Gabarito da Lista 4 Geometria Analítica e Álgebra Linear I

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GABARITO DA 4ª LISTA DE EXERCÍCIOS 
GEOMETRIA ANALÍTICA E ÁLGEBRA LINEAR I 
 
1. a) Temos ( ) ( ) ( ) .1454325,34,2vu =⋅+−⋅=−⋅=⋅ rr 
b) Temos ( ) .3453|v|e522042|u| 2222 =+−===+= rr Logo, 
o
rr
rr
5,57
170
7
arccos
170
7
3452
14
|v||u|
vu
cos ==θ⇒==
⋅
⋅
=θ . 
c) Temos ( ) 





−=−=





⋅
⋅
=
17
35
,
17
215,3
34
14
v
vv
vu
uproj v
r
rr
rr
r
r
. 
 
2. Seja ( )1,2u −=r o vetor dado e ( )y,xv =r o vetor procurado. Temos 
( ) 25x2x
25yx
x2y
25yx
0yx2
5yx
0vu
5v
vu 22
222222 =+⇒



=+
=
⇒



=+
=−
⇒




=+
=⋅
⇒



=
⊥
rr
r
rr
 
52y5x5x25x5 22 ±=⇒±=⇒=⇒=⇒ . 
Portanto, os possíveis vetores vr são ( ) ( )52,5vou52,5v −−== rr . 
3. Temos ( ) ( ) ( ) ( ) ( ) ( ) ⇒=⋅−+−⋅⇒=−⋅−⇒−⊥− 041x6204,61x,24,61x,2 222 2x4x2 ±=⇒= 
4. a)Temos ( ) ( ) ( ) ( )4,1BCBCe3,2ACAC,1,3BA,1,3ABAB −=−==−=−=−=−= →→→→ . 
Logo, ( ) ( ) 1741|BC|e1332|AC|,1013|BA||AB| 222222 =+−==+==−+== →→→→ . 
Assim, ( ) ,7,74
130
3
arccosAˆ
130
3
1310
3123
|AC||AB|
ACABAˆcos o==⇒=
⋅
⋅−+⋅
=
⋅
⋅
=
→→
→→
 
 
v
r
 
u
r
 
uprojv
r
r
 
 
2 
( ) ( ) o5,57
170
7
arccosBˆ
170
7
1710
4113
|BC||BA|
BCBABˆcos =





=⇒=
⋅
⋅+−⋅−
=
⋅
⋅
=
→→
→→
 e 
.8,47BˆAˆ180Cˆ oo =−−= 
b) Temos ( ) 





−=−=










⋅
⋅
=
→
→→
→→
→
→
10
3
,
10
91,3
10
3AB
ABAB
ABACACproj
AB
 e 
( ) 





−=−−=










⋅
⋅
=
→
→→
→→
→
→
10
7
,
10
211,3
10
7AB
ABAB
ABBCBCproj
AB
. 
c) Seja ( )y,xP o pé da altura relativa ao vértice C. Logo, .
10
3
,
10
9ACprojAP
AB






−==
→→
→ 
Mas, ( ) .
10
7
,
10
19P1,1
10
3
,
10
9AAPPAPAP 





⇒+





−=+=⇒−=
→→
 
 
5. a) Temos ( ) 





==





⋅
⋅
==
5
6
,
5
83,4
25
10
u
uu
uv
vprojv u1
r
rr
rr
rr
r
 e ( ) .
5
4
,
5
3
5
6
,
5
82,1vvv 12 





−=





−=−=
rrr
 
 
A 
C 
B 
P 
u
r
 
1v
r
 
v
r
 
2v
r
 
 
3 
b) Temos ( ) 





==





⋅
⋅
==
2
7
,
2
71,1
2
7
u
uu
uv
vprojv u1
r
rr
rr
rr
r
 e ( ) .
2
3
,
2
3
2
7
,
2
75,2vvv 12 





−=





−=−=
rrr
 
6. ( ) .42154ve14312u)a 222222 =++==+−+= rr 
( )b u v) .r r⋅ = ⋅ + − ⋅ + ⋅ =2 4 1 5 3 1 6 
( ) ( )c u w u w m m) .r r r r⊥ ⇔ ⋅ = ⇔ ⋅ + − ⋅ − + ⋅ = ⇔ = −0 2 3 1 1 3 0 73 
d u v
u v
) cos | || | .θ =
⋅
= =
r r
r r
6
14 42
3
7
 
7. a) Temos ( ) ,9242111vu =⋅+⋅+−⋅=⋅ rr ( ) e3221|v| 222 =++−=r 23411|u| 222 =++=r . 
Logo, .
4
,totanpor,e
2
2
233
9
|v||u|
vu
cos
pi
=θ=
⋅
=
⋅
=θ rr
rr
 
b) Temos ( ) ( )r r ru v u e⋅ = ⋅ + ⋅ + − ⋅ = − = + + − =2 1 0 1 3 1 1 2 0 3 132 2 2, | | | |rv = + + =1 1 1 32 2 2 . 
Logo, 
39
39
313
1
|v||u|
vu
cos −=
⋅
−
=
⋅
=θ rr
rr
 e, portanto, o2,99=θ . 
c) Temos ( ) 00200110101vu =⋅+⋅+−⋅=⋅ rr . Logo, 
2
,sejaou,vu pi=θ⊥ rr . 
8. .9x033x01x0vuvu)a −=⇔=⋅+⋅+⋅⇔=⋅⇔⊥ vrvr 
⇔=⋅+⋅+⋅⇔=⋅⇔⊥ 014xx4x0vuvu)b vrvr .2x04x4x2 −=⇔=++ 
( ) ( ) ( ) ( ) ⇔=−⋅+−⋅+−⋅+⇔=⋅⇔⊥ 022111x1x0vuvu)c vrvr .6x06x2 ±=⇔=− 
9. Temos ( ) 





−=−=





⋅
⋅
==
9
14
,
9
7
,
9
142,1,2
9
7
u
uu
uv
vprojv u1
r
rr
rr
rr
r
 e 
( ) .
9
41
,
9
38
,
9
22
9
14
,
9
7
,
9
143,5,4vvv 12 





=





−−=−=
rrr
 
10. a) 1º modo: Temos .0|uu|0
213
213
kji
uu =∧⇒=
−−
−−=∧
rrr
rrr
rr
 
2º modo: Temos .00senpois,00sen|u||u||uu| ==⋅⋅=∧ rrrr 
b) Temos .k14ji9j3i8k2k12j4i
142
213
kji
vu
rrrrrrrrr
rrr
rr
+−=++++−=
−
−−=∧ 
 
4 
Portanto, ( ) .kj23ij9kj14i
101
1419
kji
wvu
rrrrrrr
rrr
rrr
−−−=−−−−=
−
−=∧∧ 
c) Temos .k4ji4j2k4ji4
101
142
kji
wv
rrrrrrr
rrr
rr
+−=−++=
−
−=∧ 
Portanto, ( ) ( ) ( ) ( ) .5421143wvu =⋅−+−⋅−+⋅=∧⋅ rrr 
11. a) Temos .k10j4i4
214
232
kji
vu
rrr
rrr
rr
++−=
−
−=∧ 
b) Se θ é o ângulo de ,veu rr então θ⋅⋅=∧ sen|v||u||vu| rrrr . Mas, 
( ) ( ) e17232|u|,3321044|vu| 222222 =+−+==++−=∧ rrr 
( ) .
357
332
sen,Logo.21214|v| 222 =θ=+−+=r 
12. a) Temos ( ) ( )2,0,1ACACe4,0,1ABAB −−=−=−=−= →→ e j6
201
401
kji
ACAB
r
rrr
=
−−
−=∧
→→
. 
Portanto, a área do triângulo ABC é dada por 3
2
|ACAB|S =∧=
→→
 e um vetor ortogonal ao plano 
determinado pelos ponto A, B e C é o próprio vetor ( )0,6,0j6ACAB ==∧ →→ r . 
b) Temos ( ) ( )1,0,1ACACe1,1,0ABAB −=−=−=−= →→ e kji
101
110
kji
ACAB
rrr
rrr
−−−=
−
−=∧
→→
. 
Portanto, a área do triângulo ABC é dada por 
2
3
2
|ACAB|S =∧=
→→
 e um vetor ortogonal ao plano 
determinado pelos ponto A, B e C é o próprio vetor ( )1,1,1kjiACAB −−−=−−−=∧ →→ rrr .