Matrizes Vetores e Geometria Analitica
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Matrizes Vetores e Geometria Analitica


DisciplinaGeometria Analítica e Álgebra Linear3.400 materiais33.549 seguidores
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. O resultado abaixo mostra
como calcular o produto misto usando as componentes dos vetores.
Teorema 3.7. Sejam \ud448 = \ud4621\u20d7\ud456+ \ud4622\ud457\u20d7 + \ud4623\ud458\u20d7, \ud449 = \ud4631\u20d7\ud456+ \ud4632\ud457\u20d7 + \ud4633\ud458\u20d7 e \ud44a = \ud4641\u20d7\ud456+ \ud4642\ud457\u20d7 + \ud4643\ud458\u20d7. Enta\u2dco,
(\ud449 ×\ud44a ) \u22c5 \ud448 = det
\u23a1
\u23a3 \ud4631 \ud4632 \ud4633\ud4641 \ud4642 \ud4643
\ud4621 \ud4622 \ud4623
\u23a4
\u23a6 .
Marc¸o 2010 Reginaldo J. Santos
202 Vetores no Plano e no Espac¸o
Demonstrac¸a\u2dco. Segue do Teorema 3.2 na pa´gina 187, do Teorema 3.6 na pa´gina 199 e da definic¸a\u2dco
de determinante de uma matriz que
(\ud449 ×\ud44a ) \u22c5 \ud448 = (\ud4621, \ud4622, \ud4623) \u22c5
(
det
[
\ud4632 \ud4633
\ud4642 \ud4643
]
,\u2212 det
[
\ud4631 \ud4633
\ud4641 \ud4643
]
, det
[
\ud4631 \ud4632
\ud4641 \ud4642
])
= \ud4621 det
[
\ud4632 \ud4633
\ud4642 \ud4643
]
\u2212 \ud4622 det
[
\ud4631 \ud4633
\ud4641 \ud4643
]
+ \ud4623 det
[
\ud4631 \ud4632
\ud4641 \ud4642
]
= det
\u23a1
\u23a3 \ud4631 \ud4632 \ud4633\ud4641 \ud4642 \ud4643
\ud4621 \ud4622 \ud4623
\u23a4
\u23a6 ;
\u25a0
Exemplo 3.13. O produto misto dos vetores \ud448 = 2\u20d7\ud456\u2212 \ud457\u20d7+3\ud458\u20d7, \ud449 = \u2212\u20d7\ud456+ 4\u20d7\ud457+ \ud458\u20d7 e \ud44a = 5\u20d7\ud456+ \ud457\u20d7\u2212 2\ud458\u20d7 e´
(\ud449 ×\ud44a ) \u22c5 \ud448 = det
\u23a1
\u23a3 \ud4631 \ud4632 \ud4633\ud4641 \ud4642 \ud4643
\ud4621 \ud4622 \ud4623
\u23a4
\u23a6 = det
\u23a1
\u23a3 \u22121 4 15 1 \u22122
2 \u22121 3
\u23a4
\u23a6 = \u221284.
Teorema 3.8. Dados tre\u2c6s vetores no espac¸o, \ud448, \ud449 e \ud44a ,
\u2223(\ud449 ×\ud44a ) \u22c5 \ud448 \u2223
e´ numericamente igual ao volume do paralelep\u131´pedo determinado por \ud448, \ud449 e \ud44a .
Matrizes Vetores e Geometria Anal\u131´tica Marc¸o 2010
3.2 Produtos de Vetores 203
Demonstrac¸a\u2dco. O volume do paralelep\u131´pedo determinado por \ud448, \ud449 e \ud44a e´ igual ao produto da a´rea
da base pela altura, ou seja, pela definic¸a\u2dco do produto vetorial, o volume e´ dado por
Volume = \u2223\u2223\ud449 ×\ud44a \u2223\u2223\u210e .
Mas, como vemos na Figura 3.25 a altura e´ \u210e = \u2223\u2223\ud448 \u2223\u2223\u2223 cos \ud703\u2223, o que implica que
Volume = \u2223\u2223\ud449 ×\ud44a \u2223\u2223 \u2223\u2223\ud448 \u2223\u2223\u2223 cos \ud703\u2223 = \u2223(\ud449 ×\ud44a ) \u22c5 \ud448 \u2223 .
\u25a0
Exemplo 3.14. Sejam \ud449 = 4\u20d7\ud456, \ud44a = 2\u20d7\ud456+ 5\u20d7\ud457 e \ud448 = 3\u20d7\ud456+ 3\u20d7\ud457 + 4\ud458\u20d7. O volume do paralelep\u131´pedo com
um ve´rtice na origem e arestas determinadas por \ud448, \ud449 e \ud44a e´ dado por
volume = \u2223(\ud449 ×\ud44a ) \u22c5 \ud448 \u2223 = \u2223 det
\u23a1
\u23a3 4 0 02 5 0
3 3 4
\u23a4
\u23a6 \u2223 = \u222380\u2223 = 80 .
Segue imediatamente do Teorema 3.7 e do Teorema 3.8 um crite´rio para saber se tre\u2c6s vetores sa\u2dco
paralelos a um mesmo plano.
Marc¸o 2010 Reginaldo J. Santos
204 Vetores no Plano e no Espac¸o
Corola´rio 3.9. Sejam \ud448 = \ud4621\u20d7\ud456 + \ud4622\ud457\u20d7 + \ud4623\ud458\u20d7, \ud449 = \ud4631\u20d7\ud456 + \ud4632\ud457\u20d7 + \ud4633\ud458\u20d7 e \ud44a = \ud4641\u20d7\ud456 + \ud4642\ud457\u20d7 + \ud4643\ud458\u20d7. Estes
vetores sa\u2dco coplanares (isto e´, sa\u2dco paralelos a um mesmo plano) se, e somente se,
(\ud449 ×\ud44a ) \u22c5 \ud448 = det
\u23a1
\u23a3 \ud4631 \ud4632 \ud4633\ud4641 \ud4642 \ud4643
\ud4621 \ud4622 \ud4623
\u23a4
\u23a6 = 0 .
Exemplo 3.15. Vamos verificar que os pontos \ud443 = (0, 1, 1), \ud444 = (1, 0, 2), \ud445 = (1,\u22122, 0) e
\ud446 = (\u22122, 2,\u22122) sa\u2dco coplanares, isto e´, pertencem a um mesmo plano. Com estes pontos podemos
construir os vetores \u2212\u2192
\ud443\ud444= (1\u2212 0, 0\u2212 1, 2\u2212 1) = (1,\u22121, 1),
\u2212\u2192
\ud443\ud445= (1\u2212 0,\u22122\u2212 1, 0\u2212 1) = (1,\u22123,\u22121) e
\u2212\u2192
\ud443\ud446= (\u22122\u2212 0, 2\u2212 1,\u22122\u2212 1) = (\u22122, 1,\u22123)
Os pontos \ud443,\ud444,\ud445 e \ud446 pertencem a um mesmo plano se, e somente se, os vetores
\u2212\u2192
\ud443\ud444,
\u2212\u2192
\ud443\ud445 e
\u2212\u2192
\ud443\ud446 sa\u2dco coplanares. E isto acontece se, e somente se, o produto misto deles e´ igual zero.
(
\u2212\u2192
\ud443\ud445 ×
\u2212\u2192
\ud443\ud446) \u22c5
\u2212\u2192
\ud443\ud444= det
\u23a1
\u23a3 1 \u22123 \u22121\u22122 1 \u22123
1 \u22121 1
\u23a4
\u23a6 = 0.
Assim, \ud443,\ud444,\ud445 e \ud446 sa\u2dco coplanares.
Matrizes Vetores e Geometria Anal\u131´tica Marc¸o 2010
3.2 Produtos de Vetores 205
O resultado a seguir sera´ usado no pro´ximo cap\u131´tulo para deduzir as equac¸o\u2dces parame´tricas do
plano.
Corola´rio 3.10. Sejam \ud448, \ud449 e \ud44a vetores coplanares na\u2dco nulos no espac¸o.
(a) Enta\u2dco a equac¸a\u2dco vetorial
\ud465\ud449 + \ud466\ud44a + \ud467\ud448 = 0¯
tem soluc¸a\u2dco na\u2dco trivial, em que \ud465, \ud466 e \ud467 sa\u2dco escalares.
(b) Enta\u2dco um dos vetores \ud448, \ud449 ou \ud44a e´ combinac¸a\u2dco linear (soma de mu´ltiplos escalares) dos outros
dois.
(c) Se \ud449 e \ud44a sa\u2dco na\u2dco paralelos, enta\u2dco \ud448 e´ combinac¸a\u2dco linear de \ud449 e \ud44a .
Demonstrac¸a\u2dco. (a) Seja\ud434 a matriz cujas colunas sa\u2dco \ud449 , \ud44a e \ud448 escritos como vetores colunas. A
equac¸a\u2dco \ud465\ud449 + \ud466\ud44a + \ud467\ud448 = 0¯ e´ equivalente ao sistema \ud434\ud44b = 0¯. Se \ud448, \ud449 e \ud44a sa\u2dco coplanares,
enta\u2dco
det(\ud434) = det(\ud434\ud461) = (\ud449 ×\ud44a ) \u22c5 \ud448 = 0.
Logo a equac¸a\u2dco tem soluc¸a\u2dco na\u2dco trivial.
(b) Pelo item anterior a equac¸a\u2dco \ud465\ud448 + \ud466\ud449 + \ud467\ud44a = 0¯ possui soluc¸a\u2dco na\u2dco trivial. Mas, se isto
acontece, enta\u2dco um dos escalares \ud465 ou \ud466 ou \ud467 pode ser diferente de zero. Se \ud467 \u2215= 0, enta\u2dco
\ud448 = (\u2212\ud465/\ud467)\ud449 + (\u2212\ud466/\ud467)\ud44a , ou seja, o vetor \ud448 e´ combinac¸a\u2dco linear de \ud449 e \ud44a . De forma
semelhante, se \ud465 \u2215= 0, enta\u2dco \ud449 e´ combinac¸a\u2dco linear de \ud448 e \ud44a e se \ud466 \u2215= 0, enta\u2dco \ud44a e´
combinac¸a\u2dco linear de \ud448 e \ud449 .
Marc¸o 2010 Reginaldo J. Santos
206 Vetores no Plano e no Espac¸o
(c) Como \ud448, \ud449 e \ud44a sa\u2dco coplanares, enta\u2dco a equac¸a\u2dco \ud465\ud448 + \ud466\ud449 + \ud467\ud44a = 0¯ possui soluc¸a\u2dco na\u2dco
trivial com \ud465 \u2215= 0. Pois, caso contra´rio \ud466\ud449 + \ud467\ud44a = 0¯ com \ud466 ou \ud467 na\u2dco simultaneamente nulos o
que implicaria que \ud449 e \ud44a seriam paralelos (por que?). Logo \ud448 = (\u2212\ud466/\ud465)\ud449 + (\u2212\ud467/\ud465)\ud44a .
\u25a0
Exemplo 3.16. Considere os vetores
\ud448 =
\u2212\u2192
\ud443\ud444= (1,\u22121, 1),
\ud449 =
\u2212\u2192
\ud443\ud445= (1,\u22123,\u22121) e
\ud44a =
\u2212\u2192
\ud443\ud446= (\u22122, 1,\u22123)
do Exemplo 3.15 na pa´gina 204. A equac¸a\u2dco
\ud465\ud448 + \ud466\ud449 + \ud467\ud44a = 0¯
e´ equivalente ao sistema \u23a7\u23a8
\u23a9
\ud465 + \ud466 \u2212 2\ud467 = 0
\u2212\ud465 \u2212 3\ud466 + \ud467 = 0
\ud465 \u2212 \ud466 \u2212 3\ud467 = 0
Escalonando a matriz do sistema obtemos\u23a1
\u23a3 1 1 \u22122\u22121 \u22123 1
1 \u22121 \u22123
\u23a4
\u23a6 \u223c
\u23a1
\u23a3 1 1 \u221220 \u22122 \u22121
0 \u22122 \u22121
\u23a4
\u23a6 \u223c
\u23a1
\u23a3 1 1 \u221220 \u22122 \u22121
0 0 0
\u23a4
\u23a6
Matrizes Vetores e Geometria Anal\u131´tica Marc¸o 2010
3.2 Produtos de Vetores 207
A u´ltima matriz corresponde ao sistema{
\ud465 + \ud466 \u2212 2\ud467 = 0
\u2212 2\ud466 \u2212 \ud467 = 0
Assim
5\ud6fc
2
\ud448 \u2212 \ud6fc
2
\ud449 + \ud6fc\ud44a = 0¯.
Logo
\ud44a = \u22125
2
\ud448 \u2212 1
2
\ud449.
Verifique que realmente vale esta relac¸a\u2dco entre os vetores \ud448, \ud449 e \ud44a .
Marc¸o 2010 Reginaldo J. Santos
208 Vetores no Plano e no Espac¸o
Exerc\u131´cios Nume´ricos (respostas na pa´gina 588)
3.2.1. Determine a equac¸a\u2dco da reta no plano que e´ perpendicular ao vetor \ud441 = (2, 3) e passa pelo
ponto \ud4430 = (\u22121, 1).
3.2.2. Seja \ud442 = (0, 0, 0). Qual o lugar geome´trico dos pontos \ud443 = (\ud465, \ud466, \ud467) tais que \u2223\u2223
\u2212\u2192
\ud442\ud443 \u2223\u22232 = 4?
Qual figura e´ representada pela equac¸a\u2dco \ud4652 + \ud4662 = 4?
3.2.3. Sejam \ud449 = \ud456\u20d7+ 2\u20d7\ud457 \u2212 3\ud458\u20d7 e \ud44a = 2\u20d7\ud456+ \ud457\u20d7 \u2212 2\ud458\u20d7. Determine vetores unita´rios paralelos aos vetores
(a) \ud449 +\ud44a ; (b) \ud449 \u2212\ud44a ; (c) 2\ud449 \u2212 3\ud44a .
3.2.4. Determine o valor de \ud465 para o qual os vetores \ud449 = \ud465\u20d7\ud456 + 3\u20d7\ud457 + 4\ud458\u20d7 e \ud44a = 3\u20d7\ud456 + \ud457\u20d7 + 2\ud458\u20d7 sa\u2dco
perpendiculares.
3.2.5. Demonstre que na\u2dco existe \ud465 tal que os vetores \ud449 = \ud465\u20d7\ud456 + 2\u20d7\ud457 + 4\ud458\u20d7 e \ud44a = \ud465\u20d7\ud456 \u2212 2\u20d7\ud457 + 3\ud458\u20d7 sa\u2dco
perpendiculares.
3.2.6. Ache o a\u2c6ngulo entre os seguintes pares de vetores:
(a) 2\u20d7\ud456+ \ud457\u20d7 e \ud457\u20d7 \u2212 \ud458\u20d7; (b) \ud456\u20d7+ \ud457\u20d7 + \ud458\u20d7 e \u22122\u20d7\ud457 \u2212 2\ud458\u20d7; (c) 3\u20d7\ud456+ 3\u20d7\ud457 e 2\u20d7\ud456+ \ud457\u20d7 \u2212 2\ud458\u20d7.
3.2.7. Decomponha \ud44a = \u2212\u20d7\ud456\u2212 3\u20d7\ud457 + 2\ud458\u20d7 como a soma de dois vetores \ud44a1 e \ud44a2, com \ud44a1 paralelo ao
vetor \ud457\u20d7 + 3\ud458\u20d7 e \ud44a2 ortogonal a este u´ltimo. (Sugesta\u2dco: revise o Exemplo 3.10 na pa´gina 193)
3.2.8. Ache o vetor unita´rio da bissetriz do a\u2c6ngulo entre os vetores \ud449 = 2\u20d7\ud456+2\u20d7\ud457+\ud458\u20d7 e\ud44a = 6\u20d7\ud456+2\u20d7\ud457\u22123\ud458\u20d7.
(Sugesta\u2dco: observe que a soma de dois vetores esta´ na direc¸a\u2dco da bissetriz se, e somente se,
os dois tiverem o mesmo comprimento. Portanto, tome mu´ltiplos escalares de \ud449 e \ud44a de forma
que eles tenham o mesmo comprimento e tome o vetor unita´rio na direc¸a\u2dco da soma deles.)
Matrizes Vetores e Geometria Anal\u131´tica Marc¸o 2010
3.2 Produtos de Vetores 209
3.2.9. Verifique se os seguintes pontos pertencem a um mesmo plano:
(a) \ud434 = (2, 2, 1), \ud435 = (3, 1, 2), \ud436 = (2, 3, 0) e \ud437 = (2, 3, 2);
(b) \ud434 = (2, 0, 2), \ud435 = (3, 2, 0), \ud436 = (0, 2, 1) e \ud437 = (10,\u22122, 1);
3.2.10. Calcule o volume do paralelep\u131´pedo que tem um dos ve´rtices no ponto \ud434 = (2, 1, 6) e os tre\u2c6s
ve´rtices adjacentes nos pontos \ud435 = (4, 1, 3), \ud436 = (1, 3, 2) e \ud437 = (1, 2, 1).
3.2.11. Calcule a a´rea do paralelogramo em que tre\u2c6s ve´rtices consecutivos sa\u2dco \ud434 = (1, 0, 1), \ud435 =
(2, 1, 3) e \ud436 = (3, 2, 4).
3.2.12. Calcule a a´rea do tria\u2c6ngulo com ve´rtices \ud434 = (1, 2, 1), \ud435 = (3, 0, 4) e \ud436 = (5, 1, 3).
3.2.13. Ache \ud44b tal que \ud44b × (\u20d7\ud456+ \ud458\u20d7) = 2(\u20d7\ud456+ \ud457\u20d7 \u2212 \ud458\u20d7) e \u2223\u2223\ud44b\u2223\u2223 = \u221a6.
3.2.14. Sabe-se que o vetor \ud44b e´ ortogonal a \ud456\u20d7+ \ud457\u20d7 e a \u2212\u20d7\ud456+ \ud458\u20d7, tem norma\u221a3 e sendo \ud703 o a\u2c6ngulo entre
\ud44b e \ud457\u20d7, tem-se cos \ud703 > 0. Ache \ud44b .
3.2.15. Mostre que\ud434 = (3, 0, 2), \ud435 = (4, 3, 0) e\ud436 = (8, 1,\u22121) sa\u2dco ve´rtices de um tria\u2c6ngulo reta\u2c6ngulo.
Em qual dos ve´rtices esta´ o a\u2c6ngulo reto?
3.2.16. Considere dois vetores \ud449 e \ud44a tais que \u2223\u2223\ud449 \u2223\u2223 = 5, \u2223\u2223\ud44a \u2223\u2223 = 2 e o a\u2c6ngulo