Multivariable Calculus A First Step Yunzhi Zou

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Yunzhi Zou
Multi-Variable Calculus

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Yunzhi Zou
Multi-Variable
Calculus
|
A First Step

Mathematics Subject Classification 2010
Primary: 26B12, 26B20, 26B15; Secondary: 26B05, 26B10
Author
Prof. Yunzhi Zou
Department of Mathematics
Sichuan University
610065 Chengdu
People’s Republic of China
zouyz@scu.edu.cn
ISBN 978-3-11-067414-9
e-ISBN (PDF) 978-3-11-067437-8
e-ISBN (EPUB) 978-3-11-067443-9
Library of Congress Control Number: 2019953764
Bibliographic information published by the Deutsche Nationalbibliothek
The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie;
detailed bibliographic data are available on the Internet at http://dnb.dnb.de.
© 2020 Walter de Gruyter GmbH, Berlin/Boston
Cover image: shulz/iStock/Getty Images Plus
Typesetting: VTeX UAB, Lithuania
Printing and binding: CPI books GmbH, Leck
www.degruyter.com

Introduction
Calculus has been widely applied to an incredible number of disciplines since its
inception in the seventeenth century. In particular, the marvelous Maxwell equa-
tions revealed the laws that govern electric and magnetic fields, which led to the
forecasting of the existence of the electromagnetic waves. The industrial revolution
witnessed the many applications of calculus. The power of calculus never diminishes,
even in today’s scientific world. For this reason, there is no doubt that calculus is one
of the most important courses for undergraduate students at any university in the
world.
On the other hand, during the past century, especially since the 2000s, many Chi-
nese and other non-English speaking people have gone to English speaking countries
to further their studies, and more are on their way. Also, as global cooperations and
communications become important for people to tackle big problems, there are needs
for people to know and understand each other better. Fortunately, Sichuan University
has a long history of global connections. Its summer immersion program is well known
for its size and popularity. Each year, it sends and hosts thousands of students from
different parts of the world. We believe that there are other similar situations where
students come and go to different places or countries without disrupting their studies.
For those students, a suitable textbook is helpful.
However, there are many challenges in developing such a suitable book. First of
all, for most freshmen whose English is not their first language, the textbook should
employ English as plain as possible. Second, the textbook should take into account
what students have learned in high school and what they need in a calculus course.
Third, there must be a smooth transition from the local standards to those globally
accepted. Furthermore, such a book must have some new insights to inject new energy
into the many already existing texts. This includes, but is not limited to, addressing
discovery over rote learning; being as concise as possible while covering the essential
content required by most local and global universities; and being printed in color as
most texts in English are. The book Single Variable Calculus: A First Step, which was
the first such calculus text in China, has provided a response to these challenges since
it was published by the World Publishing Company in 2015 and by De Gruyter in 2018.
The present book, Multivariable calculus: a first step, makes sure that these efforts
continue.
With more than 10 years in teaching calculus courses to students at the Wu Yu-
zhang honors college at Sichuan University, I have had the chance to work with local
students using books and resource materials in English. We adopted or referred stu-
dents to many calculus books, for example Thomas’ Calculus, 10th edition, by Finney,
Weir, and Giordano; Calculus, 5th edition, by Stewart; Calculus, by Larson, Edwards,
and Hosteltle; Calculus: Ideas and Applications, by Himonnas and Howard; Calculus:
Early Transcendentals, 2nd edition, by Briggs, Cochran, and Gillett; and other books in
https://doi.org/10.1515/9783110674378-201

X | Introduction
Chinese, including the calculus books published by the Mathematics departments at
Sichuan University and Tongji University. They are all good textbooks, and I acknowl-
edge that I was inspired much by them, not only by their structure that builds the
contents, but also by the nice problems that enhance understanding. Most of the ex-
ercises in this text were developed over the past decade. My many teaching assistants
contributed a lot by helping create or collecting resources in the past years. Among
them are Zengbao Wu, Liang Li, Mengxin Li, Bo Qian, Xi Zhu, and Yang Yang. Most of
the exercises were inspired by the books mentioned before. Some are original, while
others we think original may be similar to those found in existing books or other re-
sources. Those are usually classic problems and can be found in many places.
My thanks also go to students in the years of 2015, 2016, 2017, and 2018 at Wu
Yuzhang College who helped proofread the manuscript or notes and provided use-
ful feedback. I also appreciate my wonderful colleagues Wengui Hu, Li Ren, and Hao
Wang, who worked with me teaching calculus using the early versions of this book. In
particular, Wengui contributed many good problems. I received valuable suggestions
from my dear friend Dr. Harold Reiter at UNCC and Dr. Wenyuan Liao at the University
of Calgary, who always lends me a hand in solving problems arising in using LaTeX or
MATLAB. Professor Xiaozhan Xu provided me with many excellent PPTs and anima-
tions for teaching the course.
My special thanks go to my dear friend Dr. Jonathan Kane, whose talents in math-
ematics and English improved the manuscript a lot, with thorough and professional
edits. Also the ideas of adding contents such as moment of inertia, the torus problem,
and solving simple PDEs were due to Jon.
I enjoyed very much working with the wonderful people mentioned above. I am
sure that without them this book would not have achieved this level. The following
list is in alphabetical order.
1. Dr. Wengui Hu, Associate Professor of Mathematics
Sichuan University, China
2. Dr. Jonathan Kane, Emeritus Professor of Mathematics
University of Wisconsin, Whitewalter, USA
3. Dr. Wenyuan Liao, Associate Professor of Mathematics
University of Calgary, Canada
4. Dr. Harold Reiter, Professor of Mathematics,
University of North Carolina at Charlotte, USA
5. Dr. Li Ren, Associate Professor of Mathematics
6. Dr. Hao Wang, Associate Professor of Mathematics
7. Mr. Xiaozhan Xu, Professor of Mathematics

Introduction | XI
8. Mr. Zengbao Wu, Mr. Liang Li, Ms. Mengxin Li, Mr. Bo Qian, Mr. Xi Zhu, Mr. Yang
Yang, and Mr. Yi Guo
Graduate students working as teacher assistants
I also would like to thank my Mathematics department and the academic affairs of-
fice at Sichuan University. I always have their encouragement and generous support,
which make me happy to devote time and energy in writing this book and make the
publication of the work possible.
We have been working hard on this version; however, there might still be typos
and even mistakes. The responsibility for those errors in this book lie entirely with
me. I will be happy to receive comments and feedback anytime whenever they arise.
I can be reached via zouyz@scu.edu.cn.
Sincerely,
Yunzhi Zou
Professor of Mathematics
Sichuan University
Chengdu, P.R. China
zouyz@scu.edu.cn
610065

1 Vectors and the geometry of space
In this chapter we introduce vectors and coordinate systems for three-dimensional
space. They are very helpful in our study of multivariable calculus. In particular, vec-
tors provide simple descriptions and insight concerning curves and planes. We also
introduce some surfaces in space. The graph of a function of two variables is a surface
in space which gives additional insight into the properties of the function.
1.1 Vectors
1.1.1 Concepts of vectors
The term vector is used to indicate a quantity that has both a magnitude and a di-
rection, for instance, displacement, acceleration, velocity, and force. Scientists often
represent a vector geometrically by an arrow (a directed line segment). The arrow of
the directed line segment points in the direction of the vector, while the length of the
arrow represents the magnitude of the vector. We denote vectors by letters that have
an arrow overbar, such as 󳨀
→
a,
󳨀
→
b,
󳨀
→
i ,
󳨀
→
k , 󳨀
→
v . For example, suppose an object moves along
a straight line from point A to point B. The vector ⃗
s representing this displacement geo-
metrically has initial point A (the tail) and terminal point B (the head), and we indicate
this by writing ⃗
s =
󳨀
󳨀
→
AB (as shown in Figure 1.1(a)). We also denote vectors by printing
the letters in boldface, such as a, b, i, k, v. In this book, we use both notations. We de-
note the magnitude (also called the length) of a vector ⃗
a (or a) by | ⃗
a| (or |a|). If | ⃗
a| = 1,
then we say that ⃗
a is a unit vector.
(a) (b) (c) (d)
Figure 1.1: Vectors, addition, scalar multiplication, and subtraction.
We say that two vectors ⃗
a and ⃗
b are equivalent (or equal) if they have the same length
and the same direction, and we write ⃗
a = ⃗
b. Note that two vectors with the same length
and direction are considered equal even when the vectors are in two different loca-
tions. The zero vector, denoted by ⃗
0 or 0, has length 0, and, consequently, it is the
https://doi.org/10.1515/9783110674378-001

2 | 1 Vectors and the geometry of space
only vector with no specific direction. If nonzero vectors ⃗
a and ⃗
b have the same direc-
tion or if ⃗
a has exactly the opposite direction to that of ⃗
b, then we say that they are
parallel, and we write ⃗
a ‖ ⃗
b.
1.1.2 Linear operations involving vectors
We assume that vectors considered here can be represented by directed line segments
or arrows in two-dimensional space, ℝ2
, or three-dimensional space, ℝ3
. However,
vectors can be defined much more generally without reference to the directed line seg-
ments.
Definition 1.1.1 (Vector addition). If ⃗
a and ⃗
b are vectors positioned so the initial point of ⃗
b is at the
terminal point of ⃗
a, then the sum ⃗
a+ ⃗
b is the vector from the initial point of ⃗
a to the terminal point of ⃗
b.
This definition of vector addition is illustrated in Figure 1.1(b), and you can see why
this definition is sometimes called the triangle law or parallelogram law.
Note. If the initial point of ⃗
b is not at the terminal point of ⃗
a, then a copy of ⃗
b (same
length and direction) can be made with its initial point at the terminal point of ⃗
a, and
the sum can be created using ⃗
a and this copy of ⃗
b.
Vector addition satisfies the following laws for any three vectors ⃗
a, ⃗
b, ⃗
c:
(1) Commutative law: ⃗
a + ⃗
b = ⃗
b + ⃗
a.
(2) Associative law: ( ⃗
a + ⃗
b) + ⃗
c = ⃗
a + ( ⃗
b + ⃗
c).
Definition 1.1.2 (Scalar multiplication, negative of a vector). If λ is a scalar (a number) and ⃗
a is a vec-
tor, then the scalar multiple λ ⃗
a is also a vector. If λ > 0, then λ ⃗
a has the same direction as the vector
⃗
a and has length λ times the length of 󳨀
→
a . If λ < 0, then λ ⃗
a has the reverse direction to the direction
of ⃗
a and has length that is |λ| times the length of 󳨀
→
a . If λ = 0 or ⃗
a = ⃗
0 (zero vector), then λ ⃗
a = ⃗
0.
In particular, the vector − ⃗
a is called the negative of ⃗
a, and it means the scalar multiple (−1) ⃗
a has the
same length as ⃗
a but points in the opposite direction.
Scalar multiplication satisfies the following laws for any two vectors ⃗
a, ⃗
b and any
two scalars λ, μ:
(3) Associative law: λ(μ ⃗
a) = (λμ) ⃗
a = μ(λ ⃗
a).
(4) Distributive laws: (λ + μ) ⃗
a = λ ⃗
a + μ ⃗
a and λ( ⃗
a + ⃗
b) = λ ⃗
a + λ ⃗
b.
By the distributive law (4) ⃗
b + (− ⃗
b) = 1 ⃗
b + (−1) ⃗
b = (1 − 1) ⃗
b = ⃗
0, so ⃗
b and − ⃗
b act as
negatives of each other. Also, we can see that two nonzero vectors are parallel to each
other if they are scalar multiples of one another. The zero vector is considered to be
parallel to all other vectors. It is easy to establish the following theorem.

1.1 Vectors | 3
Theorem 1.1.1. Suppose ⃗
a and ⃗
b are two nonzero vectors. Then ⃗
a ‖ ⃗
b if and only if there exists a number
λ ̸
= 0 such that ⃗
a = λ ⃗
b.
The difference or subtraction ⃗
a − ⃗
b of two vectors is defined in terms of a vector sum as
⃗
a − ⃗
b = ⃗
a + (− ⃗
b).
Hence, we can construct ⃗
a − ⃗
b geometrically by first drawing the negative − ⃗
b of ⃗
b,
and then adding − ⃗
b to ⃗
a using the parallelogram law as in Figure 1.1(d). This shows
that the vector ⃗
a − ⃗
b is the vector from the head of ⃗
b to the head of ⃗
a. The operation
of subtracting two vectors does not satisfy the commutative law (1) or the associative
law (2), but it does satisfy the distributive law (4), λ( ⃗
a − ⃗
b) = λ ⃗
a − λ ⃗
b.
1.1.3 Coordinate systems in three-dimensional space
To locate a point in a plane in a two-dimensional Cartesian coordinate system with
perpendicular x- and y-axes, two numbers or coordinates are necessary, and this is
why a plane is called two-dimensional. That is, the point can be represented as an
ordered pair (a, b) of real numbers where the x-coordinate, a, is the directed distance
from the y-axis to the point, and the y-coordinate, b, is the directed distance from the
x-axis to the point.
To locate a point in three-dimensional space, three coordinates are required. We
start with a fixed point, O, called the origin. We then draw three number lines that
all pass through O and are perpendicular to each other. Usually, we put two number
lines: one horizontal and one vertical. We call the three number lines the coordinate
axes and label them as the x-axis, the y-axis, and the z-axis in a way that satisfies the
right-hand rule. This rule helps determine the direction of the z-axis. If you curl your
right-hand fingers naturally in a 90° rotation from the positive x-axis to the positive
y-axis, then the direction that your thumb points is the positive direction of the z-axis,
as shown in Figure 1.2(a). The three axes determine three coordinate planes called
the xy-plane, the xz-plane, and the yz-plane, as shown in Figure 1.2(b). Therefore, the
(a) (b) (c) (d)
Figure 1.2: Three-dimensional coordinate system, axes, coordinate planes, and octants.

space is divided into eight octants. We label them the first octant, the second octant,
the third octant, the fourth octant, the fifth octant, the six octant, the seventh octant,
and the eighth octant in a way that is shown in Figure 1.2(c).
To locate a point P in space, we project the point onto the three coordinate planes.
If the directed distance from the yz-plane to the point P is a, the directed distance
from the xz-plane to the point P is b, and the directed distance from the xy-plane to
the point P is c, then we say that the point P has x-coordinate a, y-coordinate b, and
z-coordinate c, and we use the ordered triple (a, b, c) to represent these coordinates.
This can be seen by drawing a rectangular box where O and P are two end points of the
main diagonal, as shown in Figure 1.3(a). This coordinate system is called the three-
dimensional Cartesian coordinate system. For example, to locate the point with coor-
dinates (1, 2, −1), we start from the origin and go along the x-axis for 1 unit; then turn
left and go parallel to the y-axis for 2 units; then go downward for 1 unit arriving at
(1, 2, −1), which is in the fifth octant as shown in Figure 1.3(b).
(a) (b) (c)
Figure 1.3: Three-dimensional coordinate system, coordinates, points, distance between two points.
Note that there is a one-to-one correspondence between points in the space and the
set of all ordered triples (a, b, c). Sometimes, we use ℝ3
to denote the Cartesian product
ℝ × ℝ × ℝ = {(x, y, z)|x, y, z ∈ ℝ}.
Distance between two points in space
In a two-dimensional plane, by using the Pythagorean theorem, we have the following
formula for the distance between two points (x1, y1) and (x2, y2) in the plane:
distance between two points d = √(x1 − x2)2 + (y1 − y2)2.
In three-dimensional space, for any two points P(x1, y1, z1) and Q(x2, y2, z2), we have a
rectangular box with P and Q as the two endpoints of a main diagonal, as shown in
Figure 1.3(c). Then we apply the Pythagorean theorem twice to get
distance between P and Q = √(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2. (1.1)

1.1 Vectors | 5
1.1.4 Representing vectors using coordinates
It is extremely useful to represent vectors using coordinates. First, we have three stan-
dard basis vectors called ⃗
i, ⃗
j, and ⃗
k, which are three unit vectors in the positive direc-
tions of the x-, y-, and z-axes, respectively. If those vectors have their tails at the origin
O, then their heads will be the points (1, 0, 0), (0, 1, 0), (0, 0, 1), respectively, as shown
in Figure 1.4(a).
(a) (b)
Figure 1.4: Three-dimensional coordinate system, basis vectors, position vectors.
Definition 1.1.3. A vector
󳨀
→
OP with initial point O, the origin, and terminal point P(x, y, z) is called the
position vector of the point P(x, y, z).
By the definition of vector addition, we must have
󳨀
󳨀
→
OP = x ⃗
i+y ⃗
j+z ⃗
k. This follows from
the box determined by the vector
󳨀
󳨀
→
OP (see Figure 1.4(b)), because the parallelogram
rule for addition gives
󳨀
󳨀
→
OP =
󳨀
󳨀
→
OQ +
󳨀
󳨀
→
QP
=
󳨀
󳨀
→
OT +
󳨀
󳨀
→
TQ +
󳨀
󳨀
→
QP,
where
󳨀
󳨀
→
OT is along the x-axis with length x and is x ⃗
i,
󳨀
󳨀
→
TQ is parallel to the y-axis with
length y and is y ⃗
j, and
󳨀
󳨀
→
QP is parallel to the z-axis with length z and is z ⃗
k. The numbers
x, y, and z are referred to as the components of the vector
󳨀
󳨀
→
OP.
If we add two vectors expressed in the ⃗
i, ⃗
j, ⃗
k format, then the commutative and as-
sociative laws of vector addition show that adding two vectors can be done by adding
their components, i. e.,
(x1
⃗
i + y1
⃗
j + z1
⃗
k) + (x2
⃗
i + y2
⃗
j + z2
⃗
k) = (x1 + x2) ⃗
i + (y1 + y2) ⃗
j + (z1 + z2) ⃗
k. (1.2)

By the distributive law one can see that multiplying a vector by a scalar λ is the same
as multiplying each component by λ, i. e.,
λ(x ⃗
i + y ⃗
j + z ⃗
k) = λx ⃗
i + λy ⃗
j + λz ⃗
k. (1.3)
Example 1.1.1. If ⃗
a = 5 ⃗
i + 2 ⃗
j − 3 ⃗
k and ⃗
b = 4 ⃗
i − 9 ⃗
k, express the vector 2 ⃗
a + 3 ⃗
b in terms of ⃗
i, ⃗
j, and ⃗
k.
Solution. Using properties of vectors, we have
2 ⃗
a + 3 ⃗
b = 2(5 ⃗
i + 2 ⃗
j − 3 ⃗
k) + 3(4 ⃗
i − 9 ⃗
k)
= 10 ⃗
i + 4 ⃗
j − 6 ⃗
k + 12 ⃗
i − 27 ⃗
k
= 22 ⃗
i + 4 ⃗
j − 33 ⃗
k.
Now we use the notation ⟨x, y, z⟩ to denote a position vector with its head at the
point (x, y, z), and this is the coordinate representation of this position vector. Since
any vector in space can be translated so that its initial point is the origin, any vector
in space can be represented in the form ⟨x, y, z⟩. We now give definitions for vector
operations using its coordinates representation as follows.
Definition 1.1.4. If ⃗
a = ⟨x1, y1, z1⟩ and ⃗
b = ⟨x2, y2, z2⟩ are two position vectors and λ is a real number,
then
⃗
a + ⃗
b = ⟨x1 + x2, y1 + y2, z1 + z2⟩,
⃗
a − ⃗
b = ⟨x1 − x2, y1 − y2, z1 − z2⟩,
λ ⃗
a = ⟨λx1, λy1, λz1⟩.
Note that those operations also work for two-dimensional vectors; the only difference
is that there is no z-component (or the z-component is always 0). Also, from the defi-
nition, we know that
⃗
a = ⃗
b ⇐⇒ x1 = x2, y1 = y2, and z1 = z2, (1.4)
that is, their corresponding components are identical.
Example 1.1.2. Consider any vector
󳨀
→
PQ, where the initial point is P(x1, y1, z1) and the terminal point is
Q(x2, y2, z2). Then find coordinates of the midpoint of the line segment PQ.
Solution. Since
󳨀
󳨀
→
PQ =
󳨀
󳨀
→
OQ −
󳨀
󳨀
→
OP = ⟨x2 − x1, y2 − y1, z2 − z1⟩,

1.1 Vectors | 7
if M(x, y, z) is the midpoint of the line segment PQ, then 2
󳨀
󳨀
→
PM =
󳨀
󳨀
→
PQ, so we have
⟨2(x − x1), 2(y − y1), 2(z − z1)⟩ = ⟨x2 − x1, y2 − y1, z2 − z1⟩.
This means
2⟨x, y, z⟩ = ⟨x2 − x1, y2 − y1, z2 − z1⟩ + 2⟨x1, y1, z1⟩
= ⟨x2 + x1, y2 + y1, z2 + z1⟩.
Hence, we can deduce that the formula for the midpoint M is
M(
x1 + x2
2
,
y1 + y2
2
,
z1 + z2
2
). (1.5)
1.1.5 Lengths, direction angles
Length and distance formula
The length of a vector is the length of the line segment whose endpoints are the head
and tail of the vector. By using the Pythagorean theorem, we have the following theo-
rem.
Theorem 1.1.2. If a vector is represented by ⃗
a = ⟨x, y, z⟩, then
| ⃗
a| = √x2 + y2 + z2.
If | ⃗
a| = 1, then ⃗
a is a unit vector. If ⃗
a is not the zero vector, 1
| ⃗
a|
⃗
a is the unit vector in the
direction of ⃗
a.
Example 1.1.3. Find the unit vector in the direction of
(1) ⃗
a = ⟨1, 2, −1⟩ and (2)b = 4i − j − 8k.
Solution.
1. The length of ⃗
a is | ⃗
a| = √12 + 22 + (−1)2 = √6. So the unit vector ⃗
e in the direction
of ⃗
a is
⃗
e =
1
| ⃗
a|
⃗
a =
1
√6
⟨1, 2, −1⟩ = ⟨
1
√6
,
2
√6
, −
1
√6
⟩.
2. The given vector has length
|4i − j − 8k| = √42 + (−1)2 + (−8)2 = √81 = 9.
So the unit vector with the same direction is
1
9
(4i − j − 8k) =
4
9
i −
1
9
j −
8
9
k.

We have seen the distance formula before. Now we can derive it from the length
of a vector as well. The distance between the two points P(x1, y1, z1) and Q(x2, y2, z2) is,
therefore, the length of the vector
󳨀
󳨀
→
PQ, so it is
|
󳨀
󳨀
→
PQ| = |
󳨀
󳨀
→
OQ −
󳨀
󳨀
→
OP| = √(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2. (1.6)
Example 1.1.4. Find a point P on the y-axis such that |PA| = |PB|, where A(−4, 1, 7) and B(3, 5, 2) are
two points.
Solution. We assume the point P has the coordinates (0, y, 0). From the distance for-
mula, we have
√(−4 − 0)2 + (1 − y)2 + (7 − 0)2 = √(3 − 0)2 + (5 − y)2 + (2 − 0)2.
Solving for y, we obtain y = −7
2
. Therefore, the point P has coordinates (0, −7
2
, 0).
Direction angles and direction cosines
Let ⃗
a =
󳨀
󳨀
→
OA and ⃗
b =
󳨀
󳨀
→
OB be two vectors in a plane or space as in Figure 1.5(a) and (b).
(a) (b) (c) (d)
Figure 1.5: Angle between two vectors, perpendicular vectors, and direction angles.
Definition 1.1.5 (Angle between two vectors, direction angle, and direction cosines). If ⃗
a and ⃗
b are
two vectors with a common tail, then:
1. The angle between the vectors ⃗
a and ⃗
b is the angle θ between 0 and π formed using the two
vectors as sides.
2. The two vectors ⃗
a and ⃗
b are called perpendicular (orthogonal) if and only if the angle between
them is π
2
.
3. The angle between a vector ⃗
a and the x-axis is the angle between ⃗
a and the unit base vector ⃗
i.
a and the y-axis is the angle between ⃗
j.
a and the z-axis is the angle between ⃗
k.
6. The direction angles α, β, and γ of a vector ⃗
a are the angles between ⃗
a and the x-, y-, and z-axes,
respectively; cos α, cos β, and cos γ are called direction cosines of ⃗
a.

From Figure 1.5(d), if the vector ⃗
a = ⟨x, y, z⟩ has direction angles α, β, γ, then we have
cos α =
x
| ⃗
a|
, cos β =
y
| ⃗
a|
, and cos γ =
z
| ⃗
a|
. (1.7)
Since
cos2
α + cos2
β + cos2
γ =
x2
| ⃗
a|2
+
y2
| ⃗
a|2
+
z2
| ⃗
a|2
=
x2
+ y2
+ z2
| ⃗
a|2
= 1,
it follows that
⟨cos α, cos β, cos γ⟩ =
⟨x, y, z⟩
| ⃗
a|
(1.8)
is the unit vector in the direction of ⃗
a.
Example 1.1.5. If A(2, 2, √2) and B(1, 3, 0) are two points, find the length, direction cosines, and direc-
tion angles of the vector
󳨀
→
AB.
Solution. Because
󳨀
󳨀
→
AB = ⟨1 − 2, 3 − 2, 0 − √2⟩ = ⟨−1, 1, −√2⟩, we have
|
󳨀
󳨀
→
AB| = √(−1)2 + 12 + (−√2)2 = 2.
The unit vector in the direction of
󳨀
󳨀
→
AB is
1
2
⟨−1, 1, −√2⟩ = ⟨
−1
2
,
1
2
, −
√2
2
⟩.
Hence,
cos α = −
1
2
, cos β =
1
2
, and cos γ = −
√2
2
are the three direction cosines, and
α =
2π
3
, β =
π
3
, and γ =
3π
4
are the three direction angles with the positive x-, y- and z-axes, respectively.
1.2 Dot product, cross product, and triple product
1.2.1 The dot product
So far we have introduced the two operations on vectors: addition and multiplica-
tion by a scalar. Now the following questions arise: How about multiplication? Can

Figure 1.6: Work done by a force, dot product.
we multiply two vectors to obtain a useful quantity? In fact, there are two commonly
used useful products of vectors called the dot product and the cross product.
As shown in Figure 1.6, you may already know from physics that the work done,
W, by a force F applied during a displacement along the vector s is
W = |F||s| cos θ,
where θ is the angle between the two vectors F and s. It is, therefore, useful to define
a product of two vectors in this way.
Definition 1.2.1 (Dot/scalar/inner product). The dot product a ⋅ b of the two vectors a and b is defined
by
a ⋅ b = |a||b| cos θ,
where θ is the angle between vectors a and b.
Example 1.2.1. If the two vectors a and b have length 3 and 4, and the angle between them is π/3, find
a ⋅ b.
Solution. Using the definition, we have
a ⋅ b = |a||b| cos(π/3) = 3 ⋅ 4 ⋅
1
2
= 6.
Well, this definition looks good as it has a physical basis. However, mathemati-
cally, it is not easy to find the dot product directly as we first need to know the angle
between the vectors. Using the coordinate representation of a vector, it turns out that
there is a remarkable way to compute the dot product, as we will see in the following
theorem.
Theorem 1.2.1. If a = ⟨a1, a2, a3⟩ and b = ⟨b1, b2, b3⟩, then
a ⋅ b = a1b1 + a2b2 + a3b3.

Proof. Suppose the angle between a and b is θ. Note that the three vectors, a, b, and
c = b − a form the three sides of a triangle. By the cosine law, we have
|c|2
= |a|2
+ |b|2
− 2|a||b| cos θ.
Since
|c|2
= |b − a|2
= (b1 − a1)2
+ (b2 − a2)2
+ (b3 − a3)2
= b2
1 − 2b1a1 + a2
1 + b2
2 − 2b2a2 + a2
2 + b2
3 − 2b3a3 + a2
3,
|a|2
= a2
1 + a2
2 + a2
3, and
|b|2
= b2
1 + b2
2 + b2
3,
substituting these values into the cosine law equation and canceling out all the
squares gives
−2b1a1 − 2b2a2 − 2b3a3 = −2|a||b| cos θ.
Therefore, we have
a ⋅ b = a1b1 + a2b2 + a3b3.
In view of this theorem, we give the following alternative definition of the dot
product.
Definition 1.2.2 (Alternative definition of the dot product). If a = ⟨a1, a2, a3⟩, b = ⟨b1, b2, b3⟩, and θ is
the angle between the two vectors, then the dot product is defined by
a ⋅ b = |a||b| cos θ = a1b1 + a2b2 + a3b3.
Finding the dot product of a and b is incredibly easy by using coordinates. We just
multiply corresponding components and add. Using this definition, we can deduce
the following properties of the dot product.
Theorem 1.2.2 (Properties of the dot product). If a, b, and c are any three vectors and λ is any scalar,
then the dot product satisfies:
1. a ⋅ a = |a|2
,
2. a ⋅ b = b ⋅ a,
3. if a and b are two nonzero vectors, then a ⋅ b = 0 means that a and b are perpendicular to each
other,
4. (a + b) ⋅ c = a ⋅ c + b ⋅ c,
5. (λa) ⋅ b = λ(a ⋅ b) = a ⋅ (λb),
6. 0 ⋅ a = 0.

These properties are similar to the rules for real numbers and can be easily proved
by using either of the two definitions of the dot product. However, some properties of
real number multiplication do not apply to the dot product. For example, if two real
numbers satisfy ab = 0, then either a = 0 or b = 0 or both. This is not true for the dot
product. If a and b are two nonzero vectors, then a ⋅ b = 0 indicates the two vectors
are perpendicular to each other, and it is not necessary that either a = 0 or b = 0.
By using the dot product, we can find the angle between two vectors, as shown in
the following example.
Example 1.2.2. Find the angle between the two vectors i + 2j − k and 2j − k.
Solution. By the definition of the dot product, u ⋅ v = |u| ⋅ |v| cos θ, so cos θ = u⋅v
|u|⋅|v|
.
Thus,
cos θ =
(i + 2j − k) ⋅ (2j − k)
|i + 2j − k||2j − k|
=
1 ⋅ 0 + 2 ⋅ 2 + (−1) ⋅ (−1)
√12 + 22 + (−1)2√22 + (−1)2
≈ 0.913.
So the angle θ ≈ cos−1
(0.913) ≈ 0.42 radians (about 24°).
1.2.2 Projections
Suppose that a =
󳨀
󳨀
→
OA and b =
󳨀
󳨀
→
OB are two vectors with the same tail O. If S is the foot
of the perpendicular from B to the line containing
󳨀
󳨀
→
OA, then the vector
󳨀
→
OS is called the
vector projection of the vector b onto the vector a, written as Projab. If e = a
|a|
is the
unit vector in the direction of
󳨀
󳨀
→
OA, then the vector projection is λe, where λ = |b| cos θ
is the size (positive or negative) of the projection vector and θ is the angle between the
two vectors, as shown in Figure 1.7. Hence, the projection of vector b onto vector a is
Projab =
|b| cos θ
|a|
a.
The scalar projection of vector b onto vector a is defined as
ProjScalab = |b| cos θ.
Figure 1.7: Vector projections.

By using the dot product a ⋅ b = |a||b| cos θ, we have
Projab =
a ⋅ b
|a|2
a and ProjScalab =
a ⋅ b
|a|
.
Example 1.2.3. Show that any vector r = ⟨x, y, z⟩ can be written as
r = Projir + Projjr + Projkr.
Solution. For Projir, since |i| = 1, we have
Projir =
r ⋅ i
|i|2
i = (r ⋅ i)i = ⟨x, y, z⟩ ⋅ ⟨1, 0, 0⟩i = xi.
Similarly, Projjr = yj and Projkr = zk. Therefore,
r = Projir + Projjr + Projkr.
1.2.3 The cross product
In mechanics, the moment of a force ⃗
F acting on a rod
󳨀
󳨀
→
OP is the vector with magnitude
| ⃗
F||
󳨀
󳨀
→
OP| sin θ, where θ is the angle between the vectors ⃗
F and
󳨀
󳨀
→
OP. The direction of the
moment vector is perpendicular to ⃗
F and
󳨀
󳨀
→
OP (see Figure 1.8(a)) and satisfies the right-
hand rule: if you curl your right fingers naturally from vector ⃗
F to vector
󳨀
󳨀
→
OP, then your
thumbs points in the direction of the moment vector, as shown in Figure 1.8(b) and (c).
Therefore, it makes sense to define a product of two vectors ⃗
a and ⃗
b as follows.
(a) (b) (c)
Figure 1.8: Cross product, moment/torque.
Definition 1.2.3 (Cross/vector/outer product). The cross product denoted by a × b of vector a and vec-
tor b in ℝ3
is a new vector which is perpendicular to both vector a and vector b. The length of a × b
is
|a × b| = |a||b| sin θ
and a, b, a × b, in that order, satisfy the right-hand rule.

According to the above definition and using Figure 1.4(a), we can see that
i × i = 0, i × j = k, i × k = −j, j × j = 0, j × i = −k,
j × k = i, k × i = j, k × j = −i, and k × k = 0.
But in general, how can we compute the cross product? If we try to compute
a × b = (a1i + a2j + a3k) × (b1i + b2j + b3k)
by using the normal rules for numbers, such as the commutative, associative, and
distributive rules, we may find an interesting vector
c = ⟨a2b3 − a3b2, a3b1 − b3a1, a1b2 − a2b1⟩.
This vector, in fact, satisfies conditions that we have set for a cross product, as we will
see in the following theorem.
Theorem 1.2.3. If a = ⟨a1, a2, a3⟩, b = ⟨b1, b2, b3⟩, and
c = ⟨a2b3 − a3b2, a3b1 − b3a1, a1b2 − a2b1⟩,
then:
1. c is perpendicular to both a and b.
2. |c| = |a||b| sin θ, where θ is the angle between a and b.
Proof. We compute the dot product to show they are perpendicular. We have
a ⋅ c = ⟨a1, a2, a3⟩ ⋅ ⟨a2b3 − a3b2, a3b1 − b3a1, a1b2 − a2b1⟩
= a1a2b3 − a1a3b2 + a2a3b1 − a2b3a1 + a3a1b2 − a3a2b1
= 0.
Similarly b ⋅ c = 0. Therefore, we claim that c is perpendicular to both a and b.
Furthermore,
|c|2
= (a2b3 − a3b2)2
+ (a3b1 − b3a1)2
+ (a1b2 − a2b1)2
= a2
2b2
3 + a2
3b2
2 − 2a2b3a3b2 +
a2
3b2
1 + b2
3a2
1 − 2a3b1b3a1 + a2
1b2
2 + a2
2b2
1 − 2a1b2a2b1
= (a2
1 + a2
2 + a2
3)(b2
1 + b2
2 + b2
3) − (a1b1 + a2b2 + a3b3)2
= |a|2
|b|2
− (a ⋅ b)2
= |a|2
|b|2
(1 − cos2
θ) = |a|2
|b|2
sin2
θ.
So |c| = |a||b| ⋅ sin θ.

Now the only issue that remains is whether a, b, and c, in that order, satisfy the
right-hand rule. This can be seen in a simple case where a and b are in the first quad-
rant of the xy-plane with tails at the origin. Then the sign of the term a2
a1
− b2
b1
determines
the relative positions of a and b, and the sign of the z-component of c, a1b2 − a2b1, de-
termines whether c points upward or downward. This is exactly the right-hand rule:
when you curl your right fingers from a to b, then your thumb points in the direction
of c.
In light of the above discussion, we now give an alternative definition of the cross
product.
Definition 1.2.4 (Alternative definition of the cross product). Let a = ⟨a1, a2, a3⟩ and b = ⟨b1, b2, b3⟩.
Then the cross product (also vector product) a × b is defined by
a × b = ⟨a2b3 − a3b2, a3b1 − b3a1, a1b2 − a2b1⟩.
Using the knowledge of determinants, we have
a × b = ⟨a2b3 − a3b2, a3b1 − b3a1, a1b2 − a2b1⟩
= ⟨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
a2 a3
b2 b3
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
, −
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
a1 a3
b1 b3
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
,
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
a1 a2
b1 b2
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
⟩ =
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
i j k
a1 a2 a3
b1 b2 b3
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
, (1.9)
where
󵄨
󵄨
󵄨
󵄨
a b
c d
󵄨
󵄨
󵄨
󵄨 = ad − bc. This is much better for remembering the cross product.
Using the definition of the vector product, we have the following theorem.
Theorem 1.2.4 (Properties of the cross product for three-dimensional vectors). For any three vectors
a, b, and c in ℝ3
and a scalar λ, we have:
1. a × a = 0,
2. if a and b are nonzero vectors, then a × b = 0 if and only if a ‖ b,
3. b × a = −(a × b),
4. a × (b + c) = a × b + a × c,
5. (a + b) × c = a × c + b × c,
6. (λa) × b = λ(a × b) = a × (λb),
7. a ⋅ (b × c) = (a × b) ⋅ c,
8. a × (b × c) = (a ⋅ c)b − (a ⋅ b)c.
Using one of the definitions of the cross product, we can prove these properties by
writing the vectors in their components form. Note that the cross product fails to obey
most of the laws satisfied by real number multiplication, such as the commutative and
associative laws. Check for yourself that a × (b × c) ̸
= (a × b) × c for most vectors a, b,
and c.
Example 1.2.4. Find a vector that is perpendicular to the plane containing the three points P(1, 0, 6),
Q(2, 5, −1), and R(−1, 3, 7).

Solution. The cross product of the two vectors
󳨀
󳨀
→
PQ and
󳨀
→
PR is such a vector. This is be-
cause the cross product is perpendicular to both
󳨀
󳨀
→
PQ and
󳨀
→
PR and is, thus, perpendicular
to the plane through the three points P, Q, and R. Since
󳨀
󳨀
→
PQ = (2 − 1) ⃗
i + (5 − 0) ⃗
j + (−1 − 6) ⃗
k = ⃗
i + 5 ⃗
j − 7 ⃗
k,
󳨀
→
PR = (−1 − 1) ⃗
i + (3 − 0) ⃗
j + (7 − 6) ⃗
k = −2 ⃗
i + 3 ⃗
j + ⃗
k,
we evaluate the cross product of these two vectors using the determinant approach,
i. e.,
󳨀
󳨀
→
PQ ×
󳨀
→
PR =
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
⃗
i ⃗
j ⃗
k
1 5 −7
−2 3 1
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
= (5 + 21) ⃗
i − (1 − 14) ⃗
j + (3 + 10) ⃗
k
= 26 ⃗
i + 13 ⃗
j + 13 ⃗
k.
So the vector ⟨26, 13, 13⟩ is perpendicular to the plane passing through the three points
P, Q, and R. In fact, any nonzero scalar multiple of this vector, such as ⟨2, 1, 1⟩, is also
perpendicular to the plane. Figure 1.9 illustrates the vector perpendicular to the plane.
Figure 1.9: Cross product, Example 1.2.4.
Note that the length of the vector |a × b| = |a||b| sin θ is equal to the area of the paral-
lelogram determined by a and b, assuming they have the same initial point, as shown
in Figure 1.8(d). Therefore, we have the following theorem.
Theorem 1.2.5. Given two nonzero vectors a and b with a common tail, we have
area of a parallelogram with adjacent sides a and b = |a × b|,
area of a triangle with adjacent sides a and b =
1
2
|a × b|.

Example 1.2.5. Find the area of the triangle with vertices P(1, 0, 6), Q(2, 5, −1), and R(−1, 3, 7).
Solution. In the previous example, we already computed that
󳨀
󳨀
→
PQ ×
󳨀
→
PR = ⟨26, 13, 13⟩.
The area of the parallelogram with adjacent sides PQ and PR is the magnitude of the
cross product, i. e.,
|
󳨀
󳨀
→
PQ ×
󳨀
→
PR| = √(26)2 + (13)2 + (13)2 = 13√6.
Thus, the area of the triangle PQR is 13√6
2
.
1.2.4 Scalar triple product
Suppose three nonplanar vectors a, b, and c, have a common tail. What is the volume
of the parallelepiped determined by these three vectors as shown in Figure 1.10?
Figure 1.10: Triple product, volume of a parallelepiped.
Consider the base parallelogram; its area is A = |b × c|. Let θ be the angle between
a and b × c. Noting that b × c is perpendicular to b and c and the height h of the
parallelepiped is
h = |a|| cos θ|
(we should use | cos θ| instead of cos θ to ensure that we obtain a positive result when
θ > π
2
), we conclude that the volume V of the parallelepiped is given as follows:
V = Ah = |b × c||a|| cos θ| =
󵄨
󵄨
󵄨
󵄨a ⋅ (b × c)
󵄨
󵄨
󵄨
󵄨.
Thus, we have proved that the volume of the parallelepiped determined by the three
vectors a, b, and c with a common tail is given as follows:
V =
󵄨
󵄨
󵄨
󵄨a ⋅ (b × c)
󵄨
󵄨
󵄨
󵄨. (1.10)

A product like a ⋅ (b × c) is called a scalar triple product of the three vectors ⃗
a,
⃗
b, and ⃗
c. Note that we can write this scalar triple product as a 3 × 3 determinant as
follows:
a ⋅ (b × c) = a1
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
b2 b3
c2 c3
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
− a2
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
b1 b3
c1 c3
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
+ a3
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
b1 b2
c1 c2
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
=
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
a1 a2 a3
b1 b2 b3
c1 c2 c3
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
.
If the above scalar triple product is 0, then it means that the volume of the paral-
lelepiped determined by the three vectors a, b, and c is 0. Then, we can conclude
that the three vectors must be coplanar (that is, they lie in the same plane). In terms
of linear algebra, they are linearly dependent.
Example 1.2.6. Use the scalar triple product to determine whether the vectors a = ⟨2, 0, −7⟩, b =
⟨1, −1, −3⟩, and c = ⟨1, 1, −1⟩ are coplanar.
Solution. Since
a ⋅ (b × c) =
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
2 0 −7
1 −1 −3
1 1 −1
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
= 2
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
−1 −3
1 −1
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
− 0
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
1 −3
1 −1
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
− 7
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
1 −1
1 1
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
= 8 − 0 − 7 × 2 = −6
is not 0, the vectors a, b, and c are not coplanar.
1.3 Equations of lines and planes
1.3.1 Lines
A line in the two-dimensional xy-plane is determined by a point on the line and the
direction of the line (its slope, or angle of inclination, or a vector parallel to the line).
The equation of the line can be written by using the usual slope-intercept form y =
mx + b.
A line L in ℝ3
is also determined once we know a point P(x0, y0, z0) on L and the
direction of L. However, we do not have the concept of “slope of a line” as we do in ℝ2
.
In three-dimensional space, the direction of a line L can be conveniently described by
a vector v = ⟨m, n, p⟩ parallel to L. If P(x, y, z) is an arbitrary point on L, then the vector
󳨀
󳨀
󳨀
→
P0P is parallel to v exactly when the point P is on the line, as shown in Figure 1.11, so
for some real number t we have
󳨀
󳨀
󳨀
→
P0P = tv,
⟨x − x0, y − y0, z − z0⟩ = ⟨tm, tn, tp⟩.

(a) (b)
Figure 1.11: Lines in space.
Equating the components, we have
x − x0 = tm, y − y0 = tn, and z − z0 = tp,
or
x = x0 + tm, y = y0 + tn and z = z0 + tp, (1.11)
or
{
{
{
{
{
x = x0 + tm,
y = y0 + tn,
z = z0 + tp.
(1.12)
Equations (1.11) and (1.12) are called parametric equations of the line passing
through the point (x0, y0, z0) with the direction vector v = ⟨m, n, p⟩. Note that equa-
tion (1.11) can be rewritten as
x − x0
m
=
y − y0
n
=
z − z0
p
, (1.13)
which is called symmetric equations of the line. If one of m, n, and p is 0, say, m = 0,
then we can still use the notation symbolically, i. e.,
x − x0
0
=
y − y0
n
=
z − z0
p
,
but this should be interpreted as x = x0 and
y−y0
n
=
z−z0
p
.
Note. In general, if a vector v = ⟨m, n, p⟩ is used to describe the direction of a line L,
then the numbers m, n, and p are called direction numbers of L. Since there are many
vectors parallel to L, any of them could be used to describe the direction of L. We can

also see that any three numbers proportional to m, n, and p are also direction num-
bers for L. The three direction numbers determine the three direction angles; they are
“angles of inclination” with respect to the three coordinate axes. If v = ⟨m, n, p⟩ is a
unit vector, then the three direction numbers are actually its three direction cosines.
Also, equation (1.11) can be written in a vector form,
(
x
y
z
) = (
x0
y0
z0
) + t (
m
n
p
) , (1.14)
or
r = ⟨x0, y0, z0⟩ + t⟨m, n, p⟩, (1.15)
or
r = r0 + tv. (1.16)
Equations (1.14)–(1.16) are all called vector equations for the line L passing through the
point (x0, y0, z0) with direction v.
Example 1.3.1. Find parametric equations, a vector equation, and symmetric equations of the line L
which passes through the points A(1, 2, −1) and B(0, 1, 3).
Solution. The vector
󳨀
󳨀
→
AB = ⟨0 − 1, 1 − 2, 3 − (−1)⟩ = ⟨−1, −1, 4⟩ is a direction vector of the
line L. Hence, a vector equation of L is
r = ⟨1, 2, −1⟩ + t⟨−1, −1, 4⟩
or
(
x
y
z
) = (
1
2
−1
) + t (
−1
−1
4
) .
This gives the parametric equations of line L
x = 1 − t, y = 2 − t, z = −1 + 4t.
Symmetric equations of L are obtained by eliminating the parameter t, i. e.,
x − 1
−1
=
y − 2
−1
=
z + 1
4
.
The graph of the line is shown in Figure 1.12(a).

(a) (b)
Figure 1.12: Lines in space, Examples 1.3.1 and 1.3.2.
Example 1.3.2. Show that the lines L1 and L2 with parametric equations
x = 1 + 2t, y = 2 − t, z = −3 + 4t,
x = 2 + s, y = 4 − s, z = 4 + 2s
are skew lines. That is, L1 and L2 do not intersect in a point and are not parallel to each other and,
therefore, do not lie in the same plane.
Solution. The lines are not parallel because the corresponding direction vectors v1 =
⟨2, −1, 4⟩ and v2 = ⟨1, −1, 2⟩ are not parallel because there is no scalar λ such that
⟨1, −1, 2⟩ = λ⟨2, −1, 4⟩. In other words, their components are not proportional. We at-
tempt to solve the system of equations in t and s to find any intersection points. We
have
1 + 2t = 2 + s,
2 − t = 4 − s,
−3 + 4t = 4 + 2s.
Solving the first two equations for t and s gives t = 3 and s = 5, but these values do not
satisfy the third equation. Therefore, there are no values of t and s that satisfy all three
equations, so the system of equations is inconsistent. Thus, L1 and L2 do not intersect
and are skew lines. The graphs of the two lines are shown in Figure 1.12(b).
The angle between two lines is the angle between their direction vectors. There-
fore, we can use the dot product to find the angle, as shown in the following example.
Example 1.3.3. Find the acute angle between two lines
L1 :
x − 1
1
=
y
−4
=
z + 3
1
and L2 : x = 2t, y = −2 − 2t, z = −t.

Solution. A direction vector of L1 is v1 = ⟨1, −4, 1⟩ and of L2 is v2 = ⟨2, −2, −1⟩. If θ is
the angle between the two lines, we have
cos θ =
v1 ⋅ v2
|v1||v2|
=
1 ⋅ 2 + (−4) ⋅ (−2) + 1 ⋅ (−1)
√12 + (−4)2 + 12√22 + (−2)2 + (−1)2
=
1
√2
.
The desired angle is, therefore, θ = π/4.
Example 1.3.4. Find symmetric equations of the line L that passes through (2, 1, 14) and perpendicu-
larly intersects the line L0 : x−3
2
= y
1
= z−1
1
.
Solution. Suppose that the line L intersects L0 at the point P(x, y, z). Then the coordi-
nates of P must have the form
x = 3 + 2t, y = t, and z = 1 + t, for some t.
The vector parallel to L with initial point (2, 1, 14) and terminal point P is
⟨3 + 2t − 2, t − 1, 1 + t − 14⟩ = ⟨1 + 2t, t − 1, −13 + t⟩.
Since the two lines intersect perpendicularly, the direction of L0 is also perpen-
dicular to this vector, so
⟨1 + 2t, t − 1, −13 + t⟩ ⋅ ⟨2, 1, 1⟩ = 0.
Solving for t, we have t = 2. Hence, P has coordinates (7, 2, 3) and a vector parallel
to L is ⟨7, 2, 3⟩ − ⟨2, 1, 14⟩ = ⟨5, 1, −11⟩. Therefore, symmetric equations of L are
x − 7
5
=
y − 2
1
=
z − 3
−11
.
Note. This example shows how to find the foot of the perpendicular of a point onto a
given line. This can be used to find the distance from a given point to a given line, as
shown in the following example.
Example 1.3.5. Find the perpendicular distance from the point Q(1, 2, 3) to the straight line with para-
metric equations x = 3 + t, y = 4 − 2t, z = −2 + 2t.
Solution. Let t be the value such that the point on the line N(3 + t, 4 − 2t, −2 + 2t)
is the foot of the perpendicular from the point Q to the line. The vector
󳨀
󳨀
→
NQ must be
perpendicular to the direction of the line, so
󳨀
󳨀
→
NQ ⋅ ⟨1, −2, 2⟩ = 0. This means
⟨1 − (3 + t), 2 − (4 − 2t), 3 − (−2 + 2t)⟩ ⋅ ⟨1, −2, 2⟩ = 0,
⟨−2 − t, −2 + 2t, 5 − 2t⟩ ⋅ ⟨1, −2, 2⟩ = 0,

−2 − t − 2(−2 + 2t) + 2(5 − 2t) = 0,
t =
4
3
.
So, the foot of the perpendicular from the point Q to the line is N(3+t, 4−2t, −2+2t) =
N(13
3
, 4
3
, 2
3
), and the distance from the point Q to the line is
|NQ| = √(
13
3
− 1)
2
+ (
4
3
− 2)
2
+ (
2
3
− 3)
2
= √17.
Note. The distance can also be obtained by minimizing the function d(t) = √|NQ|.
Also, one can show that the distance from a point P to a line r = r0 + vt is
distance from P to a line =
|
󳨀
󳨀
→
MP × v|
|v|
, where M is any point on the line. (1.17)
1.3.2 Planes
A plane is a surface that is determined by a point M0(x0, y0, z0) and a normal vector n.
That is, there is a unique plane that passes through the given point M0 and is perpen-
dicular to a given direction n. How do you find an equation for this plane? Assume
that M(x, y, z) is a point in space. Then M is in the plane if and only if the vector
󳨀
󳨀
󳨀
󳨀
→
M0M
is orthogonal to the normal vector n (see Figure 1.13), that is,
n ⋅
󳨀
󳨀
󳨀
󳨀
→
M0M = 0 or n ⋅ (
󳨀
󳨀
→
OM −
󳨀
󳨀
󳨀
→
OM0) = 0.
If n = ⟨a, b, c⟩, then expanding the dot product gives
n ⋅ (
󳨀
󳨀
→
OM −
󳨀
󳨀
󳨀
→
OM0) = ⟨a, b, c⟩ ⋅ (⟨x, y, z⟩ − ⟨x0, y0, z0⟩).
Thus,
a(x − x0) + b(y − y0) + c(z − z0) = 0. (1.18)
Figure 1.13: Planes in space.

This is called the Cartesian equation/linear equation of the plane through M0(x0, y0, z0)
with normal vector n = ⟨a, b, c⟩. By collecting terms in the equation, we can write the
equation as
ax + by + cz + d = 0, (1.19)
where d = −(ax0 + by0 + cz0). A point (x, y, z) is in the plane if and only if it satisfies
this equation.
Example 1.3.6. The plane x = 0 is the yz-coordinate plane, the plane y = 0 is the xz-coordinate plane,
and the plane z = 0 is the xy-coordinate plane; z = 3 is the plane parallel to the xy-plane with distance
3 units from it.
Example 1.3.7. Find an equation of the plane that passes through the point (2, 2, −1) with normal vec-
tor ⃗
n = ⟨1, 2, 3⟩. Also, find the intercepts of the plane with the three coordinate axes and then sketch
the plane.
Solution. Plug a = 1, b = 2, c = 3 and x0 = 2, y0 = 2, z0 = −1 into the equation (1.18).
We get an equation of the plane
1(x − 2) + 2(y − 2) + 3(z + 1) = 0,
or
x + 2y + 3z = 3.
In order to find the x-intercept, we set y = z = 0 in this equation and solve for x to
get x = 3. Similarly, the y-intercept is 3/2 and the z-intercept is 1. The plane is shown
in Figure 1.14(a).
(a) (b)
Figure 1.14: Planes, Examples 1.3.7 and 1.3.8.

Example 1.3.8. Find an equation of the plane through the three points P(−1, −3, 2), Q(0, −1, 7), and
R(3, 2, −1).
Solution. The vectors
󳨀
󳨀
→
PQ and
󳨀
→
PR are
󳨀
󳨀
→
PQ = ⟨0, −1, 7⟩ − ⟨−1, −3, 2⟩ = ⟨1, 2, 5⟩
and
󳨀
→
PR = ⟨3, 2, −1⟩ − ⟨−1, −3, 2⟩ = ⟨4, 5, −3⟩.
Their cross product
󳨀
󳨀
→
PQ ×
󳨀
→
PR is orthogonal to the desired plane and, thus, ⃗
n =
󳨀
󳨀
→
PQ ×
󳨀
→
PR
is a normal vector to the plane. Hence, an equation of the plane is
󳨀
󳨀
→
PM ⋅ ⃗
n =
󳨀
󳨀
→
PM ⋅ (
󳨀
󳨀
→
PQ ×
󳨀
→
PR) = 0,
where M(x, y, z) is an arbitrary point in the plane. Using the triple product formula
gives
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
x − (−1) y − (−3) z − 2
1 2 5
4 5 −3
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
= 0.
Simplifying this, we obtain
23y − 31x − 3z + 44 = 0.
The graph of the plane is shown in Figure 1.14(b).
In general, an equation of the plane passing through three points P1(x1, y1, z1),
P2(x2, y2, z2), and P3(x3, y3, z3) is
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
x − x1 y − y1 z − z1
x2 − x1 y2 − y1 z2 − z1
x3 − x1 y3 − y1 z3 − z1
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
= 0.
In particular, if the three points are three intercepts with the x-, y-, and z-axes
given by P1(a, 0, 0), P2(0, b, 0), and P3(0, 0, c), then an equation of the plane is
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
x − a y z
−a b 0
−a 0 c
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
= 0,
and this simplifies to (provided, a, b, and c are all nonzero)
x
a
+
y
b
+
z
c
= 1.

Figure 1.15: Angle between two planes.
We can define the angle between two planes using their normal vectors as shown in
Figure 1.15.
Definition 1.3.1. The angle between two planes is defined as the acute angle between the normal
vectors of the two planes. Two planes are considered to be perpendicular if their normal vectors are
orthogonal.
Example 1.3.9. Find the angle between the planes x − y − 2z = 1 and x + y − 3z = 1.
Solution. The normal vectors of these two planes are
⃗
n1 = ⟨1, −1, −2⟩ and ⃗
n2 = ⟨1, 1, −3⟩,
respectively. Let θ be the angle between the two planes. Then
cos θ =
⃗
n1 ⋅ ⃗
n2
| ⃗
n1|| ⃗
n2|
=
1(1) + (−1)(1) + (−2)(−3)
√12 + (−1)2 + (−2)2√12 + 12 + (−3)2
≈ 0.73855.
So the acute angle between the given planes is cos−1
(0.73855) ≈ 42°.
Example 1.3.10. Find a formula for the perpendicular distance D from the point P(x0, y0, z0) to the
plane ax + by + cz + d = 0.
Solution. Let P1(x1, y1, z1) be any point in the given plane. Then
󳨀
󳨀
→
P1P = ⟨x0 − x1, y0 − y1, z0 − z1⟩.
The vector ⃗
n = ⟨a, b, c⟩ is a normal vector of the plane. Then, as shown in Figure 1.16,
the distance D from P to the plane is
D =
󵄨
󵄨
󵄨
󵄨|
󳨀
󳨀
→
P1P| cos θ
󵄨
󵄨
󵄨
󵄨.

Figure 1.16: Distance from a point P to a plane ax + by + cz + d = 0.
Thus,
D =
󵄨
󵄨
󵄨
󵄨|
󳨀
󳨀
→
P1P| cos θ
󵄨
󵄨
󵄨
󵄨 =
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
|
󳨀
󳨀
→
P1P| ⋅ cos θ ⋅
| ⃗
n|
| ⃗
n|
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
=
1
| ⃗
n|
󵄨
󵄨
󵄨
󵄨|
󳨀
󳨀
→
P1P| ⋅ cos θ ⋅ | ⃗
n|
󵄨
󵄨
󵄨
󵄨 =
|
󳨀
󳨀
→
P1P ⋅ ⃗
n|
| ⃗
n|
=
|a(x0 − x1) + b(y0 − y1) + c(z0 − z1)|
√a2 + b2 + c2
=
|ax0 + by0 + cz0 − (ax1 + by1 + cz1)|
√a2 + b2 + c2
=
|ax0 + by0 + cz0 + d|
√a2 + b2 + c2
, (1.20)
since −(ax1 + by1 + cz1) = d.
Example 1.3.11. Find the distance between the two parallel planes x +2y −2z = 5 and 2x +4y −4z = 3.
Solution. The two planes are parallel to each other since their normal vectors ⟨1, 2, −2⟩
and ⟨2, 4, −4⟩ are parallel. In order to find the distance D between the two planes, we
can, instead, find the distance from any point in one plane to the other plane. For
example, we can put y = z = 0 in the equation of the first plane, to get x = 5, so
(5, 0, 0) is a point in the first plane. Using formula (1.20) from Example 1.3.10,
D =
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
2(5) + 4(0) − 4(0) − 3
√22 + 42 + (−4)2
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
=
7
6
.
So the distance between the two planes is 7/6.

The intersection of two planes that are not parallel is of course a line. So a line L
can be described as the line of intersection of two planes in the form
L : {
A1x + B1y + C1z = D1,
A2x + B2y + C2z = D2.
(1.21)
This is a general equation of the line L. The symmetric equations of a line are an exam-
ple of this form. There will, of course, be infinitely many possible choices for the two
planes that intersect in a given line L.
Example 1.3.12. Rewrite the line L determined by the equations below in the form of parametric equa-
tions and then in the form of symmetric equations:
{
x + y − z = 1,
2x + y + 3z = 4.
Solution. First of all, we find a point on the line by choosing z = 0 and solving the
equations for x and y,
{
x + y = 1,
2x + y = 4,
obtaining x = 3 and y = −2. Therefore, the point (3, −2, 0) lies on line L. Note that the
direction vector v of line L is perpendicular to both normal vectors of the given planes,
so it is given by the cross product
v = n1 × n2 =
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
⃗
i ⃗
j ⃗
k
1 1 −1
2 1 3
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
= 4 ⃗
i − 5 ⃗
j − ⃗
k.
Hence, parametric equations of line L are x = 3 + 4t, y = −2 − 5t, z = −t. Symmetric
equations of line L are
x − 3
4
=
y + 2
−5
=
z
−1
.
Vector equations of planes
SupposetwononparallelvectorsaandblieintheplanewiththeirtailsatM0(x0, y0, z0).
Then, any vector with its head a point in the plane and tail at M0 can be given by a lin-
ear combination of the two vectors a, b. Assume M0 is the head of the position vector
r0. Then, for any position vector r with head at a point in the plane (see Figure 1.17),
there must be two scalars λ and u such that
r − r0 = λa + ub.

Figure 1.17: Vector equations of a plane.
Therefore, the vector equation
r = r0 + λa + ub (1.22)
describes a plane in space. Suppose r = ⟨x, y, z⟩, a = ⟨a1, a2, a3⟩, and b = ⟨b1, b2, b3⟩.
Then
(
x
y
z
) = (
x0
y0
z0
) + λ (
a1
a2
a3
) + u (
b1
b2
b3
) (1.23)
is a vector equation of the plane. This can be rewritten as
{
{
{
{
{
x = x0 + λa1 + ub1,
y = y0 + λa2 + ub2,
z = z0 + λa3 + ub3.
(1.24)
These are parametric equations of the plane. Note that this can be written in the form
r(λ, u) = ⟨x(λ, u), y(λ, u), z(λ, u)⟩, (1.25)
which is a vector-valued function with two parameters.
Example 1.3.13. Rewrite the equation of the plane 2x − y − 3z = 10 in a vector form ⃗
r = ⃗
r0 + λ ⃗
a + μ ⃗
b.
Solution. We must find a position vector ⃗
r0 whose terminal point is a point in the
plane and two nonparallel vectors ⃗
a and ⃗
b which are both parallel to the plane. To find
three such vectors, we find three points in the plane. It is easy to check that (5, 0, 0),
(0, −10, 0), and (2, 0, −2) are three points in the plane and, therefore,
⃗
a = (
0
−10
0
) − (
5
0
0
) = (
−5
−10
0
) and ⃗
b = (
2
0
−2
) − (
5
0
0
) = (
−3
0
−2
)

are two vectors parallel the plane, and ⃗
a 󳠑 ⃗
b, thus, a vector equation of the plane is
given by
⃗
r = (
5
0
0
) + λ (
−5
−10
0
) + μ (
−3
0
−2
) .
Also, we can solve the equation 2x−y−3z = 10 to find a general solution. For example,
we let y = λ and z = u be two free variables. Then a general solution to the equation
(
x
y
z
) = (
10+λ+3u
2
λ
u
) = (
5
0
0
) + λ (
1
2
1
0
) + u (
3
2
0
1
)
is in the desired vector form.
1.4 Curves and vector-valued functions
A line is a special curve in space. As seen from the previous section, a line in space
has parametric equations
{
{
{
{
{
x = x0 + mt,
y = y0 + nt,
z = z0 + pt,
where (x0, y0, z0) is a point on the line and ⟨m, n, p⟩ is the direction of the line. We can
rewrite this in a vector form
r = r0 + vt
with r = ⟨x, y, z⟩, r0 = ⟨x0, y0, z0⟩, and v = ⟨m, n, p⟩ is the direction vector. This can be
written as
r(t) = ⟨x0 + mt, y0 + nt, z0 + pt⟩,
which is a vector-valued function of t with each component being a linear function of
t. In general, the graph of a vector-valued function r(t) = ⟨x(t), y(t), z(t)⟩ is a curve in
space. Its parametric form is x = x(t), y = y(t), and z = z(t). You can imagine that this
curve is the trajectory of a moving object: at each specific time t, its position vector
is r(t).
Example 1.4.1 (A helix). The graph of the vector-valued function r(t) = 2 cos ti + 2 sin tj + 0.5tk, t ≥ 0,
is called a helix. The curve is shown in Figure 1.18.

1.4 Curves and vector-valued functions | 31
Figure 1.18: A picture of a helix.
Example 1.4.2 (Slinky curve). A slinky curve is defined as r(t) = ⟨a(t) cos t, a(t) sin t, 1.2 sin 20t⟩. The
graph of the curve when a(t) = 5 + cos 20t and 0 ≤ t ≤ 2π is shown in Figure 1.19.
Figure 1.19: A picture of a slinky curve.
Sometimes it is helpful to visualize a curve in space by projecting the curve onto one
of the coordinate planes. If a curve has the vector equation r(t) = ⟨x(t), y(t), z(t)⟩, then
its view from above is its projection onto the xy-plane, and when it is projected, its
x- and y-coordinates remain unchanged, but the z-coordinate becomes 0. Thus, the
projection of the curve onto the xy-plane has the equation
r(t) = ⟨x(t), y(t), 0⟩.
In Example 1.4.1, the projection of the helix onto the xy-plane has the equation
r(t) = ⟨2 cos t, 2 sin t, 0⟩
or x2
+ y2
= 4 and z = 0. It is a circle in the xy-plane.

Similarly,toobtainanequationfortheprojectionofthecurver(t) = ⟨x(t), y(t), z(t)⟩
onto the xz-plane, we set the y-coordinate to be 0. To obtain an equation for the projec-
tion curve of the curve r(t) = ⟨x(t), y(t), z(t)⟩ onto the yz-plane, we set the x-coordinate
to be 0. For instance, the projection of the curve r(t) = ⟨2 cos t, 2 sin t, 0.5t⟩ onto the
yz-plane has an x-coordinate equal to 0, giving
r(t) = ⟨0, 2 sin t, 0.5t⟩
or x = 0 and y = 2 sin(2z). It is the graph of y = 2 sin(2z) in the plane x = 0.
1.5 Calculus of vector-valued functions
1.5.1 Limits, derivatives, and tangent vectors
We can also define the limit of a vector-valued function r(t) = ⟨x(t), y(t), z(t)⟩ at a point
t0. Similar to a scalar function, if t → t0 implies r(t) → L, then we say limt→t0
r(t) = L,
where L = ⟨a, b, c⟩ is a constant vector. More precisely, it is defined as follows.
Definition 1.5.1. Let L = ⟨a, b, c⟩ be a constant vector and r(t) = ⟨x(t), y(t), z(t)⟩ be a vector-valued
function. Then limt→t0
r(t) = L if and only if for any given ε > 0, there is a number δ > 0 such that
󵄨
󵄨
󵄨
󵄨
󵄨r(t) − L
󵄨
󵄨
󵄨
󵄨
󵄨 < ε whenever 0 < |t − t0| < δ.
Since r(t) = x(t)i + y(t)j + z(t)k and
󵄨
󵄨
󵄨
󵄨r(t) − L
󵄨
󵄨
󵄨
󵄨 = √(x(t) − a)
2
+ (y(t) − b)
2
+ (z(t) − c)
2
< ε,
using the above definition and applying the limit laws for scalar functions, we have
the following theorem.
Theorem 1.5.1. Let L = ⟨a, b, c⟩ be a constant vector and let r(t) = ⟨x(t), y(t), z(t)⟩ be a vector-valued
function. Then
lim
t→t0
r(t) = L ⇐⇒ lim
t→t0
x(t) = a, lim
t→t0
y(t) = b, and lim
t→t0
z(t) = c.
Therefore, limt→t0
r(t) = ⟨limt→t0
x(t), limt→t0
y(t), limt→t0
z(t)⟩. That is, to evaluate the
limit of a vector-valued function, we evaluate the limit of each component of the func-
tion, given that all limits exist.
Example 1.5.1. Given the vector-valued function r(t) = ⟨1−cos t
t2 , e−t
, tan−1
t⟩, evaluate the limits
(a) limt→0 r(t) and (b) limt→∞ r(t).

Solution.
(a) We have limt→0 r(t) = ⟨limt→0
1−cos t
t2 , limt→0 e−t
, limt→0 tan−1
t⟩ = ⟨limt→0
sin t
2t
,
1, 0⟩ = ⟨1
2
, 1, 0⟩.
(b) We have limt→∞ r(t) = ⟨limt→∞
1−cos t
t2 , limt→∞ e−t
, limt→∞ tan−1
t⟩ = ⟨0, 0, π
2
⟩.
Intuitively, we know that if each component of r(t) is continuous, then the curve r(t)
must be continuous, which means that you can draw the curve continuously, without
lifting your pencil. The formal definition of continuity is given below.
Definition 1.5.2. A vector-valued function r(t) is continuous at t0 if and only if limt→t0
r(t) = r(t0).
This means that each component of r(t) must be continuous at t = t0.
We now consider the trajectory of a moving object, where at any instant its posi-
tion vector is given by r(t). Its displacement over the time Δt is r(t) − r(t0) = r(t0 + Δt) −
r(t0). Therefore,
Δr
Δt
=
r(t0 + Δt) − r(t0)
Δt
is the average velocity of the object during this time interval. The limit as Δt → 0, if
it exists, is the instantaneous velocity at that t0. This is the definition of a derivative.
Thus,
r󸀠
(t0) = lim
Δt→0
Δr
Δt
= lim
Δt→0
r(t0 + Δt) − r(t0)
Δt
.
If x(t), y(t), and z(t) are differentiable one-variable functions, then
r󸀠
(t0) = lim
Δt→0
r(t0 + Δt) − r(t0)
Δt
= lim
Δt→0
(x(t0 + Δt) − x(t0))i + (y(t0 + Δt) − y(t0))j+(z(t0 + Δt) − z(t0))k
Δt
= lim
Δt→0
(x(t0 + Δt) − x(t0))i
Δt
+ lim
Δt→0
(y(t0 + Δt) − x(t0))j
Δt
+ lim
Δt→0
(y(t0 + Δt) − y(t0))k
Δt
= x󸀠
(t0)i + y󸀠
(t0)j + z󸀠
(t0)k.
Or this can be written as r󸀠
(t0) = ⟨x󸀠
(t0), y󸀠
(t0), z󸀠
(t0)⟩. If this vector is not 0, then
it is a vector tangent (tangent vector) to the curve r(t) at t0. Figure 1.20(a) illustrates
this idea. We can extend this idea to define the derivative as a function of t as follows.
Definition 1.5.3. If x(t), y(t), and z(t) are three differentiable functions on the interval (a, b), then the
derivative of the vector-valued function r(t) = ⟨x(t), y(t), z(t)⟩ is
r
󸀠
(t) = ⟨x
󸀠
(t), y
󸀠
(t), z
󸀠
(t)⟩.
If this vector is not 0, then it is a vector tangent to the curve r(t).

(a) (b)
Figure 1.20: Tangent vector/line and normal plane.
In light of the above definition, we are now able to derive an equation for the tangent
line to the curve r(t) at any point t = t0. Since the curve at point (x(t0), y(t0), z(t0)) has
tangent vector r󸀠
(t0) = ⟨x󸀠
(t0), y󸀠
(t0), z󸀠
(t0)⟩, the symmetric equations of the tangent
line, provided r󸀠
(t0) ̸
= 0, are
x − x(t0)
x󸀠(t0)
=
y − y(t0)
y󸀠(t0)
=
z − z(t0)
z󸀠(t0)
. (1.26)
Parametric equations of the tangent line at t = t0 are
{
{
{
{
{
x = x(t0) + x󸀠
(t0)t,
y = y(t0) + y󸀠
(t0)t,
z = z(t0) + z󸀠
(t0)t,
(1.27)
and a vector equation of the tangent line at t = t0 is
r(t) = r(t0) + r󸀠
(t0)t, (1.28)
where r󸀠
(t0) is a tangent vector. The unit tangent vector at t = t0 is T =
r󸀠
(t0)
|r󸀠(t0)|
.
Note that the plane passing through the curve at t = t0 with a normal vector paral-
lel to the tangent vector to the curve at t = t0 is the normal plane to the curve at t = t0,
as shown in Figure 1.20(b). The normal plane to the curve at t = t0 has the equation
x󸀠
(t0)(x − x0) + y󸀠
(t0)(y − y0) + z󸀠
(t0)(z − z0) = 0. (1.29)
Example 1.5.2. Find an equation for the tangent line and normal plane to the curve
r(t) = ⟨sin t, cos t, sin 2t⟩ at t = π/6.
Solution. The point is (sin π/6, cos π/6, sin(2 × π/6)) = (1/2, √3/2, √3/2), and since
r󸀠
(t) = ⟨cos t, − sin t, 2 cos 2t⟩,
r󸀠
(π/6) = ⟨cos(π/6), − sin(π/6), 2 cos(2 ⋅ π/6)⟩ = ⟨√3/2, −1/2, 1⟩.

So, the parametric equations for the desired tangent line are
{
{
{
{
{
{
{
{
{
x = 1
2
+
√3
2
t,
y =
√3
2
− 1
2
t,
z =
√3
2
+ t.
An equation for the normal plane at t = π/6 is
√3
2
(x −
1
2
) −
1
2
(y −
√3
2
) + (z −
√3
2
) = 0.
Figure 1.21 shows the tangent line and normal plane at t = π/6.
(a) (b) (c) (d)
Figure 1.21: Tangent line and normal plane, Example 1.5.2.
By using the above definition of the derivative for a vector-valued function, we can
deduce the following theorem, the proof of which is omitted here.
Theorem 1.5.2. Let u(t) and v(t) be two differentiable vector-valued functions and f (t) be a differen-
tiable scalar-valued function over a < t < b. Let c be a constant vector. Then at any t in (a, b), we
have:
1. d
dt
(c) = 0,
2. d
dt
(u(t) ± v(t)) = d
dt
u(t) ± d
dt
v(t) (sum or difference rule),
3. d
dt
(f (t)u(t)) = ( d
dt
f (t))u(t) + f (t) d
dt
u(t) (constant multiple rule),
4. d
dt
u(f (t)) = u󸀠
(f (t))f 󸀠
(t) (chain rule),
5. d
dt
(u(t) ⋅ v(t)) = u󸀠
(t) ⋅ v(t) + u(t) ⋅ v󸀠
(t) (dot product rule),
6. d
dt
(u(t) × v(t)) = u󸀠
(t) × v(t) + u(t) × v󸀠
(t) (cross product rule).
1.5.2 Antiderivatives and definite integrals
Similar to scalar functions, if R󸀠
(t) = r(t), then we say R(t) is an antiderivative of r(t),
and we write the indefinite integral of r(t) as
∫ r(t)dt = R(t) + C,

where C is an arbitrary constant vector. For definite integrals, we write ∫
b
a
r(t)dt =
R(b) − R(a). In light of the previous definition for derivative, we have the following
formal definition using the components of r(t).
Definition 1.5.4. If r(t) = ⟨x(t), y(t), z(t)⟩ is continuous for a ≤ t ≤ b, then we define
∫ r(t)dt = ⟨∫ x(t)dt, ∫ y(t)dt, ∫ z(t)dt⟩ and
b
∫
a
r(t)dt = ⟨
b
∫
a
x(t)dt,
b
∫
a
y(t)dt,
b
∫
a
z(t)dt⟩.
Example 1.5.3. If r󸀠
(t) = e2t
i + sec2
tj + sin tk,
1. find r(t).
2. furthermore, if r(0) =< 1, 1, 2 >, determine r(t).
3. find ∫
π/4
0
r(t)dt.
Solution.
1. Since r󸀠
(t) = ⟨e2t
, sec2
t, sin t⟩,
r(t) = ∫ r󸀠
(t)dt = ⟨∫ e2t
dt, ∫ sec2
tdt, ∫ sin tdt⟩
= ⟨
1
2
e2t
+ c1, tan t + c2, − cos t + c3⟩
= ⟨
1
2
e2t
, tan t, − cos t⟩ + ⟨c1, c2, c3⟩.
2. Since r(0) = ⟨1, 1, 2⟩, we have
⟨1, 1, 2⟩ = ⟨
1
2
e0
, tan 0, − cos 0⟩ + ⟨c1, c2, c3⟩
so ⟨c1, c2, c2⟩ = ⟨
1
2
, 1, 3⟩.
Then, r(t) = ⟨1
2
e2t
+ 1
2
, tan t + 1, − cos t + 3⟩.
3. By definition
π/4
∫
0
r(t)dt = ⟨
π/4
∫
0
(
1
2
e2t
+
1
2
)dt,
π/4
∫
0
(tan t + 1)dt,
π/4
∫
0
(− cos t + 3)dt⟩
= ⟨
1
4
e2t
+
t
2
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
π/4
0
, − ln | cos t|π/4
0 + π/4, − sin t + 3t|π/4
0 ⟩
= ⟨(
eπ/2
4
+
π
8
−
1
4
),
ln 2
2
+
π
4
, −
√2
2
+
3π
4
⟩.

1.5.3 Length of curves, curvatures, TNB frame
As seen before, s, the arc length, or length of a plane curve ⟨x(t), y(t)⟩ for a ≤ t ≤ b, is
s =
b
∫
a
√[x󸀠(t)]
2
+ [y󸀠(t)]
2
dt.
The analog for a curve in space is the length of a curve r(t) = ⟨x(t), y(t), z(t)⟩ for a ≤
t ≤ b, which is
s =
b
∫
a
√[x󸀠(t)]
2
+ [y󸀠(t)]
2
+ [z󸀠(t)]
2
dt,
provided that the integrand is integrable. The integrand is always integrable when the
curve is smooth, that is, x󸀠
(t), y󸀠
(t), and z󸀠
(t) are continuous on [a, b].
Again, thinking of a moving object along the curve, the length of the curve is in-
deed the distance traveled by the object over time interval [a, b]. Since the derivative
of position with respect to time is the velocity, v(t), and the derivative of distance trav-
eled with respect the time t is the speed, we have
v(t) = r󸀠
(t) and
ds
dt
=
󵄨
󵄨
󵄨
󵄨v(t)
󵄨
󵄨
󵄨
󵄨.
It is not a surprise that the length of the curve is s = ∫
b
a
|v(t)|dt.
We conclude this in the following definition.
Definition 1.5.5. If r󸀠
(t) is continuous, the curve r(t) is a smooth curve, and the length of this curve for
a ≤ t ≤ b is defined as
b
∫
a
󵄨
󵄨
󵄨
󵄨
󵄨r
󸀠
(t)
󵄨
󵄨
󵄨
󵄨
󵄨dt =
b
∫
a
√[x󸀠(t)]
2
+ [y󸀠(t)]
2
+ [z󸀠(t)]
2
dt.
Example 1.5.4. Find the length of the curve r(t) = ⟨3 cos t, 4 cos t, 5 sin t⟩ for 0 ≤ t ≤ 2π.
Solution. The length of the curve s is given by the integral
s =
2π
∫
0
√((3 cos t)󸀠)
2
+ ((4 cos t)󸀠)
2
+ ((5 sin t)󸀠)
2
dt
=
2π
∫
0
√9 sin2
t + 16 sin2
t + 25 cos2 tdt =
2π
∫
0
5dt = 10π.

Parameterization by arc length
Now we consider a vector-valued function r(t) = ⟨x(t), y(t)⟩ with the following para-
metric equations representations:
(a) {
x = R cos t,
y = R sin t,
0 ≤ t ≤ 2π , (b) {
x = R cos u
2
,
y = R sin u
2
,
0 ≤ u ≤ 4π ,
(c) {
x = R cos 2t,
y = R sin 2t,
0 ≤ t ≤ π , (d) {
x = R sin 3θ,
y = R cos 3θ,
0 ≤ θ ≤
2π
3
.
They actually describe the same curve. In this case, it is a circle centered at (0, 0)
with radius R. The name of a parameter, of course, does not matter. However, how
the curve evolves as the parameter increases does make a difference. For example,
in (a), the circle is formed counterclockwise while in (d) it is formed clockwise. The
positive orientation of a curve is the direction in which the curve is generated as the
parameter increases. So the positive orientation of (a) is counterclockwise, while the
positive orientation of (d) is clockwise.
A curve may be parameterized in many ways, as shown above. In some ways, the
parameter may have a nice geometric interpretation. For example, in (a), at each point
onthe circle,the correspondingvalueof theparametert isexactlytheangle(measured
in radians) formed by the corresponding radius and the positive x-axis. We now intro-
duce a very natural way for describing a curve where its parameter represents the arc
length. We first investigate the following curve:
{
x = 2 cos t
2
,
y = 2 sin t
2
,
for 0 ≤ t ≤ 4π.
The initial point is (2, 0). When t = π, the corresponding arc length is also π. When t =
2π, the corresponding arc length is also 2π. In general, the length of the interval [0, t] is
equal to the length of the curve generated. We say that the curve r(t) = ⟨2 cos t
2
, 2 sin t
2
⟩
is parameterized by arc length. In this case we also write
r(s) = ⟨2 cos
s
2
, 2 sin
s
2
⟩
with s being the arc length parameter.
But how do we know whether a curve r(t) is parameterized by arc length? First,
we note that by definition
s =
t
∫
a
󵄨
󵄨
󵄨
󵄨r󸀠
(t)
󵄨
󵄨
󵄨
󵄨dt,
and so ds
dt
= |r󸀠
(t)|. This means ds = |r󸀠
(t)|dt. Therefore, the change in t is equal to
the change in s if and only of if |r󸀠
(t)| = 1. In particular, if the curve starts at r(a) and
|r󸀠
(t)| = 1 for all t, then when t = a, we have s = 0, and when t ̸
= a, we have s = t − a.

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knowledge of the criminal aspects of the Rascher experiments. He
was notified that Dachau inmates were to be used. He himself
inspected the experiments. Sievers admitted that Rascher told him
that several died as a result of the high-altitude experiments.
Under these facts Sievers is specially chargeable with the criminal
aspects of these experiments.
FREEZING EXPERIMENTS
Before the high-altitude experiments had actually been
completed, freezing experiments were ordered to be performed at
Dachau. They were conducted from August 1942 to the early part of
1943 by Holzloehner, Finke and Rascher, all of whom were officers in
the Medical Services of the Luftwaffe. Details of the freezing
experiments have been given elsewhere in this judgment.
In May 1943 Rascher was transferred to the Waffen SS and then
proceeded alone to conduct freezing experiments in Dachau until
May 1945. Rascher advised the defendant Rudolf Brandt that Poles
and Russians had been used as subjects.
The witness Neff testified that the defendant Sievers visited the
experimental station quite frequently during the freezing
experiments. He testified further that in September 1942 he received
orders to take the hearts and lungs of 5 experimental subjects killed
in the experiments to Professor Hirt in Strasbourg for further
scientific study; that the travel warrant for the trip was made out by
Sievers; and that the Ahnenerbe Society paid the expenses for the
transfer of the bodies. One of the 5 experimental subjects killed was
a Dutch citizen.
Neff’s testimony is corroborated in large part by the affidavits of
the defendants Rudolf Brandt and Becker-Freyseng, by the testimony
of the witnesses Lutz, Michalowsky and Vieweg, and by the
documentary evidence in the record. In the Sievers’ diary, there are
numerous instances of Sievers’ activities in the aid of Rascher. On 1
February 1943 Sievers noted efforts in obtaining apparatus,
implements and chemicals for Rascher’s experiments. On 6 and 21

January 1944 Sievers noted the problem of location. Rascher
reported to Sievers periodically concerning the status and details of
the freezing experiments.
It is plain from the record that the relationship of Sievers and
Rascher in the performance of freezing experiments required Sievers
to make the preliminary arrangements for the performance of the
experiments to familiarize himself with the progress of the
experiments by personal inspection, to furnish necessary equipment
and material, including human beings used during the freezing
experiments, to receive and make progress reports concerning
Rascher, and to handle the matter of evaluation and publication of
such reports. Basically, such activities constituted a performance of
his duties as defined by Sievers in his letter of 28 January 1943 to
Rudolf Brandt, in which he stated that he smoothed the way for
research workers and saw to it that Himmler’s orders were carried
out.
Under these facts Sievers is chargeable with the criminal
activities in these experiments.
MALARIA EXPERIMENTS
Details of these experiments are given elsewhere in this
judgment. These experiments were performed at Dachau by
Schilling and Ploetner. The evidence shows that Sievers had
knowledge of the nature and purpose of these criminal enterprises
and supported them in his official position.
LOST GAS EXPERIMENTS
These experiments were conducted in the Natzweiler
concentration camp under the supervision of Professor Hirt of the
University of Strasbourg. The Ahnenerbe Society and the defendant
Sievers supported this research on behalf of the SS. The
arrangement for the payment of the research subsidies of the
Ahnenerbe was made by Sievers. The defendant Sievers participated
in these experiments by actively collaborating with the defendants

Karl Brandt and Rudolf Brandt and with Hirt and his principal
assistant, Dr. Wimmer. The record shows that Sievers was in
correspondence with Hirt at least as early as January 1942, and that
he established contact between Himmler and Hirt.
In a letter of 11 September 1942 to Gluecks, Sievers wrote that
the necessary conditions existed in Natzweiler “for carrying out our
military scientific research work”. He requested that Gluecks issue
the necessary authorization for Hirt, Wimmer, and Kieselbach to
enter Natzweiler, and that provision be made for their board and
accommodations. The letter also stated:
“The experiments which are to be performed on
prisoners are to be carried out in four rooms of an already
existing medical barrack. Only slight changes in the
construction of the building are required, in particular the
installation of the hood which can be produced with very
little material. In accordance with attached plan of the
construction management at Natzweiler, I request that
necessary orders be issued to same to carry out the
reconstruction. All the expenses arising out of our activity
at Natzweiler will be covered by this office.”
In a memorandum of 3 November 1942 to the defendant Rudolf
Brandt, Sievers complained about certain difficulties which had
arisen in Natzweiler because of the lack of cooperation from the
camp officials. He seemed particularly outraged by the fact that the
camp officials were asking that the experimental prisoners be paid
for. A portion of the memorandum follows:
“When I think of our military research work conducted
at the concentration camp Dachau, I must praise and call
special attention to the generous and understanding way in
which our work was furthered there and to the cooperation
we were given. Payment of prisoners was never discussed.
It seems as if at Natzweiler they are trying to make as
much money as possible out of this matter. We are not
conducting these experiments, as a matter of fact, for the

sake of some fixed scientific idea, but to be of practical help
to the armed forces and beyond that, to the German people
in a possible emergency.”
Brandt was requested to give his help in a comradely fashion in
setting up the necessary conditions at Natzweiler. The defendant
Rudolf Brandt replied to this memorandum on 3 December 1942 and
told Sievers that he had had occasion to speak to Pohl concerning
these difficulties, and that they would be remedied.
The testimony of the witness Holl was that approximately 220
inmates of Russian, Polish, Czech, and German nationality were
experimented upon by Hirt and his collaborators, and that
approximately 50 died. None of the experimental subjects
volunteered. During the entire period of these experiments, Hirt was
associated with the Ahnenerbe Society.
In early 1944 Hirt and Wimmer summarized their findings from
the Lost experiments in a report entitled “Proposed Treatment of
Poisoning Caused by Lost.” The report was described as from the
Institute for Military Scientific Research, Department H of the
Ahnenerbe, located at the Strasbourg Anatomical Institute. Light,
medium, and heavy injuries due to Lost gas are mentioned. Sievers
received several copies of this report. On 31 March 1944, after Karl
Brandt had received a Fuehrer Decree giving him broad powers in
the field of chemical-warfare, Sievers informed Brandt about Hirt’s
work and gave him a copy of the report. This is proved by Sievers’
letter to Rudolf Brandt on 11 April 1944. Karl Brandt admitted that
the wording of the report made it clear that experiments had been
conducted on human beings.
Sievers testified that on 25 January 1943, he went to Natzweiler
concentration camp and consulted with the camp authorities
concerning the arrangements to be made for Hirt’s Lost experiments.
These arrangements included the obtaining of laboratories and
experimental subjects. Sievers testified that the Lost experiments
were harmful. On the visit of 25 January 1943, Sievers saw ten
persons who had been subjected to Lost experiments and watched
Hirt change the bandages on one of the persons. Sievers testified

that in March 1943 he asked Hirt whether any of the experimental
subjects had suffered harm from the experiments and was told by
Hirt that two of the experimental subjects had died due to other
causes.
It is evident that Sievers was criminally connected with these
experiments.
SEA-WATER EXPERIMENTS
These experiments were conducted at Dachau from July through
September 1944. Details of these experiments are explained
elsewhere in the judgment.
The function of the Ahnenerbe in the performance of sea-water
experiments conducted at Dachau from July through September
1944 was chiefly in connection with the furnishing of space and
equipment for the experiments. Sievers made these necessary
arrangements on behalf of the Ahnenerbe. As a result of Schroeder’s
request to Himmler through Grawitz for permission to perform the
sea-water experiments on inmates in Dachau, Himmler directed on 8
July 1944 that the experiments be made on gypsies and three other
persons with other racial qualities as control subjects. Sievers was
advised by Himmler’s office of the above authorization for
experiments at the Rascher station at Dachau.
On 27 June 1944, Rascher was replaced by Ploetner as head of
the Ahnenerbe Institute for Military Scientific Research at Dachau.
Sievers, on 20 July, went to Dachau and conferred with Ploetner of
the Ahnenerbe Institute and the defendant Beiglboeck, who was to
perform the experiments, concerning the execution of the sea-water
experiments and the availability of working space for them. Sievers
agreed to supply working space in Ploetner’s department and at the
Ahnenerbe Entomological Institute.
On 26 July 1944, Sievers made a written report to Grawitz
concerning details of his conference at Dachau. Sievers wrote that
40 experimental persons could be accommodated at “our” research
station, that the Ahnenerbe would supply a laboratory, and that Dr.
Ploetner would give his assistance, help, and advice to the Luftwaffe

physicians performing the experiments. Sievers also stated the
number and assignment of the personnel to be employed, estimating
that the work would cover a period of three weeks and designated
23 July 1944 as the date of commencement, provided that
experimental persons were available and the camp commander had
received the necessary order from Himmler. In conclusion, Sievers
expressed his hope that the arrangements which he had made
would permit a successful conduct of the experiments and requested
that acknowledgment be made to Himmler as a participant in the
experiments.
In his testimony Sievers admitted that he had written the above
letter and had conferred with Beiglboeck at Dachau. As the letter
indicates, Sievers knew that concentration camp inmates were to be
used.
Sievers had knowledge of and criminally participated in sea-water
experiments.
TYPHUS EXPERIMENTS
Detailed description of these experiments is contained elsewhere
in this judgment. Sievers participated in the criminal typhus
experiments conducted by Haagen on concentration camp inmates
at Natzweiler by making the necessary arrangements in connection
with securing experimental subjects, handling administrative
problems incident to the experiments, and by furnishing the
Ahnenerbe station with its equipment in Natzweiler for their
performance.
On 16 August 1943, when Haagen was preparing to transfer his
typhus experiments from Schirmeck to Natzweiler, he requested
Sievers to make available a hundred concentration camp inmates for
his research. This is seen from a letter of 30 September 1943 from
Sievers to Haagen in which he states that he will be glad to assist,
and that he is accordingly contacting the proper source to have the
“desired personnel” placed at Haagen’s disposal. As a result of
Sievers’ efforts, a hundred inmates were shipped from Auschwitz to
Natzweiler for Haagen’s experiments. These were found to be unfit

for experimentation because of their pitiful physical condition. A
second group of one hundred was then made available. Some of
these were used by Haagen as experimental subjects.
That the experiments were carried out in the Ahnenerbe
experimental station in Natzweiler is proved by excerpts from
monthly reports of the camp doctor in Natzweiler. A number of
deaths occurred among non-German experimental subjects as a
direct result of the treatment to which they were subjected.
POLYGAL EXPERIMENTS
Evidence has been introduced during the course of the trial to
show that experiments to test the efficacy of a blood coagulant
“polygal” were conducted on Dachau inmates by Rascher. The
Sievers’ diary shows that the defendant had knowledge of activities
concerning the production of polygal, and that he lent his support to
the conduct of the experiments.
JEWISH SKELETON COLLECTION
Sievers is charged under the indictment with participation in the
killing of 112 Jews who were selected to complete a skeleton
collection for the Reich University of Strasbourg.
Responding to a request by the defendant Rudolf Brandt, Sievers
submitted to him on 9 February 1942 a report by Dr. Hirt of the
University of Strasbourg on the desirability of securing a Jewish
skeleton collection. In this report, Hirt advocated outright murder of
“Jewish Bolshevik Commissars” for the procurement of such a
collection. On 27 February 1942, Rudolf Brandt informed Sievers that
Himmler would support Hirt’s work and would place everything
necessary at his disposal. Brandt asked Sievers to inform Hirt
accordingly and to report again on the subject. On 2 November 1942
Sievers requested Brandt to make the necessary arrangements with
the Reich Main Security Office for providing 150 Jewish inmates from
Auschwitz to carry out this plan. On 6 November, Brandt informed
Adolf Eichmann, the Chief of Office IV B/4 (Jewish Affairs) of the

Reich Main Security Office to put everything at Hirt’s disposal which
was necessary for the completion of the skeleton collection.
From Sievers’ letter to Eichmann of 21 June 1943, it is apparent
that SS Hauptsturmfuehrer Beger, a collaborator of the Ahnenerbe
Society, carried out the preliminary work for the assembling of the
skeleton collection in the Auschwitz concentration camp on 79 Jews,
30 Jewesses, 2 Poles, and 4 Asiatics. The corpses of the victims
were sent in three shipments to the Anatomical Institute of Hirt in
the Strasbourg University.
When the Allied Armies were threatening to overrun Strasbourg
early in September 1944, Sievers dispatched to Rudolf Brandt the
following teletype message:
“Subject: Collection of Jewish Skeletons.
“In conformity with the proposal of 9 February 1942 and
with the consent of 23 February 1942 * * * SS
Sturmbannfuehrer Professor Hirt planned the hitherto
missing collection of skeletons. Due to the extent of the
scientific work connected herewith, the preparation of the
skeletons is not yet concluded. Hirt asks with respect to the
time needed for 80 specimens, and in case the endangering
of Strasbourg has to be reckoned with, how to proceed with
the collection situated in the dissecting room of the
anatomical institute. He is able to carry out the maceration
and thus render them irrecognizable. Then, however, part
of the entire work would have been partly done in vain, and
it would be a great scientific loss for this unique collection,
because hominit casts could not be made afterwards. The
skeleton collection as such is not conspicuous. Viscera could
be declared as remnants of corpses, apparently left in the
anatomical institute by the French and ordered to be
cremated. Decision on the following proposals is requested:
“1. Collection can be preserved.
“2. Collection is to be partly dissolved.
“3. Entire collection is to be dissolved.

“Sievers”
The pictures of the corpses and the dissecting rooms of the
Institute, taken by the French authorities after the liberation of
Strasbourg, point up the grim story of these deliberate murders to
which Sievers was a party.
Sievers knew from the first moment he received Hirt’s report of 9
February 1942 that mass murder was planned for the procurement
of the skeleton collection. Nevertheless he actively collaborated in
the project, sent an employee of the Ahnenerbe to make the
preparatory selections in the concentration camp at Auschwitz, and
provided for the transfer of the victims from Auschwitz to Natzweiler.
He made arrangements that the collection be destroyed.
Sievers’ guilt under this specification is shown without question.
Sievers offers two purported defenses to the charges against him
(1) that he acted pursuant to superior orders; (2) that he was a
member of a resistance movement.
The first defense is wholly without merit. There is nothing to
show that in the commission of these ghastly crimes, Sievers acted
entirely pursuant to orders. True, the basic policies or projects which
he carried through were decided upon by his superiors, but in the
execution of the details Sievers had an unlimited power of discretion.
The defendant says that in his position he could not have refused an
assignment. The fact remains that the record shows the case of
several men who did, and who have lived to tell about it.
Sievers’ second matter of defense is equally untenable. In
support of the defense, Sievers offered evidence by which he hoped
to prove that as early as 1933 he became a member of a secret
resistance movement which plotted to overthrow the Nazi
Government and to assassinate Hitler and Himmler; that as a leading
member of the group, Sievers obtained the appointment as Reich
Business Manager of the Ahnenerbe so that he could be close to
Himmler and observe his movements; that in this position he
became enmeshed in the revolting crimes, the subject matter of this
indictment; that he remained as business manager upon advice of
his resistance leader to gain vital information which would hasten

the day of the overthrow of the Nazi Government and the liberation
of the helpless peoples coming under its domination.
Assuming all these things to be true, we cannot see how they
may be used as a defense for Sievers. The fact remains that murders
were committed with cooperation of the Ahnenerbe upon countless
thousands of wretched concentration camp inmates who had not the
slightest means of resistance. Sievers directed the program by which
these murders were committed.
It certainly is not the law that a resistance worker can commit no
crime, and least of all, against the very people he is supposed to be
protecting.
MEMBERSHIP IN A CRIMINAL ORGANIZATION
Under count four of the indictment, Wolfram Sievers is charged
with being a member of an organization declared criminal by the
judgment of the International Military Tribunal, namely, the SS. The
evidence shows that Wolfram Sievers became a member of the SS in
1935 and remained a member of that organization to the end of the
war. As a member of the SS he was criminally implicated in the
commission of war crimes and crimes against humanity, as charged
under counts two and three of the indictment.
CONCLUSION
Military Tribunal I finds and adjudges the defendant Wolfram
Sievers guilty under counts two, three and four of the indictment.
ROSE
The defendant Rose is charged under counts two and three of
the indictment with special responsibility for, and participation in
Typhus and Epidemic Jaundice Experiments.
The latter charge has been abandoned by the prosecution.
Evidence was offered concerning Rose’s criminal participation in
malaria experiments at Dachau, although he was not named in the

indictment as one of the defendants particularly charged with
criminal responsibility in connection with malaria experiments.
Questions presented by this situation will be discussed later.
The defendant Rose is a physician of large experience, for many
years recognized as an expert in tropical diseases. He studied
medicine at the Universities of Berlin and Breslau and was admitted
to practice in the fall of 1921. After serving as interne in several
medical institutes, he received an appointment on the staff of the
Robert Koch Institute in Berlin. Later he served on the staff of
Heidelberg University and for three years engaged in the private
practice of medicine in Heidelberg. In 1929 he went to China, where
he remained until 1936, occupying important positions as medical
adviser to the Chinese Government. In 1936 he returned to Germany
and became head of the Department for Tropical Medicine at the
Robert Koch Institute in Berlin. Late in August 1939 he joined the
Luftwaffe with the rank of first lieutenant in the Medical Corps. In
that service he was commissioned brigadier general in the reserve
and continued on active duty until the end of the war. He was
consultant on hygiene and tropical medicine to the Chief of the
Medical Service of the Luftwaffe. From 1944 he was also consultant
on the staff of defendant Handloser and was medical adviser to Dr.
Conti in matters pertaining to tropical diseases. During the war Rose
devoted practically all of his time to his duties as consultant to the
Chief of the Medical Service of the Luftwaffe, Hippke, and after 1
January 1944, the defendant Schroeder.
MALARIA EXPERIMENTS
Medical experiments in connection with malaria were carried on
at Dachau concentration camp from February 1942 until the end of
the war. These experiments were conducted under Dr. Klaus Schilling
for the purpose of discovering a method of establishing immunity
against malaria. During the course of the experiments probably as
many as 1,000 inmates of the concentration camp were used as
subjects of the experiments. Very many of these persons were
nationals of countries other than Germany who did not volunteer for

the experiments. By credible evidence it is established that
approximately 30 of the experimental subjects died as a direct result
of the experiments and that many more succumbed from causes
directly following the experiments, including non-German nationals.
With reference to Rose’s participation in these experiments, the
record shows the following: The defendant Rose had been
acquainted with Schilling for a number of years, having been his
successor in a position once held by Schilling in the Robert Koch
Institute. Under date 3 February 1941, Rose, writing to Schilling,
then in Italy, referred to a letter received from Schilling, in which the
latter requested “malaria spleens” (spleens taken from the bodies of
persons who had died from malaria). Rose in reply asked for
information concerning the exact nature of the material desired.
Schilling wrote 4 April 1942 from Dachau to Rose at Berlin, stating
that he had inoculated a person intracutaneously with sporocoides
from the salivary glands of a female anopheles which Rose had sent
him. The letter continues:
“For the second inoculation I miss the sporocoides
material because I do not possess the ‘Strain Rose’ in the
anopheles yet. If you could find it possible to send me in
the next days a few anopheles infected with ‘Strain Rose’
(with the last consignment two out of ten mosquitoes were
infected) I would have the possibility to continue this
experiment and I would naturally be very thankful to you
for this new support of my work.
“The mosquito breeding and the experiments proceed
satisfactorily and I am working now on six tertiary strains.”
The letter bears the handwritten endorsement “finished 17 April
1942. L. g. RO 17/4,” which evidence clearly reveals that Rose had
complied with Schilling’s request for material.
Schilling again wrote Rose from Dachau malaria station 5 July
1943, thanking Rose for his letter and “the consignment of
atroparvus eggs.” The letter continues:

“Five percent of them brought on water went down and
were therefore unfit for development; the rest of them
hatched almost 100 percent.
“Thanks to your solicitude, achieved again the
completion of my breed.
“Despite this fact I accept with great pleasure your offer
to send me your excess of eggs. How did you dispatch this
consignment? The result could not have been any better!
“Please tell Fraeulein Lange, who apparently takes care
of her breed with greater skill and better success than the
prisoner August, my best thanks for her trouble.
“Again my sincere thanks to you!”
The “prisoner August” mentioned in the letter was doubtless the
witness August Vieweg, who testified before this Tribunal concerning
the malaria experiments.
Rose wrote Schilling 27 July 1943 in answer to the latter’s letter
of 5 July 1943, stating he was glad the shipment of eggs had arrived
in good order and had proved useful. He also gave the information
that another shipment of anopheles eggs would follow.
In the fall of 1942 Rose was present at the “Cold Conference”
held at Nuernberg and heard Holzloehner deliver his lecture on the
freezing experiments which had taken place at Dachau. Rose
testified that after the conference he talked with Holzloehner, who
told him that the carrying out of physiological experiments on
human beings imposed upon him a tremendous mental burden,
adding that he hoped he never would receive another order to
conduct such experiments.
It is impossible to believe that during the years 1942 and 1943
Rose was unaware of malaria experiments on human beings which
were progressing at Dachau under Schilling, or to credit Rose with
innocence of knowledge that the malaria research was not confined
solely to vaccinations designed for the purpose of immunizing the
persons vaccinated. On the contrary, it is clear that Rose well knew
that human beings were being used in the concentration camp as
subjects for medical experimentation.

However, no adjudication either of guilt or innocence will be
entered against Rose for criminal participation in these experiments
for the following reason: In preparing counts two and three of its
indictment the prosecution elected to frame its pleading in such a
manner as to charge all defendants with the commission of war
crimes and crimes against humanity, generally, and at the same time
to name in each sub-paragraph dealing with medical experiments
only those defendants particularly charged with responsibility for
each particular item.
In our view this constituted, in effect, a bill of particulars and
was, in essence, a declaration to the defendants upon which they
were entitled to rely in preparing their defenses, that only such
persons as were actually named in the designated experiments
would be called upon to defend against the specific items. Included
in the list of names of those defendants specifically charged with
responsibility for the malaria experiments the name of Rose does not
appear. We think it would be manifestly unfair to the defendant to
find him guilty of an offense with which the indictment affirmatively
indicated he was not charged.
This does not mean that the evidence adduced by the
prosecution was inadmissible against the charges actually preferred
against Rose. We think it had probative value as proof of the fact of
Rose’s knowledge of human experimentation upon concentration
camp inmates.
TYPHUS EXPERIMENTS
These experiments were carried out at Buchenwald and
Natzweiler concentration camps, over a period extending from 1942
to 1945, in an attempt to procure a protective typhus vaccine.
In the experimental block at Buchenwald, with Dr. Ding in
charge, inmates of the camp were infected with typhus for the
purpose of procuring a continuing supply of fresh blood taken from
persons suffering from typhus. Other inmates, some previously
immunized and some not, were infected with typhus to demonstrate

the efficacy of the vaccines. Full particulars of these experiments
have been given elsewhere in the judgment.
Rose visited Buchenwald in company with Gildemeister of the
Robert Koch Institute in the spring of 1942. At this time Dr. Ding was
absent, suffering from typhus as the result of an accidental infection
received while infecting his experimental subjects. Rose inspected
the experimental block where he saw many persons suffering from
typhus. He passed through the wards and looked at the clinical
records “of * * * persons with severe cases in the control cases and
* * * lighter cases among those vaccinated.”
The Ding diary, under dates 19 August-4 September 1942,
referring to use of vaccines for immunization, states that 20 persons
were inoculated with vaccine from Bucharest, with a note “this
vaccine was made available by Professor Rose, who received it from
Navy Doctor Professor Ruegge from Bucharest.” Rose denied that he
had ever sent vaccine to Mrugowsky or Ding for use at Buchenwald.
Mrugowsky, from Berlin, under date 16 May 1942, wrote Rose as
follows:
“Dear Professor:
“The Reich Physician SS and Police has consented to the
execution of experiments to test typhus vaccines. May I
therefore ask you to let me have the vaccines.
“The other question which you raised, as to whether the
louse can be infected by a vaccinated typhus patient, will
also be dealt with. In principle, this also has been
approved. There are, however, still some difficulties at the
moment about the practical execution, since we have at
present no facilities for breeding lice.
“Your suggestion to use Olzscha has been passed on to
the personnel department of the SS medical office. It will
be given consideration in due course.”
From a note on the letter, it appears that Rose was absent from
Berlin and was not expected to return until June. The letter,
however, refers to previous contact with Rose and to some

suggestions made by him which evidently concern medical
experiments on human beings. Rose in effect admitted that he had
forwarded the Bucharest vaccine to be tested at Buchenwald.
At a meeting of consulting physicians of the Wehrmacht held in
May 1943, Ding made a report in which he described the typhus
experiments he had been performing at Buchenwald. Rose heard the
report at the meeting and then and there objected strongly to the
methods used by Ding in conducting the experiments. As may well
be imagined, this protest created considerable discussion among
those present.
The Ding diary shows that, subsequent to this meeting,
experiments were conducted at Buchenwald at the instigation of the
defendant Rose. The entry under date of 8 March 1944, which refers
to “typhus vaccine experimental series VIII”, appears as follows:
“Suggested by Colonel M. C. of the Air Corps, Professor
Rose (Oberstarzt), the vaccine ‘Kopenhagen’ (Ipsen-Murine-
vaccine) produced from mouse liver by the National Serum
Institute in Copenhagen was tested for its compatibility on
humans. 20 persons were vaccinated for immunization by
intramuscular injection * * *. 10 persons were
contemplated for control and comparison. 4 of the 30
persons were eliminated before the start of the artificial
injection because of intermittent sickness * * *. The
remaining experimental persons were infected on 16 April
44 by subcutaneous injection of 1/20 cc. typhus sick fresh
blood * * *. The following fell sick: 17 persons immunized:
9 medium, 8 seriously; 9 persons control: 2 medium, 7
seriously * * *. 2 June 44: The experimental series was
concluded 13 June 44: Chart and case history completed
and sent to Berlin. 6 deaths (3 Copenhagen) (3 control). Dr.
Ding.”
When on the witness stand Rose vigorously challenged the
correctness of this entry in the Ding diary and flatly denied that he
had sent a Copenhagen vaccine to Mrugowsky or Ding for use at
Buchenwald. The prosecution met this challenge by offering in

evidence a letter from Rose to Mrugowsky dated 2 December 1943,
in which Rose stated that he had at his disposal a number of
samples of a new murine virus typhus vaccine prepared from mice
livers, which in animal experiments had been much more effective
than the vaccine prepared from the lungs of mice. The letter
continued:
“To decide whether this first-rate murine vaccine should
be used for protective vaccination of human beings against
lice typhus, it would be desirable to know if this vaccine
showed in your and Ding’s experimental arrangement at
Buchenwald an effect similar to that of the classic virus
vaccines.
“Would you be able to have such an experimental series
carried out? Unfortunately I could not reach you over the
phone. Considering the slowness of postal communications
I would be grateful for an answer by telephone * * *.”
The letter shows on its face that it was forwarded by Mrugowsky to
Ding, who noted its receipt by him 21 February 1944.
On cross-examination, when Rose was confronted with the letter
he admitted its authorship, and that he had asked that experiments
be carried out by Mrugowsky and Ding at Buchenwald.
The fact that Rose contributed actively and materially to the
Mrugowsky-Ding experiments at Buchenwald clearly appears from
the evidence.
The evidence also shows that Rose actively collaborated in the
typhus experiments carried out by Haagen at the Natzweiler
concentration camp for the benefit of the Luftwaffe.
From the exhibits in the record, it appears that Rose and Haagen
corresponded during the month of June 1943 concerning the
production of a vaccine for typhus. Under date 5 June 1943 Haagen
wrote to Rose amplifying a telephone conversation between the two
and referring to a letter from a certain Giroud with reference to a
vaccine which had been used on rabbits. A few days later Rose
replied, thanking him for his letters of 4 and 5 June and for “the
prompt execution of my request.” The record makes it plain that by

use of the phrase “the prompt execution of my request” was meant
a request made by Rose to the Chief of the Medical Service of the
Wehrmacht for an order to produce typhus vaccine to be used by the
armed forces in the eastern area.
Under date 4 October 1943 Haagen again wrote Rose concerning
his plans for vaccine production, making reference in the letter to a
report made by Rose on the Ipsen vaccine. Haagen stated that he
had already reported to Rose on the results of experiments with
human beings and expressed his regret that, up to the date of the
letter, he had been unable to “perform infection experiments on the
vaccinated persons.” He also stated that he had requested the
Ahnenerbe to provide suitable persons for vaccination but had
received no answer; that he was then vaccinating other human
beings and would report results later. He concluded by expressing
the wish and need for experimental subjects upon whom to test
vaccinations, and suggested that when subjects were procured,
parallel tests should be made between the vaccine referred to in the
letter and the Ipsen tests.
We think the only reasonable inference which can be drawn from
this letter is that Haagen was proposing to test the efficacy of the
vaccinations which he had completed, which could only be
accomplished by infecting the vaccinated subjects with a virulent
pathogenic virus.
In a letter written by Rose and dated “in the field, 29 September
1943”, directed to the Behring Works at Marburg/Lahn, Rose states
that he is enclosing a memorandum regarding reports by Dr. Ipsen
on his experience in the production of typhus vaccine. Copy of the
report which Rose enclosed is in evidence, Rose stating therein that
he had proposed, and Ipsen had promised, that a number of Ipsen’s
liver vaccine samples should be sent to Rose with the object of
testing its protective efficacy on human beings whose lives were in
special danger. Copies of this report were forwarded by Rose to
several institutions, including that presided over by Haagen.
In November 1943, 100 prisoners were transported to Natzweiler,
of whom 18 had died during the journey. The remainder were in
such poor health that Haagen found them worthless for his

experiments and requested additional healthy prisoners through Dr.
Hirt, who was a member of the Ahnenerbe.
Rose wrote to Haagen 13 December 1943, saying among other
things “I request that in procuring persons for vaccination in your
experiment, you request a corresponding number of persons for
vaccination with Copenhagen vaccine. This has the advantage, as
also appeared in the Buchenwald experiments, that the test of
various vaccines simultaneously gives a clearer idea of their value
than the test of one vaccine alone.”
There is much other evidence connecting Rose with the series of
experiments conducted by Haagen but we shall not burden the
judgment further. It will be sufficient to say that the evidence proves
conclusively that Rose was directly connected with the criminal
experiments conducted by Haagen.
Doubtless at the outset of the experimental program launched in
the concentration camps, Rose may have voiced some vigorous
opposition. In the end, however, he overcame what scruples he had
and knowingly took an active and consenting part in the program.
He attempts to justify his actions on the ground that a state may
validly order experiments to be carried out on persons condemned to
death without regard to the fact that such persons may refuse to
consent to submit themselves as experimental subjects. This defense
entirely misses the point of the dominant issue. As we have pointed
out in the case of Gebhardt, whatever may be the condition of the
law with reference to medical experiments conducted by or through
a state upon its own citizens, such a thing will not be sanctioned in
international law when practiced upon citizens or subjects of an
occupied territory.
We have indulged every presumption in favor of the defendant,
but his position lacks substance in the face of the overwhelming
evidence against him. His own consciousness of turpitude is clearly
disclosed by the statement made by him at the close of a vigorous
cross-examination in the following language:
“It was known to me that such experiments had earlier
been carried out, although I basically objected to these

experiments. This institution had been set up in Germany
and was approved by the state and covered by the state. At
that moment I was in a position which perhaps corresponds
to a lawyer who is, perhaps, a basic opponent of execution
or death sentence. On occasion when he is dealing with
leading members of the government, or with lawyers during
public congresses or meetings, he will do everything in his
power to maintain his opinion on the subject and have it
put into effect. If, however, he does not succeed, he stays
in his profession and in his environment in spite of this.
Under circumstances he may perhaps even be forced to
pronounce such a death sentence himself, although he is
basically an opponent of that set-up.”
The Tribunal finds that the defendant Rose was a principal in,
accessory to, ordered, abetted, took a consenting part in, and was
connected with plans and enterprises involving medical experiments
on non-German nationals without their consent, in the course of
which murders, brutalities, cruelties, tortures, atrocities, and other
inhuman acts were committed. To the extent that these crimes were
not war crimes they were crimes against humanity.
CONCLUSION
Military Tribunal I finds and adjudges the defendant Gerhard
Rose guilty under counts two and three of the indictment.
RUFF, ROMBERG, AND WELTZ
The defendants Ruff, Romberg, and Weltz are charged under
counts two and three of the indictment with special responsibility for,
and participation in, High-Altitude Experiments.
The defendant Weltz is also charged under counts two and three
with special responsibility for, and participation in, Freezing
Experiments.

To the extent that the evidence in the record relates to the high-
altitude experiments, the cases of the three defendants will be
considered together.
Defendant Ruff specialized in the field of aviation medicine from
the completion of his medical education at Berlin and Bonn in 1932.
In January 1934 he was assigned to the German Experimental
Institute for Aviation, a civilian agency, in order to establish a
department for aviation medicine. Later he became chief of the
department.
Defendant Romberg joined the NSDAP in May 1933. From April
1936 until 1938 he interned as an assistant physician at a Berlin
hospital. On 1 January 1938 he joined the staff of the German
Experimental Institution for Aviation as an associate assistant to the
defendant Ruff. He remained as a subordinate to Ruff until the end
of the war.
Defendant Weltz for many years was a specialist in X-ray work. In
the year 1935 he received an assignment as lecturer in the field of
aviation medicine at the University of Munich. At the same time he
instituted a small experimental department at the Physiological
Institute of the University of Munich. Weltz lectured at the University
until 1945; at the same time he did research work at the Institute.
In the summer of 1941 the experimental department at the
Physiological Institute, University of Munich, was taken over by the
Luftwaffe and renamed the “Institute for Aviation Medicine in
Munich.” Weltz was commissioned director of this Institute by
Hippke, then Chief of the Medical Inspectorate of the Luftwaffe. In
his capacity as director of this Institute, Weltz was subordinated to
Luftgau No. VII in Munich for disciplinary purposes. In scientific
matters he was subordinated directly to Anthony, Chief of the
Department for Aviation Medicine in the Office of the Medical
Inspectorate of the Luftwaffe.
HIGH-ALTITUDE EXPERIMENTS
The evidence is overwhelming and not contradicted that
experiments involving the effect of low air pressure on living human

beings were conducted at Dachau from the latter part of February
through May 1942. In some of these experiments great numbers of
human subjects were killed under the most brutal and senseless
conditions. A certain Dr. Sigmund Rascher, Luftwaffe officer, was the
prime mover in the experiments which resulted in the deaths of the
subjects. The prosecution maintains that Ruff, Romberg, and Weltz
were criminally implicated in these experiments.
The guilt of the defendant Weltz is said to arise by reason of the
fact that, according to the prosecution’s theory, Weltz, as the
dominant figure proposed the experiments, arranged for their
conduct at Dachau, and brought the parties Ruff, Romberg, and
Rascher together. The guilt of Ruff and Romberg is charged by
reason of the fact that they are said to have collaborated with
Rascher in the conduct of the experiments. The evidence on the
details of the matter appears to be as follows:
In the late summer of 1941 soon after the Institute Weltz at
Munich was taken over by the Luftwaffe, Hippke, Chief of the
Medical Service of the Luftwaffe, approved, in principle, a research
assignment for Weltz in connection with the problem of rescue of
aviators at high altitudes. This required the use of human
experimental subjects. Weltz endeavored to secure volunteer
subjects for the research from various sources; however, he was
unsuccessful in his efforts.
Rascher, one of Himmler’s minor satellites, was at the time an
assistant at the Institute. He, Rascher, suggested the possibility of
securing Himmler’s consent to conducting the experiments at
Dachau. Weltz seized upon the suggestion, and thereafter
arrangements to that end were completed, Himmler giving his
consent for experiments to be conducted on concentration camp
inmates condemned to death, but only upon express condition that
Rascher be included as one of the collaborators in the research.
Rascher was not an expert in aviation medicine. Ruff was the
leading German scientist in this field, and Romberg was his principal
assistant. Weltz felt that before he could proceed with his research
these men should be persuaded to come into the undertaking. He
visited Ruff in Berlin and explained the proposition. Thereafter Ruff

and Romberg came to Munich, where a conference was held with
Weltz and Rascher to discuss the technical nature of the proposed
experiments.
According to the testimony of Weltz, Ruff, and Romberg, the
basic consideration which impelled them to agree to the use of
concentration camp inmates as subjects was the fact that the
inmates were to be criminals condemned to death who were to
receive some form of clemency in the event they survived the
experiments. Rascher, who was active in the conference, assured the
defendants that this also was one of the conditions under which
Himmler had authorized the use of camp inmates as experimental
subjects.
The decisions reached at the conference were then made known
to Hippke, who gave his approval to the institution of experiments at
Dachau and issued an order that a mobile low-pressure chamber
which was then in the possession of Ruff at the Department for
Aviation Medicine, Berlin, should be transferred to Dachau for use in
the project.
A second meeting was held at Dachau, attended by Ruff,
Romberg, Weltz, Rascher, and the camp commander, to make the
necessary arrangements for the conduct of the experiments. The
mobile low-pressure chamber was then brought to Dachau, and on
22 February 1942 the first series of experiments was instituted.
Weltz was Rascher’s superior; Romberg was subordinate to Ruff.
Rascher and Romberg were in personal charge of the conduct of the
experiments. There is no evidence to show that Weltz was ever
present at any of these experiments. Ruff visited Dachau one day
during the early part of the experiments, but thereafter remained in
Berlin and received information concerning the progress of the
experiments only through his subordinate, Romberg.
There is evidence from which it may reasonably be found that at
the outset of the program personal friction developed between Weltz
and his subordinate Rascher. The testimony of Weltz is that on
several occasions he asked Rascher for reports on the progress of
the experiments and each time Rascher told Weltz that nothing had
been started with reference to the research. Finally Weltz ordered

Rascher to make a report; whereupon Rascher showed his superior a
telegram from Himmler which stated, in substance, that the
experiments to be conducted by Rascher were to be treated as top
secret matter and that reports were to be given to none other than
Himmler. Because of this situation Weltz had Rascher transferred out
of his command to the DVL branch at Dachau. Defendant Romberg
stated that these experiments had been stopped soon after their
inception by the adjutant of the Reich War Ministry, because of
friction between Weltz and Rascher, and that the experiments were
resumed only after Rascher had been transferred out of Weltz
Institute.
While the evidence is convincingly plain that Weltz participated in
the initial arrangements for the experiments and brought all parties
together, it is not so clear that illegal experiments were planned or
carried out while Rascher was under Weltz command, or that he
knew that experiments which Rascher might conduct in the future
would be illegal and criminal.
There appear to have been two distinct groups of prisoners used
in the experimental series. One was a group of 10 to 15 inmates
known in the camp as “exhibition patients” or “permanent
experimental subjects”. Most, if not all, of these were German
nationals who were confined in the camp as criminal prisoners.
These men were housed together and were well-fed and reasonably
contented. None of them suffered death or injury as a result of the
experiments. The other group consisted of 150 to 200 subjects
picked at random from the camp and used in the experiments
without their permission. Some 70 or 80 of these were killed during
the course of the experiments.
The defendants Ruff and Romberg maintain that two separate
and distinct experimental series were carried on at Dachau; one
conducted by them with the use of the “exhibition subjects”, relating
to the problems of rescue at high altitudes, in which no injuries
occurred; the other conducted by Rascher on the large group of
nonvolunteers picked from the camp at random, to test the limits of
human endurance at extremely high altitudes, in which experimental
subjects in large numbers were killed.

The prosecution submits that no such fine distinction may be
drawn between the experiments said to have been conducted by
Ruff and Romberg, on the one hand, and Rascher on the other, or in
the prisoners who were used as the subjects of these experiments;
that Romberg—and Ruff as his superior—share equal guilt with
Rascher for all experiments in which deaths to the human subjects
resulted.
In support of this submission the members of the prosecution
cite the fact that Rascher was always present when Romberg was
engaged in work at the altitude chamber; that on at least three
occasions Romberg was at the chamber when deaths occurred to
the so-called Rascher subjects, yet elected to continue the
experiments. They point likewise to the fact that, in a secret
preliminary report made by Rascher to Himmler which tells of
deaths, Rascher mentions the name of Romberg as being a
collaborator in the research. Finally they point to the fact that, after
the experiments were concluded, Romberg was recommended by
Rascher and Sievers for the War Merit Cross, because of the work
done by him at Dachau.
The issue on the question of the guilt or innocence of these
defendants is close; we would be less than fair were we not to
concede this fact. It cannot be denied that there is much in the
record to create at least a grave suspicion that the defendants Ruff
and Romberg were implicated in criminal experiments at Dachau.
However, virtually all of the evidence which points in this direction is
circumstantial in its nature. On the other hand, it cannot be gainsaid
that there is a certain consistency, a certain logic, in the story told by
the defendants. And some of the story is corroborated in significant
particulars by evidence offered by the prosecution.
The value of circumstantial evidence depends upon the
conclusive nature and tendency of the circumstances relied on to
establish any controverted fact. The circumstances must not only be
consistent with guilt, but they must be inconsistent with innocence.
Such evidence is insufficient when, assuming all to be true which the
evidence tends to prove, some other reasonable hypothesis of
innocence may still be true; for it is the actual exclusion of every

other reasonable hypothesis but that of guilt which invests mere
circumstances with the force of proof. Therefore, before a court will
be warranted in finding a defendant guilty on circumstantial
evidence alone, the evidence must show such a well-connected and
unbroken chain of circumstances as to exclude all other reasonable
hypotheses but that of the guilt of the defendant. What
circumstances can amount to proof can never be a matter of general
definition. In the final analysis the legal test is whether the evidence
is sufficient to satisfy beyond a reasonable doubt the understanding
and conscience of those who, under their solemn oaths as officers,
must assume the responsibility for finding the facts.
On this particular specification, it is the conviction of the Tribunal
that the defendants Ruff, Romberg, and Weltz must be found not
guilty.
FREEZING EXPERIMENTS
In addition to the high-altitude experiments, the defendant Weltz
is charged with freezing experiments, likewise conducted at Dachau
for the benefit of the German Luftwaffe. These began at the camp at
the conclusion of the high-altitude experiments and were performed
by Holzloehner, Finke, and Rascher, all of whom were officers in the
medical services of the Luftwaffe. Non-German nationals were killed
in these experiments.
We think it quite probable that Weltz had knowledge of these
experiments, but the evidence is not sufficient to prove that he
participated in them.
CONCLUSION
Military Tribunal I finds and adjudges that the defendant
Siegfried Ruff is not guilty under either counts two or three of the
indictment, and directs that he be released from custody under the
indictment when this Tribunal presently adjourns; and
Military Tribunal I finds and adjudges that the defendant Hans
Wolfgang Romberg is not guilty under either counts two or three of

the indictment, and directs that he be released from custody under
the indictment when this Tribunal presently adjourns; and
Military Tribunal I finds and adjudges that the defendant Georg
August Weltz is not guilty under either counts two or three of the
indictment; and directs that he be released from custody under the
indictment when this Tribunal presently adjourns.
BRACK
The defendant Brack is charged under counts two and three of
the indictment with personal responsibility for, and participation in,
Sterilization Experiments and the Euthanasia Program of the German
Reich. Under count four the defendant is charged with membership
in an organization declared criminal by the judgment of the
International Military Tribunal, namely, the SS.
The defendant Brack enlisted in an artillery unit of an SA
regiment in 1923, and became a member of the NSDAP and the SS
in 1929. Throughout his career in the Party he was quite active in
high official circles. He entered upon full-time service in the Braune
Haus, the Nazi headquarters at Munich, in the summer of 1932. The
following year he was appointed to the Staff of Bouhler, business
manager of the NSDAP in Munich. When in 1934 Bouhler became
Chief of the Chancellery of the Fuehrer of the NSDAP, Brack was
transferred from the Braune Haus to Bouhler’s Berlin office. In 1936
Brack was placed in charge of office 2 (Amt 2) in the Chancellery of
the Fuehrer in Berlin, that office being charged with the
examinations of complaints received by the Fuehrer from all parts of
Germany. Later, he became Bouhler’s deputy in office 2. As such he
frequently journeyed to the different Gaue for the purpose of gaining
first-hand information concerning matters in which Bouhler was
interested.
Brack was promoted to the rank of Sturmbannfuehrer in the SS in
1935, and in April 1936 to the rank of Obersturmbannfuehrer. The
following September he became a Standartenfuehrer in the SS, and
was transferred to the staff of the Main Office of the SS in

November. In November 1940 he was promoted to the grade of
Oberfuehrer.
In 1942 Brack joined the Waffen SS, and during the late summer
of that year was ordered to active duty with a Waffen SS division. He
apparently remained on active duty until the close of the war.
STERILIZATION EXPERIMENTS
The persecution of the Jews had become a fixed Nazi policy very
soon after the outbreak of World War II. By 1941 that persecution
had reached the stage of the extermination of Jews, both in
Germany and in the occupied territories. This fact is confirmed by
Brack himself, who testified that he had been told by Himmler that
he, Himmler, had received a personal order to that effect from Hitler.
The record shows that the agencies organized for the so-called
euthanasia of incurables were used for this bloody pogrom. Later,
because of the urgent need for laborers in Germany, it was decided
not to kill Jews who were able to work but, as an alternative, to
sterilize them.
With this end in view Himmler instructed Brack to inquire of
physicians who were engaged in the Euthanasia Program about the
possibility of a method of sterilizing persons without the victim’s
knowledge. Brack worked on the assignment, with the result that in
March 1941 he forwarded to Himmler his signed report on the
results of experiments concerning the sterilization of human beings
by means of X-rays. In the report a method was suggested by which
sterilization with X-ray could be effected on groups of persons
without their being aware of the operation.
On 23 June 1942 Brack wrote the following letter to Himmler:
“Dear Reichsfuehrer:
“* * * Among 10 millions of Jews in Europe, there are, I
figure, at least 2-3 millions of men and women who are fit
enough to work. Considering the extraordinary difficulties
the labor problem presents us with I hold the view that
those 2-3 millions should be specially selected and

preserved. This can however only be done if at the same
time they are rendered incapable to propagate. About a
year ago I reported to you that agents of mine have
completed the experiments necessary for this purpose. I
would like to recall these facts once more. Sterilization, as
normally performed on persons with hereditary diseases is
here out of the question, because it takes too long and is
too expensive. Castration by X-ray however is not only
relatively cheap, but can also be performed on many
thousands in the shortest time. I think, that at this time it is
already irrelevant whether the people in question become
aware of having been castrated after some weeks or
months, once they feel the effects.
“Should you, Reichsfuehrer, decide to choose this way in
the interest of the preservation of labor, then Reichsleiter
Bouhler would be prepared to place all physicians and other
personnel needed for this work at your disposal. Likewise
he requested me to inform you that then I would have to
order the apparatus so urgently needed with the greatest
speed.
“Heil Hitler!
“Yours
“Viktor Brack.”
Brack testified from the witness stand that at the time he wrote
this letter he had every confidence that Germany would win the war.
Brack’s letter was answered by Himmler on 11 August 1942. In
the reply Himmler directed that sterilization by means of X-rays be
tried in at least one concentration camp in a series of experiments,
and that Brack place at his disposal expert physicians to conduct the
operation.
Blankenburg, Brack’s deputy, replied to Himmler’s letter and
stated that Brack had been transferred to an SS division, but that he,
Blankenburg, as Brack’s permanent deputy would “immediately take
the necessary measures and get in touch with the chiefs of the main
offices of the concentration camps.”

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