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Unformatted text preview: qazi (kaq87) – HW05 – berg – (55290)
This printout should have 18 questions.
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before answering.
001 to determine a × b.
002 10.0 points Which of the following expressions are welldeﬁned for all vectors a, b, c, and d? 10.0 points Find the cross product of the vectors
a = i + 2j + 3k , 1 b = 3i + 3j + 3k . I a × ( b · c) , II (a × b) · (c × d) , III a × (b × c) .
1. a × b = −2i + 7j − 4k
2. a × b = −3i + 6j − 3k correct 1. I and II only
2. I only 3. a × b = −2i + 6j − 4k 3. none of them 4. a × b = −2i + 6j − 3k 4. I and III only 5. a × b = −3i + 6j − 4k 5. III only 6. a × b = −3i + 7j − 3k
Explanation:
One way of computing the cross product 6. II and III only correct
7. II only (i + 2j + 3k) × (3i + 3j + 3k)
8. all of them is to use the fact that
i ×j = k, j× k = i, k ×i = j, j ×j = 0, k× k = 0. while
i×i = 0,
For then
a × b = −3i + 6j − 3k .
Alternatively, we can use the deﬁnition
ij
12
33 a×b = = 2
3 k
3
3 1
3
i−
3
3
+ 1
3 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.
For the three given expressions, therefore,
we see that
I is not welldeﬁned because the second
term in the cross product is a dot product,
hence not a vector.
II is welldeﬁned because both terms in the
dot product are cross products, hence vectors.
III is welldeﬁned because it is the cross
product of two vectors. 3
j
3
2
k
3 keywords: vectors, dot product, cross product, T/F, length,
003 10.0 points qazi (kaq87) – HW05 – berg – (55290) 2 Determine all unit vectors v orthogonal to
a = 3 i + 4 j + k, 1. v = b = 3i +6j +2k. 004 2
3
6
i− j+ k
7
7
7 P (2, 2) , 2
6
3
i+ j− k
7
7
7 6. v = −2 i − 6 j + 3 k 3. area = 8
4. area = 7
5. area = 9 Explanation:
The nonzero vectors orthogonal to a and b
are all of the form
λ = 0, with λ a scalar. The only unit vectors orthogonal to a, b are thus
v=± a×b
.
a × b Explanation:
To use vectors we shall identify a line segment with the corresponding directed line segment.
Since the area of the parallelogram having
adjacent edges P Q and P R is given by
−
−
→−
→
P Q × P R ,
∆P QR has But for the given vectors a and b,
a×b = ij
34
36 R(3, −2) . 17
2
15
correct
2. area =
2 2
3
6
i − j + k correct
7
7
7 v = λ(a × b) , Q(−2, 3) , 1. area = 4. v = 2 i − 3 j + 6 k
5. v = ± 10.0 points Find the area of the triangle having vertices 2
6
3
2. v = − i − j + k
7
7
7
3. v = ± keywords: vector product, cross product, unit
vector, orthogonal, area = k
1
2 3
31
41
j+
i−
=
3
32
62 1−
−
→−
→
P Q × P R .
2 Now
−
−
→
P Q = −4, 1, 0 ,
4
k
6 = 2i− 3j+ 6k. −
→
PR = 1, −4, 0 . But then
i
−
−
→−
→
P Q × P R = −4
1 j
1
−4 k
−4
0=
1
0 1
k.
−4 Consequently, ∆P QR has In this case,
a × b2 = 49 . area = Consequently,
v=± 2
3
6
i− j+ k
7
7
7 . 15
2 . keywords: vectors, cross product area, triangle, parallelogram qazi (kaq87) – HW05 – berg – (55290)
005 10.0 points 3 3. volume = 31 Find a vector v orthogonal to the plane
through the points 4. volume = 28 P (5, 0, 0), Q(0, 2, 0), R(0, 0, 4) . 5. volume = 27 1. v = 4, 20, 10 Explanation:
For the parallelopiped determined by vectors a, b, and c its 2. v = 8, 5, 10 volume = a · (b × c) . 3. v = 8, 20, 10 correct But
3 −4 4. v = 8, 4, 10 −1 3 5. v = 2, 20, 10
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. =3 12
14 1 2 3 a · ( b × c) = 1 4 +4 32
34 − 3 1 3 1 . Consequently, Here volume = 30 . −
−
→
P Q = −5, 2, 0 , −
→
P R = −5, 0, 4 .
keywords: determinant, cross product, scalar
triple product, parallelopiped, volume, Consequently,
−
−
→−
→
v = P Q × P R = 8, 20, 10
is othogonal to the plane through P, Q and
R.
006 10.0 points Compute the volume of the parallelopiped
determined by the vectors
a = 3, −4, −1 , 007 10.0 points Which of the following statements are true for
all vectors a, b = 0?
A. a × b2 + a · b2 = a2b2 ,
B. a × b = b × a,
C. If a × b = 0, then a ⊥ b. b = 3, 1, 2 ,
1. C only and
c = 3, 1, 4 . 2. all of them 1. volume = 30 correct 3. B and C only 2. volume = 29 4. A only correct qazi (kaq87) – HW05 – berg – (55290) 4 so if a = 0 and b = 0, then
5. A and C only a × b = 0 6. none of them =⇒ sin θ = 0 . Thus θ = 0, π . In this case, a is parallel to b,
not perpendicular. 7. B only
keywords:
8. A and B only
008 Explanation:
A. TRUE: if θ, 0 ≤ θ ≤ π , is the angle
between a, and b, then 10.0 points Compute the volume of the parallelopiped
with adjacent edges
PQ, a × b = ab sin θ , PR, PS determined by vertices while P (1, −1, 1) , Thus
a × b2 + a · b2 Q(3, −5, −1) , R(2, 0, −1) , a · b = ab cos θ . S (2, 1, 0) . 1. volume = 8 correct = a2 b2(sin2 θ + cos2 θ ) = a2 b2 .
B. FALSE: if 2. volume = 7
3. volume = 5 a = a1 , a2 , a3 , b = b1 , b2 , b3 ,
4. volume = 6 then
a2
b2 a×b = a3
b3 i− a1 a3 b1 b1 j+ a1 a3 b1 b2 b1 b2 a1 a2 5. volume = 4
k=0 when a, b = 0, while
b2 b3 a2 b×a = a3 i+ b1 b3 a1 a3 j+ k=0 when a, b = 0.
On the other hand, for a 2 × 2 determinant,
a b c d = ad − cb = − c d a b Explanation:
The parallelopiped is determined by the
vectors
−
−
→
a = P Q = 2, −4, −2 ,
−
→
b = PR =
−
→
c = PS = 1, 2, −1 . Thus its volume is given in terms of a scalar
triple product by . Consequently, 1, 1, −2 , V =  a · ( b × c)  .
But
2 −4
a · ( b × c) = C. FALSE: if θ, 0 ≤ θ ≤ π , is the angle
between a, and b, then
a × b = ab sin θ , =2 1 −2
2 −1 1 1 −2 1 a × b = −b × a . −2 2 −1 +4 1 −2
1 −1 −2 1 1 1 2 . qazi (kaq87) – HW05 – berg – (55290) 5 of the third line, so must be scalar multiples
of each other. Consequently,
volume = 8 . 010
keywords: determinant, cross product, vector
product, scalar triple product, parallelopiped,
volume,
009 10.0 points Which equation has the surface
z 10.0 points Which of the following statements are true for
all lines and planes in 3space?
I. two planes parallel to a third plane are
parallel,
y II. two lines perpendicular to a plane are
parallel,
III. two lines parallel to a third line are
parallel. x
as its graph in the ﬁrst octant?
1. 1. II and III only
2. all of them correct 2. 3. III only 3. 4. I and III only 4. 5. I and II only 5. 6. II only
6.
7. none of them
8. I only
Explanation:
I. 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.
II. 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.
III. TRUE: each of the two lines has a direction vector parallel to the direction vector x
3
x
3
x
5
x
4
x
4
x
5 +
+
+
+
+
+ y
4
y
5
y
4
y
3
y
5
y
3 +
+
+
+
+
+ z
5
z
4
z
3
z
5
z
3
z
4 =1
=1
=1
=1
= 1 correct
=1 Explanation:
As the surface is a plane, it must be the
graph of a linear function which can be written in intercept form as
xyz
+ + = 1.
ab
c
But by inspection we see that the xintercept
is x = 4, the y intercept is y = 5 and the z intercept is z = 3. Consequently, the surface
is the graph in the ﬁrst octant of the equation
xyz
+ + =1.
453 qazi (kaq87) – HW05 – berg – (55290)
011 1. Q(5, 0, 1) 10.0 points Find parametric equations for the line passing through the point P (4, −3, 4) and parallel
to the vector 2, 4, −1 .
1. x = 4 − 2t, y = 3 − 4t, z = 4 − t
2. x = 2 − 4t, y = −4 + 3t, z = −1 + 4t
3. x = 2 + 4t, y = 4 + 3t, z = 1 − 4t
4. x = 2 + 4t, y = 4 − 3t, z = −1 + 4t
5. x = −4 + 2t, y = 3 + 4t, z = −4 − t
6. x = 4 + 2t, y = −3 + 4t, z = 4 − t
correct
Explanation:
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 . 2. Q(3, 0, 5)
3. Q(−1, 0, 3)
4. Q(−1, 1, 0) correct
5. Q(1, −1, 0)
6. Q(5, 3, 0)
Explanation:
Since the xy plane is given by z = 0, we
have to ﬁnd an equation for 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 , 4, −3, 4 , 3x + y + 2z = 2 .
v = 2, 4, −1 . Thus Thus
a = 2, 2, 2 , r(t) = 4 + 2t, −3 + 4t, 4 − t . Consequently, are parametric equations for the line. A line passes through the point P (2, 2, 2)
and is perpendicular to the plane
3x + y + 2z = 2 .
At what point Q does
plane? intersect the xy  3, 1, 2 , 2 + 3t, 2 + t, 2 + 2t . In this case, z = 0 when t = −1. Consequently, intersects the xy plane at
Q(−1, 1, 0) . keywords: line, parametric equations, direction vector, point on line
10.0 points v= and so
r(t) = x = 4 + 2t, y = −3 + 4t, z = 4 − t 012 a = a, b, c . But for , its direction vector will be parallel
to the normal to the plane Now for the given line,
a= 6 keywords: line, parametric equations, direction vector, point on line, intercept, coordinate plane
013 10.0 points Find parametric equations for the line
through the point P (9, 5, 1) that is parallel qazi (kaq87) – HW05 – berg – (55290)
to the plane x + y + z = 2 and perpendicular
to the line
x = 1 + t , y = 1 − t , z = 3t .
1. x = 9 − 4t, y = 5 + 2t, z = 1 + t
2. x = 9 + t, y = a + t, z = 1 − 2t
3. x = 9 − 4t,
correct y = 5 + 2t, z = 1 + 2t 7 3. P (−7, 9, 6)
4. P (7, 9, 3)
5. P (7, 9, 6) correct
Explanation:
To determine where the lines intesect it is
convenient ﬁrst to convert the lines from equations in symmetric form to ones in parametric
form: 4. x = 9 + 4t, y = 5 + 2t, z = 1 + t x = 3 + 4t , y = 4 + 5t , z = 5 + t , 5. x = 9 + 4t, y = 5 + 2t, z = 1 + 2t x = 2 + 5s , y = 6 + 3s , z = 4 + 2s . Explanation:
Two vectors which are perpendicular to the
required line are the normal, 1, 1, 1 , of the
given plane and a direction vector, 1, −1, 3 ,
for the given line. So a direction vector for
the required line is
1, 1, 1 × 1, −1, 3 = −4, 2, 2 .
Thus L is given by
x, y, z = 9, 5, 1 + t −4, 2, 2 ,
which can be written in parametric form as For then the lines intersect when the equations
3 + 4t = 2 + 5s ,
4 + 5t = 6 + 3s ,
5 + t = 4 + 2s ,
are satisﬁed simultaneously. Solving the ﬁrst
two equations gives t = 1, s = 1 and a check
shows that these values then satisfy the third
equation.
Consequently, the lines intersect when s =
t = 1, i.e., at the point x = 9 − 4t , y = 5 + 2t , z = 1 + 2t .
P (7, 9, 6) .
keywords: parametric equations, line, plane,
parallel perpendicular, direction vector, cross
product
014 10.0 points Find the point of intersection, P , of the
lines
y−4
z−5
x−3
=
=
,
4
5
1
x−2
y−6
z−4
=
=
.
5
3
2 015 10.0 points Find an equation for the plane passing
through the point P (1, 2, 5) and perpendicular to the y axis.
1. z = −5
2. y = −2
3. x = 1 1. P (7, −9, 6) 4. x = −1 2. P (7, −9, 4) 5. z = 5 qazi (kaq87) – HW05 – berg – (55290) 8 lies in the plane, so
6. y = 2 correct
Explanation:
If the plane is perpendicular to the y axis,
then it is of the form y = d. But if this
plane also passes through P (1, 2, 5), then it
has equation v ·n = 0.
Now
i k 3 1 0 −2 y=2 . j
0 3 = 3, −9, 2 . But then
016 10.0 points Find an equation for the plane passing
through the points
Q(−1, −2, −1) , R(2, −1, −1) , v · n = 3(x + 1) − 9(y + 2) + 2(z + 1)
= 3x − 9y + 2z − 13 = 0 .
Consequently, the plane S (−3, −2, 2) .
3x − 9y + 2z − 13 = 0
1. 3x + 2y + 9z + 13 = 0
passes through Q, R and S . 2. 3x + 9y − 2z − 13 = 0 keywords: plane, cross product, plane determined by three points, dot product 3. 3x − 9y + 2z + 13 = 0
4. 3x − 9y + 2z − 13 = 0 correct
5. 3x − 2y + 9z + 13 = 0 Explanation:
Since the points Q, R, and S lie in the
plane, the displacement vectors
−
−
→
QR = 3, 1, 0 , 6x + 4y − z = 1 .
1. x + 6y − 4z + 1 = 0
2. 4x − y − 4z + 1 = 0 −2, 0, 3 , lie in the plane. Thus the cross product
i j k 3 1 0 −2 0 −
−
→−
→
n = QR × QS = 10.0 points Find an equation for the plane passing
through the origin and parallel to the plane 6. 3x + 2y + 9z − 13 = 0 −
→
QS = 017 3 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 = P Q = x + 1, y + 2, z + 1 3. 6x − 4y + z + 1 = 0
4. x − 6y + 4z = 0
5. 6x + 4y − z = 0 correct
6. 4x + y + 4z = 0
Explanation:
The scalar equation for the plane through
P (a, b, c) with normal vector
n = Ai + B j + C k qazi (kaq87) – HW05 – berg – (55290)
is
A(x − a) + B (y − b) + C (z − c) = 0 .
But P (a, b, c) = (0, 0, 0) if the plane passes
through the origin, while
n = 6i + 4j − k
if the plane is parallel to
6x + 4y − z = 1
since parallel planes have the same normal
vector.
Consequently, the plane has equation
6x + 4y − z = 0 .
018 10.0 points Find the point at which the line
x = 3 + 5t , y = −1 , z = 3t , intersects the plane
4x + 6y − 6z = 12 .
1. P (18, −1, −9)
2. P (18, 1, −9)
3. P (−18, 1, −9)
4. P (−18, 1, 9)
5. P (18, −1, 9) correct
6. P (−18, −1, 9)
Explanation:
The line
x = 3 + 5t , y = −1 z = 3t intersects the plane
4x + 6y − 6z = 12
when
4(3 + 5t) + 6(−1) − 6(3t) = 12 ,
i.e. when 2t = 6. Thus the point of intersection occurs at t = 3, which corresponds to the
point
P (18, −1, 9) . 9 ...
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 Spring '07
 Gilbert
 Multivariable Calculus

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