Homework 4

# Homework 4 - Vincent(jaw ~ HVVO4 w Gilbert ~—(57495 I 1...

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

Homework 4 - Vincent(jaw ~ HVVO4 w Gilbert ~—(57495 I 1...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online