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MATH 275: Calculus III

Lecture Notes by Angel V. Kumchev

• Lecture 1. Three-Dimensional Coordinate Systems Preface iii
• Lecture 2. Vectors
• Lecture 3. Lines and Planes
• Lecture 4. Vector Functions
• Lecture 5. Space Curves
• Lecture 6. Multivariable Functions
• Lecture 7. Surfaces
• Lecture 8. Partial Derivatives
• Lecture 9. Directional Derivatives and Gradients
• Lecture 10. Tangent Planes and Normal Vectors
• Lecture 11. Extremal Values of Multivariable Functions
• Lecture 12. Lagrange Multipliers*
• Lecture 13. Double Integrals over Rectangles
• Lecture 14. Double Integrals over General Regions
• Lecture 15. Double Integrals in Polar Coordinates
• Lecture 16. Triple Integrals
• Lecture 17. Triple Integrals in Cylindrical and Spherical Coordinates
• Lecture 18. Applications of Double and Triple Integrals
• Lecture 19. Change of Variables in Multiple Integrals*
• Lecture 20. Vector Fields
• Lecture 21. Line Integrals
• Lecture 22. The Fundamental Theorem for Line Integrals
• Lecture 23. Green’s Theorem
• Lecture 24. Surface Integrals
• Lecture 25. Stokes’ Theorem
• Lecture 26. Gauss’ Divergence Theorem
• Appendix A. Determinants
• Appendix B. Definite Integrals of Single-Variable Functions
• Appendix C. Polar Coordinates
• Appendix D. Introduction to Mathematica
• Appendix E. Answers to Selected Exercises

Preface

These lecture notes originated from a set of class notes that I used to suplement my lectures for M408D Sequences, Series, and Multivariable Calculus at UT-Austin, and later for MATH 275 here, at TU. About 2006, the TU mathematics department set a goal to include in MATH 275 some topics that had previously been excluded “for lack of class time.” I found that difficult to achieve within the constraints imposed by the organization of the standard calculus text that we were using at the time, yet quite manageable if I moved some material around and kept the applications more focused on geometry. Those changes were significant enough to diminish the value of the textbook as a primary reference for the course—it became a (rather expensive) source of homework assign- ments and pretty pictures. That motivated me to add some exercises and illustrations to the class notes and to turn them in a self-contained resource—these lecture notes. I have not written these notes in vacuum, so there are several people whose help and suggestions I gratefully acknowledge. The earliest version of my UT-Austin class notes benefited greatly from numerous exchanges with Michael Tehranchi and Vrej Zarikian, and my good friend and colleague Alexei Kolesnikov helped me with the addition of some new material to the current version. My colleagues John Chollet and Tatyana Sorokina have used these notes in their own teaching and in the process have not only helped me find various typos, but also have provided me with insights regarding possible improvements—as have many of their and my own former students. The present version of these notes differs from earlier versions in two main ways. First, I added a couple of new topics: the method of Lagrange multipliers (Lecture #12) and the general theory of change of variables in multiple integrals (Lecture #19). While those topics are not officially a part of the MATH 275 curriculum, I was convinced over time that they are too important not to be mentioned. The second major addition to this version is the inclusion of more visual illustrations and of several Mathematica tutorials. Among the latter, I want to mention specifically Appendix D, which is a version of the introduction to Mathematica that is part of the computer laboratories used in calculus courses at TU. To that end, I thank my colleagues Raouf Boules, Geoffrey Goodson, Ohoe Kim and Michael O’Leary for allowing me to use their work.

iii

LECTURE 1

Three-Dimensional Coordinate Systems

You are familiar with coordinate systems in the plane from single-variable calculus. The first coordinate system in the plane that you have seen is most likely the Cartesian system, which consists of two perpendicular axes. These axes are usually positioned so that one of them, called the x-axis, is horizontal and oriented from left to right and the other, called the y-axis, is vertical and oriented upward (see Figure 1.1). The intersection point of the two axes is called the origin and is commonly denoted by O.

x

y

O

Figure 1.1. Two-dimensional Cartesian coordinate system

1.1. Cartesian coordinates in space The three-dimensional Cartesian coordinate system consists of a point O in space, called the origin, and of three mutually perpendicular axes through O, called the coordinate axes and labeled the x-axis, the y-axis, and the z-axis. The labeling of the axes is usually done according to the right-hand rule: if one places one’s right hand in space so that the origin lies on one’s palm, the y-axis points toward one’s arm, and the x-axis points toward one’s curled fingers, then the z-axis points toward one’s thumb. There are several standards for drawing two-dimensional models of three-dimensional objects. In particular, there are three somewhat standard ways to draw a three-dimensional coordinate sys- tem. These are shown on Figure 1.2. Model (c) is the one commonly used in manual sketches. Model (a) is a version of model (c) that is commonly used in textbooks and will be the one we shall use in these notes; unlike model (c), it is based on real mathematics that transforms three- dimensional coordinates to two-dimensional ones. Model (b) is the default model preferred by computer algebra systems such as Mathematica. The coordinate axes determine several notable objects in space. First, we have the coordinate planes: these are the unique planes determined by the three different pairs of coordinate axes. The coordinate planes determined by the x- and y-axes, by the x- and z-axes, and by the y- and z-axes are called the xy-plane, the xz-plane, and the yz-plane, respectively. The three coordinate planes divide space into eight “equal” parts, called octants; the octants are similar to the four quadrants

y

z

x (a)

x

z

y

(b)

y

z

x (c)

Figure 1.2. Three-dimensional Cartesian coordinate systems

that the two coordinate axes divide the xy-plane into. The octants are labeled I through VIII, so that octants I through IV lie above the respective quadrants in the xy-plane and octants V through VIII lie below those quadrants. Next, we define the Cartesian coordinates of a point P in space. We start by finding its orthog- onal projections onto the yz-, the xz-, and the xy-planes. Let us denote those points by Px, Py, and Pz, respectively (see Figure 1.3). Let a be the directed distance from Px to P: that is,

a =

|PPx| if the direction from Px to P is the same as that of the x-axis, −|PPx| if the direction from Px to P is opposite to that of the x-axis.

Further, let b be the directed distance from Py to P, and let c be the directed distance from Pz to P. The three-dimensional Cartesian coordinates of P are (a, b, c) and we write P(a, b, c).

Example 1.1. Plot the points A(4, 3 , 0), B(2, − 2 , 2), and C(2, 2 , −2).

Example 1.2. Determine the sets of points described by the equations: z = 1; x = −2; and y = 3 , x = −1.

Answer. The equation z = 1 represents a plane parallel to the xy-plane. The equation x = − 2 represents a plane parallel to the yz-plane. The equations y = 3 , x = −1 represent a vertical line perpendicular to the xy-plane and passing through the point (3, − 1 , 0). 

y

z

x (^) P z

Py^ Px P c

b a

Figure 1.3. Cartesian coordinates of a point in space

y

z

x

P(x, y, z)

P 0 (x, y)

z

θ^ r

Figure 1.4. Cylindrical coordinates of a point in space

by passing between Cartesian and polar coordinates in the xy-plane and keeping the z-coordinate unchanged. The formulas for conversion from cylindrical to Cartesian coordinates are

x = r cos θ, y = r sin θ, z = z, (1.3)

and those for conversion from Cartesian to cylindrical coordinates are

r =

x^2 + y^2 , θ = arctan(y/x), z = z. (1.4)

The formula θ = arctan(y/x) in (1.4) comes with a caveat: unless the point (x, y) lies in the first quadrant of the xy-plane, this formula yields a reference angle which needs to be adjusted to obtain the actual value of θ. (This is similar to the conversion from Cartesian to polar coordinates in the plane: see §C.2 in Appendix C).

Example 1.6. Find the Cartesian coordinates of the points P(2, π, e) and Q(4, arctan(−2), 2 .7). Solution. The Cartesian coordinates of P are (x, y, e), where x = 2 cos π = − 2 , y = 2 sin π = 0.

That is, P(− 2 , 0 , e). The Cartesian coordinates of Q are (x, y, 2 .7), where

x = 4 cos(arctan(−2)) =

, y = 4 sin(arctan(−2)) = −

That is, Q( √^45 , − √^85 , 2 .7). 

Example 1.7. Find the cylindrical coordinates of the points P(3, 4 , 5) and Q(− 2 , 2 , −1). Solution. The cylindrical coordinates of P are (r, θ, 5), where

r =

32 + 42 = 5 , tan θ =

and θ lies in the first quadrant. Hence, θ = arctan(^43 ) and the cylindrical coordinates of P are

(5, arctan(^43 ), 5). The cylindrical coordinates of Q are (r, θ, −1), where

r =

22 + (−2)^2 = 2

2 , tan θ = − 1 ,

and θ lies in the second quadrant. Hence, θ = 34 π and the cylindrical coordinates of P are

(

2 , 34 π , −1). 

z

y x

y

x (^12 , 0)

Figure 1.5. The cylinder x^2 + y^2 = x and its cross-section

Example 1.8. Find the Cartesian equation of the surface r = cos θ. What kind of surface is this?

Solution. We have r = cos θ ⇐⇒ r^2 = r cos θ ⇐⇒ x^2 + y^2 = x.

The last equation is the equation of a circle of radius 12 centered at the point (^12 , 0) in the xy-plane:

x^2 + y^2 = x ⇐⇒ x^2 − x + 14 + y^2 = 14 ⇐⇒ (x − 12 )^2 + y^2 = 14.

Here, however, we use the equation x^2 + y^2 = x to describe a surface in space, so we view it as an equation in x, y, z. Since z does not appear in the equation, if the x- and y-coordinates of a point satisfy the equation x^2 + y^2 = x, then the point is on the surface. In other words, if a^2 + b^2 = a, then all points (a, b, z), whose x-coordinate is a and y-coordinate is b, are on the surface. We also know that all the points (x, y, 0), with x^2 + y^2 = x, are on the surface: these are the points on the circle in the xy-plane with the same equation. Putting these two observations together, we conclude that the given surface is the cylinder that consists of all the vertical lines passing through the points of the circle x^2 + y^2 = x in the xy-plane (see Figure 1.5). 

1.4. Spherical coordinates We now introduce the so-called “spherical coordinates” in space. In order to define this set of coordinates, we choose a special point O, the origin, a plane that contains O, and a pair of mutually perpendicular axes passing through O—one that lies in the chosen plane and another that is orthogonal to the chosen plane. Since we want to relate the spherical coordinates of a point to its Cartesian coordinates, it is common to choose O to be the origin O(0, 0 , 0) of the Cartesian coordinate system, the special plane to be the xy-plane, and the two axes to be the x-axis and the z-axis. With these choices, the spherical coordinates of a point P(x, y, z) in space are (ρ, θ, φ), where ρ = |OP|, θ is the angular polar coordinate of the point P 0 (x, y) in the xy-plane, and φ is the angle between OP and the positive direction of the z-axis (see Figure 1.6); it is common to restrict φ to the range 0 ≤ φ ≤ π. The angles θ and φ are often called the longitude and the latitude of P, because with the right choice of the spherical coordinate system, the angles θ and 90◦^ − φ for a point on the surface of the Earth are its geographic longitude and latitude.

Hence, θ = 3 π/4 and the spherical coordinates of Q are (3, 3 π/ 4 , arccos(− 1 /3)). 

Example 1.11. Find the Cartesian equation of the surface ρ^2 (sin^2 φ − 2 cos^2 φ) = 1.

Solution. Since

x^2 + y^2 = ρ^2 sin^2 φ cos^2 θ + ρ^2 sin^2 φ sin^2 θ = ρ^2 sin^2 φ,

we have

ρ^2 sin^2 φ − 2 ρ^2 cos^2 φ = 1 ⇐⇒ x^2 + y^2 − 2 z^2 = 1.

The last equation represents a surface known as “one-sheet hyperboloid”; it is shown on Figure 1.7. 

y x

z

Figure 1.7. The hyperboloid x^2 + y^2 − 2 z^2 = 1

y

x

z

Figure 1.8. The region 2 ≤ ρ ≤ 3, π/ 2 ≤ φ ≤ π

Example 1.12. Describe the solid defined by the inequalities

2 ≤ ρ ≤ 3 , π/ 2 ≤ φ ≤ π.

Solution. The second inequality restricts the points of the solid to the lower half-space. The inequality ρ ≤ 3 describes the points in space that are within distance of 3 from the origin, that is, the interior of the sphere x^2 + y^2 + z^2 = 9. Similarly the inequality ρ ≥ 2 describes the exterior of the sphere x^2 + y^2 + z^2 = 4. Thus, the given solid is obtained from the lower half of the interior of the sphere x^2 + y^2 + z^2 = 9 by removing from it the interior of the smaller sphere x^2 + y^2 + z^2 = 4 (see Figure 1.8). 

Example 1.13. Find the cylindrical and spherical equations of the sphere x^2 + y^2 + z^2 + 4 z = 0.

Solution. By (1.4), x^2 + y^2 = r^2 , so the cylindrical form of the given equation is

x^2 + y^2 + z^2 + 4 z = 0 ⇐⇒ r^2 + z^2 + 4 z = 0.

By (1.6), x^2 + y^2 + z^2 = ρ^2 , so the spherical form of the given equation is

ρ^2 + 4 ρ cos φ = 0 ⇐⇒ ρ = −4 cos φ. 

Exercises

1.1. Which of the points P(1, 3 , 2), Q(0, − 2 , −1), and R(5, 3 , −3): is closest to the xy-plane; lies in the xz-plane; lies in the yz-plane; is farthest from the origin?

1.2. Describe the set in R^3 represented by the equation x + 2 y = 2.

1.3. Find the side lengths of the triangle with vertices A(0, 6 , 2), B(3, 4 , 1) and C(1, 3 , 4). Is this a special kind of a triangle: equilateral, isosceles, right?

1.4. Find an equation of the sphere that passes through P(4, 2 , 1) and has center C(1, 1 , 1).

1.5. Show that the equation x^2 + y^2 + z^2 = x + 2 y + 4 z represents a sphere and find its center and radius.

Describe verbally the set in R^3 represented by the given equation or inequality. 1.6. x = − 3 1.7. z ≤ 2

1.8. − 1 ≤ y ≤ 2 1.9. (x − 1)^2 + y^2 + z^2 ≥ 4

1.10. 4 < x^2 + y^2 + z^2 < 9 1.11. xy = 0

1.12. Write inequalities to describe the half-space consisting of all points to the right of the plane y = −2.

1.13. Write inequalities to describe the interior of the upper hemisphere of radius 2 centered at the origin.

Find the Cartesian coordinates of the point with the given cylindrical coordinates. 1.14. (2, 0 , 3) 1.15. (3, π/ 4 , −3) 1.16. (3, 7 π/ 6 , 2) 1.17. (3, 3 π/ 2 , −1)

Find the cylindrical coordinates of the point with the given Cartesian coordinates. 1.18. (1, − 1 , 4) 1.19. (− 2 , 2 , 2) 1.20. (

√ 3 , 1 , −2) 1.21. (3, 3 , 2)

Find the Cartesian coordinates of the point with the given spherical coordinates. 1.22. (1, 0 , 0) 1.23. (3, π/ 4 , 3 π/4) 1.24. (3, 7 π/ 6 , π/2) 1.25. (2, π/ 2 , π/4)

Find the spherical coordinates of the point with the given Cartesian coordinates. 1.26. (1, 1 , 2) 1.27. (− 1 , 1 , −

√

2. 1.28. (

√ 3 , 1 , 0) 1.29. (

√ 2 , 1 , −1)

Find the cylindrical and spherical equations of the surface represented by the given Cartesian equation. 1.30. x^2 + y^2 = 2 z 1.31. x^2 + 2 x + y^2 + z^2 = 1 1.32. z =

√ x^2 + y^2

Find the Cartesian equation of the surface represented by the given cylindrical or spherical equation. If possible, describe the surface verbally. 1.33. ρ cos φ = − 2 1.34. r = 2 sin θ

1.35. r^2 + z^2 = 4 1.36. ρ = 2 sin φ sin θ

1.37. ρ sin^2 φ − 2 cos φ = 0 1.38. r cos θ − 2 r sin θ = 3

Describe verbally the set in R^3 represented by the given conditions. 1.39. θ = π/ 4 1.40. ρ = 2

1.41. r = 1 , 0 ≤ z ≤ 1 1.42. 2 ≤ ρ ≤ 4 , 0 ≤ φ ≤ π/ 2

1.43. ρ ≤ 1 , −π/ 2 ≤ θ ≤ π/ 2 1.44. ρ sin φ ≤ 1 , π/ 2 ≤ φ ≤ π

2.3. Multiplication by scalars Let v be a vector and let c be a real number (called also a scalar). The scalar multiple cv is the vector whose length is |c| times the length of v and whose direction is the same as the direction of v when c > 0, or opposite to the direction of v when c < 0. If c = 0 or v = 0 , then cv = 0. The vector (−1)v has the same length as v but the opposite direction. We call this vector the negative (or opposite) of v and denote it by −v. For two vectors u and v, we define their difference u − v by

u − v = u + (−v). We visualize the sum and the difference of two vectors u and v by drawing them so that they share the same initial point. We then consider the parallelogram they define. The sum u + v is represented by the diagonal of the parallelogram that passes through the common initial point of u and v and the difference u − v is represented by the other diagonal (see Figure 2.2).

u

v

u^

+^

v

(a)

u

v

u (^) − (^) v

(b)

Figure 2.2. The sum and the difference of u and v

2.4. Coordinate representation of vectors. Algebraic vectors Often, it is convenient to introduce a coordinate system and to work with vectors algebraically

instead of geometrically. If we represent a (geometric) vector a by the arrow

OA, where O is the origin of a Cartesian coordinate system, the terminal point A of a has coordinates (a 1 , a 2 ) or (a 1 , a 2 , a 3 ), depending on the dimension. We call these coordinates the coordinates or the compo- nents of the algebraic vector a and write

a = 〈a 1 , a 2 〉 or a = 〈a 1 , a 2 , a 3 〉.

We will use the same algebraic vector to represent all the geometric vectors equal to

OA. For example, if A(1, −1), B(− 1 , 3), and C(0, 2), then

−−→ BC =

OA = 〈 1 , − 1 〉.

Thus, for any two points A(a 1 , a 2 , a 3 ) and B(b 1 , b 2 , b 3 ), the algebraic vector v representing the arrow −−→ AB is

v = 〈b 1 − a 1 , b 2 − a 2 , b 3 − a 3 〉.

From now on, we shall state all the formulas for three-dimensional vectors. One usually obtains the respective two-dimensional definition or result simply by dropping the last component of the vectors; any exceptions to this rule will be mentioned explicitly when they occur.

2.5. Operations with algebraic vectors Note that when we write a vector a algebraically, the distance formula provides a convenient tool for computing the length (also called the norm) ‖a‖ of that vector: if a = 〈a 1 , a 2 , a 3 〉, then

‖a‖ =

a^21 + a^22 + a^23

Furthermore, if a = 〈a 1 , a 2 , a 3 〉 and b = 〈b 1 , b 2 , b 3 〉 are vectors and c is a scalar, we have the following coordinate formulas for a + b, a − b, and ca:

a + b = 〈a 1 + b 1 , a 2 + b 2 , a 3 + b 3 〉, (2.2) a − b = 〈a 1 − b 1 , a 2 − b 2 , a 3 − b 3 〉, (2.3) ca = 〈ca 1 , ca 2 , ca 3 〉. (2.4)

In other words, we add vectors, subtract vectors, and multiply vectors by scalars by simply per- forming the respective operations componentwise.

Example 2.1. Consider the vectors a = 〈 1 , − 2 , 3 〉 and b = 〈 0 , 1 , − 2 〉. Then

2 a + b = 〈 2 , − 4 , 6 〉 + 〈 0 , 1 , − 2 〉 = 〈 2 , − 3 , 4 〉 , ‖ 2 a + b‖ =

22 + (−3)^2 + 42 =

The algebraic operations with vectors have several properties that resemble familiar properties of the algebraic operations with numbers.

Theorem 2.1. Let a, b, and c denote vectors and let c, c 1 , c 2 denote scalars. Then: i) a + b = b + a ii) a + (b + c) = (a + b) + c iii) a + 0 = a iv) a + (−a) = 0 v) c(a + b) = ca + cb vi) (c 1 + c 2 )a = c 1 a + c 2 a vii) (c 1 c 2 )a = c 1 (c 2 a) viii) 1a = a ix) ‖a‖ ≥ 0 , with ‖a‖ = 0 if and only if a = 0 x) ‖ca‖ = |c| · ‖a‖ xi) ‖a + b‖ ≤ ‖a‖ + ‖b‖ (the triangle inequality)

Proof. The proofs of parts i)–viii) are all similar: they use (2.2)–(2.4) to deduce i)–viii) from the respective properties of numbers. For example, to prove the distributive law vi), we write a = 〈a 1 , a 2 , a 3 〉 and argue as follows:

(c 1 + c 2 )a = 〈(c 1 + c 2 )a 1 , (c 1 + c 2 )a 2 , (c 1 + c 2 )a 3 〉 by (2.4) = 〈c 1 a 1 + c 2 a 1 , c 1 a 2 + c 2 a 2 , c 1 a 3 + c 2 a 3 〉 by the distributive law for numbers = 〈c 1 a 1 , c 1 a 2 , c 1 a 3 〉 + 〈c 2 a 1 , c 2 a 2 , c 2 a 3 〉 by (2.2) = c 1 a + c 2 a by (2.3).

Definition. Given vectors a = 〈a 1 , a 2 , a 3 〉 and b = 〈b 1 , b 2 , b 3 〉, we define their dot product a · b to be a · b = a 1 b 1 + a 2 b 2 + a 3 b 3. (2.5) Notice that the dot product of two vectors is a scalar, not a vector. Sometimes, the dot product is called also the inner product or the scalar product of a and b. Because of this, you should avoid the temptation to refer to the scalar multiple ca as the “scalar product of c and a”.

Example 2.3. The dot product of a = 〈 4 , 2 , − 1 〉 and b = 〈− 2 , 2 , 1 〉 is a · b = (4)(−2) + (2)(2) + (−1)(1) = − 5.

The dot product of u = 2 e 1 + 3 e 2 + e 3 and v = e 1 + 4 e 2 is

u · v = (2)(1) + (3)(4) + (1)(0) = 14. The next theorem summarizes some important algebraic properties of the dot product. Theorem 2.2. Let a, b, and c denote vectors and let c 1 , c 2 denote scalars. Then: i) a · a = ‖a‖^2 ii) a · 0 = 0 iii) a · b = b · a iv) (c 1 a) · (c 2 b) = c 1 c 2 (a · b) v) (a + b) · c = a · c + b · c Proof. These follow from the definition of the dot product and from the properties of the real numbers. For example, if a = 〈a 1 , a 2 , a 3 〉 and b = 〈b 1 , b 2 , b 3 〉, we have

a · b = a 1 b 1 + a 2 b 2 + a 3 b 3 by (2.5) = b 1 a 1 + b 2 a 2 + b 3 a 3 by the commutative law for real numbers = b · a by (2.5).

That is, iii) holds. 

We said earlier that our definition of the dot product is motivated by applications. The next theorem relates the dot product of two vectors to their geometric representations.

Theorem 2.3. If a and b are two vectors, then a · b = ‖a‖ ‖b‖ cos θ,

where θ is the angle between a and b.

Corollary 2.4. If a and b are two nonzero vectors and θ is the angle between them, then

cos θ =

a · b ‖a‖ ‖b‖

Example 2.4. Find the angle between a = 〈 1 , 2 , − 1 〉 and b = 〈 0 , 2 , 1 〉. Solution. The angle θ between a and b satisfies

cos θ =

12 + 22 + (−1)^2

so θ = arccos

√ 30

u

a

projua

a^

proj

au

Figure 2.3. The projection of a vector onto a unit vector

We say that two vectors a and b are orthogonal (or perpendicular) if a·b = 0. By the corollary, a · b = 0 implies that the angle between a and b is 90◦, so this definition makes perfect sense.

Note that if a = 〈a 1 , a 2 , a 3 〉, then we can use the dot products of a and the standard basis vectors ej to extract the components of a:

a 1 = a · e 1 , a 2 = a · e 2 , a 3 = a · e 3.

More generally, suppose that u is a unit vector. Then a · u is a scalar and

projua = (a · u)u

is a vector that is parallel to u. This vector is called the (orthogonal) projection of a onto u. It has the property that the difference between it and the original vector a is perpendicular to u: ( a − projua

· u = 0.

The two-dimensional case of this is illustrated on Figure 2.3. Note that the size of the dot product a·u equals the length of the projection; the sign of that dot product indicates whether the projection and u have the same direction (positive sign) or opposite directions (negative sign).

2.8. The cross product The dot product of two vectors is a scalar. In this section, we define another product of two vectors: their “cross product”, which is a vector. Since our discussion requires basic familiarity with 2 × 2 and 3 × 3 determinants, a quick look at Appendix A may be helpful before proceeding further.

Definition. Let a = 〈a 1 , a 2 , a 3 〉 and b = 〈b 1 , b 2 , b 3 〉 be two three-dimensional vectors. Then their cross product a × b is the vector

a × b =

a 2 a 3 b 2 b 3

∣ e^1 −

a 1 a 3 b 1 b 3

∣ e^2 +

a 1 a 2 b 1 b 2

∣ e^3.

Remarks. There are a couple of things we should point out:

1. The cross product is defined only for three-dimensional vectors. There is no analog of this operation for two-dimensional vectors.
2. The defining formula may be confusing at first. On the other hand, following the exact labeling is important because of the properties of the cross product (see Theorem 2.