Chapter 2 Supplementary Exercises
3.2 Norm, Dot Product, and Distance in R n
In this section we will be concerned with the notions of length and distance as they relate to vectors. We will first discuss these ideas inR2andR3and then extend them algebraically toRn.
Norm of a Vector In this text we will denote the length of a vectorvby the symbolv, which is read as thenormofv, thelengthofv, or themagnitudeofv(the term “norm” being a common mathematical synonym for length). As suggested in Figure 3.2.1a, it follows from the
x y
||v||
||v||
(v1, v2)
v1
v2
P(v1, v2, v3) y z
x O
(a)
(b) Q
R S
Figure 3.2.1
Theorem of Pythagoras that the norm of a vector(v1, v2)inR2is v =√
v21+v22 (1)
Similarly, for a vector(v1, v2, v3)inR3, it follows from Figure 3.2.1band two applica- tions of the Theorem of Pythagoras that
v2=(OR)2+(RP )2=(OQ)2+(QR)2+(RP )2=v21+v22+v23 and hence that
v =√
v21+v22+v32 (2)
Motivated by the pattern of Formulas (1) and (2), we make the following definition.
DEFINITION1 Ifv=(v1, v2, . . . , vn)is a vector inRn, then thenormofv(also called thelengthofvor themagnitudeofv) is denoted byv, and is defined by the formula
v ="
v21+v22+ · · · +vn2 (3)
3.2 Norm, Dot Product, and Distance inRn 143
E X A M P L E 1 Calculating Norms
It follows from Formula (2) that the norm of the vectorv=(−3,2,1)inR3is v ="
(−3)2+22+12=√ 14
and it follows from Formula (3) that the norm of the vectorv=(2,−1,3,−5)inR4is v ="
22+(−1)2+32+(−5)2=√ 39
Our first theorem in this section will generalize toRn the following three familiar facts about vectors inR2andR3:
• Distances are nonnegative.
• The zero vector is the only vector of length zero.
• Multiplying a vector by a scalar multiplies its length by the absolute value of that scalar.
It is important to recognize that just because these results hold inR2andR3does not guarantee that they hold in Rn—their validity inRn must beproved using algebraic properties ofn-tuples.
THEOREM3.2.1 Ifvis a vector inRn,and ifkis anyscalar,then:
(a) v ≥0
(b) v =0if and onlyifv=0 (c) kv = |k|v
We will prove part (c) and leave (a) and (b) as exercises.
Proof(c) Ifv=(v1, v2, . . . , vn), thenkv=(kv1, kv2, . . . , kvn), so kv ="
(kv1)2+(kv2)2+ · · · +(kvn)2
=#
(k2)(v12+v22+ · · · +v2n)
= |k|#
v12+v22+ · · · +v2n
= |k|v
Unit Vectors A vector of norm 1 is called a unit vector. Such vectors are useful for specifying a direction when length is not relevant to the problem at hand. You can obtain a unit vector in a desired direction by choosing anynonzerovectorvin that direction and multiplying vby the reciprocal of its length. For example, ifvis a vector of length 2 inR2orR3, then 12vis a unit vector in the same direction asv. More generally, ifvis any nonzero vector inRn, then
u= 1
vv (4)
defines a unit vector that is in the same direction asv. We can confirm that (4) is a unit
WARNING Sometimes you will see Formula (4) expressed as
u= vv
This is just a more compact way of writing that formula and isnotintended to convey thatvis being divided byv.
vector by applying part (c) of Theorem 3.2.1 withk=1/vto obtain u = kv = |k|v =kv = 1
vv =1
The process of multiplying a nonzero vector by the reciprocal of its length to obtain a unit vector is callednormalizingv.
E X A M P L E 2 Normalizing a Vector
Find the unit vectoruthat has the same direction asv=(2,2,−1).
Solution The vectorvhas length v ="
22+22+(−1)2=3 Thus, from (4)
u= 13(2,2,−1)=2
3,23,−13 As a check, you may want to confirm thatu =1.
The Standard Unit Vectors When a rectangular coordinate system is introduced inR2orR3, the unit vectors in the positive directions of the coordinate axes are called thestandard unit vectors. InR2these vectors are denoted by
i=(1,0) and j=(0,1) and inR3by
i=(1,0,0), j=(0,1,0), and k=(0,0,1)
(Figure 3.2.2). Every vectorv=(v1, v2)inR2and every vectorv=(v1, v2, v3)inR3
x y
(0, 1)
(1, 0) i j
(a)
(b) x
y z
(0, 1, 0) (1, 0, 0)
(0, 0, 1)
j i k
Figure 3.2.2
can be expressed as a linear combination of standard unit vectors by writing
v=(v1, v2)=v1(1,0)+v2(0,1)=v1i+v2j (5) v=(v1, v2, v3)=v1(1,0,0)+v2(0,1,0)+v3(0,0,1)=v1i+v2j+v3k (6) Moreover, we can generalize these formulas toRnby defining thestandard unit vectors inRnto be
e1=(1,0,0, . . . ,0), e2=(0,1,0, . . . ,0), . . . , en=(0,0,0, . . . ,1) (7) in which case every vectorv=(v1, v2, . . . , vn)inRncan be expressed as
v=(v1, v2, . . . , vn)=v1e1+v2e2+ · · · +vnen (8)
E X A M P L E 3 Linear Combinations of Standard Unit Vectors (2,−3,4)=2i−3j+4k
(7,3,−4,5)=7e1+3e2−4e3+5e4
Distance in Rn IfP1andP2are points inR2orR3, then the length of the vector−−→
P1P2is equal to the distancedbetween the two points (Figure 3.2.3). Specifically, ifP1(x1, y1)andP2(x2, y2) are points inR2, then Formula (4) of Section 3.1 implies that
d = −−→
P1P2 ="
(x2−x1)2+(y2−y1)2 (9)
3.2 Norm, Dot Product, and Distance inRn 145 This is the familiar distance formula from analytic geometry. Similarly, the distance between the pointsP1(x1, y1, z1)andP2(x2, y2, z2)in 3-space is
d = ||P1P2||
P2 d
P1
Figure 3.2.3
d(u,v)= −−→
P1P2 ="
(x2−x1)2+(y2−y1)2+(z2−z1)2 (10) Motivated by Formulas (9) and (10), we make the following definition.
DEFINITION2 Ifu=(u1, u2, . . . , un)andv=(v1, v2, . . . , vn)are points inRn, then we denote thedistancebetweenuandvbyd(u,v)and define it to be
d(u,v)= u−v ="
(u1−v1)2+(u2−v2)2+ · · · +(un−vn)2 (11) We noted in the previous
section that n-tuples can be viewed either as vectors or points in Rn. In Definition 2 we chose to describe them as points, as that seemed the more natural interpretation.
E X A M P L E 4 Calculating Distance inRn If
u=(1,3,−2,7) and v=(0,7,2,2) then the distance betweenuandvis
d(u,v)="
(1−0)2+(3−7)2+(−2−2)2+(7−2)2=√ 58
Dot Product Our next objective is to define a useful multiplication operation on vectors inR2andR3 and then extend that operation toRn. To do this we will first need to define exactly what we mean by the “angle” between two vectors inR2orR3. For this purpose, letuand vbe nonzero vectors inR2 orR3that have been positioned so that their initial points coincide. We define theangle betweenuandvto be the angleθdetermined byuandv that satisfies the inequalities 0≤θ≤π(Figure 3.2.4).
Figure 3.2.4 The angle θ between u and v satisfies 0 ≤ θ ≤ π.
θ v u
θ v
u θ
u v
θ v u
DEFINITION3 Ifuandvare nonzero vectors inR2orR3, and ifθis the angle between uandv, then thedot product(also called theEuclidean inner product) ofuandvis denoted byu·vand is defined as
u·v= uvcosθ (12)
Ifu=0orv=0, then we defineu·vto be 0.
The sign of the dot product reveals information about the angleθthat we can obtain by rewriting Formula (12) as
cosθ= u·v
uv (13)
Since 0≤θ≤π, it follows from Formula (13) and properties of the cosine function studied in trigonometry that
• θis acute ifu·v>0. • θis obtuse ifu·v<0. • θ=π/2 ifu·v=0.
E X A M P L E 5 Dot Product
Find the dot product of the vectors shown in Figure 3.2.5.
z
y
x (0, 0, 1)
(0, 2, 2) v
u θ = 45°
Figure 3.2.5
Solution The lengths of the vectors are
u =1 and v =√
8=2√ 2 and the cosine of the angleθbetween them is
cos(45◦)=1/√ 2 Thus, it follows from Formula (12) that
u·v= uvcosθ=(1)(2√ 2)(1/√
2)=2 Component Form of the
Dot Product
For computational purposes it is desirable to have a formula that expresses the dot product of two vectors in terms of components. We will derive such a formula for vectors in 3-space; the derivation for vectors in 2-space is similar.
Letu=(u1, u2, u3)andv=(v1, v2, v3)be two nonzero vectors. If, as shown in Figure 3.2.6,θis the angle betweenuandv, then the law of cosines yields
v u
θ z
y x
P(u1, u2, u3)
Q(v1, v2, v3)
Figure 3.2.6
−→
PQ2= u2+ v2−2uvcosθ (14)
Since−→
PQ=v−u, we can rewrite (14) as
uvcosθ =12(u2+ v2− v−u2) or
u·v=12(u2+ v2− v−u2) Substituting
u2=u21+u22+u23, v2=v21+v22+v23 and
v−u2=(v1−u1)2+(v2−u2)2+(v3−u3)2 we obtain, after simplifying,
u·v=u1v1+u2v2+u3v3 (15) The companion formula for vectors in 2-space is
Although we derived Formula (15) and its 2-space compan- ion under the assumption that uandvare nonzero, it turned out that these formulas are also applicable if u=0 or
v=0(verify). u·v=u1v1+u2v2 (16)
Motivated by the pattern in Formulas (15) and (16), we make the following definition.
Josiah Willard Gibbs (1839–1903)
Historical Note The dot product notation was first in- troduced by the American physicist and mathemati- cian J. Willard Gibbs in a pamphlet distributed to his students at Yale University in the 1880s. The prod- uct was originally written on the baseline, rather than centered as today, and was referred to as thedirect product. Gibbs’s pamphlet was eventually incorpo- rated into a book entitledVector Analysisthat was pub- lished in 1901 and coauthored with one of his students.
Gibbs made major contributions to the fields of ther- modynamics and electromagnetic theory and is gen- erally regarded as the greatest American physicist of the nineteenth century.
[Image: SCIENCE SOURCE/Photo Researchers/
Getty Images]
3.2 Norm, Dot Product, and Distance inRn 147 DEFINITION4 Ifu=(u1, u2, . . . , un)andv=(v1, v2, . . . , vn)are vectors inRn, then thedot product(also called theEuclidean inner product) ofuandvis denoted byu·v and is defined by
u·v=u1v1+u2v2+ · · · +unvn (17) In words, to calculate the
dot product (Euclidean inner product) multiplycorrespond- ing components and add the resulting products.
E X A M P L E 6 Calculating Dot Products Using Components
(a) Use Formula (15) to compute the dot product of the vectorsuandvin Example 5.
(b) Calculateu·vfor the following vectors inR4:
u=(−1,3,5,7), v=(−3,−4,1,0)
Solution(a) The component forms of the vectors areu=(0,0,1)andv=(0,2,2).
Thus,
u·v=(0)(0)+(0)(2)+(1)(2)=2 which agrees with the result obtained geometrically in Example 5.
Solution(b)
u·v=(−1)(−3)+(3)(−4)+(5)(1)+(7)(0)= −4
E X A M P L E 7 A Geometry Problem Solved Using Dot Product Find the angle between a diagonal of a cube and one of its edges.
Solution Letkbe the length of an edge and introduce a coordinate system as shown in Figure 3.2.7. If we letu1=(k,0,0),u2=(0, k,0), andu3=(0,0, k), then the vector
u3
u2
u1 (0, k, 0)
(k, k, k)
(k, 0, 0) (0, 0, k)
d
θ z
y
x
Figure 3.2.7 d=(k, k, k)=u1+u2+u3
is a diagonal of the cube. It follows from Formula (13) that the angleθbetweendand the edgeu1satisfies
cosθ= u1·d
u1d = k2 (k)(√
3k2)= 1
√3 With the help of a calculator we obtain
Note that the angleθobtained in Example 7 does not involve k. Why was this to be ex- pected?
θ=cos−1 1
√3
!
≈54.74◦
Algebraic Properties of the Dot Product
In the special case whereu=vin Definition 4, we obtain the relationship
v·v=v12+v22+ · · · +v2n= v2 (18) This yields the following formula for expressing the length of a vector in terms of a dot product:
v =√
v·v (19)
Dot products have many of the same algebraic properties as products of real numbers.
THEOREM3.2.2 Ifu,v,andware vectors inRn,and ifkis a scalar,then:
(a) u·v=v·u [ Symmetry property ]
(b) u·(v+w)=u·v+u·w [ Distributive property ]
(c) k(u·v)=(ku)·v [ Homogeneity property ]
(d) v·v≥0andv·v=0if and onlyifv=0 [ Positivity property ]
We will prove parts (c) and (d) and leave the other proofs as exercises.
Proof(c) Letu=(u1, u2, . . . , un)andv=(v1, v2, . . . , vn). Then k(u·v)=k(u1v1+u2v2+ · · · +unvn)
=(ku1)v1+(ku2)v2+ · · · +(kun)vn=(ku)·v
Proof(d) The result follows from parts (a) and (b) of Theorem 3.2.1 and the fact that v·v=v1v1+v2v2+ · · · +vnvn=v12+v22+ · · · +vn2= v2
The next theorem gives additional properties of dot products. The proofs can be obtained either by expressing the vectors in terms of components or by using the algebraic properties established in Theorem 3.2.2.
THEOREM3.2.3 Ifu,v,andware vectors inRn,and ifkis a scalar,then:
(a) 0·v=v·0=0
(b) (u+v)·w=u·w+v·w (c) u·(v−w)=u·v−u·w (d) (u−v)·w=u·w−v·w (e) k(u·v)=u·(kv)
We will show how Theorem 3.2.2 can be used to prove part (b) without breaking the vectors into components. The other proofs are left as exercises.
Proof(b)
(u+v)·w=w·(u+v) [By symmetry]
=w·u+w·v [By distributivity]
=u·w+v·w [By symmetry]
Formulas (18) and (19) together with Theorems 3.2.2 and 3.2.3 make it possible to manipulate expressions involving dot products using familiar algebraic techniques.
E X A M P L E 8 Calculating with Dot Products
(u−2v)·(3u+4v)=u·(3u+4v)−2v·(3u+4v)
=3(u·u)+4(u·v)−6(v·u)−8(v·v)
=3u2−2(u·v)−8v2 Cauchy–Schwarz Inequality
and Angles in Rn
Our next objective is to extend toRnthe notion of “angle” between nonzero vectorsu andv. We will do this by starting with the formula
θ=cos−1 u·v uv
!
(20) which we previously derived for nonzero vectors inR2andR3. Since dot products and norms have been defined for vectors inRn, it would seem that this formula has all the ingredients to serve as adefinitionof the angleθbetween two vectors,uandv, inRn. However, there is a fly in the ointment, the problem being that the inverse cosine in Formula (20) is not defined unless its argument satisfies the inequalities
−1≤ u·v
uv ≤1 (21)
Fortunately, these inequalitiesdohold for all nonzero vectors inRnas a result of the following fundamental result known as theCauchy–Schwarz inequality.
3.2 Norm, Dot Product, and Distance inRn 149
THEOREM3.2.4 Cauchy–Schwarz Inequality
Ifu=(u1, u2, . . . , un)andv=(v1, v2, . . . , vn)are vectors inRn,then
|u·v| ≤ uv (22)
or in terms of components
|u1v1+u2v2+ · · · +unvn| ≤(u21+u22+ · · · +u2n)1/2(v12+v22+ · · · +v2n)1/2 (23)
We will omit the proof of this theorem because later in the text we will prove a more general version of which this will be a special case. Our goal for now will be to use this theorem to prove that the inequalities in (21) hold for all nonzero vectors inRn. Once that is done we will have established all the results required to use Formula (20) as our definitionof the angle between nonzero vectorsuandvinRn.
To prove that the inequalities in (21) hold for all nonzero vectors inRn, divide both sides of Formula (22) by the productuvto obtain
|u·v|
uv ≤1 or equivalently u·v
uv ≤1
from which (21) follows.
Geometry in Rn Earlier in this section we extended various concepts toRnwith the idea that familiar results that we can visualize inR2andR3might be valid inRnas well. Here are two fundamental theorems from plane geometry whose validity extends toRn:
• The sum of the lengths of two side of a triangle is at least as large as the third (Figure 3.2.8).
||u + v|| ≤ ||u|| + ||v||
v
u u + v
Figure 3.2.8
• The shortest distance between two points is a straight line (Figure 3.2.9).
The following theorem generalizes these theorems toRn.
d(u, v) ≤ d(u, w) + d(w, v) u
w v
Figure 3.2.9
THEOREM3.2.5 Ifu,v,andware vectors inRn,then:
(a) u+v ≤ u + v [ Triangle inequality for vectors ]
(b) d(u,v)≤d(u,w)+d(w,v) [ Triangle inequality for distances ]
Hermann Amandus Schwarz
(1843–1921)
Viktor Yakovlevich Bunyakovsky (1804–1889)
Historical Note The Cauchy–Schwarz in- equality is named in honor of the French mathematician Augustin Cauchy (see p. 121) and the German mathemati- cian Hermann Schwarz. Variations of this inequality occur in many different settings and under various names. Depending on the context in which the inequality occurs, you may find it called Cauchy’s inequal- ity, the Schwarz inequality, or sometimes even the Bunyakovsky inequality, in recog- nition of the Russian mathematician who published his version of the inequality in 1859, about 25 years before Schwarz.
[Images: © Rudolph Duehrkoop/
ullstein bild/The Image Works(Schwarz);
http://www-history.mcs.st-and.ac.uk/
Biographies/Bunyakovsky.html (Bunyakovsky)]
Proof(a)
u+v2=(u+v)·(u+v)=(u·u)+2(u·v)+(v·v)
= u2+2(u·v)+ v2
≤ u2+2|u·v| + v2 Property of absolute value
≤ u2+2uv + v2 Cauchy–Schwarz inequality
=(u + v)2
This completes the proof since both sides of the inequality in part (a) are nonnegative.
Proof(b) It follows from part (a) and Formula (11) that d(u,v)= u−v = (u−w)+(w−v)
≤ u−w + w−v =d(u,w)+d(w,v)
It is proved in plane geometry that for any parallelogram the sum of the squares of the diagonals is equal to the sum of the squares of the four sides (Figure 3.2.10). The following theorem generalizes that result toRn.
v
u
u – v u + v
Figure 3.2.10 THEOREM3.2.6 Parallelogram Equation for Vectors
Ifuandvare vectors inRn,then
u+v2+ u−v2=2
u2+ v2
(24)
Proof
u+v2+ u−v2=(u+v)·(u+v)+(u−v)·(u−v)
=2(u·u)+2(v·v)
=2
u2+ v2
We could state and prove many more theorems from plane geometry that generalize toRn, but the ones already given should suffice to convince you thatRnis not so different fromR2andR3even though we cannot visualize it directly. The next theorem establishes a fundamental relationship between the dot product and norm inRn.
THEOREM3.2.7 Ifuandvare vectors inRnwith the Euclidean inner product,then u·v= 14u+v2−14u−v2 (25)
Proof
u+v2=(u+v)·(u+v)= u2+2(u·v)+ v2 u−v2=(u−v)·(u−v)= u2−2(u·v)+ v2 from which (25) follows by simple algebra.
Note that Formula (25) ex- presses the dot product in terms of norms.
Dot Products as Matrix Multiplication
There are various ways to express the dot product of vectors using matrix notation.
The formulas depend on whether the vectors are expressed as row matrices or column matrices. Table 1 shows the possibilities.
3.2 Norm, Dot Product, and Distance inRn 151
Table 1
Form Dot Product Example
ua column matrix andva column matrix
u·v=uTv=vTu u=
⎡
⎣−13 5
⎤
⎦
v=
⎡
⎣54 0
⎤
⎦
uTv= [1 −3 5]
⎡
⎣54 0
⎤
⎦= −7
vTu= [5 4 0]
⎡
⎣−13 5
⎤
⎦= −7
ua row matrix andva column matrix
u·v=uv=vTuT
u= [1 −3 5]
v=
⎡
⎣54 0
⎤
⎦
uv= [1 −3 5]
⎡
⎣54 0
⎤
⎦= −7
vTuT = [5 4 0]
⎡
⎣−13 5
⎤
⎦= −7
ua column matrix andva row matrix
u·v=vu=uTvT u=
⎡
⎣−13 5
⎤
⎦ v= [5 4 0]
vu= [5 4 0]
⎡
⎣−13 5
⎤
⎦= −7
uTvT = [1 −3 5]
⎡
⎣54 0
⎤
⎦= −7
ua row matrix andva row matrix
u·v=uvT =vuT
u= [1 −3 5] v= [5 4 0]
uvT = [1 −3 5]
⎡
⎣54 0
⎤
⎦= −7
vuT = [5 4 0]
⎡
⎣−13 5
⎤
⎦= −7
Application of Dot Products to ISBN Numbers
Although the system has recently changed, most older books have been assigned a unique 10-digit number called anInternational Stan- dard Book Numberor ISBN. The first nine digits of this number are split into three groups—the first group representing the country or group of countries in which the book originates, the second iden- tifying the publisher, and the third assigned to the book title itself.
The tenth and final digit, called acheck digit, is computed from the first nine digits and is used to ensure that an electronic transmission of the ISBN, say over the Internet, occurs without error.
To explain how this is done, regard the first nine digits of the ISBN as a vectorbinR9, and letabe the vector
a=(1,2,3,4,5,6,7,8,9)
Then the check digitcis computed using the following procedure:
1. Form the dot producta·b.
2. Dividea·bby 11, thereby producing a remaindercthat is an integer between 0 and 10, inclusive. The check digit is taken to bec, with the proviso thatc=10 is written as X to avoid double digits.
For example, the ISBN of the brief edition ofCalculus, sixth edition, by Howard Anton is
0-471-15307-9
which has a check digit of 9. This is consistent with the first nine digits of the ISBN, since
a·b=(1,2,3,4,5,6,7,8,9)·(0,4,7,1,1,5,3,0,7)=152 Dividing 152 by 11 produces a quotient of 13 and a remainder of 9, so the check digit isc=9. If an electronic order is placed for a book with a certain ISBN, then the warehouse can use the above procedure to verify that the check digit is consistent with the first nine digits, thereby reducing the possibility of a costly shipping error.
IfAis ann×nmatrix anduandvaren×1 matrices, then it follows from the first row in Table 1 and properties of the transpose that
Au·v=vT(Au)=(vTA)u=(ATv)Tu=u·ATv u·Av=(Av)Tu=(vTAT)u=vT(ATu)=ATu·v The resulting formulas
Au·v=u·ATv (26)
u·Av=ATu·v (27)
provide an important link between multiplication by ann×nmatrixAand multiplica- tion byAT.
E X A M P L E 9 Verifying thatAu·v=u·ATv Suppose that
A=
⎡
⎢⎣
1 −2 3
2 4 1
−1 0 1
⎤
⎥⎦, u=
⎡
⎢⎣
−1 2 4
⎤
⎥⎦, v=
⎡
⎢⎣
−2 0 5
⎤
⎥⎦
Then
Au=
⎡
⎢⎣
1 −2 3
2 4 1
−1 0 1
⎤
⎥⎦
⎡
⎢⎣
−1 2 4
⎤
⎥⎦=
⎡
⎢⎣ 7 10 5
⎤
⎥⎦
ATv=
⎡
⎢⎣
1 2 −1
−2 4 0
3 1 1
⎤
⎥⎦
⎡
⎢⎣
−2 0 5
⎤
⎥⎦=
⎡
⎢⎣
−7 4
−1
⎤
⎥⎦
from which we obtain
Au·v=7(−2)+10(0)+5(5)=11 u·ATv=(−1)(−7)+2(4)+4(−1)=11
Thus,Au·v=u·ATvas guaranteed by Formula (26). We leave it for you to verify that Formula (27) also holds.
A Dot Product View of Matrix Multiplication
Dot products provide another way of thinking about matrix multiplication. Recall that ifA= [aij]is anm×rmatrix andB = [bij]is anr×nmatrix, then theijth entry of ABis
ai1b1j +ai2b2j+ · · · +airbrj
which is the dot product of theith row vector ofA [ai1 ai2 · · · air] and thejth column vector ofB ⎡
⎢⎢
⎢⎢
⎣ b1j
b2j
... brj
⎤
⎥⎥
⎥⎥
⎦
3.2 Norm, Dot Product, and Distance inRn 153 Thus, if the row vectors ofAarer1,r2, . . . ,rm and the column vectors ofB arec1, c2, . . . ,cn, then the matrix productABcan be expressed as
AB=
⎡
⎢⎢
⎢⎣
r1·c1 r1·c2 · · · r1·cn
r2·c1 r2·c2 · · · r2·cn
... ... ... rm·c1 rm·c2 · · · rm·cn
⎤
⎥⎥
⎥⎦ (28)
Exercise Set 3.2
In Exercises1–2, find the norm ofv, and a unit vector that is oppositely directed tov.
1.(a) v=(2,2,2) (b)v=(1,0,2,1,3) 2.(a) v=(1,−1,2) (b)v=(−2,3,3,−1)
In Exercises3–4, evaluate the given expression with u=(2,−2,3),v=(1,−3,4),andw=(3,6,−4).
3.(a) u+v (b)u + v
(c) −2u+2v (d)3u−5v+w
4.(a) u+v+w (b)u−v (c) 3v −3v (d)u − v In Exercises5–6, evaluate the given expression with u=(−2,−1,4,5),v=(3,1,−5,7),andw=(−6,2,1,1).
5.(a) 3u−5v+w (b)3u −5v + w (c) −uv
6.(a) u + −2v + −3w (b)$$u−vw$$
7.Letv=(−2,3,0,6).Find all scalarsksuch thatkv =5. 8.Letv=(1,1,2,−3,1).Find all scalarsksuch that
kv =4.
In Exercises9–10, findu·v,u·u,andv·v.
9.(a) u=(3,1,4), v=(2,2,−4) (b)u=(1,1,4,6), v=(2,−2,3,−2) 10.(a) u=(1,1,−2,3),v=(−1,0,5,1)
(b)u=(2,−1,1,0,−2), v=(1,2,2,2,1)
In Exercises11–12, find the Euclidean distance betweenuandv and the cosine of the angle between those vectors. State whether that angle is acute, obtuse, or 90◦.
11.(a) u=(3,3,3), v=(1,0,4)
(b)u=(0,−2,−1,1), v=(−3,2,4,4) 12.(a) u=(1,2,−3,0),v=(5,1,2,−2)
(b)u=(0,1,1,1,2), v=(2,1,0,−1,3)
13.Suppose that a vectorain thexy-plane has a length of 9 units and points in a direction that is 120◦counterclockwise from
the positivex-axis, and a vectorbin that plane has a length of 5 units and points in the positivey-direction. Finda·b. 14. Suppose that a vectorain thexy-plane points in a direction
that is 47◦ counterclockwise from the positivex-axis, and a vectorbin that plane points in a direction that is 43◦clock- wise from the positivex-axis. What can you say about the value ofa·b?
In Exercises15–16, determine whether the expression makes sense mathematically. If not, explain why.
15. (a) u·(v·w) (b)u·(v+w)
(c) u·v (d)(u·v)− u
16. (a) u·v (b)(u·v)−w (c) (u·v)−k (d)k·u
In Exercises17–18, verify that the Cauchy–Schwarz inequality holds.
17. (a) u=(−3,1,0), v=(2,−1,3) (b)u=(0,2,2,1),v=(1,1,1,1) 18. (a) u=(4,1,1),v=(1,2,3)
(b)u=(1,2,1,2,3), v=(0,1,1,5,−2)
19. Letr0=(x0, y0)be a fixed vector inR2. In each part, describe in words the set of all vectorsr=(x, y)that satisfy the stated condition.
(a) r−r0 =1 (b)r−r0 ≤1 (c) r−r0>1 20. Repeat the directions of Exercise 19 for vectorsr=(x, y, z)
andr0=(x0, y0, z0)inR3.
Exercises 21–25 The direction of a nonzero vectorvin anxyz- coordinate system is completely determined by the anglesα, β, andγ betweenvand the standard unit vectorsi,j, andk(Fig- ure Ex-21). These are called thedirection anglesofv, and their cosines are called thedirection cosinesofv.
21. Use Formula (13) to show that the direction cosines of a vector v=(v1, v2, v3)inR3are
cosα= v1
v, cosβ= v2
v, cosγ= v3
v
v
α γ β
y z
j i
k
x Figure Ex-21
22.Use the result in Exercise 21 to show that cos2α+cos2β+cos2γ =1
23.Show that two nonzero vectorsv1andv2inR3are orthogonal if and only if their direction cosines satisfy
cosα1cosα2+cosβ1cosβ2+cosγ1cosγ2=0 24.The accompanying figure shows a cube.
(a) Find the angle between the vectorsdanduto the nearest degree.
(b) Make a conjecture about the angle between the vectors dandv, and confirm your conjecture by computing the angle.
d
u v z
x
y
Figure Ex-24
25.Estimate, to the nearest degree, the angles that a diagonal of a box with dimensions 10 cm×15 cm×25 cm makes with the edges of the box.
26.Ifv =2 andw =3, what are the largest and smallest val- ues possible forv−w? Give a geometric explanation of your results.
27.What can you say about two nonzero vectors,uandv, that satisfy the equationu+v = u + v?
28.(a) What relationship must hold for the pointp=(a, b, c) to be equidistant from the origin and thexz-plane? Make sure that the relationship you state is valid for positive and negative values ofa,b, andc.
(b) What relationship must hold for the pointp=(a, b, c)to be farther from the origin than from thexz-plane? Make sure that the relationship you state is valid for positive and negative values ofa,b, andc.
29.State a procedure for finding a vector of a specified lengthm that points in the same direction as a given vectorv.
30.Under what conditions will the triangle inequality (Theo- rem 3.2.5a) be an equality? Explain your answer geometri- cally.
Exercises 31–32 The effect that a force has on an object de- pends on the magnitude of the force and the direction in which it is applied. Thus, forces can be regarded as vectors and represented as arrows in which the length of the arrow specifies the magnitude of the force, and the direction of the arrow specifies the direction in which the force is applied. It is a fact of physics that force vectors obey the parallelogram law in the sense that if two force vectors F1 andF2are applied at a point on an object, then the effect is the same as if the single forceF1+F2(called theresultant) were applied at that point (see accompanying figure). Forces are com- monly measured in units called pounds-force (abbreviated lbf) or Newtons (abbreviated N).
F1
F1 + F2 F2
The single force F1 + F2 has the same effect as the two forces F1 and F2.
31.A particle is said to be instatic equilibriumif the resultant of all forces applied to it is zero. For the forces in the accompa- nying figure, find the resultantFthat must be applied to the indicated point to produce static equilibrium. DescribeFby giving its magnitude and the angle in degrees that it makes with the positivex-axis.
32.Follow the directions of Exercise 31.
x y
60°
10 lb
8 lb
Figure Ex-31
x y
75°
45°
150 N 120 N
100 N
Figure Ex-32
Working with Proofs
33.Prove parts (a) and (b) of Theorem 3.2.1.
34.Prove parts (a) and (c) of Theorem 3.2.3.
35.Prove parts (d) and (e) of Theorem 3.2.3.
True-False Exercises
TF.In parts (a)–(j) determine whether the statement is true or false, and justify your answer.