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Math 241
Sample Exam 1
Solutions
Print your name neatly. If you forget to write your name, or write so sloppy that
I can’t read it, you can lose up to 100 points! Answer all the questions that you can.
Circle your answer.
Exam format: There will be 26 problems. Each problem will be worth 1 point.
There will be no partial credit on any problem. Do any 25 of the 26 problems. You may
do all 26, but I will stop grading after you get the first 25 correct. In other words, you
cannot earn more than 100% on this exam. You have been given 1 extra problem that
you may miss without penalty. In this practice exam I have given you 33 problems for
completeness sake. I consider every problem in this practice exam to be an elementarylevel problem. A low score on this exam indicates to me that the student is not ready
for vector calculus. Think of this exam as a right-of-passage exam.
You must show your work. You will not receive credit for lucky guesses. Show
your work as clearly as you can: if I can’t understand how you got an answer, I will not
give you credit for it.
WARNING: There are no calculators allowed on this exam.
SOME USEFUL FORMULAS
Lines: The game is to reduce your data to the case of knowing a point P0 on the line
and a direction vector m
~ to the line. Recall, to find the equation of a line you either need
two points on the line P1 and P2 , or a point P0 = (x0 , y0 , z0 ) on the line and a direction
vector m
~ = ha, b, ci(a 3-D slope). In the latter form the equation of the line is


x0 + at, x-component
~r(t) = hx(t), y(t), z(t)i; where y0 + bt, y-component


z0 + ct, z-component
Planes: The game is to reduce your data to the case of knowing a point P0 = (x0 , y0 , z0 )
on the plane and a normal vector ~n to the plane P. You can then use these pieces of
information to determine the equation of a plane as follows. Let P = (x, y, z) be any
−−→
other point on the plane, then the equation of the plane is given by e equation ~n · P P0 = 0.
Projections: The formula for the projection of F~ onto ~v = ha, b, ci is
!
Ã
~ · ~v
F
~v
proj~v (F~ ) =
||~v ||2
Math 241, Sample Exam 1
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Copyright °Wayne
Hacker 2007. All rights reserved.
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Study Guide for Exam 1: Exam 1 is based on homework sets 1, 2, and 3 from my
web site. Before exam 1 you should review your homework sets relevant to exam 1 and
work this sample exam.
DO NOT WRITE ON THIS PAGE IN THE SPACE BELOW
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Math 241, Sample Exam 1
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Recall from college algebra:(problems 1-2) If a curve g(x, y) = 0 divides the xy-plane
into two parts, then solution sets to the inequalities g(x, y) < 0 and g(x, y) > 0 are
disjoint subsets of the xy-plane with the curve serving as the boundary between these
two sets. In more detail, if we let S1 be the solution set to the equation g(x, y) < 0,
S2 be the solution set to the equation g(x, y) > 0, and S0 be the solution set to the
equation g(x, y) = 0, then S0 ∪ S1 ∪ S2 = the xy-plane, and the sets are mutually disjoint:
Si ∩ Sj = ∅ for i 6= j. Such a collection of sets is said to be a partition of the xy-plane.
Problem 1: Circle the answer that most accurately describes the set of points in space
that whose coordinates satisfy the given inequality and equation x2 + y 2 + z 2 ≤ 1, z = 0.
(a) the first octant
(b) the inside of the unit sphere
(c) the inside of a right circular cylinder
(d) the inside of a circle in the xy-plane
(e) the first quadrant
(f) the upper-half of the xy-plane
Solution: Setting z = 0 in the inequality x2 + y 2 + z 2 ≤ 1 yields x2 + y 2 ≤ 1. Since
z = 0 the solution set must be located in the xy-plane. The boundary of this region is
the unit circle centered at the origin: x2 + y 2 = 1. Since the inequality is not a strict
inequality (i.e. <, or >) the boundary must be part of the solution set. The circle is the
boundary of the solution set. The boundary (the circle) divides the xy-plane into two
disjoint sets. Clearly the solution set is either the set inside the circle or outside the circle.
To determine the solution we simply use a test point in the inequality. Because the circle
divides the plane into disjoint sets it follows that if one point inside the circle satisfies
the inequality, then every point inside the circle satisfies the inequality. In particular,
(0, 0) satisfies the inequality since 02 + 02 ≤ 1, so the solution set is the circle together
with its interior. The answer is (d).
Math 241, Sample Exam 1
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Problem 2: Describe the given set with with a single equation or pair of equations.
The circle of radius 1 centered at (−3, 4, 1) and lying in a plane parallel to the xy-plane.
Circle the correct answer.
(a) (x − 3)2 + (z + 1)2 = 1,
y=4
(b) (x + 3)2 + (z − 1)2 = 1,
y=4
(c) (x + 3)2 + (y − 4)2 = 1,
z=1
(d) (x − 3)2 + (y + 4)2 = 1,
z=1
(e) (y − 4)2 + (z − 1)2 = 1,
x = −3
(f) None of these
Solution: Since the circle lies in the a plane parallel to the xy-plane it follows that z
is fixed, say z = z0 . That is, the plane containing the circle is given by the equation
z = z0 . Since this plane also contains the point (−3, 4, 1) it follows that z = 1. Recall: if a circle is centered at (x0 , y0 ) with radius r, then the equation of the circle is
(x − x0 )2 + (y − y0 )2 = r2 . Thus the desired circle is (x − (−3))2 + (y − 4)2 = 1 with
z = 1. The answer is (c).
Math 241, Sample Exam 1
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Copyright °Wayne
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Problem 3: Find the distance between the points (7, 1, 1) and (5, −1, 2) .
(a)
√
7
(b)
√
3
(c) 9
(d) 3
(e) −9
(f) None of these
Solution: Let A = (x1 , y1 , z1 ) = (7, 1, 1) and B = (x2 , y2 , z2 ) = (5, −1, 2). The the
distance between the points A and B is
p
D(A, B) = (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2
p
= (5 − 7)2 + (−1 − 1)2 + (2 − 1)2
p
= (−2)2 + (−2)2 + (1)2 = 3
The answer is (d).
Problem 4: Find the midpoint between the points (7, 1, 2) and (5, −1, 2) .
(a) (6, 0, 2)
(b) (12, 2, 4)
(c) (12, 0, 4)
(d) (6, 2, 2)
(e) (2, 2, 0)
(f) None of these
Solution: Let A = (x1 , y1 , z1 ) = (7, 1, 2) and B = (x2 , y2 , z2 ) = (5, −1, 2). The the
midpoint between the points A and B is
µ
¶
x1 + x2 y1 + y2 z1 + z2
(mx , my , mz ) =
,
,
2
2
2
µ
¶
7+5 1−1 2+2
=
,
,
2
2
2
= (6 , 0 , 2)
The answer is (a).
Math 241, Sample Exam 1
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Problem 5: Find the center and radius of the sphere x2 + y 2 + z 2 − 2z = 0.
(a) Center = (0, 0, −1); radius = 2
(b) Center = (0, 0, 1); radius = 2
(c) Center = (0, 0, −1); radius = 4
(d) Center = (0, 0, 1); radius = 4
(e) Center = (0, 0, 1); radius = 1
(f) None of these
Solution: To find the center and radius of the circle we must complete the square.
Notice that the circle is already a perfect square in x and y. Thus we need only complete
the square in z.
x2 + y 2 + (z 2 − 2z +
+1
−−→
⇒
⇒
)=0
x2 + y 2 + (z 2 − 2z + 1) = 1 (complete the square)
x2 + y 2 + (z − 1)2 = 1
center = (0, 0, 1) and radius = 1
The answer is (e).
Problem 6: For the given center and radius, find the equation of the sphere: center is
at (1, 2, 3); Radius = 3.
(a) (x − 1)2 + (y − 2)2 + (z − 3)2 = 3
(b) (x + 1)2 + (y + 2)2 + (z + 3)2 = 3
(c) (x − 1)2 + (y − 2)2 + (z − 3)2 = 9
(d) (x + 1)2 + (y + 2)2 + (z + 3)2 = 9
(e) (x + 3)2 + (y − 2)2 + (z − 1)2 = 3
(f) None of these
Solution: The equation for a sphere with center (x0 , y0 , z0 ) and radius R is
(x − x0 )2 + (y − y0 )2 + (z−0 )2 = R2 .
The answer is (c).
Problem 7: Find the vector with initial point (tail) at (−3, 1, 2) and final point (tip)
at (1, 2, 3) .
(a) h0, 1, −1i
(b) h4, 1, −5i
(c) h0, 1, −5i
(d) h1, −2, 3i
(e) h4, 1, 1i
(f) None of these
Solution: Let P0 = (−3, 1, 2) and P = (1, 2, 3). Then
−−→
P0 P = tip - tail = h1 − (−3), 2 − 1, 3 − 2i = h4, 1, 1i
The answer is (e).
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Problem 8: Find the vector with initial point (tail) at (2, 1, 3) and final point (tip) at
(1, 2, 3) .
(a) h0, 0, −1i
(b) h1, −1, 0i
(c) h−1, 1, 0i
(d) h1, −1, 3i
(e) h−3, 1, 2i
(f) None of these
Solution: Let P0 = (2, 1, 3) and P = (1, 2, 3). Then
−−→
P0 P = tip - tail = h1 − 2, 2 − 1, 3 − 3i = h−1, 1, 0i
The answer is (c).
Problem 9: Convert the vector ~v = h1, 3, −1i into a unit vector.
(a) h1/11, 3/11, −1/11i
(b) h1, 3, −1i
√
√
√
(c) h1/ 11, 3/ 11, −1/ 11i
(d) h1/3, 1, −1/3i
(e) h1/2, 3/2, −1/2i
(f) None of these
Solution: First find the magnitude of ~v :
q
p
√
||~v || = v12 + v22 + v32 = 12 + 32 + (−1)2 = 11 .
The unit vector in the direction of ~v is
uˆv =
~v
.
||~v ||
The answer is (c).
Problem 10: Let ~u = h1, −1, 1i, ~v = h2, 1, −1i, and w
~ = h0, 1, −1i be 3-D vectors.
Compute ~u + ~v − 3w.
~
(a) h1, 0, −4i
(b) h3, 3, 2i
(c) h2, −3, −2i
(d) h3, −3, 3i
(e) h2, 1, −3i
(f) None of these
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Solution:
~u + ~v − 3w
~ = h1, −1, 1i + h2, 1, −1i − 3h0, 1, −1i
= h3, 0, 0i + h0, −3, 3i
= h3, −3, 3i
The answer is (d).
Problem 11: Circle all of the expressions below that are not vectors written in anglebracket notation.
(a) h1, 2, 3i
(b) hˆı, −3, 2i
(c) h5, −3ˆ
, −2i
(d) h1, −3, 2i
ˆ
(e) hˆı, ˆ, ki
(f) None of these
Solution: Recall that the form of a vector represented in angle-bracket notation is
hnumber, number, numberi. Thus any expression that contains a vector as one of the
components is wrong. The answer is: (b), (c), and (e).
Problem 12: Find the dot product of the vectors ~u = h2, −1, 3i and ~v = h0, 3, 2i .
(a) 0
(b) 3
(c) 9
(d) 6
(e) cos−1 ( √13 )
(f) None of these
Solution: ~u · ~v = h2, −1, 3i · h0, 3, 2i = 2(0) − 1(3) + 3(2) = 3 . The answer is (b).
Problem 13: Find the dot product of the vectors ~u = h3, −7, 5i and ~v = h1, 2, 2i .
(a) −21
(b) 1
(c) −1
(d) 13
(e) −6
(f) None of these
Solution: ~u · ~v = h3, −7, 5i · h1, 2, 2i = 3(1) − 7(2) + 5(2) = −1 . The answer is (c).