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Unformatted text preview: romasko (qrr58) – Assignment 2 – luecke – (55035)
This printout should have 20 questions.
Multiplechoice questions may continue on
the next column or page – ﬁnd all choices
before answering. Find the value of f (−2) when
1
1 f ( x) = 1
3 x2 + 1. D = −25 correct
2. D = −23
3. D = −27 10.0 points 001 1 13
11 4. D = −29
x. 5. D = −21
Explanation:
For any 3 × 3 determinant 1. f (−2) = 12 correct
2. f (−2) = 16 A B C 3. f (−2) = 14 a1 b1 c1 a2 b2 c2 4. f (−2) = 8
5. f (−2) = 10 a b c d = ad − bc . 1 1 3 x2 + 13
11 + ((1) (1) − (1) (3)) x .
Consequently, c2 −1
= 3
3 3 2 3 1 +C a1 b1 a2 b2 3 3 −1 3 −1
2 1 −2 3 2 −1 1 − = (3)(1) − (3)(2)
− 2 ((3)(1) − (−1)(2)) f ( x) = 2 x 2 − 2 x , − ((3)(3) − (−1)(3)) . and so
f (−2) = 12 . Consequently, keywords: determinant D = −25 . 10.0 points Find the value of the determinant
1 2 −1
D= c1 2 3 x = ((1) (3) − (1) (1)) x2 002 c2 a2 1
D= 1 b2 Thus Thus
f ( x) = c1 a1 −B Explanation:
For any 2 × 2 determinant b1 =A 3 3 2 −1 3 1 . keywords: determinant
003 10.0 points . romasko (qrr58) – Assignment 2 – luecke – (55035) 2 Find the cross product of the vectors
a = −2i − 3j − k , b = −3i − j − 3k . 1. a × b = 9i − 4j − 8k 1. a × b = 8, 11, 2
2. a × b = 9, 7, 2
3. a × b = 2. a × b = 9i − 3j − 8k 9, 7, 3 4. a × b = 8, 11, 3 3. a × b = 8i − 5j − 8k
4. a × b = 8i − 3j − 7k correct
5. a × b = 9i − 3j − 7k 5. a × b = 9, 11, 2
6. a × b = 8, 7, 3 correct Explanation:
By deﬁnition 6. a × b = 8i − 4j − 7k Explanation:
One way of computing the cross product
(−2i − 3j − k) × (−3i − j − 3k)
is to use the fact that
i × j = k, = j× k = i, k ×i = j, j ×j = 0, i
−3
−2 a×b = k× k = 0. 3
1 jk
31
13 −3
1
i−
−2
3 −3
1
j+
−2
3 3
k.
1 Consequently, while
i×i = 0, a × b = 8, 7, 3 . For then
a × b = 8i − 3j − 7k . keywords: vectors, cross product Alternatively, we can use the deﬁnition
a×b = i
−2
−3 −3
=
−1 j
−3
−1 k
−1
−3 −2
−1
i−
−3
−3
−2
+
−3 10.0 points Which of the following expressions are welldeﬁned for all vectors a, b, c, and d?
−1
j
−3 −3
k
−1 to determine a × b.
004 005 I  a  × ( b × c) , II a · ( b × c) , III (a · b) × (c · d) .
1. I and III only
2. all of them 10.0 points Find the cross product of the vectors
a = −3, 3, 1 , b = −2, 1, 3 . 3. none of them
4. II only correct romasko (qrr58) – Assignment 2 – luecke – (55035) 3 Explanation:
The nonzero vectors orthogonal to a and b
are all of the form 5. I and II only
6. II and III only v = λ(a × b) , 7. III only λ = 0, with λ a scalar. The only unit vectors orthogonal to a, b are thus 8. I only
Explanation:
The cross product is deﬁned only for two
vectors, and its value is a vector; on the other
hand, the dot product is deﬁned only for two
vectors, and its value is a scalar. v=± But for the given vectors a and b, For the three given expressions, therefore,
we see that
I is not welldeﬁned because the ﬁrst term
in the cross product is a scalar because it is
the length of a vector, not a vector. = i
3
3 a×b = II is welldeﬁned because it is the dot product of two vectors. a×b
.
a × b jk
14
26 3
14
i−
3
26 3
4
j+
3
6 1
k
2 = −2 i − 6 j + 3 k . III is not welldeﬁned because each term in
the cross product is a dot product, hence a
scalar. In this case, keywords: vectors, dot product, cross product, T/F, length, Consequently, 006 10.0 points a × b2 = 49 . v=± 6
3
2
i+ j− k
7
7
7 . Determine all unit vectors v orthogonal to
a = 3 i + j + 4 k, b = 3i + 2j + 6k. 6
3
2
1. v = − i − j + k
7
7
7
2. v = ±
3. v = 2
3
6
i− j+ k
7
7
7 3
6
2
i− j+ k
7
7
7 4. v = −2 i − 6 j + 3 k
5. v = ± 6
3
2
i + j − k correct
7
7
7 6. v = 2 i − 3 j + 6 k keywords: vector product, cross product, unit
vector, orthogonal,
007 10.0 points If a is a vector parallel to the xy plane and
b is a vector parallel to k, determine a × b
when a = 2 and b = 4.
1. a × b = −4
√
2. a × b = 4 2
3. a × b = 4
4. a × b = 0 romasko (qrr58) – Assignment 2 – luecke – (55035) 4 Consequently,
5. a × b = −8 −
−
→−
→
v = P Q × P R = 20, 15, 12 √ 6. a × b = −4 2
7. a × b = 8 correct is othogonal to the plane through P, Q and
R. Explanation:
For vectors a and b, 10.0 points 009 a × b = ab sin θ
when the angle between them is θ , 0 ≤ θ < π .
But θ = π/2 in the case when a is parallel
to the xy plane and b is parallel to k because k is then perpendicular to the xy plane.
Consequently, for the given vectors,
a × b = 8 . Compute the volume of the parallelopiped
with adjacent edges OP , OQ, and OR determined by vertices
P (3, −4, 2) , Q(2, −3, 2) , R(2, 4, 4) , where O is the origin in 3space.
1. volume = 17
2. volume = 14 keywords: cross product, length, angle,
008 10.0 points Find a vector v orthogonal to the plane
through the points
P (3, 0, 0), Q(0, 4, 0), R(0, 0, 5) .
1. v = 20, 5, 12 3. volume = 16 correct
4. volume = 15
5. volume = 18
Explanation:
The parallelopiped is determined by the
vectors
−
−
→
a = OP = 3, −4, 2 ,
−
−
→
b = OQ = 2. v = 4, 15, 12 −
−
→
c = OR = 3. v = 20, 3, 12 2, 4, 4 . Thus its volume is given in terms of a scalar
triple product by 4. v = 5, 15, 12 vol = a · (b × c) . 5. v = 20, 15, 12 correct
Explanation:
Because the plane through P , Q, R con−
−
→
−
→
tains the vectors P Q and P R, any vector v
orthogonal to both of these vectors (such as
their cross product) must therefore be orthogonal to the plane. But
a · ( b × c) = Here
−
−
→
P Q = −3, 4, 0 , 2, −3, 2 , −
→
P R = −3, 0, 5 . =3 −3
4 2
4 3 −4 2 2 4 2 −3 +4 4 2 22
24 +2 2 −3
2 4 . romasko (qrr58) – Assignment 2 – luecke – (55035) 5 Consequently, the parallelopiped has
10.0 points 011
volume = 16 .
Which of the following surfaces is the graph
of keywords: determinant, cross product scalar
triple product, parallelopiped, volume,
010 10.0 points
3x + 6y + 4z = 12 Which of the following statements are true for
all lines and planes in 3space?
I. two lines perpendicular to a plane are
parallel,
II. two planes parallel to a third plane are
parallel,
III. two lines perpendicular to a third line
are parallel. in the ﬁrst octant?
z
correct 1.
1. II and III only
2. I only
3. I and II only correct
y
4. III only
x 5. none of them z
6. I and III only 2. 7. all of them
8. II only
Explanation:
I. TRUE: the two lines will have direction
vectors parallel to the normal vector of the
plane, and so be parallel, hence the two lines
are parallel. y
x
z
3. II. TRUE: each of the two planes has a
normal vector parallel to the normal vector of
the third plane, and so are parallel, hence the
planes are parallel.
III. FALSE: the xaxis and y axis are perpendicular, not parallel, yet both are perpendicular to the z axis. y
x romasko (qrr58) – Assignment 2 – luecke – (55035)
z 6
z 4. y
y
x x 012 z 10.0 points Which equation has the surface 5.
z
z
y x x y
y
x z
6.
as its graph?
1. y + z = 1
x y 2. y − x = 1
3. x + z = 1 correct Explanation:
Since the equation is linear, it’s graph will
be a plane. To determine which plane, we
have only to compute the intercepts of 4. x + y = 1
5. z − y = 1
6. z − x = 1 3x + 6y + 4z = 12 . Now the xintercept occurs at y = z = 0,
i.e. at x = 4; similarly, the y intercept is at
y = 2, while the z intercept is at z = 3. By
inspection, therefore, the graph is Explanation:
As the surface is a plane, the equation must
be linear; in addition, for ﬁxed y and varying
x, the corresponding value of z stays constant.
So the equation must depend only on y and
z . To determine the equation precisely, we
thus look at the trace of the surface on the
yz plane as shown in romasko (qrr58) – Assignment 2 – luecke – (55035) 7 Consequently, the linear equation is
014 10.0 points x+z = 1 .
013 Find parametric equations for the line passing through the point P (1, −2, 4) and perpendicular to the plane 10.0 points A line ℓ passes through the point P (1, 4, 1)
and is parallel to the vector 3, 1, 1 .
At what point Q does ℓ intersect the xy plane? 4x + y − 2z = 4 .
1. x = 1 − 4t, y = 2 − t, z = 4 − 2t
2. x = 4 + t, y = 1 − 2t, z = −2 + 4t 1. Q(−2, 3, 0) correct 3. x = 1 + 4t, y = −2 + t, z = 4 − 2t
correct 2. Q(0, 4, 3)
3. Q(0, −2, 5) 4. x = −1 + 4t, y = 2 + t, z = −4 − 2t 4. Q(0, 5, 4) 5. x = 4 − t, y = −1 + 2t, z = −2 + 4t 5. Q(3, −2, 0) 6. x = 4 + t, y = 1 + 2t, z = 2 − 4t 6. Q(4, 5, 0)
Explanation:
Since the xy plane is given by z = 0, we
have to ﬁnd an equation for ℓ and then set
z = 0.
Now a line passing through a point
P (a, b, c) and having direction vector v is
given parametrically by
r(t) = a + tv , a= a, b, c . Explanation:
A line passing through a point P (a, b, c)
and having direction vector v is given parametrically by
r(t) = a + tv , Now for the given line, its direction vector will
be parallel to the normal to the plane
4x + y − 2z = 4 . But for ℓ,
a = 1, 4, 1 , v = 3, 1, 1 . Thus Thus
a = 1, 2, 4, , r(t) = a = a, b, c . 1 + 3t, 4 + t, 1 + t , so z = 0 when t = −1. Consequently, the line
ℓ intersects the xy plane at
Q(−2, 3, 0) .
keywords: line, parametric equations, direction vector, point on line, intercept, coordinate plane v = 4, 1, −2 , and so
r(t) = 1 + 4t, 2 + t, 4 + 2t . Consequently,
x = 1 + 4t, y = 2 + t, z = 4 + 2t
are parametric equations for the line. romasko (qrr58) – Assignment 2 – luecke – (55035)
1. 2x − 4y − 3z + 6 = 0 correct 10.0 points 015 8 Find parametric equations for the line
passing through the points P (4, 1, 1) and
Q(3, 5, 5). 2. 2x − 3y + 4z + 6 = 0
3. 2x + 3y + 4z − 6 = 0 1. x = −1 − 4t, y = 4 − t, z = 4 + t 4. 2x − 3y + 4z − 6 = 0 2. x = −1 − 4t, y = 4 + t, z = 4 + t 5. 2x + 4y + 3z + 6 = 0 3. x = −1 + 4t, y = 4 + t, z = 4 − t 6. 2x − 4y − 3z − 6 = 0 4. x = 4 − t, y = 1 − 4t, z = 1 + 4t Explanation:
Since the points Q, R, and S lie in the
plane, the displacement vectors 5. x = 4 + t, y = 1 + 4t, z = 1 − 4t
6. x = 4 − t,
correct y = 1 + 4t, Now −
−
→
PQ = a= is a direction vector for the given line, so
a = 4, 1, 1 , v = −1, 4, 4 . ijk
−
−
→−
→
n = QR × QS = 2 1 0
30 4 − t, 1 + 4t, 1 + 4t . 2 is normal to the plane.
On the other hand, if P (x, y, z ) is an arbitrary point on the plane, then the displacement vector
−
−
→
v = PQ = Thus
r(t) = 3, 0, 2 , lie in the plane. Thus the cross product a, b, c . −1, 4, 4 2, 1, 0 , −
→
QS = z = 1 + 4t Explanation:
A line passing through a point P (a, b, c)
and having direction vector v is given parametrically by
r(t) = a + tv , −
−
→
QR = x + 2, y + 1, z − 2 lies in the plane, so Consequently, v ·n = 0. x = 4 − t, y = 1 + 4t, z = 1 + 4t Now
i 016 Find an equation for the plane passing
through the points
Q(−2, −1, 2) , R(0, 0, 2) , S (1, −1, 4) . k
0 30 10.0 points j 21 are parametric equations for the line. 2 = 2, −4, −3 . But then
v · n = 2(x + 2) − 4(y + 1) − 3(z − 2)
= 2x − 4y − 3z + 6 = 0 . romasko (qrr58) – Assignment 2 – luecke – (55035)
Consequently, the plane
2x − 4y − 3z + 6 = 0 9 will be parallel to the plane and the point
Q(4, 2, 4) on the line obtained by setting t =
0 lies in the plane also. Thus
−
−
→
b = QP = 3, −2, −7 passes through Q, R and S . too is parallel to the plane. In this case,
keywords: plane, cross product, plane determined by three points, dot product
017 a × b = −28 − (−8), −21 + 12, 6 − 12
= 10.0 points Find an equation for the plane passing
through the point P (7, 0, −3) and containing the line
x = 4 − 3t, y = 2 + 4t, z = 4 + 4t .
1. 20x + 9y − 6z = 122 20x + 9y + 6z = 122
is an equation for the plane.
keywords: plane, normal, cross product, line,
018 3. 3x + 4y − 4z = 10
4. 20x + 9y + 6z = 122 correct Explanation:
If we can ﬁnd vectors a, b parallel to the
plane, their cross product
A, B, C will be a vector normal to the plane. In this
case
A(x − 7) + By + C (z + 3) = 0
will be an equation for the plane passing
through the point P (7, 0, −3) and containing the line x + 3y + 2z = 3 .
1. x + 3y + 2z = 5 correct
2. 2x + y + 3z = −2
3. 2x + y + 3z = 1
4. x + 3y + 2z = 2
5. 3x + 2y + z = 3
6. 3x + 2y + z = 6
Explanation:
The scalar equation for the plane through
P (a, b, c) with normal vector x = 4 − 3t, y = 2 + 4t, z = 4 + 4t .
To determine a and b, note ﬁrst that the
given line lies in the plane, so its direction
vector
a = −3, 4, 4 10.0 points Find an equation for the plane passing
through the point P (1, 2, −1) and parallel
to the plane 5. 20x − 9y + 6z = 122
6. 3x − 4y − 4z = 10 = n. Consequently, 2. 3x + 4y + 4z = 10 n = a×b = −20, −9, −6 n = Ai + B j + C k
is
A(x − a) + B (y − b) + C (z − c) = 0 . romasko (qrr58) – Assignment 2 – luecke – (55035)
In this question
P (a, b, c) = (1, 2, −1) ,
while
n = i + 3j + 2k
if the plane is parallel to 10 Explanation:
The graph is a circular cylinder whose axis
of symmetry is parallel to the xaxis, so it
will be the graph of an equation containing no
xterm. This already eliminates the equations
x2 + y 2 + 2 y = 0 , x2 + y 2 + 4 y = 0 , x + 3y + 2z = 3 x2 + y 2 − 2 x = 0 , x2 + y 2 − 4 x = 0 . since parallel planes have parallel normal vectors.
Consequently, the plane has equation On the other hand, the intersection of the
graph with the yz plane, i.e. the x = 0 plane,
is a circle centered on the y axis and passing
through the origin as shown in x + 3y + 2z = 5 . z keywords: plane, normal vector, point on
plane, scalar equation
y
019 10.0 points Which one of the following equations has
graph But this circle has radius 2 because the cylinder has radius 2, and so its equation is
(y − 2)2 + z 2 = 4
as a circle in the yz plane. Consequently,
after expansion we see that the cylinder is the
graph of the equation
when the circular cylinder has radius 2. y 2 + z 2 − 4y = 0 . 1. x2 + y 2 + 4y = 0
2. x2 + y 2 − 2x = 0
3. y 2 + z 2 − 4y = 0 correct
2 keywords: quadric surface, graph of equation,
cylinder, 3D graph, circular cylinder, trace
020 10.0 points 2 4. x + y + 2y = 0
5. y 2 + z 2 − 2y = 0 Which one of the following is the graph of
the equation 6. x2 + y 2 − 4x = 0 z − y2 = 1 ? romasko (qrr58) – Assignment 2 – luecke – (55035) 11
z z 5. 1. y
x y x z z 6. 2. x y
y
x Explanation:
The graph must be a cylinder with axis
parallel to the xaxis. In addition, its trace
on the yz plane must be a parabola opening
in the positive z direction and having positive
z intercept. Consequently, the graph of z 3. x y z − y2 = 1
is
z z
4. cory rect y x x keywords:
quadric surface, cylinder,
parabolic cylinder, 3Dgraph, surface, Surfaces, ...
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This note was uploaded on 06/15/2011 for the course M 408M taught by Professor Gilbert during the Fall '07 term at University of Texas at Austin.
 Fall '07
 Gilbert

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