8. 1.1
O Espac¸o Euclidiano Rn
Nesta se¸c˜ao, introduziremos a no¸c˜ao de espa¸co euclidiano Rn
, estabelecendo suas propri-
edades alg´ebricas e geom´etricas b´asicas. Come¸camos com sua defini¸c˜ao.
1.1.1
Definic¸˜ao Dado n ∈ N, o espa¸co euclidiano Rn
´e definido como sendo o conjunto de todas
as n-uplas de n´umeros reais X = (x1, x2, . . . , xn), isto ´e,
Rn
= {X = (x1, x2, . . . , xn); xi ∈ R, i = 1, 2, . . . n}.
1.1.2
Definic¸˜ao Dada uma n-upla X = (x1, x2, . . . , xn), os n´umeros reais x1, x2, . . . , xn s˜ao
chamados coordenadas de X.
1.1.3
Definic¸˜ao Dadas n-uplas X = (x1, x2, . . . , xn) e Y = (y1, y2, . . . , yn), diremos que X = Y
se x1 = y1, x2 = y2, . . . , xn = yn.
1.1.4
Operac¸˜oes com n-uplas
As estruturas aditiva e multiplicativa do corpo R induzem, naturalmente, uma estrutura
de espa¸co vetorial sobre Rn
. A adi¸c˜ao de n-uplas e a multiplica¸c˜ao de uma n-upla por um
n´umero real s˜ao definidas a seguir.
1.1.5
Definic¸˜ao Sejam X = (x1, x2, . . . , xn) e Y = (y1, y2, . . . , yn). A soma de X com Y , indicada
por X + Y , ´e a n-upla
X + Y = (x1 + y1, x2 + y2, . . . , xn + yn).
1.1.6
Definic¸˜ao Sejam X = (x1, x2, . . . , xn) e a ∈ R. O produto de X pelo n´umero real a,
indicada por aX, ´e a n-upla
aX = (ax1, ax2, . . . , axn).
1.1.7
Exemplo No espa¸co R4
, considere X = (1, 2, −π,
√
2), Y = (2,
√
3, π, 1) e a =
√
2. Ent˜ao,
X + Y = (3, 2 +
√
3, 0, 1 +
√
2) e aX = (
√
2, 2
√
2, −π
√
2, 2).
2
9. Vetores e Func¸˜oes Vetoriais 3
1.1.8
Exemplo Em R7
, considere X = (0, 1, 2, −1, 0,
√
2,
√
3) e Y = (2,
√
3, π, 1, 0, 2, −
√
3).
Ent˜ao, X + Y = (2, 1 +
√
3, 2 + π, 0, 0, 2 +
√
2, 0).
As proposi¸c˜oes que seguem mostram que as opera¸c˜oes com n-uplas rec´em-definidas satis-
fazem os axiomas de espa¸co vetorial. Tal fato justifica a terminologia que consiste em chamar
uma n-upla, de vetor no Rn
.
1.1.9
Proposic¸˜ao Se X, Y, Z ∈ Rn
, ent˜ao valem as seguintes propriedades:
(i) [Comutatividade] X + Y = Y + X;
(ii) [Associatividade] (X + Y ) + Z = X + (Y + Z);
(iii) [Elemento Neutro] a n-upla O = (0, 0, . . . , 0), chamada n-upla nula (ou zero), ´e a ´unica
n-upla tal que X + O = X;
(iv) [Sim´etrico] a n-upla −X = (−x1, −x2, . . . , −xn), chamada sim´etrico da n-upla X, ´e a
´unica n-upla tal que X + (−X) = O.
Demonstrac¸˜ao: Vejamos a demonstra¸c˜ao de (ii). As demais s˜ao igualmente simples, e
ser˜ao deixadas como exerc´ıcio para o leitor. Temos que
(X + Y ) + Z = (x1 + y1, x2 + y2, . . . , xn + yn) + (z1, z2, . . . , zn)
= ((x1 + y1) + z1, (x2 + y2) + z2, . . . , (xn + yn) + zn))
= (x1 + (y1 + z1), x2 + (y2 + z2), . . . , xn + (yn + zn))
= X + (Y + Z),
onde, na passagem da segunda para a terceira equa¸c˜ao, foi usada a propriedade associativa dos
n´umeros reais. Os demais pontos envolvem apenas a defini¸c˜ao 1.1.5. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
1.1.10
Proposic¸˜ao Se X, Y ∈ Rn
e a, b ∈ R, ent˜ao valem as seguintes propriedades:
(i) [Distributividade] a(X + Y ) = aX + aY ;
(ii) [Distributividade] (a + b)X = aX + bX;
(iii) [Associatividade] (ab)X = a(bX);
(iv) 1 X = X.
Demonstrac¸˜ao: Sejam X = (x1, x2, . . . , xn) e Y = (y1, y2, . . . , yn). Temos que
a(X + Y ) = a(x1 + y1, x2 + y2, . . . , xn + yn)
= (a(x1 + y1), a(x2 + y2), . . . , a(xn + yn))
= (ax1, ax2, . . . , axn) + (ay1, ay2, . . . , ayn) = aX + aY,
10. 4 O Espac¸o Euclidiano Rn
onde usamos a propriedade distributiva de R junto com a defini¸c˜ao 1.1.5, e obtemos (i). Para (ii),
a propriedade distributiva de R e a defini¸c˜ao 1.1.6 s˜ao usadas:
(a + b)X = ((a + b)x1, (a + b)x2, . . . , (a + b)xn)
= (ax1 + bx1, ax2 + bx2, . . . , axn + bxn)
= aX + bX,
como quer´ıamos. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
1.1.11
Corol´ario Rn
´e um espa¸co vetorial de dimens˜ao n.
Demonstrac¸˜ao: As proposi¸c˜oes 1.1.9 e 1.1.10 mostram que Rn
´e uma espa¸co vetorial.
Falta mostrar que dim Rn
= n. Para isto, sejam e1, e2, . . . , en, definidos por
e1 = (1, 0, 0, . . . , 0)
e2 = (0, 1, 0, . . . , 0)
...
...
...
en = (0, 0, . . . , 0, 1).
Note que se X = (x1, x2, . . . , xn), ent˜ao
X = x1e1 + x2e2 + · · · + xnen.
Logo, {e1, e2, . . . , en} gera Rn
. Agora se c1, c2, . . ., cn s˜ao n´umeros reais tais que
c1e1 + c2e2 + · · · + cnen = (0, 0, . . . , 0),
vem que c1 = c2 = · · · = cn = 0, o que completa a prova. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
1.1.12
Definic¸˜ao A base {e1, e2, . . . , en} constru´ıda acima ´e chamada base canˆonica do espa¸co Rn
.
1.1.13
Interpretac¸˜oes Geom´etricas
A interpreta¸c˜ao geom´etrica que descreve R como uma reta orientada, sobre a qual se esco-
lhe um ponto O, o qual corresponde ao n´umero zero, uma unidade de medida, que corresponde
ao n´umero 1, pode ser estendida a uma interpreta¸c˜ao geom´etrica dos espa¸cos euclidianos R2
e
R3
. Para Rn
, n ≥ 4, fica por conta da imagina¸c˜ao de cada um.
11. Vetores e Func¸˜oes Vetoriais 5
Figura 1: Os N´umeros Reais R
r
0
r
1
r√
3
r
π +∞−∞
r
−
√
2
Para visualizar o R2
, tomamos duas c´opias de R, as quais chamamos de eixos coordenados.
Estes eixos s˜ao denotados por eixo-x e eixo-y. O passo seguinte consiste em dispor os eixos
coordenados em um plano euclidiano de modo que eles se interceptem ortogonalmente ao longo
de suas origens, o que produz o ponto O, que ser´a associado `a dupla (2-upla) nula (0, 0),
conforme figura 2. Feito isso, uma dupla de R2
, digamos X = (x1, x2), ´e olhada como aquele
ponto do plano, tamb´em indicado por X, que se projeta ortogonalmente sobre eixo-x e eixo-y
naqueles pontos que correspondem a x1 e x2, respectivamente. Isto ´e feito tra¸cando-se por x1,
uma reta perpendicular ao eixo-x, e por x2, uma reta perpendicular ao eixo-y. A interse¸c˜ao
destas perpendiculares ´e exatamente o ponto do plano que representar´a X. Deste modo, fica
estabelecida uma bije¸c˜ao entre o plano euclidiano que fixamos e o espa¸co R2
.
H´a situa¸c˜oes em que ´e conveniente representar uma dupla X como um segmento orientado
localizado em O. A figura 2 exibe duas duplas X = (x1, x2) e Y = (y1, y2), de modo que cada
uma delas aparece ora como ponto, ora como segmento orientado. Neste ponto, observamos
que a no¸c˜ao geom´etrica de ˆangulo entre X e Y ´e mais adequada `a figura 2-(c), onde ambas s˜ao
olhadas como segmentos orientados localizados em O. J´a a figura 2-(b) ´e perfeita para motivar
a no¸c˜ao de reta que passa por Y e ´e paralela a X, conforme defini¸c˜ao 1.3.1.
O
s E
T
x
y
Figura 2-(a)
s
Y
y2 r
y1
r
sXx2 r
x1
r
O
s E
T
x
y
Figura 2-(b)
s
Y
y2 r
y1
r¨
¨¨¨¨¨¨¨B
Xx2 r
x1
r
O
s E
T
x
y
Figura 2-(c)
¢
¢
¢
¢
¢
¢¢
Y
y2 r
y1
r¨
¨¨¨¨¨¨¨B
Xx2 r
x1
r
Analogamente, para fazermos geometria em R3
, recorremos a trˆes retas, que formar˜ao os
eixos coordenados, indicados, respectivamente, por eixo-x, eixo-y e eixo-z, e as colocamos no
espa¸co euclidiano tridimensional de modo que elas se interceptem ortogonalmente em suas ori-
gens, produzindo o ponto O, que corresponder´a `a tripla (3-upla) nula (0, 0, 0). Feito isso, temos
em m˜aos trˆes planos especiais, chamados coordenados, e denotados por plano-xy, plano-xz e
plano-yz. Agora a uma tripla X = (x1, x2, x3) fazemos corresponder o ponto do espa¸co cujas
proje¸c˜oes ortogonais sobre os eixos coordenados eixo-x, eixo-y e eixo-z coincidem, respectiva-
mente, com os pontos destes eixos que est˜ao associados aos n´umeros reais x1, x2, x3. A figura 3
mostra como isso ´e feito: inicialmente, marcamos x1 no eixo-x, x2 no eixo-y e x3 no eixo-z.
Depois, a partir de x1, caminhamos paralelamente ao eixo-y at´e atingir a medida x2, onde en-
contramos o ponto que representa a proje¸c˜ao de X no plano-xy, que corresponde a (x1, x2, 0).
13. Vetores e Func¸˜oes Vetoriais 7
Agora podemos descrever geometricamente a adi¸c˜ao de n-uplas e a multiplica¸c˜ao de uma n-upla
por um n´umero real. Para isso, usamos a figura 4, onde marcamos as duplas X = (x1, x2) e
Y = (y1, y2) juntamente com sua soma.
Olhando atentamente o conjunto de figuras 4, observamos que a (b), que mostra X e Y
como segmentos orientados localizados em O, nos d´a uma regra geom´etrica evidente: o segmento
orientado X + Y ´e a diagonal do paralelogramo com arestas X e Y . Portanto, a dupla X + Y
´e o ponto final deste segmento. A regra geom´etrica contida na figura 4-(c) ´e a mais simples:
localizamos em Y , o segmento orientado X. O ponto final obtido ´e a dupla X + Y . F´acil, n˜ao?
Para finalizar, consideremos a figura abaixo que ilustra geometricamente como funciona
a multiplica¸c˜ao de uma n-upla por um escalar: a n-upla aX, a ∈ R, tem comprimento igual ao
comprimento de X multiplicado pelo valor absoluto de a. Seu sentido ´e o mesmo de X, quando
a 0, e lhe ´e contr´ario, quando a 0.
O
s E
T
x
y
Figura 5
¨¨¨
¨¨¨
¨¨B
¨¨¨
¨¨¨
¨¨%
X
−X
x2 r
x1
r¨¨¨
¨¨¨
¨¨¨
¨¨¨B
¨¨¨
¨B
r aX (a 1)
aX (0 a 1)
ax1
r
ax2
1.2
Produto Interno e Norma
Vimos, recentemente, que o espa¸co Rn
possui uma estrutura de espa¸co vetorial. Em
v´arias situa¸c˜oes, precisamos das no¸c˜oes de comprimento, ˆangulo, ortogonalidade, n˜ao presentes
nas opera¸c˜oes de espa¸co vetorial. Estas no¸c˜oes s˜ao obtidas a partir de um produto escalar (ou
interno), que introduziremos agora.
1.2.1
Definic¸˜ao Sejam X = (x1, x2, . . . , xn) e Y = (y1, y2, . . . , yn). O produto escalar (ou in-
terno) de X por Y , indicado por X · Y , ´e o n´umero real dado por
X · Y = x1y1 + x2y2 + · · · + xnyn.
A seguinte proposi¸c˜ao descreve as propriedades deste produto.
14. 8 Produto Interno e Norma
1.2.2
Proposic¸˜ao Se X, Y, Z ∈ Rn
e a ∈ R s˜ao arbitr´arios, ent˜ao valem as seguintes proprieda-
des:
(i) [Positividade] X · X ≥ 0, e X · X = 0 se, e somente se, X = O = (0, 0, . . . , 0);
(ii) [Comutatividade] X · Y = Y · X;
(iii) [Distributividade] X · (Y + Z) = X · Y + X · Z;
(iv) [Homogeneidade] (aX) · Y = X · (aY ) = a(X · Y ).
Demonstrac¸˜ao: Com X = (x1, x2, . . . , xn), Y = (y1, y2, . . . , yn) e Z = (z1, z2, . . . , zn),
temos que
X · (Y + Z) = x1(y1 + z1) + x2(y2 + z2) + · · · + xn(yn + zn)
= x1y1 + x1z1 + x2y2 + x2z2 + · · · + xnyn + xnzn
= (x1y1 + x2y2 + · · · + xnyn) + (x1z1 + x2z2 + · · · + xnzn)
= X · Y + X · Z.
Assim, fica provado (iii). pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
A proposi¸c˜ao 1.2.2, em seu item (i), permite-nos dar a seguinte defini¸c˜ao.
1.2.3
Definic¸˜ao Seja X = (x1, x2, . . . , xn). A norma (ou comprimento) de X ´e dada por
X =
√
X · X = x2
1 + x2
2 + · · · + x2
n.
1.2.4
Definic¸˜ao X ∈ Rn
´e dito unit´ario se X = 1.
1.2.5
Exemplo Se X = (1, 2, −1, 0) e Y = (3, 1, 0, −
√
2), ent˜ao X · Y = 5, X =
√
6 e
Y = 2
√
3.
1.2.6
Proposic¸˜ao Dados X, Y ∈ Rn
e a ∈ R, temos que
(i) X ≥ 0, e X = 0 se, e somente se, X = O;
(ii) aX = |a| X , onde |a| ´e o valor absoluto de a;
(iii) se X = O, o vetor uX = X/ X ´e unit´ario (uX ´e conhecido como vetor unit´ario na dire¸c˜ao
de X);
(iv) X + Y 2
= X 2
+ 2X · Y + Y 2
;
16. 10 Produto Interno e Norma
Tomemos agora X, Y dois vetores em R2
(ou R3
) que fazem entre si um ˆangulo θ, como
mostra a figura 7. (Note a interpreta¸c˜ao geom´etri-
ca para a diferen¸ca Y − X: o segmento orientado,
localizado em X, que representa Y − X termina
em Y .) Aplicando a lei dos cossenos ao triˆangulo
OXY , obtemos que
Y − X 2
= X 2
+ Y 2
− 2 X Y cos θ,
o que comparado com (v) da proposi¸c˜ao 1.2.6 d´a
que
X · Y = X Y cos θ.
O
s E
T
x
y
Figura 7
¢
¢
¢
¢
¢
¢¢
Yy2 r
y1
r¨¨
¨¨
¨¨
¨¨B
Xx2 r
x1
r€€€€€€i
Y − X y2 − x2 r
y1 − x1
r
€€
€€€i
θppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
Assim vemos que a no¸c˜ao de produto interno est´a bem ligada `a no¸c˜ao de ˆangulo entre vetores,
e colhemos a seguinte proposi¸c˜ao, onde ∠(X, Y ) indica o ˆangulo, no intervalo [0, π], entre os
vetores X e Y .
1.2.8
Proposic¸˜ao Se X, Y ∈ Rn
(n = 2, 3) e θ = ∠(X, Y ), ent˜ao X · Y = X Y cos θ. Em
particular, vale a desigualdade de Cauchy-Schwarz: |X · Y | ≤ X Y , a
igualdade ocorrendo apenas quando X e Y s˜ao linearmente dependentes.
1.2.9
Exemplo A proposi¸c˜ao 1.2.8 mostra que dois vetores X e Y em Rn
, n = 2, 3, s˜ao perpen-
diculares se, e somente se, X · Y = 0. De fato, X e Y s˜ao perpendiculares se, e
somente se, ∠(X, Y ) = π/2. Como caso particular disto, note que dado X = (x1, x2), o vetor
Y = (−x2, x1) ´e perpendicular a X. O vetor Y ´e obtido de X por uma rota¸c˜ao em torno de O
no sentido anti-hor´ario. Como exerc´ıcio, o leitor deve esbo¸car X e Y , para se convencer deste
fato.
Continuando com a nossa discuss˜ao geom´e-
trica construiremos, agora, o que chamamos de
proje¸c˜ao ortogonal de um vetor na dire¸c˜ao de ou-
tro n˜ao-nulo dado. Sejam, ent˜ao, X = O e Y
como na figura 8, onde uX = X/ X ´e o vetor
unit´ario na dire¸c˜ao de X, θ = ∠(X, Y ) e PXY ´e
o vetor obtido pela proje¸c˜ao ortogonal de Y sobre
X. Assim PXY = auX, onde
a = Y cos θ = uX · Y =
X · Y
X
.
O
s
e
¨
Figura 8: Proje¸c˜ao de Y sobre X
¨
¨
e
e¨¨
a
¢
¢
¢
¢
¢
¢
¢¢
e
e
e
e
eu
¨
¨¨
¨¨B
Y
¨
¨¨
¨¨¨
¨¨
¨¨B
¨
¨¨B
uX
PXY
Y − PXY
X
θppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
Logo, PXY =
X · Y
X 2 X. Da constru¸c˜ao de PXY decorre facilmente que o vetor Y − PXY ´e
ortogonal a X, o que pode ser verificado, tamb´em, analiticamente:
X · (Y − PXY ) = X · (Y −
X · Y
X 2 X) = X · Y − X · Y = 0.
17. Vetores e Func¸˜oes Vetoriais 11
´E conveniente notar aqui que a express˜ao que define PXY pode muito bem ser usada para
o espa¸co Rn
, visto que ela n˜ao cont´em nenhum apelo geom´etrico expl´ıcito. Isto ser´a parte do
conte´udo da pr´oxima subse¸c˜ao.
1.2.10
Exemplo Considere, em R2
, o triˆangulo ABC, onde
A = (1, 1), B = (3, 2) e C = (0, 4). Os veto-
res X = C − B e Y = A − B aparecem na figura 9 localiza-
dos em B. Temos que X = (−3, 2) e Y = (−2, −1). Assim,
X · Y = 4, X 2
= 13 e
PXY =
4
13
X =
4
13
(−3, 2).
Agora fica f´acil calcular a altura relativa ao lado BC, hBC,
do triˆangulo ABC. De fato, temos
hBC = Y − PXY = (−
14
13
, −
21
13
) = 7
√
13
13
.
O
s E
T
x
y
Figura 9
sBhBC
PXY
r
3
r2
s
A
X
Y
1
1
r
r
ssC4
k
k
¨¨¨¨%f
f
f
f
f
ff
r
1.2.11
A Desigualdade de Cauchy-Schwarz
Inicialmente, nos inspiramos nas no¸c˜oes geom´etricas que usamos h´a pouco, para definir
ortogonalidade entre n-uplas e construir a proje¸c˜ao ortogonal de uma n-upla sobre outra.
1.2.12
Definic¸˜ao Dados X, Y ∈ Rn
, diremos que X ´e ortogonal (perpendicular) a Y se X · Y = 0.
1.2.13
Definic¸˜ao Um subconjunto {v1, v2, . . . , vk} ⊂ Rn
´e dito ortogonal se vi · vj = 0, para
1 ≤ i, j ≤ k, i = j. {v1, v2, . . . , vk} ´e ortonormal se ´e ortogonal e seus elementos
s˜ao vetores unit´arios.
1.2.14
Exemplo Seja {e1, e2, . . . , en} a base canˆonica do espa¸co Rn
(veja defini¸c˜ao 1.1.12). ´E claro
que e1 = e2 = · · · = en = 1. Al´em disto, dados i, j ∈ {1, 2, . . . , n}, i = j,
temos que ei · ej = 0. Logo, a base canˆonica ´e uma base ortonormal do espa¸co Rn
.
1.2.15
Definic¸˜ao Dados dois vetores X, Y ∈ Rn
, X = O, o vetor
PXY =
X · Y
X 2 X
´e chamado proje¸c˜ao de Y sobre X.
18. 12 Produto Interno e Norma
1.2.16
Proposic¸˜ao Sejam X, Y ∈ Rn
com X = O. Ent˜ao, Y − PXY ´e perpendicular a X.
Portanto, ´e tamb´em perpendicular a PXY .
Demonstrac¸˜ao: X · (Y − PXY ) = X · (Y −
X · Y
X 2 X) = X · Y − X · Y = 0. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
Agora podemos estender o teorema de Pit´agoras para o Rn
.
1.2.17 [Pit´agoras]
Proposic¸˜ao Sejam X, Y ∈ Rn
com X = O. Ent˜ao, X ´e perpendicular a Y se, e
somente se, X + Y 2
= X 2
+ Y 2
.
Demonstrac¸˜ao: Resulta imediatamente de X + Y 2
= X 2
+ Y 2
+ 2X · Y , como
indica a proposi¸c˜ao 1.2.6, item (iv). pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
Enfim a desigualdade de Cauchy-Schwarz, j´a obtida via argumento geom´etricos para os
espa¸cos R2
e R3
.
1.2.18 [Cauchy-Schwarz]
Teorema Sejam X, Y ∈ Rn
. Ent˜ao, |X · Y | ≤ X Y , e a igual-
dade ´e atingida se, e somente se, X e Y s˜ao linearmente
dependentes.
Demonstrac¸˜ao: Inicialmente notamos que se X = O, a desigualdade ´e facilmente
verificada. Portanto, podemos supor X = O. Seja PXY a proje¸c˜ao ortogonal de Y sobre X.
Segue-se da proposi¸c˜ao 1.2.16 que Y − PXY ´e perpendicular a PXY . Usando a proposi¸c˜ao 1.2.17,
obtemos que
Y 2
= (Y − PXY ) + PXY 2
= (Y − PXY ) 2
+ PXY 2
≥ PXY 2
, (¶1)
e a igualdade ocorre se, e somente se, Y = PXY =
X · Y
X 2 X. Mas
PXY 2
=
X · Y
X 2 X
2
=
(X · Y )2
X 2 ,
o que combinado com a desigualdade (¶1) d´a (X · Y )2
≤ X 2
Y 2
, como quer´ıamos. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
Como corol´ario da desigualdade de Cauchy-Schwarz, obtemos as propriedades da norma
que faltavam ser apresentadas: as desigualdades triangulares.
1.2.19
Corol´ario Se X, Y ∈ Rn
e a ∈ R, ent˜ao
(i) X ≥ 0, e X = 0 se, e somente se, X = O;
(ii) aX = |a| X ;
19. Vetores e Func¸˜oes Vetoriais 13
(iii) [Desigualdade triangular] X + Y ≤ X + Y ;
(iv) [Desigualdade triangular] | X − Y | ≤ X − Y .
Demonstrac¸˜ao: Note que (i) e (ii) aparecem na proposi¸c˜ao 1.2.6. Daremos uma prova
para (iii) e (iv). Temos que
X + Y 2
= X 2
+ 2X · Y + Y 2
≤ X 2
+ 2|X · Y | + Y 2
≤ X 2
+ 2 X Y + Y 2
= ( X + Y )2
,
o que implica (iii). A segunda desigualdade triangular resulta da primeira. De fato,
X = (X − Y ) + Y ≤ X − Y + Y .
Logo,
X − Y ≤ X − Y . (¶2)
Trocando X por Y , vem que
Y − X ≤ Y − X = X − Y . (¶3)
Agora, juntando (¶2) e (¶3), segue-se que
− X − Y ≤ X − Y ≤ X − Y ,
o que ´e equivalente a
| X − Y | ≤ X − Y ,
e est´a pronto o corol´ario. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
Neste ponto, tomamos duas n-uplas n˜ao-nulas X e Y . Da desigualdade de Cauchy-
Schwarz, obtemos que
−1 ≤
X · Y
X Y
≤ 1.
Logo, existe um ´unico n´umero real θ ∈ [0, π] tal que
cos θ =
X · Y
X Y
.
Posto isto, temos a seguinte defini¸c˜ao.
1.2.20
Definic¸˜ao Dadas as n-uplas n˜ao-nulas X e Y , o n´umero real
∠(X, Y ) = arccos
X · Y
X Y
.
´e chamado ˆangulo entre X e Y .
20. 14 Produto Interno e Norma
1.2.21
Exemplo Sejam X = (1, 2, 1, 0) e Y = (1, 1, 3, 1) dois elementos do R4
. Ent˜ao, X =
√
6,
Y = 2
√
3 e X · Y = 6. Logo, ∠(X, Y ) = arccos
6
2
√
18
= arccos
√
2
2
=
π
4
.
Observac¸˜ao Seja V um espa¸co vetorial qualquer sobre R, de dimens˜ao finita ou n˜ao. Um
produto interno em V ´e definido como sendo uma forma bilinear sim´etrica e
positiva definida, que indicamos por , . Isto significa que se X, Y, Z ∈ V e a ∈ R, ent˜ao
devem valer:
(i) X, X ≥ 0, e X, X = 0 se, e somente se, X ´e o vetor nulo de V;
(ii) X, Y = Y, X ;
(iii) X, Y + Z = X, Y + X, Z ;
(iv) a X, Y = aX, Y = X, aY .
Note que o produto escalar · que definimos para o Rn
satisfaz estas propriedades, como indica a
proposi¸c˜ao 1.2.2. O que queremos chamar a aten¸c˜ao aqui ´e que todo o conte´udo desta subse¸c˜ao
poderia ser aplicado para o espa¸co V, com apenas uma mudan¸ca, a saber: a troca do produto
escalar · por , . Em particular, ter´ıamos a desigualdade de Cauchy-Schwarz:
| X, Y | ≤ X Y ,
onde, ´e claro, X = X, X . Esta norma tamb´em satisfaz as propriedades do corol´ario 1.2.19.
Para finalizar esta subse¸c˜ao, consideraremos em Rn
a distˆancia induzida por sua norma.
1.2.22
Definic¸˜ao A distˆancia entre as n-uplas X e Y , indicada por d(X, Y ), ´e o n´umero real
d(X, Y ) = Y − X = (y1 − x1)2 + (y2 − x2)2 + · · · + (yn − xn)2.
(A figura 7 sugere, tamb´em, esta defini¸c˜ao.)
1.2.23
Exemplo Se X = (1, 2, 3, −1, 2), Y = (1, 1, 2, 0, 1), ent˜ao d(X, Y ) = 2.
1.2.24
Proposic¸˜ao Sejam X, Y, Z ∈ Rn
. A distˆancia tem as seguintes propriedades.
(i) d(X, Y ) ≥ 0, e d(X, Y ) = 0 se, e somente se, X = Y ;
(ii) d(X, Y ) = d(Y, X);
21. Vetores e Func¸˜oes Vetoriais 15
(iii) d(X, Z) ≤ d(X, Y ) + d(Y, Z).
Demonstrac¸˜ao: Para (ii) basta observar que Y − X = X − Y . Vejamos (iii).
d(X, Z) = Z − X
= (Z − Y ) + (Y − X)
≤ Z − Y + Y − X
≤ d(X, Y ) + d(Y, Z),
onde a desigualdade obtida vem do corol´ario 1.2.19, item (iii). pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
1.3
Retas e Planos
Como vimos fazendo at´e aqui, para definirmos reta e plano no Rn
, usaremos alguns argu-
mentos geom´etricos no espa¸co euclidiano R2
.
A figura 10 ao lado mostra a dupla P, os
vetores V = O e N (perpendicular a V ) e a reta
l que passa por P e ´e paralela a V . Se X ´e um
ponto qualquer de l, ent˜ao o vetor X −P deve ser
um m´ultiplo de V , isto ´e, existe t ∈ R tal que
X − P = tV,
ou
X = P + tV,
equa¸c˜ao que descreve os pontos de l, e motiva a
seguinte defini¸c˜ao.
O
s E
T
x
y
Figura 10: Reta passando por P
e paralela a V
s¨¨¨
¨¨¨
¨¨¨
¨
¨¨¨¨¨¨¨¨
¨¨¨
¨¨¨
¨
¨¨¨B
¨¨¨
¨¨Bs
X
¨¨
¨¨¨
P
X − P
¨¨¨B
e
e
eu
V
N
l = P + [V ]
[V ]
1.3.1
Definic¸˜ao Dados P, V ∈ Rn
, V = O, o subconjunto l = P + [V ], onde [V ] indica o
subespa¸co gerado por V , ´e chamado reta que passa por P e ´e paralela ao vetor V .
Assim,
l = {X ∈ Rn
; X = P + tV, t ∈ R}.
A equa¸c˜ao X = P + tV ´e a equa¸c˜ao param´etrica de l.
1.3.2
Exemplo Dados P, Q ∈ Rn
, P = Q, a reta que passa por P (ou Q) e ´e paralela ao vetor
Q − P ´e a reta lPQ = P + [Q − P]. Para t = 1, obtemos X = P + t(Q − P) = Q.
22. 16 Retas e Planos
Logo, Q ∈ l, o que implica que l ´e a reta que passa por P e Q. Deixando t percorrer o intervalo
fechado [0, 1], obtemos o subconjunto [P, Q] ⊂ lPQ, o qual
chamaremos de segmento de reta ligando P a Q. Assim,
[P, Q] = {X = P + t(Q − P); 0 ≤ t ≤ 1}.
Para t = 1/2, obtemos
M = P +
1
2
(Q − P) =
P + Q
2
∈ [P, Q],
¨¨¨
¨¨
Figura 11: Segmento [P, Q]
P
QM
s
s
s
o ponto m´edio de [P, Q]. Observe que d(M, P) = d(M, Q) = M − P = M − Q = d(P, Q)/2.
1.3.3
Exemplo Tomemos, em R2
, P = (x0, y0) e V = (v1, v2) = (0, 0). Se
X = (x, y) ∈ l = P + [V ] = {X = (x, y) = (x0, y0) + t(v1, v2), t ∈ R}
´e um ponto qualquer de l, ent˜ao x = x0 + tv1 e y = y0 + tv2, t ∈ R. Donde
v2x = v2x0 + tv2v1 e v1y = v1y0 + tv1v2
e, portanto,
ax + by = c,
onde a = −v2, b = v1 e c = ax0 + by0. Esta ´e a equa¸c˜ao cartesiana de l, forma usual nos textos
elementares de Geometria Anal´ıtica, e que pode ser reescrita como
(X − P) · N = 0,
onde N = (a, b) = (−v2, v1) ´e perpendicular a V e, portanto, a l (veja o exemplo 1.2.9).
Observac¸˜ao Uma reta l = P + [V ] n˜ao determina unicamente P e V . De fato, se Q ∈ l ´e
um ponto qualquer de l e W = λV , λ = 0, ent˜ao l = Q + [W].
1.3.4
Definic¸˜ao Duas retas no Rn
, l1 = P + [V ] e l2 = Q + [W], s˜ao ditas paralelas se V e W
s˜ao linearmente dependentes.
1.3.5
Exemplo Sejam l1 = P + [V ] e l2 = Q + [W] duas retas no R2
que n˜ao s˜ao paralelas.
Logo, como nossa intui¸c˜ao espera, l1 e l2 devem se tocar num ´unico ponto (o
que pode n˜ao ocorrer em dimens˜oes maiores que 2, como mostra o exemplo 1.3.6). Com efeito,
{V, W} ´e uma base do R2
(por quˆe?) e, portanto, devem existir ´unicos t1, t2 ∈ R tais que
Q − P = t1V + t2W. Donde, Q − t2W = P + t1V . Mas P + t1V ∈ l1 e Q − t2W ∈ l2. Logo, l1
e l2 se interceptam em R = Q − t2W = P + t1V .
24. 18 Retas e Planos
cont´em os pontos Q e R. Para ver isto, ponha s = 1 e t = 0, para obter Q, e s = 0 e t = 1, para
encontrar R. Este ´e o plano que passa pelos pontos P, Q, R, que indicaremos por πPQR. Como
caso particular, tomemos, em R3
, os pontos P = (0, 0, 2), Q = (4, 1, 0) e R = (1, 1, 1). Ent˜ao,
V = (4, 1, −2) e W = (1, 1, −1), e πPQR fica
πPQR = {(x, y, z) = (0, 0, 2) + s(4, 1, −2) + t(1, 1, −1), s, t ∈ R}
= {(x, y, z) = (4s + t, s + t, 2 − 2s − t), s, t ∈ R},
Eliminando s e t na equa¸c˜ao param´etrica obtida, obtemos que
πPQR = {(x, y, z); x + 2y + 3z = 6},
que ´e a forma cartesiana de πPQR. Note que os coeficientes desta ´ultima equa¸c˜ao, a saber, 1,
2 e 3, d˜ao origem ao vetor N = (1, 2, 3) que, como ´e f´acil de ver, ´e perpendicular aos vetores
V e W. Portanto, N ´e tamb´em perpendicular a πPQR. A seguinte proposi¸c˜ao generaliza esta
situa¸c˜ao.
1.3.9
Proposic¸˜ao Seja π = P + [{V, W}] um plano do R3
. Ent˜ao existe N = (a, b, c), n˜ao-nulo,
perpendicular a V e W (e portanto perpendicular a π) tal que
π = {X ∈ R3
; (X − P) · N = 0} = {(x, y, z); ax + by + cz = d},
onde d = N · P.
Demonstrac¸˜ao: Sejam P = (p1, p2, p3), V = (v1, v2, v3) e W = (w1, w2, w3). Assim,
π = {(x, y, z) = (p1 + sv1 + tw1, p2 + sv2 + tw2, p3 + sv3 + tw3), s, t ∈ R}. (¶4)
Como V e W s˜ao linearmente independentes, a matriz
v1 w1
v2 w2
v3 w3
tem posto 2. Resulta da´ı, que pelo menos uma das matrizes
v1 w1
v2 w2
,
v1 w1
v3 w3
e
v2 w2
v3 w3
tem determinante n˜ao-nulo. Logo, podemos supor, sem perda de generalidade, que a primeira
destas matrizes tem inversa, a qual ´e dada por
v1 w1
v2 w2
−1
=
1
v1w2 − v2w1
w2 −w1
−v2 v1
.
Seja X = (x, y, z) ∈ π um ponto qualquer. De (¶4) vem que
x − p1
y − p2
=
v1 w1
v2 w2
s
t
z − p3 = (v3 w3)
s
t
.
25. Vetores e Func¸˜oes Vetoriais 19
Logo,
z − p3 = (v3 w3)
v1 w1
v2 w2
−1
x − p1
y − p2
=
1
v1w2 − v2w1
(v3 w3)
w2 −w1
−v2 v1
x − p1
y − p2
.
Donde,
(v2w3 − v3w2)(x − p1) + (v3w1 − v1w3)(y − p2) + (v1w2 − v2w1)(z − p3) = 0,
ou (X − P) · N = 0, onde
N = (v2w3 − v3w2, v3w1 − v1w3, v1w2 − v2w1).
A equa¸c˜ao (X − P) · N = 0 implica, em particular, que N ´e perpendicular aos vetores V e W.
De fato, tomando X = P + V ∈ π, temos que (X − P) · N = V · N = 0. Da mesma forma,
vemos que W · N = 0. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
A equa¸c˜ao (X −P)·N = 0, obtida para planos no R3
, serve, como vimos no exemplo 1.3.3,
tamb´em para retas em R2
. Isto sugere a seguinte defini¸c˜ao.
1.3.10
Definic¸˜ao Sejam P, N ∈ Rn
, onde N ´e um vetor n˜ao-nulo. O subconjunto
H = {X ∈ Rn
; (X − P) · N = 0}
´e chamado hiperplano que passa por P e ´e perpendicular a N.
Observac¸˜ao Os hiperplanos de R2
s˜ao retas; os do R3
s˜ao os planos. Por analogia a estes
casos ´e de se esperar que a “dimens˜ao” destes objetos dependa do ambiente no
qual eles habitam. Um hiperplano H ⊂ Rn
deve ter dimens˜ao n − 1.
1.3.11
Exemplo Seja H = {X = (x1, x2, x3, x4) ∈ R4
; x1 + x2 − 2x3 − x4 = 1}. Temos que H ´e o
hiperplano do R4
que ´e perpendicular a N = (1, 1, −2, −1) e passa, por exemplo,
por P = (0, 0, 0, −1). Agora observe que X ∈ H se, e somente se,
X = (x1, x2, x3, x1 + x2 − 2x3 − 1) = P + x1(1, 0, 0, 1) + x2(0, 1, 0, 1) + x3(0, 0, 1, −2),
o que mostra que os pontos de H s˜ao descritos por uma equa¸c˜ao param´etrica a trˆes parˆametros.
Isto basta para sentir que a dimens˜ao de H ´e 3.
1.3.12
Produto Vetorial
Devido ao seu valor geom´etrico, o vetor N constru´ıdo na proposi¸c˜ao 1.3.9 merece destaque
especial. Nesta subse¸c˜ao colocaremos as propriedades b´asicas deste vetor.
26. 20 Retas e Planos
1.3.13
Definic¸˜ao Sejam X = (x1, x2, x3) e Y = (y1, y2, y3) duas triplas em R3
. O produto vetorial
de X por Y , denotado por X × Y (ou X ∧ Y ), ´e definido por
X × Y = (x2y3 − x3y2, x3y1 − x1y3, x1y2 − x2y1),
que pode ser facilmente lembrado expandindo o determinante abaixo ao longo da primeira linha:
X × Y =
e1 e2 e3
x1 x2 x3
y1 y2 y3
= (x2y3 − x3y2)e1 + (x3y1 − x1y3)e2 + (x1y2 − x2y1)e3,
onde {e1, e2, e3} ´e a base canˆonica do R3
.
1.3.14
Exemplo Sejam X = (1, 1, 2) e Y = (3, −1, 1). O produto vetorial de X por Y ´e o vetor
X × Y =
e1 e2 e3
1 1 2
3 −1 1
= 3e1 + 5e2 − 4e3 = (3, 5, −4).
Note que (X × Y ) · X = (3, 5, −4) · (1, 1, 2) = 0. Tamb´em (X × Y ) · Y = 0, o que diz que X × Y
´e perpendicular a X e a Y . Esta propriedade ´e verdadeira em geral, como veremos a seguir.
1.3.15
Proposic¸˜ao Sejam X, Y, Z ∈ R3
e a ∈ R. As seguintes propriedades s˜ao verificadas.
(i) X × Y = −Y × X;
(ii) a(X × Y ) = (aX) × Y = X × (aY );
(iii) X × (Y + Z) = X × Y + X × Z;
(iv) (X × Y ) · Z = det (X, Y, Z);
(v) X × Y 2
= X 2
Y 2
− (X · Y )2
,
onde (X, Y, Z) indica a matriz cujas colunas (ou linhas) s˜ao as triplas X, Y e Z, olhadas como
matrizes 3 × 1. Assim,
det (X, Y, Z) =
x1 y1 z1
x2 y2 z2
x3 y3 z3
=
x1 x2 x3
y1 y2 y3
z1 z2 z3
.
Demonstrac¸˜ao: A demonstra¸c˜ao destas propriedades ´e feita via computa¸c˜ao direta,
usando a defini¸c˜ao 1.3.13. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
28. 22 Retas e Planos
1.3.19
Corol´ario Se X, Y ∈ R3
s˜ao ortonormais (unit´arios e ortogonais), ent˜ao {X, Y, X × Y } ´e
uma base ortonormal do R3
.
Demonstrac¸˜ao: Falta verificar que X × Y ´e tamb´em unit´ario. Isto segue-se facilmente
de (v) da proposi¸c˜ao 1.3.15. Com efeito,
X × Y 2
= X 2
Y 2
− (X · Y )2
= 1 − 0 = 1. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
1.3.20 [Duplo Produto Vetorial]
Corol´ario Se X, Y, Z ∈ R3
, ent˜ao vale a f´ormula do duplo
produto vetorial:
(X × Y ) × Z = (X · Z) Y − (Y · Z) X.
Demonstrac¸˜ao: Suponhamos, inicialmente, que X e Y sejam ortonormais. Logo,
{X, Y, X × Y } ´e uma base ortonormal do R3
, e isto implica que existem (e s˜ao ´unicos) n´umeros
reais c1, c2 e c3 tais que Z = c1X + c2Y + c3X × Y . Na realidade, c1 = X · Z, c2 = Y · Z e
c3 = (X ×Y )·Z. Como (X ×Y )×Z ´e perpendicular a X ×Y , vem que ele deve ser combina¸c˜ao
linear de X e Y . Portanto,
(X × Y ) × Z = aX + bY,
onde a = ((X × Y ) × Z) · X e b = ((X × Y ) × Z) · Y . Mas
((X × Y ) × Z) · X = det (X × Y, Z, X)
= det (X × Y, c1X + c2Y + c3X × Y, X)
= c2 det (X × Y, Y, X)
= −c2 det (X, Y, X × Y )
= −c2 X × Y 2
= −c2 = −Y · Z.
Analogamente, vemos que ((X × Y ) × Z) · Y = c1 = X · Z. Logo, a = −Y · Z e b = X · Z, o
que prova a f´ormula do duplo produto vetorial para o caso onde X e Y s˜ao ortonormais. Para
estendˆe-la para o caso onde X e Y s˜ao apenas ortogonais, tomamos os unit´arios uX e uY . Logo
(uX × uY ) × Z = (uX · Z) uY − (uY · Z) uX,
ou, equivalentemente,
1
X Y
(X × Y ) × Z =
1
X Y
(X · Z) Y −
1
X Y
(Y · Z) X,
que simplificada d´a
(X × Y ) × Z = (X · Z) Y − (Y · Z) Y,
e a f´ormula do duplo produto vetorial funciona tamb´em quando X e Y s˜ao ortogonais. Para o
caso geral, onde X e Y s˜ao linearmente independentes, recorremos `a proje¸c˜ao de Y sobre X,
29. Vetores e Func¸˜oes Vetoriais 23
PXY . Da proposi¸c˜ao 1.2.16 vem que PXY = λX, onde λ = (X · Y )/ X 2
, e que Y − PXY ´e
perpendicular a X. Logo,
(X × (Y − PXY )) × Z = (X · Z) (Y − PXY ) − ((Y − PXY ) · Z) X. (¶5)
Mas
X × (Y − PXY ) = X × Y − X × PXY = X × Y,
visto que PXY ´e paralelo a X, e
(X · Z)(Y − PXY ) − ((Y − PXY ) · Z)X = (X · Z)Y − (X · Z)PXY − (Y · Z)X + (PXY · Z)X
= (X · Z)Y − (Y · Z)X − λ((X · Z)X − (X · Z)X)
= (X · Z)Y − (Y · Z)X.
Assim, (¶5) fica:
(X × Y ) × Z = (X · Z) Y − (Y · Z) X.
A f´ormula do duplo produto vetorial agora vale sempre que X e Y s˜ao linearmente independentes.
O caso onde X e Y s˜ao linearmente dependente (Y = kX) ´e trivial: a f´ormula tem ambos os
membros nulos, como pode ser facilmente verificado pelo leitor. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
1.3.21
Corol´ario Sejam T e N dois vetores ortonormais do R3
. Se B = T ×N, ent˜ao B × T = N
e N × B = T.
Demonstrac¸˜ao: A f´ormula do duplo produto vetorial d´a que
B × T = (T × N) × T = (T · T)N − (N · T)T = N,
visto que T = 1 e T · N = 0. Agora convidamos o leitor a provar que N × B = T. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
1.3.22
Exemplo Sejam e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1). Um c´alculo direto mostra que
e1 × e2 = e3. Logo, e2 × e3 = e1 e e3 × e1 = e2. Em particular,
(e1 × e2) × e2 = e3 × e2 = −e1
e1 × (e2 × e2) = e1 × O = O,
o que implica que (e1 × e2) × e2 = e1 × (e2 × e2) e mostra que o produto vetorial, em geral, n˜ao
´e associativo. O pr´oximo corol´ario mostra quando o produto vetorial ´e associativo.
30. 24 Retas e Planos
1.3.23
Corol´ario Dados vetores X, Y e Z em R3
, ent˜ao X × (Y × Z) = (X × Y ) × Z se, e
somente se, (X · Y ) Z = (Y · Z) X.
Demonstrac¸˜ao: Usando o corol´ario 1.3.20 temos que
X × (Y × Z) = −(Y × Z) × X = −((X · Y ) Z − (X · Z) Y ) = (X · Z) Y − (X · Y ) Z,
que comparado com
(X × Y ) × Z = (X · Z) Y − (Y · Z) X,
mostra que X × (Y × Z) = (X × Y ) × Z se, e somente se, (X · Y ) Z = (Y · Z) X. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
1.3.24
Distˆancia de um Ponto a uma Reta
Seja l = P + [V ] uma reta no Rn
. Dado
Q ∈ Rn
um ponto qualquer, seja Y = Q − P,
como mostra a figura 15. A proje¸c˜ao de Y sobre
V ´e dada por
PV Y = λV, onde λ =
Y · V
V 2 .
Como a figura 15 mostra ´e bastante razo´avel se
esperar que a distˆancia de Q a l, que indicaremos
por d(Q, l), definida como sendo o m´ınimo das
distˆancias de Q a pontos de l, isto ´e,
d(Q, l) = min{d(Q, X), X ∈ l},
¨¨
¨¨¨
¨¨¨
¨¨¨
¨¨
¨
¨¨¨
Figura 15
¢
¢
¢
¢
¢
¢
¢¢
e
e
e
e
eu
¨¨
¨¨¨B
Y = Q − P
Qs
¨¨
¨¨¨
¨¨¨
¨¨B
PV Y
Y − PV Y
V
l = P + [V ]
sQ
P
s
seja atingida no ponto Q ∈ l, a proje¸c˜ao ortogonal de Q sobre l, dado por
Q = P + PV Y = P + λV = P +
(Q − P) · V
V 2 V.
Portanto,
(d(Q, l))2
= Y − PV Y 2
= (Q − P) −
(Q − P) · V
V 2 V
2
= Q − P 2
− 2(Q − P) · (
(Q − P) · V
V 2 V ) +
((Q − P) · V )2
V 4 V 2
=
Q − P 2
V 2
− ((Q − P) · V )2
V 2 ,
o que produz o seguinte resultado.
31. Vetores e Func¸˜oes Vetoriais 25
1.3.25
Proposic¸˜ao Seja l = P +[V ] a reta do Rn
que passa por P e ´e paralela a V . Dado Q ∈ Rn
a distˆancia de Q a l ´e dada por
d(Q, l) =
Q − P 2
V 2
− ((Q − P) · V )2
V
.
Al´em disto, o ponto Q ∈ l onde esta distˆancia ´e atingida ´e dado por
Q = P +
(Q − P) · V
V 2 V.
A partir desta proposi¸c˜ao obtemos as f´ormulas usuais da distˆancia de um ponto a uma
reta, em R3
e R2
, como mostram os corol´arios 1.3.27 e 1.3.28.
1.3.26
Exemplo Sejam P = (1, 0, −2, 3), Q = (1, 1, , 0, 2) e V = (1, 1, −1, 1), e consideremos a
reta l = P + [V ]. Temos que Q − P 2
= 6, (Q − P) · V = −2 e V = 2. Logo,
usando a proposi¸c˜ao 1.3.25, a distˆancia de Q a l ´e
d(Q, l) =
Q − P 2
V 2
− ((Q − P) · V )2
V
=
√
5.
O ponto Q ∈ l onde d(Q, l) ´e atingida ´e dado por
Q = P +
(Q − P) · V
V 2 V = (1, 0, −2, 3) −
1
2
(1, 1, −1, 1) =
1
2
(1, −1, −3, 5).
Sugerimos ao leitor o c´alculo de d(Q, Q ) que, claro, deve produzir
√
5.
1.3.27
Corol´ario Seja l = P + [V ] a reta do R3
que passa por P e ´e paralela a V . Dado Q ∈ R3
a distˆancia de Q a l ´e dada por
d(Q, l) =
(Q − P) × V
V
.
Demonstrac¸˜ao: Resulta de (v), proposi¸c˜ao 1.3.15, junto com a proposi¸c˜ao 1.3.25. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
1.3.28
Corol´ario Seja l = P + [V ] a reta do R2
que passa por P = (x1, x2) e ´e paralela a
V = (v1, v2). Dado Q = (x0, y0) a distˆancia de Q a l ´e dada por
d(Q, l) =
|ax0 + by0 − c|
√
a2 + b2
,
32. 26 Retas e Planos
onde N = (a, b) = (−v2, v1) ´e normal a l e c = ax1 + bx2. (Neste caso, a equa¸c˜ao cartesiana de
l ´e: ax + by = c.)
Demonstrac¸˜ao: Visando utilizar o corol´ario 1.3.27, mergulharemos R2
em R3
, isto ´e,
olharemos uma dupla X = (x1, x2), como sendo a tripla X = (x1, x2, 0). Assim,
d(Q, l) =
(x0 − x1, y0 − y1, 0) × (v1, v2, 0)
(v1, v2, 0)
,
que expressa em termos de a = −v2, b = v1 e c = ax1 + bx2 fica:
d(Q, l) =
(0, 0, −ax0 − by0 + c)
(b, −a, 0)
=
|ax0 + by0 − c|
√
a2 + b2
. pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
1.3.29
Distˆancia de um Ponto a um Hiperplano
Seja π o plano do R3
que ´e perpendicular a N e passa por P, conforme mostra a figura 16.
Dado Q ∈ R3
, a distˆancia de Q a π, d(Q, π), ´e definida como sendo o m´ınimo das distˆancias de
Q a pontos de π, isto ´e,
d(Q, π) = min{d(Q, X); X ∈ π}.
Seja l a reta que passa por Q e ´e paralela a N. Temos que l intercepta (ortogonalmente) π no
ponto
Q = Q − PN (Q − P) = Q −
(Q − P) · N
N 2 N,
onde PN (Q − P) ´e a proje¸c˜ao de Q − P sobre N.
Fixemos X ∈ π um ponto arbitr´ario. Como Q − Q ´e perpendicular a π, ele ´e perpendicular a
X − Q . Usando o teorema de Pit´agoras (veja proposi¸c˜ao 1.2.17), vem que
X − Q 2
= X − Q − Q + Q 2
= X − Q 2
+ Q − Q 2
≥ Q − Q 2
e, portanto, obtemos
Q − Q ≤ X − Q , ∀X ∈ π.
Segue-se, ent˜ao, que d(Q, π) ´e atingida em Q e
d(Q, π) = Q − Q =
(Q − P) · N
N 2 N =
|(Q − P) · N|
N
.
Isto prova a seguinte proposi¸c˜ao.
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ˆ
¨¨¨¨
¨¨¨¨¨¨¨
ˆˆˆˆˆˆˆˆ
d
d
d
d
d
d
d
d
g
g
g
g
g
g
g
g
Q
d(Q, π) d(Q, X)
Q
π
Q − PT
N
P
s
sXs
Figura 16
33. Vetores e Func¸˜oes Vetoriais 27
1.3.30
Proposic¸˜ao Seja H o hiperplano do Rn
que ´e perpendicular a N e passa por P. Dado
Q ∈ Rn
, a distˆancia de Q a H ´e dada por
d(Q, H) =
|(Q − P) · N|
N
.
Mais ainda, d(Q, H) ´e atingida no ponto de H
Q = Q −
(Q − P) · N
N 2 N.
Em particular, se n = 3, Q = (x0, y0, z0), N = (a, b, c) e d = N · P, a distˆancia de Q ao plano H
fica:
d(Q, H) =
|ax0 + by0 + cz0 − d|
√
a2 + b2 + c2
.
1.3.31
Exemplo A distˆancia de Q = (3, −2, 1) ao plano π de equa¸c˜ao cartesiana 2x − 2y − z = −9
vale
d(Q, π) =
|(3, −2, 1) · (2, −2, −1) + 9|
3
= 6.
Agora se X = (x, y, z) ´e um ponto qualquer de π, ent˜ao
X = (x, y, 2x − 2y + 9) = (0, 0, 9) + x(1, 0, 2) + y(0, 1, −2)
= P + xV + yW,
onde P = (0, 0, 9), V = (1, 0, 2) e W = (0, 1, −2). Logo, π = P + [{V, W}], e obtemos uma
representa¸c˜ao param´etrica para π.
1.3.32
Exemplo Sejam l1 = P + [V ] e l2 = Q + [W] duas retas em Rn
. Se V e W s˜ao linearmente
independentes, h´a duas possibilidades para a interse¸c˜ao l1 ∩ l2, a saber:
(i) l1 ∩ l2 ´e um ponto, digamos l1 ∩ l2 = {R};
(ii) l1 ∩ l2 ´e vazio.
No primeiro caso, o plano π = R + [{V, W}] cont´em l1 e l2, e dizemos que l1 e l2 s˜ao retas
coplanares. J´a em (ii), n˜ao existe um plano que contenha ambas as retas, e diremos que l1 e l2
s˜ao retas reversas. O leitor deve observar que as retas do exemplo 1.3.6 s˜ao retas reversas em
R3
. Suponhamos, agora, que os vetores V e W sejam linearmente dependentes. Obtemos, outra
vez, duas alternativas:
(iii) l1 e l2 s˜ao coincidentes, isto ´e, l1 = l2;
(iv) l1 e l2 s˜ao paralelas e l1 ∩ l2 = ∅.
Em (iv), l1 e l2 s˜ao coplanares, visto que o plano π = P + [{V, W}] = Q + [{V, W}], onde
W = Q − P, as cont´em.
34. 28 Func¸˜oes Vetoriais
1.4
Func¸˜oes Vetoriais
Nesta se¸c˜ao, estudaremos as no¸c˜oes b´asicas relacionadas com aplica¸c˜oes entre espa¸cos
euclidianos de dimens˜oes quaisquer.
1.4.1
Definic¸˜ao Uma fun¸c˜ao vetorial ´e uma fun¸c˜ao com dom´ınio D ⊂ Rn
e contradom´ınio Rm
,
isto ´e, uma fun¸c˜ao do tipo
f : D ⊂ Rn
−−−−−→ Rm
X −−−−−→ f(X) = (f1(X), f2(X), . . . , fm(X)),
onde X = (x1, x2, . . . , xn) ∈ D. Quando m = 1, diremos que f ´e uma fun¸c˜ao real. J´a quando
n = 1, f ´e dita uma fun¸c˜ao vetorial de uma vari´avel real. A imagem de f, denotada por Im(f),
ou por f(D), ´e o conjunto
Im(f) = f(D) = {Y ∈ Rm
; Y = f(X), X ∈ D}.
Dizemos, tamb´em, que f parametriza o conjunto Im(f), ou que Im(f) ´e o conjunto parametri-
zado por f.
1.4.2
Definic¸˜ao Dada uma fun¸c˜ao vetorial
f : D ⊂ Rn
−−−−−→ Rm
X −−−−−→ f(X) = (f1(X), f2(X), . . . , fm(X)),
as m fun¸c˜oes reais
f1 : D ⊂ Rn
−−−−−→ R
X −−−−−→f1(X)
f2 : D ⊂ Rn
−−−−−→ R
X −−−−−→f2(X)
...
fm : D ⊂ Rn
−−−−−→ R
X −−−−−→fm(X)
s˜ao as fun¸c˜oes coordenadas de f.
1.4.3
Exemplo Seja f : R2
−→ R definida por f(x, y) = x2
+ y2
. Temos que f ´e uma fun¸c˜ao real
(de duas vari´aveis) cuja imagem coincide com o intervalo [0, ∞).
35. Vetores e Func¸˜oes Vetoriais 29
1.4.4
Exemplo Seja f(t) = (x0 + a cos t, y0 + b sen t), t ∈ [0, 2π], onde a 0, b 0, x0 e y0 s˜ao
n´umeros reais fixados. A imagem de f,
Im(f) = {(x, y) ∈ R2
; x = x0 + a cos t, y = y0 + b sen t, t ∈ R}, (¶6)
coincide com a elipse de semi-eixos a e b, centrada em C = (x0, y0), que denotaremos por
E(C, a, b). De fato, se x e y s˜ao como em (¶6), ent˜ao
(x − x0)2
a2 +
(y − y0)2
b2 = cos2
t + sen2
t = 1.
E
f
O
C
s E
T
x
E(C, a, b)
y
r
q
qy0
x0
a
b
0 2πt
r
Figura 17: Elipse
(x − x0)2
a2 +
(y − y0)2
b2 = 1
¨¨¨¨B
r
f(t)
r
f1(t)
rf2(t)
t
r
P2
rP1
Assim f parametriza E(C, a, b). A figura 17 mostra, em particular, a constru¸c˜ao geom´etrica
da fun¸c˜ao f: pelo ponto C = (x0, y0) tra¸camos a semi-reta que faz o ˆangulo t com o eixo-x.
Esta semi-reta intercepta os c´ırculos, centrados em C e de raios a e b, nos pontos P1 e P2,
respectivamente. Agora, por P1 tra¸camos uma
reta paralela ao eixo-y, e por P2 tra¸camos uma
reta paralela ao eixo-x. A interse¸c˜ao destas retas
´e exatamente o ponto da elipse
(x − x0)2
a2 +
(y − y0)2
b2 = 1
que indicamos por f(t). Em particular, se b = a,
obtemos que f parametriza o c´ırculo de centro
C = (x0, y0) e raio a, denotado por S1
(C, a). Ob-
serve que, neste caso, as fun¸c˜oes coordenadas de
f ficam assim:
f1(t) = x(t) = x0 + a cos t
f2(t) = y(t) = y0 + a sen t,
onde t ∈ [0, 2π].
O
C
s E
T
x
S1
(C, a)
y
r
q
qy0
x0
a
Figura 18: (x − x0)2
+ (y − y0)2
= a2
ppppppppppppppppppppppppppppppppp ppppppppppp pppppppppp pppppppppp ppppppppp ppppppppp pppppppp pppppp pppppp ppppp ppppp pppp pppp pppppp ppppp ppppp pppp pppp ppppp pppp ppppp ppppp pppp ppp pp pp ppp ppp ppp pppp pppp pppp ppp ppp ppp pp pp ppp ppp pp ppp ppp pp ppp ppp pp pp pp ppp ppp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp ppp ppp pp pp pp ppp ppp pp ppp ppp ppp ppp pp pp ppp ppp ppp pppp pppp pppp ppp ppp ppp pp pp ppp pppp ppppp ppppp pppp ppppp pppp pppp ppppp ppppp pppppp pppp pppp ppppp ppppp pppppp pppppp pppppppp ppppppppp ppppppppp ppppppppp ppppppppp pppppppppppp pppppppppppppppppppppppppppppppppp
ppppppppppppppppppppppppppppppppp ppppppppppp pppppppppp pppppppppp ppppppppp ppppppppp pppppppp pppppp pppppp ppppp ppppp pppp pppp pppppp ppppp ppppp pppp pppp ppppp pppp ppppp ppppp pppp ppp pp pp ppp ppp ppp pppp pppp pppp ppp ppp ppp pp pp ppp ppp pp ppp ppp pp ppp ppp pp pp pp ppp ppp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp ppp ppp pp pp pp ppp ppp pp ppp ppp ppp ppp pp pp ppp ppp ppp pppp pppp pppp ppp ppp ppp pp pp ppp pppp ppppp ppppp pppp ppppp pppp pppp ppppp ppppp pppppp pppp pppp ppppp ppppp pppppp pppppp pppppppp ppppppppp ppppppppp ppppppppp ppppppppp pppppppppppp pppppppppppppppppppppppppppppppppp
r
f1(t)
rf2(t)
t
rf(t)
36. 30 Func¸˜oes Vetoriais
1.4.5
Exemplo A fun¸c˜ao vetorial
g : R2
−−−−−→ R3
(u, v) −−−−−→ g(u, v) = (1 + u, 2 + v, u + v)
tem fun¸c˜oes coordenadas
g1(u, v) = x(u, v) = 1 + u
g2(u, v) = y(u, v) = 2 + v
g3(u, v) = z(u, v) = u + v,
(u, v) ∈ R2
, e sua imagem coincide com o plano que passa por (1, 2, 0) e ´e paralelo aos vetores
(1, 0, 1) e (0, 1, 1).
1.4.6
Conjuntos Associados a Func¸˜oes Vetoriais
Estudaremos, agora, alguns subconjuntos do espa¸co euclidiano que desempenham papel
de grande relevˆancia na descri¸c˜ao das fun¸c˜oes vetoriais.
1.4.7 [Gr´aficos]
Definic¸˜ao Seja f : D ⊂ Rn
−→ Rm
, f(X) = (f1(X), f2(X), . . . , fm(X)), uma fun-
¸c˜ao vetorial. O gr´afico de f, indicado por G(f), ´e definido por
G(f) = {(X, Y ) ∈ Rn+m
; Y = f(X), X ∈ D}
= {(x1, x2, . . . , xn, f1(X), f2(X), . . . , fm(X)); X = (x1, x2, . . . , xn) ∈ D}.
Diremos, tamb´em, que G(f) ´e o conjunto definido explicitamente por f.
Observac¸˜ao Na defini¸c˜ao acima, introduzimos uma nova nota¸c˜ao, que ser´a ´util em outras
situa¸c˜oes: dados X = (x1, x2, . . . , xn) ∈ Rn
e Y = (y1, y2, . . . , yn) ∈ Rm
, escre-
vemos (X, Y ) para representar a (n + m)-upla (x1, x2, . . . , xn, y1, y2, . . . , ym).
1.4.8
Exemplo Seja f : [−1, 1] −→ R definida por
f(x) = x2
. Temos que f ´e uma fun-
¸c˜ao real de uma vari´avel real cuja imagem coincide
com o intervalo [0, 1], e cujo gr´afico ´e o subcon-
junto do R2
dado por
G(f) = {(x, y); y = x2
, x ∈ [−1, 1]},
que coincide com o arco da par´abola y = x2
que
se projeta sobre o intervalo [−1, 1].
−1 1O
E
T
x
y
Figura 19: Par´abola y = x2
s
39. Vetores e Func¸˜oes Vetoriais 33
Observac¸˜ao ´E muito comum na pr´atica nos referirmos a uma equa¸c˜ao Y = f(x1, x2, . . . , xn),
onde o lado direito indica uma m-upla envolvendo os s´ımbolos x1, x2, . . ., xn,
como uma fun¸c˜ao. O que estamos pensando, na realidade, ´e na fun¸c˜ao f : D −→ Rm
, onde D ´e
o maior subconjunto de Rn
, onde tal express˜ao faz sentido.
1.4.12
Exemplo A express˜ao
z = 4 − x2 − y2 log (x2
+ y2
− 1)
define uma fun¸c˜ao f no anel de R2
dado por
D = {(x, y); 1 x2
+ y2
≤ 4}.
De fato, se x2
+y2
≤ 1, ent˜ao n˜ao existe log (x2
+ y2
− 1),
e se x2
+ y2
4 n˜ao tem sentido 4 − x2 − y2.
pppppppppppppppppppppppppppppppp ppppppppppppp ppppppppppp pppppppppp ppppppppp ppppppppp pppppppp ppppp ppppp pppppp pppppp ppppp ppppp pppppp ppppp ppppp ppppp pppp pppp pppp pppp pppp pppp pppp ppp ppp ppp ppp pppp ppp ppp pppp ppp ppp ppp pp ppp ppp pp pp ppp ppp pp ppp ppp pp pp pp pp pp pp ppp ppp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pppppppppp pp p pppppp pp pppp pppp pp ppp pppp pp pppp pppp pp ppp ppp pp p pp pppp pp p pp ppp pp p pp ppp pp p p pp p pp pp pp ppp pp p pp p pp pp pp p pp p pp p pp pp pp p pp p p pp p p pp pp pp pp pp p pp pp p pp p p pp pp p pp pp p pp pp p pp p p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp p p pp pp pp pp p p pp p p pp p p pp p pp p pp p pp p pp p pp ppp pp p pp pp pp p p pp ppp pp pp pp ppp pp p pp ppp pp p pp ppp ppp pp ppppp ppppp pp pppp pppp pp pppp pppp pp pp pppppp pp pppppppppp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp pp ppp ppp pp pp pp pp pp pp ppp ppp pp ppp ppp pp pp ppp ppp pp ppp ppp ppp pppp ppp ppp ppp pp ppp ppp ppp pppp pppp pppp pppp pppp pppp pppp ppppp ppppp ppppp pppppp ppppp ppppp ppppppp ppppppp ppppp ppppp pppppppp ppppppppp ppppppppp pppppppppp ppppppppppp ppppppppppppp pppppppppppppppppppppppppppppppp
r
1
r
2
E
x
T
y
Figura 24: Anel
1.4.13 [Conjuntos de N´ıvel]
Definic¸˜ao Sejam f : D ⊂ Rn
−→ Rm
uma fun¸c˜ao vetorial e K
um vetor em Rm
fixado. O conjunto de n´ıvel K de f
´e o subconjunto de D definido por
f−1
(K) = {X ∈ D; f(X) = K}.
Isto ´e, se f1, f2, . . ., fm s˜ao as fun¸c˜oes coordenadas de f, e K = (k1, k2, . . . , km), ent˜ao f−1
(K)
´e o conjuntos das solu¸c˜oes em D do sistema de m-equa¸c˜oes e n-inc´ognitas:
f1(x1, x2, . . . , xn) = k1
f2(x1, x2, . . . , xn) = k2
...
...
...
fm(x1, x2, . . . , xn) = km.
Os conjuntos de n´ıvel f−1
(K) s˜ao, tamb´em, denominados conjuntos definidos implicitamente
por f.
Observac¸˜ao f−1
(K) = ∅ se, e somente se, K ∈ Im(f).
1.4.14
Exemplo Os conjuntos de n´ıvel f−1
(a), a ∈ R, de z = f(x, y) = x2
+ y2
, (x, y) ∈ R2
, s˜ao
os seguintes:
(i) f−1
(a) = ∅, se a 0, pois Im(f) = [0, +∞);
(ii) f−1
(0) = {(0, 0)};
43. Vetores e Func¸˜oes Vetoriais 37
Agora, se a ∈ R, vem que
T(aX) = T(ax1, ax2)
= (ax1 + ax2, ax2 − ax1)
= a(x1 + x2, x2 − x1)
= aT(X).
Logo, T ´e linear. Um modo eficiente de ver que T ´e linear, ´e introduzindo a seguinte identifica-
¸c˜ao: uma dupla X = (x1, x2) passar´a a ser olhada como a matriz-coluna X =
x1
x2
. Isto posto,
vem que
T(X) = T(x1, x2) =
x1 + x2
x2 − x1
=
1 1
−1 1
x1
x2
,
ou T(X) = M(T)X, onde
M(T) =
1 1
−1 1
.
Agora, usando propriedades da multiplica¸c˜ao de matrizes, segue-se facilmente a linearidade de T.
De fato,
e
T(X + Y ) = M(T)(X + Y ) = M(T)X + M(T)Y = T(X) + T(Y )
T(aX) = M(T)(aX) = aM(T)X = aT(X).
A identifica¸c˜ao feita no exemplo anterior pode ser usada com uma k-upla qualquer, o que
facilitar´a a compreens˜ao das fun¸c˜oes lineares definidas em Rn
. Dado X = (x1, x2, . . . , xk) ∈ Rk
,
identificaremos, sempre que for preciso, X com a matriz-coluna (vetor-coluna)
X =
x1
x2
...
xk
.
Assim sendo, sejam T : Rn
−→ Rm
uma aplica¸c˜ao linear, e X ∈ Rn
. Temos que
X = (x1, x2, . . . , xn) =
x1
x2
...
xn
= x1
1
0
...
0
+ x2
0
1
...
0
+ · · · + xn
0
0
...
1
.
Logo,
T(X) = x1T(e1) + x2T(e2) + · · · + xnT(en) = x1T
1
0
...
0
+ x2T
0
1
...
0
+ · · · + xnT
0
0
...
1
. (¶8)
44. 38 Func¸˜oes Vetoriais Especiais
Como Im(T) ⊂ Rm
, vem que
T(e1) =
a11
a21
...
am1
, T(e2) =
a12
a22
...
am2
, . . . , T(en) =
a1n
a2n
...
amn
,
para alguns n´umeros reais aij, 1 ≤ i ≤ m e 1 ≤ j ≤ n, o que posto em (¶8) d´a que
T(X) = x1
a11
a21
...
am1
+ x2
a12
a22
...
am2
+ · · · + xn
a1n
a2n
...
amn
=
x1a11
x1a21
...
x1am1
+
x2a12
x2a22
...
x2am2
+ · · · +
xna1n
xna2n
...
xnamn
=
a11x1 + a12x2 + · · · + a1nxn
a21x1 + a22x2 + · · · + a2nxn
...
am1x1 + am2x2 + · · · + amnxn
=
a11 a12 . . . a1n
a21 a22 . . . a2n
...
...
...
am1 am2 . . . amn
x1
x2
...
xn
.
Isto prova o seguinte teorema.
1.5.3
Teorema Sejam T : Rn
−→ Rm
uma aplica¸c˜ao linear, e M(T) a matriz de ordem m × n
cujas colunas s˜ao os vetores T(e1), T(e2), . . ., T(en), nesta ordem. Temos que
(i) T(X) = M(T)X;
(ii) Im(T) ´e gerado pelas colunas de M(T);
(iii) posto T = posto M(T),
onde posto T indica a dimens˜ao de Im(T), e posto M(T) indica o posto da matriz M(T), isto
´e, o n´umero m´aximo de colunas linearmente independentes que ela possui.
45. Vetores e Func¸˜oes Vetoriais 39
1.5.4
Definic¸˜ao A matriz M(T) = (aij) ´e conhecida como a matriz de T com rela¸c˜ao `as bases
canˆonicas do Rn
e Rm
. Por simplicidade, chamaremos M(T) de matriz de T.
Um conjunto de n´ıvel especial de uma fun¸c˜ao linear T : Rn
−→ Rm
´e o seu n´ucleo,
N(T) = T−1
(0, 0, . . . , 0) = {X ∈ Rn
; T(X) = (0, 0, . . . , 0)}.
Os outros conjuntos definidos implicitamente por T, quando n˜ao-vazios, s˜ao determinados a
partir dele, como mostra seguinte proposi¸c˜ao.
1.5.5
Proposic¸˜ao Seja T : Rn
−→ Rm
uma aplica¸c˜ao linear. Se K = T(P), P ∈ Rn
, ent˜ao
T−1
(K) = P + N(T) = {X ∈ Rn
; X = P + V, V ∈ N(T)}.
Demonstrac¸˜ao: Seja X ∈ T−1
(K). Logo, T(X) = K = T(P). Como T ´e linear,
vem que T(X − P) = O, isto ´e, V = X − P ∈ N(T). Assim, X = P + V , o que prova
que T−1
(K) ⊂ P + N(T). Por outro lado, se X = P + V , para algum V ∈ N(T), ent˜ao
T(X) = T(P) + T(V ) = T(P) = K. Donde, P + N(T) ⊂ T−1
(K)). pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
ppppppppppppppppppppp
1.5.6
Exemplo Seja
T : R3
−−−−−→ R3
(x, y, z) −−−−−→ T(x, y, z) = (x + y, x + 2y + z, −x + 3y + 4z).
Temos que T(e1) = (1, 1, −1), T(e2) = (1, 2, 3) e T(e3) = (0, 1, 4). Logo, a matriz de T ´e
M(T) =
1 1 0
1 2 1
−1 3 4
,
e, usando o teorema 1.5.3, obtemos
T
x
y
z
=
1 1 0
1 2 1
−1 3 4
x
y
z
.
Sugerimos ao leitor que verifique diretamente esta identidade. Como det M(T) = 0, segue-se
que posto M(T) ≤ 2. Como, por exemplo, as duas primeiras colunas de M(T) s˜ao linearmente
independentes, devemos ter posto T = 2. (Conv´em observar, que esta informa¸c˜ao pode ser
obtida, tamb´em, usando opera¸c˜oes elementares sobre as linhas (colunas) de M(T), o que ´e mais
conveniente para matrizes de ordem alta.) Logo, Im(T) tem dimens˜ao dois e ´e gerado pelos