Sets and Functions
1.1 Sets
Denition. A set is a collection of objects. We call the objects, elements. A set is denoted by listing the elements
between braces. For example: fe; {; ; 1g. We use ellipses to indicate patterns. The set of positive integers is
f1; 2; 3; : : :g. We also denote a sets with the notation fxjconditions on xg for sets that are more easily described than
enumerated. This is read as \the set of elements x such that x satises . . . ". x 2 S is the notation for \x is an
element of the set S." To express the opposite we have x 62 S for \x is not an element of the set S."
Examples. We have notations for denoting some of the commonly encountered sets.
; = fg is the empty set, the set containing no elements.
Z = f: : : ;1; 0; 1 : : :g is the set of integers. (Z is for \Zahlen", the German word for \number".)
Q = fp=qjp; q 2 Z; q 6= 0g is the set of rational numbers. (Q is for quotient.)
R = fxjx = a1a2 an:b1b2 g is the set of real numbers, i.e. the set of numbers with decimal expansions. 1
1Guess what R is for.
2
C = fa + {bja; b 2 R; {2 = 1g is the set of complex numbers. { is the square root of 1. (If you haven't seen
complex numbers before, don't dismay. We'll cover them later.)
Z+, Q+ and R+ are the sets of positive integers, rationals and reals, respectively. For example, Z+ = f1; 2; 3; : : :g.
Z0+, Q0+ and R0+ are the sets of non-negative integers, rationals and reals, respectively. For example, Z0+ =
f0; 1; 2; : : :g.
(a : : : b) denotes an open interval on the real axis. (a : : : b) fxjx 2 R; a < x < bg
We use brackets to denote the closed interval. [a : : : b] fxjx 2 R; a x bg
The cardinality or order of a set S is denoted jSj. For nite sets, the cardinality is the number of elements in the
set. The Cartesian product of two sets is the set of ordered pairs:
X Y f(x; y)jx 2 X; y 2 Y g:
The Cartesian product of n sets is the set of ordered n-tuples:
X1 X2 Xn f(x1; x2; : : : ; xn)jx1 2 X1; x2 2 X2; : : : ; xn 2 Xng:
Equality. Two sets S and T are equal if each element of S is an element of T and vice versa. This is denoted,
S = T. Inequality is S 6= T, of course. S is a subset of T, S T, if every element of S is an element of T. S is a
proper subset of T, S T, if S T and S 6= T. For example: The empty set is a subset of every set, ; S. The
rational numbers are a proper subset of the real numbers, Q R.
Operations. The union of two sets, S [ T, is the set whose elements are in either of the two sets. The union of n
sets,
[nj
=1Sj S1 [ S2 [ [ Sn
is the set whose elements are in any of the sets Sj . The intersection of two sets, S \ T, is the set whose elements are
in both of the two sets. In other words, the intersection of two sets in the set of elements that the two sets have in
common. The intersection of n sets,
\nj
=1Sj S1 \ S2 \ \ Sn
3
is the set whose elements are in all of the sets Sj . If two sets have no elements in common, S \ T = ;, then the sets
are disjoint. If T S, then the di erence between S and T, S n T, is the set of elements in S which are not in T.
S n T fxjx 2 S; x 62 Tg
The di erence of sets is also denoted S T.
Properties. The following properties are easily veried from the above denitions.
S [ ; = S, S \ ; = ;, S n ; = S, S n S = ;.
Commutative. S [ T = T [ S, S \ T = T \ S.
Associative. (S [ T) [ U = S [ (T [ U) = S [ T [ U, (S \ T) \ U = S \ (T \ U) = S \ T \ U.
Distributive. S [ (T \ U) = (S [ T) \ (S [ U), S \ (T [ U) = (S \ T) [ (S \ U).
1.2 Single Valued Functions
Single-Valued Functions. A single-valued function or single-valued mapping is a mapping of the elements x 2 X
into elements y 2 Y . This is expressed as f : X ! Y or X
f!
Y . If such a function is well-dened, then for each
x 2 X there exists a unique element of y such that f(x) = y. The set X is the domain of the function, Y is the
codomain, (not to be confused with the range, which we introduce shortly). To denote the value of a function on a
particular element we can use any of the notations: f(x) = y, f : x 7! y or simply x 7! y. f is the identity map on
X if f(x) = x for all x 2 X.
Let f : X ! Y . The range or image of f is
f(X) = fyjy = f(x) for some x 2 Xg:
The range is a subset of the codomain. For each Z Y , the inverse image of Z is dened:
f1(Z) fx 2 Xjf(x) = z for some z 2 Zg:
4
Examples.
Finite polynomials and the exponential function are examples of single valued functions which map real numbers
to real numbers.
The greatest integer function, bc, is a mapping from R to Z. bxc in the greatest integer less than or equal to x.
Likewise, the least integer function, dxe, is the least integer greater than or equal to x.
The -jectives. A function is injective if for each x1 6= x2, f(x1) 6= f(x2). In other words, for each x in the domain
there is a unique y = f(x) in the range. f is surjective if for each y in the codomain, there is an x such that y = f(x).
If a function is both injective and surjective, then it is bijective. A bijective function is also called a one-to-one mapping.
Examples.
The exponential function y = ex is a bijective function, (one-to-one mapping), that maps R to R+. (R is the set
of real numbers; R+ is the set of positive real numbers.)
f(x) = x2 is a bijection from R+ to R+. f is not injective from R to R+. For each positive y in the range, there
are two values of x such that y = x2.
f(x) = sin x is not injective from R to [1::1]. For each y 2 [1; 1] there exists an innite number of values of
x such that y = sin x.
1.3 Inverses and Multi-Valued Functions
If y = f(x), then we can write x = f1(y) where f1 is the inverse of f. If y = f(x) is a one-to-one function, then
f1(y) is also a one-to-one function. In this case, x = f1(f(x)) = f(f1(x)) for values of x where both f(x) and
f1(x) are dened. For example log x, which maps R+ to R is the inverse of ex. x = elog x = log(ex) for all x 2 R+.
(Note the x 2 R+ ensures that log x is dened.)
5
Injective Surjective Bijective
Figure 1.1: Depictions of Injective, Surjective and Bijective Functions
If y = f(x) is a many-to-one function, then x = f1(y) is a one-to-many function. f1(y) is a multi-valued function.
We have x = f(f1(x)) for values of x where f1(x) is dened, however x 6= f1(f(x)). There are diagrams showing
one-to-one, many-to-one and one-to-many functions in Figure 1.2.
domain range domain range domain range
one-to-one many-to-one one-to-many
Figure 1.2: Diagrams of One-To-One, Many-To-One and One-To-Many Functions
Example 1.3.1 y = x2, a many-to-one function has the inverse x = y1=2. For each positive y, there are two values of
x such that x = y1=2. y = x2 and y = x1=2 are graphed in Figure 1.3.
6
Figure 1.3: y = x2 and y = x1=2
We say that there are two branches of y = x1=2: the positive and the negative branch. We denote the positive
branch as y =
p
x; the negative branch is y =
p
x. We call
p
x the principal branch of x1=2. Note that
p
x is a
one-to-one function. Finally, x = (x1=2)2 since (
p
x)2 = x, but x 6= (x2)1=2 since (x2)1=2 = x. y =
p
x is graphed
in Figure 1.4.
Figure 1.4: y =
p
x
Now consider the many-to-one function y = sin x. The inverse is x = arcsin y. For each y 2 [1; 1] there are an
innite number of values x such that x = arcsin y. In Figure 1.5 is a graph of y = sin x and a graph of a few branches
of y = arcsin x.
Example 1.3.2 arcsin x has an innite number of branches. We will denote the principal branch by Arcsin x which
maps [1; 1] to
2 ;
2
. Note that x = sin(arcsin x), but x 6= arcsin(sin x). y = Arcsin x in Figure 1.6.
7
Figure 1.5: y = sin x and y = arcsin x
Figure 1.6: y = Arcsin x
Example 1.3.3 Consider 11=3. Since x3 is a one-to-one function, x1=3 is a single-valued function. (See Figure 1.7.)
11=3 = 1.
Figure 1.7: y = x3 and y = x1=3
8
Example 1.3.4 Consider arccos(1=2). cos x and a few branches of arccos x are graphed in Figure 1.8. cos x = 1=2
Figure 1.8: y = cos x and y = arccos x
has the two solutions x = =3 in the range x 2 [; ]. Since cos(x + ) = cos x,
arccos(1=2) = f=3 + ng:
1.4 Transforming Equations
We must take care in applying functions to equations. It is always safe to apply a one-to-one function to an equation,
(provided it is dened for that domain). For example, we can apply y = x3 or y = ex to the equation x = 1. The
equations x3 = 1 and ex = e have the unique solution x = 1.
If we apply a many-to-one function to an equation, we may introduce spurious solutions. Applying y = x2 and
y = sin x to the equation x =
2 results in x2 = 2
4 and sin x = 1. The former equation has the two solutions x =
2 ;
the latter has the innite number of solutions x =
2 + 2n, n 2 Z.
We do not generally apply a one-to-many function to both sides of an equation as this rarely is useful. Consider the
equation
sin2 x = 1:
9
Applying the function f(x) = x1=2 to the equation would not get us anywhere
(sin2 x)1=2 = 11=2:
Since (sin2 x)1=2 6= sin x, we cannot simplify the left side of the equation. Instead we could use the denition of
f(x) = x1=2 as the inverse of the x2 function to obtain
sin x = 11=2 = 1:
Then we could use the denition of arcsin as the inverse of sin to get
x = arcsin(1):
x = arcsin(1) has the solutions x = =2 + 2n and x = arcsin(1) has the solutions x = =2 + 2n. Thus
x =
2
+ n; n 2 Z:
Note that we cannot just apply arcsin to both sides of the equation as arcsin(sin x) 6= x.
10
1.5 Exercises
Exercise 1.1
The area of a circle is directly proportional to the square of its diameter. What is the constant of proportionality?
Hint, Solution
Exercise 1.2
Consider the equation
x + 1
y 2
=
x2 1
y2 4
:
1. Why might one think that this is the equation of a line?
2. Graph the solutions of the equation to demonstrate that it is not the equation of a line.
Hint, Solution
Exercise 1.3
Consider the function of a real variable,
f(x) =
1
x2 + 2
:
What is the domain and range of the function?
Hint, Solution
Exercise 1.4
The temperature measured in degrees Celsius 2 is linearly related to the temperature measured in degrees Fahrenheit 3.
Water freezes at 0 C = 32 F and boils at 100 C = 212 F. Write the temperature in degrees Celsius as a function
of degrees Fahrenheit.
2 Originally, it was called degrees Centigrade. centi because there are 100 degrees between the two calibration points. It is now
called degrees Celsius in honor of the inventor.
3 The Fahrenheit scale, named for Daniel Fahrenheit, was originally calibrated with the freezing point of salt-saturated water to
be 0. Later, the calibration points became the freezing point of water, 32, and body temperature, 96. With this method, there are
64 divisions between the calibration points. Finally, the upper calibration point was changed to the boiling point of water at 212.
This gave 180 divisions, (the number of degrees in a half circle), between the two calibration points.
11
Hint, Solution
Exercise 1.5
Consider the function graphed in Figure 1.9. Sketch graphs of f(x), f(x + 3), f(3 x) + 2, and f1(x). You may
use the blank grids in Figure 1.10.
Figure 1.9: Graph of the function.
Hint, Solution
Exercise 1.6
A culture of bacteria grows at the rate of 10% per minute. At 6:00 pm there are 1 billion bacteria. How many bacteria
are there at 7:00 pm? How many were there at 3:00 pm?
Hint, Solution
Exercise 1.7
The graph in Figure 1.11 shows an even function f(x) = p(x)=q(x) where p(x) and q(x) are rational quadratic
polynomials. Give possible formulas for p(x) and q(x).
Hint, Solution
12
Figure 1.10: Blank grids.
Exercise 1.8
Find a polynomial of degree 100 which is zero only at x = 2; 1; and is non-negative.
Hint, Solution
Exercise 1.9
Hint, Solution
13
1 2
1
2
2 4 6 8 10
1
2
Figure 1.11: Plots of f(x) = p(x)=q(x).
Exercise 1.10
Hint, Solution
Exercise 1.11
Hint, Solution
Exercise 1.12
Hint, Solution
Exercise 1.13
Hint, Solution
Exercise 1.14
Hint, Solution
Exercise 1.15
Hint, Solution
Exercise 1.16
Hint, Solution
14
1.6 Hints
Hint 1.1
area = constant diameter2.
Hint 1.2
A pair (x; y) is a solution of the equation if it make the equation an identity.
Hint 1.3
The domain is the subset of R on which the function is dened.
Hint 1.4
Find the slope and x-intercept of the line.
Hint 1.5
The inverse of the function is the re
ection of the function across the line y = x.
Hint 1.6
The formula for geometric growth/decay is x(t) = x0rt, where r is the rate.
Hint 1.7
Note that p(x) and q(x) appear as a ratio, they are determined only up to a multiplicative constant. We may take the
leading coecient of q(x) to be unity.
f(x) =
p(x)
q(x)
=
ax2 + bx + c
x2 + x +
Use the properties of the function to solve for the unknown parameters.
Hint 1.8
Write the polynomial in factored form.
15
1.7 Solutions
Solution 1.1
area = radius2
area =
4
diameter2
The constant of proportionality is
4 .
Solution 1.2
1. If we multiply the equation by y2 4 and divide by x + 1, we obtain the equation of a line.
y + 2 = x 1
2. We factor the quadratics on the right side of the equation.
x + 1
y 2
=
(x + 1)(x 1)
(y 2)(y + 2)
:
We note that one or both sides of the equation are undened at y = 2 because of division by zero. There are
no solutions for these two values of y and we assume from this point that y 6= 2. We multiply by (y2)(y+2).
(x + 1)(y + 2) = (x + 1)(x 1)
For x = 1, the equation becomes the identity 0 = 0. Now we consider x 6= 1. We divide by x + 1 to obtain
the equation of a line.
y + 2 = x 1
y = x 3
Now we collect the solutions we have found.
f(1; y) : y 6= 2g [ f(x; x 3) : x 6= 1; 5g
The solutions are depicted in Figure /refg not a line.
16
-6 -4 -2 2 4 6
-6
-4
-2
2
4
6
Figure 1.12: The solutions of x+1
y2 = x21
y24 .
Solution 1.3
The denominator is nonzero for all x 2 R. Since we don't have any division by zero problems, the domain of the
function is R. For x 2 R,
0 <
1
x2 + 2
2:
Consider
y =
1
x2 + 2
: (1.1)
For any y 2 (0 : : : 1=2], there is at least one value of x that satises Equation 1.1.
x2 + 2 =
1
y
x =
r
1
y
2
Thus the range of the function is (0 : : : 1=2]
17
Solution 1.4
Let c denote degrees Celsius and f denote degrees Fahrenheit. The line passes through the points (f; c) = (32; 0) and
(f; c) = (212; 100). The x-intercept is f = 32. We calculate the slope of the line.
slope =
100 0
212 32
=
100
180
=
5
9
The relationship between fahrenheit and celcius is
c =
5
9
(f 32):
Solution 1.5
We plot the various transformations of f(x).
Solution 1.6
The formula for geometric growth/decay is x(t) = x0rt, where r is the rate. Let t = 0 coincide with 6:00 pm. We
determine x0.
x(0) = 109 = x0
11
10
0
= x0
x0 = 109
At 7:00 pm the number of bacteria is
109
11
10
60
=
1160
1051
3:04 1011
At 3:00 pm the number of bacteria was
109
11
10
180
=
10189
11180
35:4
18
Figure 1.13: Graphs of f(x), f(x + 3), f(3 x) + 2, and f1(x).
Solution 1.7
We write p(x) and q(x) as general quadratic polynomials.
f(x) =
p(x)
q(x)
=
ax2 + bx + c
x2 + x +
We will use the properties of the function to solve for the unknown parameters.
19
Note that p(x) and q(x) appear as a ratio, they are determined only up to a multiplicative constant. We may take
the leading coecient of q(x) to be unity.
f(x) =
p(x)
q(x)
=
ax2 + bx + c
x2 + x +
f(x) has a second order zero at x = 0. This means that p(x) has a second order zero there and that 6= 0.
f(x) =
ax2
x2 + x +
We note that f(x) ! 2 as x ! 1. This determines the parameter a.
lim
x!1
f(x) = lim
x!1
ax2
x2 + x +
= lim
x!1
2ax
2x +
= lim
x!1
2a
2
= a
f(x) =
2x2
x2 + x +
Now we use the fact that f(x) is even to conclude that q(x) is even and thus = 0.
f(x) =
2x2
x2 +
Finally, we use that f(1) = 1 to determine .
f(x) =
2x2
x2 + 1
20
Solution 1.8
Consider the polynomial
p(x) = (x + 2)40(x 1)30(x )30:
It is of degree 100. Since the factors only vanish at x = 2; 1; , p(x) only vanishes there. Since factors are nonnegative,
the polynomial is non-negative.
21
Chapter 2
Vectors
2.1 Vectors
2.1.1 Scalars and Vectors
A vector is a quantity having both a magnitude and a direction. Examples of vector quantities are velocity, force
and position. One can represent a vector in n-dimensional space with an arrow whose initial point is at the origin,
(Figure 2.1). The magnitude is the length of the vector. Typographically, variables representing vectors are often
written in capital letters, bold face or with a vector over-line, A; a;~a. The magnitude of a vector is denoted jaj.
A scalar has only a magnitude. Examples of scalar quantities are mass, time and speed.
Vector Algebra. Two vectors are equal if they have the same magnitude and direction. The negative of a vector,
denoted a, is a vector of the same magnitude as a but in the opposite direction. We add two vectors a and b by
placing the tail of b at the head of a and dening a + b to be the vector with tail at the origin and head at the head
of b. (See Figure 2.2.)
The di erence, a b, is dened as the sum of a and the negative of b, a + (b). The result of multiplying a by
a scalar is a vector of magnitude j j jaj with the same/opposite direction if is positive/negative. (See Figure 2.2.)
22
x
z
y
Figure 2.1: Graphical Representation of a Vector in Three Dimensions
a+b
a
b
-a
a
2a
Figure 2.2: Vector Arithmetic
Here are the properties of adding vectors and multiplying them by a scalar. They are evident from geometric
considerations.
a + b = b + a a = a commutative laws
(a + b) + c = a + (b + c) (a) = ( )a associative laws
(a + b) = a + b ( + )a = a + a distributive laws
23
Zero and Unit Vectors. The additive identity element for vectors is the zero vector or null vector. This is a vector
of magnitude zero which is denoted as 0. A unit vector is a vector of magnitude one. If a is nonzero then a=jaj is a
unit vector in the direction of a. Unit vectors are often denoted with a caret over-line, ^n.
Rectangular Unit Vectors. In n dimensional Cartesian space, Rn, the unit vectors in the directions of the
coordinates axes are e1; : : : en. These are called the rectangular unit vectors. To cut down on subscripts, the unit
vectors in three dimensional space are often denoted with i, j and k. (Figure 2.3).
x
z
y
j
k
i
Figure 2.3: Rectangular Unit Vectors
Components of a Vector. Consider a vector a with tail at the origin and head having the Cartesian coordinates
(a1; : : : ; an). We can represent this vector as the sum of n rectangular component vectors, a = a1e1 + + anen.
(See Figure 2.4.) Another notation for the vector a is ha1; : : : ; ani. By the Pythagorean theorem, the magnitude of
the vector a is jaj =
p
a21
+ + a2
n.
24
x
z
y
a
a
a
1
3
i
k
a2 j
Figure 2.4: Components of a Vector
2.1.2 The Kronecker Delta and Einstein Summation Convention
The Kronecker Delta tensor is dened
ij =
(
1 if i = j;
0 if i 6= j:
This notation will be useful in our work with vectors.
Consider writing a vector in terms of its rectangular components. Instead of using ellipses: a = a1e1+ +anen, we
could write the expression as a sum: a =
Pn
i=1 aiei. We can shorten this notation by leaving out the sum: a = aiei,
where it is understood that whenever an index is repeated in a term we sum over that index from 1 to n. This is the
Einstein summation convention. A repeated index is called a summation index or a dummy index. Other indices can
take any value from 1 to n and are called free indices.
25
Example 2.1.1 Consider the matrix equation: A x = b. We can write out the matrix and vectors explicitly.
0
B@
a11 a1n
...
. . .
...
an1 ann
1
CA
0
B@
x1
...
xn
1
CA
=
0
B@
b1
...
bn
1
CA
This takes much less space when we use the summation convention.
aijxj = bi
Here j is a summation index and i is a free index.
2.1.3 The Dot and Cross Product
Dot Product. The dot product or scalar product of two vectors is dened,
a b jajjbj cos ;
where is the angle from a to b. From this denition one can derive the following properties:
a b = b a, commutative.
(a b) = ( a) b = a ( b), associativity of scalar multiplication.
a (b + c) = a b + a c, distributive.
eiej = ij . In three dimension, this is
i i = j j = k k = 1; i j = j k = k i = 0:
a b = aibi a1b1 + + anbn, dot product in terms of rectangular components.
If a b = 0 then either a and b are orthogonal, (perpendicular), or one of a and b are zero.
26
The Angle Between Two Vectors. We can use the dot product to nd the angle between two vectors, a and
b. From the denition of the dot product,
a b = jajjbj cos :
If the vectors are nonzero, then
= arccos
a b
jajjbj
:
Example 2.1.2 What is the angle between i and i + j?
= arccos
i (i + j)
jijji + jj
= arccos
1
p
2
=
4
:
Parametric Equation of a Line. Consider a line that passes through the point a and is parallel to the vector t,
(tangent). A parametric equation of the line is
x = a + ut; u 2 R:
Implicit Equation of a Line. Consider a line that passes through the point a and is normal, (orthogonal, perpendicular),
to the vector n. All the lines that are normal to n have the property that x n is a constant, where x is
any point on the line. (See Figure 2.5.) x n = 0 is the line that is normal to n and passes through the origin. The
line that is normal to n and passes through the point a is
x n = a n:
27
=0
=1 =a n
n a
=-1
x n
x n
x n
x n
Figure 2.5: Equation for a Line
The normal to a line determines an orientation of the line. The normal points in the direction that is above the
line. A point b is (above/on/below) the line if (b a) n is (positive/zero/negative). The signed distance of a point
b from the line x n = a n is
(b a)
n
jnj
:
Implicit Equation of a Hyperplane. A hyperplane in Rn is an n1 dimensional \sheet" which passes through
a given point and is normal to a given direction. In R3 we call this a plane. Consider a hyperplane that passes through
the point a and is normal to the vector n. All the hyperplanes that are normal to n have the property that x n is a
constant, where x is any point in the hyperplane. x n = 0 is the hyperplane that is normal to n and passes through
the origin. The hyperplane that is normal to n and passes through the point a is
x n = a n:
The normal determines an orientation of the hyperplane. The normal points in the direction that is above the
hyperplane. A point b is (above/on/below) the hyperplane if (b a) n is (positive/zero/negative). The signed
28
distance of a point b from the hyperplane x n = a n is
(b a)
n
jnj
:
Right and Left-Handed Coordinate Systems. Consider a rectangular coordinate system in two dimensions.
Angles are measured from the positive x axis in the direction of the positive y axis. There are two ways of labeling the
axes. (See Figure 2.6.) In one the angle increases in the counterclockwise direction and in the other the angle increases
in the clockwise direction. The former is the familiar Cartesian coordinate system.
x y
y x
q
q
Figure 2.6: There are Two Ways of Labeling the Axes in Two Dimensions.
There are also two ways of labeling the axes in a three-dimensional rectangular coordinate system. These are called
right-handed and left-handed coordinate systems. See Figure 2.7. Any other labelling of the axes could be rotated into
one of these congurations. The right-handed system is the one that is used by default. If you put your right thumb in
the direction of the z axis in a right-handed coordinate system, then your ngers curl in the direction from the x axis
to the y axis.
Cross Product. The cross product or vector product is dened,
a b = jajjbj sin n;
where is the angle from a to b and n is a unit vector that is orthogonal to a and b and in the direction such that a,
b and n form a right-handed system.
29
x
z
j y
i
k
z
k
j
i
y
x
Figure 2.7: Right and Left Handed Coordinate Systems
You can visualize the direction of a b by applying the right hand rule. Curl the ngers of your right hand in the
direction from a to b. Your thumb points in the direction of a b. Warning: Unless you are a lefty, get in the habit
of putting down your pencil before applying the right hand rule.
The dot and cross products behave a little di erently. First note that unlike the dot product, the cross product is not
commutative. The magnitudes of a b and b a are the same, but their directions are opposite. (See Figure 2.8.)
a
b
b a
a b
Figure 2.8: The Cross Product is Anti-Commutative.
30
Let
a b = jajjbj sin n and b a = jbjjaj sin m:
The angle from a to b is the same as the angle from b to a. Since fa; b; ng and fb; a;mg are right-handed systems,
m points in the opposite direction as n. Since a b = b a we say that the cross product is anti-commutative.
Next we note that since
ja bj = jajjbj sin ;
the magnitude of a b is the area of the parallelogram dened by the two vectors. (See Figure 2.9.) The area of the
triangle dened by two vectors is then 1
2
ja bj.
b
sin
b
b
a
q
a
Figure 2.9: The Parallelogram and the Triangle Dened by Two Vectors
From the denition of the cross product, one can derive the following properties:
a b = b a, anti-commutative.
(a b) = ( a) b = a ( b), associativity of scalar multiplication.
a (b + c) = a b + a c, distributive.
(a b) c 6= a (b c). The cross product is not associative.
i i = j j = k k = 0.
31
i j = k, j k = i, k i = j.
a b = (a2b3 a3b2)i + (a3b1 a1b3)j + (a1b2 a2b1)k =
i j k
a1 a2 a3
b1 b2 b3
;
cross product in terms of rectangular components.
If a b = 0 then either a and b are parallel or one of a or b is zero.
Scalar Triple Product. Consider the volume of the parallelopiped dened by three vectors. (See Figure 2.10.)
The area of the base is jjbjjcj sin j, where is the angle between b and c. The height is jaj cos , where is the angle
between b c and a. Thus the volume of the parallelopiped is jajjbjjcj sin cos .
f
q
b c
a
b
c
Figure 2.10: The Parallelopiped Dened by Three Vectors
Note that
ja (b c)j = ja (jbjjcj sin n)j
= jjajjbjjcj sin cos j :
32
Thus ja (b c)j is the volume of the parallelopiped. a (bc) is the volume or the negative of the volume depending
on whether fa; b; cg is a right or left-handed system.
Note that parentheses are unnecessary in a bc. There is only one way to interpret the expression. If you did the
dot product rst then you would be left with the cross product of a scalar and a vector which is meaningless. a bc
is called the scalar triple product.
Plane Dened by Three Points. Three points which are not collinear dene a plane. Consider a plane that
passes through the three points a, b and c. One way of expressing that the point x lies in the plane is that the vectors
x a, b a and c a are coplanar. (See Figure 2.11.) If the vectors are coplanar, then the parallelopiped dened by
these three vectors will have zero volume. We can express this in an equation using the scalar triple product,
(x a) (b a) (c a) = 0:
b
c
x
a
Figure 2.11: Three Points Dene a Plane.
2.2 Sets of Vectors in n Dimensions
Orthogonality. Consider two n-dimensional vectors
x = (x1; x2; : : : ; xn); y = (y1; y2; : : : ; yn):
33
The inner product of these vectors can be dened
hxjyi x y =
Xn
i=1
xiyi:
The vectors are orthogonal if x y = 0. The norm of a vector is the length of the vector generalized to n dimensions.
kxk =
p
x x
Consider a set of vectors
fx1; x2; : : : ; xmg:
If each pair of vectors in the set is orthogonal, then the set is orthogonal.
xi xj = 0 if i 6= j
If in addition each vector in the set has norm 1, then the set is orthonormal.
xi xj = ij =
(
1 if i = j
0 if i 6= j
Here ij is known as the Kronecker delta function.
Completeness. A set of n, n-dimensional vectors
fx1; x2; : : : ; xng
is complete if any n-dimensional vector can be written as a linear combination of the vectors in the set. That is, any
vector y can be written
y =
Xn
i=1
cixi:
34
Taking the inner product of each side of this equation with xm,
y xm =
Xn
i=1
cixi
!
xm
=
Xn
i=1
cixi xm
= cmxm xm
cm =
y xm
kxmk2
Thus y has the expansion
y =
Xn
i=1
y xi
kxik2xi:
If in addition the set is orthonormal, then
y =
Xn
i=1
(y xi)xi:
35
2.3 Exercises
The Dot and Cross Product
Exercise 2.1
Prove the distributive law for the dot product,
a (b + c) = a b + a c:
Exercise 2.2
Prove that
a b = aibi a1b1 + + anbn:
Exercise 2.3
What is the angle between the vectors i + j and i + 3j?
Exercise 2.4
Prove the distributive law for the cross product,
a (b + c) = a b + a b:
Exercise 2.5
Show that
a b =
i j k
a1 a2 a3
b1 b2 b3
Exercise 2.6
What is the area of the quadrilateral with vertices at (1; 1), (4; 2), (3; 7) and (2; 3)?
Exercise 2.7
What is the volume of the tetrahedron with vertices at (1; 1; 0), (3; 2; 1), (2; 4; 1) and (1; 2; 5)?
36
Exercise 2.8
What is the equation of the plane that passes through the points (1; 2; 3), (2; 3; 1) and (3; 1; 2)? What is the distance
from the point (2; 3; 5) to the plane?
37
2.4 Hints
The Dot and Cross Product
Hint 2.1
First prove the distributive law when the rst vector is of unit length,
n (b + c) = n b + n c:
Then all the quantities in the equation are projections onto the unit vector n and you can use geometry.
Hint 2.2
First prove that the dot product of a rectangular unit vector with itself is one and the dot product of two distinct
rectangular unit vectors is zero. Then write a and b in rectangular components and use the distributive law.
Hint 2.3
Use a b = jajjbj cos .
Hint 2.4
First consider the case that both b and c are orthogonal to a. Prove the distributive law in this case from geometric
considerations.
Next consider two arbitrary vectors a and b. We can write b = b? + bk where b? is orthogonal to a and bk is
parallel to a. Show that
a b = a b?:
Finally prove the distributive law for arbitrary b and c.
Hint 2.5
Write the vectors in their rectangular components and use,
i j = k; j k = i; k i = j;
and,
i i = j j = k k = 0:
38
Hint 2.6
The quadrilateral is composed of two triangles. The area of a triangle dened by the two vectors a and b is 1
2
ja bj.
Hint 2.7
Justify that the area of a tetrahedron determined by three vectors is one sixth the area of the parallelogram determined
by those three vectors. The area of a parallelogram determined by three vectors is the magnitude of the scalar triple
product of the vectors: a b c.
Hint 2.8
The equation of a line that is orthogonal to a and passes through the point b is a x = a b. The distance of a point
c from the plane is (c b)
a
jaj
39
2.5 Solutions
The Dot and Cross Product
Solution 2.1
First we prove the distributive law when the rst vector is of unit length, i.e.,
n (b + c) = n b + n c: (2.1)
From Figure 2.12 we see that the projection of the vector b+c onto n is equal to the sum of the projections b n and
c n.
b
c
n b
n c
b+c
n
n (b+c)
Figure 2.12: The Distributive Law for the Dot Product
Now we extend the result to the case when the rst vector has arbitrary length. We dene a = jajn and multiply
Equation 2.1 by the scalar, jaj.
jajn (b + c) = jajn b + jajn c
a (b + c) = a b + a c:
Solution 2.2
First note that
ei ei = jeijjeij cos(0) = 1:
40
Then note that that dot product of any two distinct rectangular unit vectors is zero because they are orthogonal. Now
we write a and b in terms of their rectangular components and use the distributive law.
a b = aiei bjej
= aibjei ej
= aibjij
= aibi
Solution 2.3
Since a b = jajjbj cos , we have
= arccos
a b
jajjbj
when a and b are nonzero.
= arccos
(i + j) (i + 3j)
ji + jjji + 3jj
= arccos
4
p
2
p
10
= arccos
2
p
5
5
!
0:463648
Solution 2.4
First consider the case that both b and c are orthogonal to a. b + c is the diagonal of the parallelogram dened by
b and c, (see Figure 2.13). Since a is orthogonal to each of these vectors, taking the cross product of a with these
vectors has the e ect of rotating the vectors through =2 radians about a and multiplying their length by jaj. Note
that a(b+c) is the diagonal of the parallelogram dened by ab and ac. Thus we see that the distributive law
holds when a is orthogonal to both b and c,
a (b + c) = a b + a c:
Now consider two arbitrary vectors a and b. We can write b = b? + bk where b? is orthogonal to a and bk is
parallel to a, (see Figure 2.14).
By the denition of the cross product,
a b = jajjbj sin n:
41
b
b+c c
a c
a
a b
a (b+c)
Figure 2.13: The Distributive Law for the Cross Product
a
b
b
q
b
Figure 2.14: The Vector b Written as a Sum of Components Orthogonal and Parallel to a
Note that
jb?j = jbj sin ;
42
and that a b? is a vector in the same direction as a b. Thus we see that
a b = jajjbj sin n
= jaj(sin jbj)n
= jajjb?jn = jajjb?j sin(=2)n
a b = a b?:
Now we are prepared to prove the distributive law for arbitrary b and c.
a (b + c) = a (b? + bk + c? + ck)
= a ((b + c)? + (b + c)k)
= a ((b + c)?)
= a b? + a c?
= a b + a c
a (b + c) = a b + a c
Solution 2.5
We know that
i j = k; j k = i; k i = j;
and that
i i = j j = k k = 0:
Now we write a and b in terms of their rectangular components and use the distributive law to expand the cross
product.
a b = (a1i + a2j + a3k) (b1i + b2j + b3k)
= a1i (b1i + b2j + b3k) + a2j (b1i + b2j + b3k) + a3k (b1i + b2j + b3k)
= a1b2k + a1b3(j) + a2b1(k) + a2b3i + a3b1j + a3b2(i)
= (a2b3 a3b2)i (a1b3 a3b1)j + (a1b2 a2b1)k
43
Next we evaluate the determinant.
i j k
a1 a2 a3
b1 b2 b3
= i
a2 a3
b2 b3
j
a1 a3
b1 b3
+ k
a1 a2
b1 b2
= (a2b3 a3b2)i (a1b3 a3b1)j + (a1b2 a2b1)k
Thus we see that,
a b =
i j k
a1 a2 a3
b1 b2 b3
Solution 2.6
The area area of the quadrilateral is the area of two triangles. The rst triangle is dened by the vector from (1; 1) to
(4; 2) and the vector from (1; 1) to (2; 3). The second triangle is dened by the vector from (3; 7) to (4; 2) and the
vector from (3; 7) to (2; 3). (See Figure 2.15.) The area of a triangle dened by the two vectors a and b is 1
2
ja bj.
The area of the quadrilateral is then,
1
2
j(3i + j) (i + 2j)j +
1
2
j(i 5j) (i 4j)j =
1
2
(5) +
1
2
(19) = 12:
Solution 2.7
The tetrahedron is determined by the three vectors with tail at (1; 1; 0) and heads at (3; 2; 1), (2; 4; 1) and (1; 2; 5).
These are h2; 1; 1i, h1; 3; 1i and h0; 1; 5i. The area of the tetrahedron is one sixth the area of the parallelogram
determined by these vectors. (This is because the area of a pyramid is 1
3 (base)(height). The base of the tetrahedron is
half the area of the parallelogram and the heights are the same. 1
2
1
3 = 1
6 ) Thus the area of a tetrahedron determined
by three vectors is 1
6
ja b cj. The area of the tetrahedron is
1
6
jh2; 1; 1i h1; 3; 1i h1; 2; 5ij =
1
6
jh2; 1; 1i h13;4;1ij =
7
2
44
x
y (3,7)
(4,2)
(2,3)
(1,1)
Figure 2.15: Quadrilateral
Solution 2.8
The two vectors with tails at (1; 2; 3) and heads at (2; 3; 1) and (3; 1; 2) are parallel to the plane. Taking the cross
product of these two vectors gives us a vector that is orthogonal to the plane.
h1; 1;2i h2;1;1i = h3;3;3i
We see that the plane is orthogonal to the vector h1; 1; 1i and passes through the point (1; 2; 3). The equation of the
plane is
h1; 1; 1i hx; y; zi = h1; 1; 1i h1; 2; 3i;
x + y + z = 6:
Consider the vector with tail at (1; 2; 3) and head at (2; 3; 5). The magnitude of the dot product of this vector with
the unit normal vector gives the distance from the plane.
h1; 1; 2i
h1; 1; 1i
jh1; 1; 1ij
=
4
p
3
=
4
p
3
3
45
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment