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Unformatted text preview: Vincent (jrnv77'7) — HVVOEi — Gilbert ~ (57495) 1 This printmout should have 19 questions.
Multipiewchoice questions may continue on
the next column or page W ﬁnd all choices
before answering. 001 10.0 points Find a vector v orthogonal to the plane
through the points P(3,o,0), Q(e,2,0), wear). 1. v m .(4,12,6)
2. v = (8,4,6)
av = (2, 12,6)
4. v m (6,3,6) 5. v = {8,12,6) correct Explanation: Because the plane through P, Q, R con—
tains the vectors 336 and $73, any vector v
orthogonal to both of these vectors (such as
their cross product) must therefore be orth0g~
onal to the plane. Here
1972’ w (—3,'2,U), 133% = («»3,0,4). Consequently, v m F‘Q’xP‘E = (8,12,6) is othogonai to the plane through P, Q and
R. 002 10.0 points As r varies the equation
(rma)v(r~b) m 0 deﬁnes a sphere. .
Determine the center of this sphere when an: (2,4,1), b: (4,6,3). 1. center 2 (1, 5, 2)
2. center 3 (3, 3, 0)
3. center = (3, 5, 0)
4. center = (1, 3, 0) 5. center m {1, 3, 2)
6. center = {3, 5, 2) correct Explanation:
The sphere deﬁned by the vector equation (rma)(r—b) m 0 has 1 — b
center 2 §(a+b), radius = [a 2 I. For the given a and b, therefore, the sphere
has center 2 (3, 5, 2) . 003 10.0 points Which of the foilOwéng surfaces is the graph
of 12 6$+4y+32 in the ﬁrst octant? Vincent (ij777) w HWO6 w Giibert w (57495) 3 6. $=m1+3t, y=1+t, 2mm4m4t
Explanation: A line passing through a point P(o, b, c)
and having direction vector v is given para—
metri'caily by 1‘05) = a+tv, a = (a, b, c).
Now for the given line,
a = (1, 41,4), v = (3,1,m4>_
Thus
r(t) = (1+3t, m1+t, was). ‘ Consequently, $=1+3t, y=wi+t, z=4m~4tt are parametric equations for the iine. keywords: line, parametric equations, direc—
tion vector, point on line 005 10.0 points . A line 6 passes through the point P{1, 3, Si)
and is perpendicular to the piano 4x+4y~z m 5. At what point Q does I? intersect the my—
plane? 1. ohms, mi, 0) 2. Q(—1,0, a3) 3. ow, 5, e) 4. Q(5, 7, 0) correct
5. Q(—3, e, 7) 6.. 9(5, 0, m1) Explanation: Since the atywplane is given by 2: == 0, we
have to find an equation for I? and then set
2 = 0. Now a line passing through a point
PM, 39, c) and having (iirection vector v is given parametrically by
r(t) = a+tv, a z (a, b, 0). But for 6, its direction vector will be parallel '
to the normal to the plane 4m+4ymz = 5.
Thus
a=(l,3,l), v=(4,4,—1),
andso r05) = (1+4t,3+4t,1~t). In this case, 2 r; 0 when t r: 1. Consequently,
E intersects the zap—plane at Q(5, 7, O) . keywords: line, parametric equations, direc—
tion vector, point on line, intercept, coordi—
nate plane 006 10.0 points Fimi an equation for the plane passing
through the point P(2, ml, ——1) and paral»«
lel to the plane 73$m2y+z = 2. 1. ct—3y+22 m1
2. m~3y+2z = 3
3. ESC—Zy—é—z m 5 4. 2x~y+3z 2: e H
to 5. 2mwy+32 6. 33: — 2g; + z m 7 correct Vincent (jmv777) — HWOG m Gilbert —— (57495) 5 .5. elliptic paraboloid 6. hyperboiic paraboloid Explanation: The trace on any horizontal plane, via, any
plane parallel to the coordinate plane z = c,
is always a circle for all icl 5 (:0 for some
choice of co, whiie the trace on any piano
parallel to the coordinate planes :2: = 0 and
y = 0 is alway a hyperbolas. Consequently
the quadric surface is a i two—sheeted hyperboloid 1. keywords: quadric surface, ellipsoid, ellip—
tic paraboloid, hyperbolic paraboloid, one—
sheeted hyperboloid, twowsheeted hyperw
boloid, ‘ 009 10.8 points Which of the following quadratic equations
has graph a one—sheeted hyperboloid? 2 2
$ y
1 =m— m
Z 9+16
22
2.m2+y2—§— m 1correct
2 2
3.z=§ww~yv~m
16 25
2 2
y 2:
4. 2 m_._=1
C“Halal
2
5.%~—m2——y2wl
6 z2=fimyf
25 9
Explanation; In every case, the graph of the given
quadratic equation wili be a quadric surface
in standard position. Now if a one—sheeted hyperboloid is in stan—
dard position and its trace on the mywplane is a circle 2
23 +92 w r2a then it will be the graph of a quadratic equa— tion 2
Z 2 332+ 2—? :17”
a while the trace on the yz— and (cc—planes are
the hyperbolas
2 2
2 N EN = 2 2 m i = 2
y a2 r , m e2 r ,
respectively. Consequently, only the graph of
the quadratic equation 272 2 2
.._m1
x+y 9 is a one—S'li‘eeted hyperboloid. keywords: 010 Which one of the following equations has
graph 10.0 points 2 when the circular cylinder has radius 1.
1. y2+22—2y :IO
2. 252+.932w22 = 0
3 gig—rag +22 w 0 0 correct ll 4. m2+y2+2y
5. $2+y2m2x = 0 e. z2+sc2+2x 2:: s Explanation: Vincent (jmv’???) — HWOG — Gilbert m (57495) the trace in the mz plane is also a parabola: a: = z2, and equation for surface is ' y2+z2=m. 013 10.0 points
Find an equation for the surface consisting
of ail points P(m, y, z) equidistant from the
point PU), 2, O) and the plane a m 1. 1. ,y2+z2+2xw4y+3 : Ocorrect 2. xgizg+2ym4z+3 = 0
3. $2+y2+4m~2z+3 ﬂ 0
4. $2+22w2y+4z+3 m 0 ,5. y2+z2~2x+4y+3 = 0 6. $2+y2~4x+23+3 x U . Explanation:
The distance from P($, y, 2:) to PG), 2, 0) is .
if$2+ (y: 2)2 +32, While the distance from P(m,y, z) to the
plane :1: m l is 156 —~ 1. Thus P lies on the
graph or" Imm1=«./:r2l~l~(yw2)2+z2. After squaring and expanding both sides this
becomes wQeZm—i—i m m2+y2—4y4~4+z2. Consequently, an equation for the surface is y2+22+2$~4y+3 == 0. 014 10.0 points Find parametric equations for the line
passing through the points P(3,1,2) and
Q(1,3,5). 7
1. mm——2m3t, 1122—4: 2331—275
2. mm~2~3t, ymZ—i—t, zn3+2t
3. m=3w2t, y=1—2t, z=2mi3t
4. $=3+2t, y=1+2t, z=2w3t
5. mm3~2t, mid—2t, zm2+3£
correct 6. mw2t3t, ym2+t, Zm3—2t Explanation: i A line passing through a point P(a, b, c)
and having direction vector v is given para»
metrically by r05) m a+tv, a m (a, b, c).
Now
m 2 (W21 2: 3)
is a direction vector for the given line, so
am(3,1,2), VW<—2,2,3>.
Thus r(t) m (3—2t,1+2t,2+3t). Consequently, 33:3—2t, yml—l—Zt, zm2+3t are parametric equations for the iine. 015 10.0 points _
Find the point at which the line :1; m4+4t, y m —3, z m 215,
intersects the plane . 4x+5y~62 m 13‘ 1. P(16, 9,6) 2. P(16, 3,6) vincent (jrnv777) m HW06 w Gilbert m— (57495) keywords: distance, distance from line, cross
product, vector product, vector, length, 10.0 points 017 Which one of the following is the graph of the equation m;
o
C reet Explanation: In addition its trace on the icy—plane must be a parabola opening
in the positive y—direction and having positive The graph must be a cylinder with axis
ywintereept. Consequently, the graph of parailel to the z—axis. is equation Z2 is "113 019 Which one of the following equations has graph 1. w+z2m4
2. m—y2+4 3. z»~—:172+4 ll H Vincent (jaw???) w HVVOG ——Gilbert w (57495) 11 2 10.0 points MN 0 0 correct 6. y—z2+4 = 0 Explanation:  The graph is a parabolic cylincier that has
constant vaer on any line parallel to the 3:—
axis, so it will be the graph of an equation
containing no xwterrn. This already eliminates
the equations m+zzm4 0, 3—3324—4}; 0, $m92+42 O, y+332~4 m 0. On the other hand, the intersection of the
graph with the erg—plane, ale. the z = 0 plane,
is a parabola opening downwards on the z—
axis as shown in Consequently, the graph is that of the equa— tion
z+y2—4 m 0 . keywords: quadric surface, graph of equation,
cylinder, 31) graph, parabolic cylinder, trace ...
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 Fall '09
 Gilbert

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