: (mm1)2+(y+352+(z+3)2. While 1%]? = (mm4)2+(y—1)2+(z+2)2. After expansion therefore, 11mg: x2e2m+y2+6y+z2+62+19,‘ While [Wig x2—8x+y2—2y+z2+4z+21. Thus IFI= when 2 2 2 a: ~2x+y +6y+z +6z+19 : x2—8m+y2—2y+zg+4z+21. Consequently, the set of all points equidistant from A and B satisfies the equation 3m+4y+z—1 : 0 . Notice that this is a plane perpendicular to the line segment joining A and B (since it must contain the perpendicular biseotor of the line Segment keywords: plane, locus points, equidistant two points ' 011 10.0 points Which, if any, of the following A- in, 13- “P5. are representations of vectors when P, Q are points in 3nspace? 1. B only 2. both of them Vincent (jmv’???) w HWOél -~ Gilbert m (57495) 7 3. a m 3 4. a m “2 5. a = 2 Explanation: Since addition and scalar muitiplication of vectors is carried out componentwise, we see that u = (-«5,1)= a,(—1,1)+b(3,—1) = (-a+~3l),owb>. Thus ma+35 = W5, amb =1. Consequently, after solving these for a we see ' that a z: —1 015 (part 2 of 2) 10.0 points 2. Determine b so that the vector u = {M5, 1 ) is a linear combination u m av + bw of vectors 'v=(—m1,1), w:(3,m1). Correct answer: m2. Explanation: By Part 1, b satisfies the equation —a + 3?) = ——5 with a 2 W1. Consequently, b=w2 keywords: vectors, vector sum, linear combi~ nation 016 10.0 points The parallelopiped in 3—Spece shown in P is determined by its vertices (2(3) "33 —1)3 Sons, 0, —2). P(»-2, w3, w3), RU), 1, 0), Find the vector v represented by the directed line segment 136 . 1. v m (9,2,1) 2. v = (6,7,6) 3. v = (1, 7,4) correct Jive (7,8,6) 5. v : (7,4,5) 6. v = (4,3,3) Explanation: As a vector sum,‘ 13? m Fifi—1‘33. But Pit: (2,4,3), 13?: (4,3,1). Consequently, vmfic’: (1,7,4). Vincent (jmv???) w HW04 w Gilbert W (57495) 9 and so a vector equivalent to u —— v is given by keywords: 018 10.0 points Determine the vector c m a — 21) when a w 8i+2j+k, b m —i+33'+k. 1. c we 4i+5§~k 2. c 2—: 5i—4j+2k 3. c = 4iwém5j+2k 4. c H 5i—4j—-kcorrect 5.c 5i+5jm~k 6. c = 4i~4j+2k Explanation: The sum of vectors a W ali+o2j+a3k, b w 01i+023+b3k is defined componentwise: a—l—b = (a;+b1)i+(o2 +bg)j+(a3+b3)k; similarly, multiplication by a scalar A also is defined componentwise: Aa = (Aofli + 0022).} —i~ (Aafik. When a=3i+2jmk, b = “1—3j+k, therefore, we see that c m (me) "— mm): + (mm) m (mom +((1>(1>—(2>c>)k. Consequently, c 2 5i w 4j m k . 019 10.0 points Find all scalars /\ so that A(—2a + 2b} is a unit vector when ' a = (2,1), b m (—3,3). 1 A -— z}: 1 correct ' 2m 1 . A m ~—- 2 116 1 . A = ———— 3 116 1 4. A 2 mm 2m 1 5. A : WWW-m 2m 1 6- A — Explanation: Avector C = (61:62) is said to be a unit vector when [c] = Magdrcg = 1. ...
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Homework 4 - Vincent (jaw???) ~ HVVO4 w Gilbert ~—...

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