GEOMETRIA SOLIDA (1)

GEOMETRÍA SOLIDA
Univ. Cecilia Tola Pacheco
10 de mayo de 2022
1.
EJERCICIOS
1. Hallar las ecuaciones vectorial y cartesiana de la recta, que pasa por el punto P0 con vector
direccional V , cuando.
(a) P0 = (0, 0, 0), V = (1, 1, 1)
Solución:
Ecuacion vectorial es
X = P0 + tV
X = (0, 0, 0) + t(1, 1, 1)
X = t(1, 1, 1)
Con X = (x, y, z),igualando componentes obtenemos la ecuación cartesiana
(x, y, z) = (0, 0, 0) + t(1, 1, 1)
x=t
y=t
z=t
(b) P0 = (5, 3, −2),
Solución:
V = (2, −3, 2)
Ecuacion vectorial es
X = P0 + tV
X = (5, 3, −2) + t(2, −3, 2)
Con X = (x, y, z),igualando componentes obtenemos la ecuacion cartesiana
(x, y, z) = (5, 3, −2) + t(2, −3, 2)
x = 5 + 2t
y = 3 − 3t
z = −2 + 2t
1
Cecilia Tola Pacheco
(c) P0 = (−1, −3, −5),
Solución:
Calculo II
MAT 102
V = (−2, 7, 3)
Ecuacion vectorial es
X = P0 + tV
X = (−1, −3, −5) + t(−2, 7, 3)
Con X = (x, y, z),igualando componentes obtenemos la ecuacion cartesiana
(x, y, z) = (−1, −3, −5) + t(−2, 7, 3)
x = −1 − 2t
y = −3 + 7t
z = −5 + 3t
(d) P0 = (3, 2, 1) V = (1, 5, −4)
Solución:
Ecuacion vectorial es
X = P0 + tV
X = (3, 2, 1) + t(1, 5, −4)
Con X = (x, y, z),igualando componentes obtenemos la ecuacion cartesiana
(x, y, z) = (3, 2, 1) + t(1, 5, −4)
x = 3+t
y = 2 + 5t
z = 1 − 4t
2. Hallar la ecuación de la recta que pasa por los puntos:
(a) (0, 0, 0) y (1, 1, 1)
Solución:
X = P0 + tV V = P1 − P0
P0 = (0, 0, 0) y P1 = (1, 1, 1)
V = P1 − P0 = (1, 1, 1) − (0, 0, 0) = (1, 1, 1)
X = P0 + tV = (0, 0, 0) + t(1, 1, 1)
X =t(1, 1, 1)
(b) (8, −3, 2)
Solución:
y
(5, 0, 0)
V = P1 − P0 = (5, 0, 0) − (8, −3, 2) = (−3, 3, −2)
X = P0 + tV = (8, −3, 2) + t(−3, 3, −2)
(c)(5, 8, 1) y (2, 6, −1)
2
Cecilia Tola Pacheco
Calculo II
MAT 102
Solución:
V = P1 − P0 = (2, 6, −1) − (5, 8, 1) = (−3, −2, −2)
X = P0 + tV = (5, 8, 1) + t(−3, −2, −2)
(d) (1, 1, 1) y −3, 2, −1
Solución:
V = P1 − P0 = (−3, 2, −1) − (1, 1, 1) = (−4, 1, −2)
X = P0 + tV = (1, 1, 1) + t(−4, 1, −2)
3. Calcular la distancia del punto P1 a la recta dada
(a)P1 = (1, 1, 1); X = (1, 2, −1) + t(3, 4, 1)
Solución:
Con P1 =(1, 1, 1); P0 = (1, 2, −1), V = (3, 4, 1)
P1 − P0 = (1, 1, 1) − (1, 2, −1) = (0, −1, 2)
(P1 − P0 ) · V = (0, −1, 2) · (3, 4, 1) = 0 − 4 + 2 = −2
√
| V |2 = ( 32 + 42 + 12 )2 = 9 + 16 + 1 = 26
d =| (P1 − P0 ) −
(P1 − P0 ) · V
(−2)
1
(3, 4, 1) | =| (0, −1, 2) + (3, 4, 1) |
· V |=| (0, −1, 2) −
2
26
13
|V |
3 4 1
=| (0, −1, 2) + ( , , ) |
13 13 13
3
9 27
=| ( , − , ) |
r13 13 13
9
27
3
= ( )2 + (− )2 + ( )2
13
13
13
r
r
√
819
63 3 7
=
=√
=
169
13
13
r
7
=3
13
(b) P1 = (0, 0, −1);X = (1, 0, −1) + t(1, 2, 3)
Solución:
P0 =(1, 0, −1); P1 = (0, 0, −1), V = (1, 2, 3)
P1 − P0 = (0, 0, −1) − (1, 0, −1) = (−1, 0, 0)
(P1 − P0 ) · V = (−1, 0, 0) · (1, 2, 3) = −1
√
| V |2 = ( 32 + 42 + 12 )2 = 14
3
Cecilia Tola Pacheco
d =| (P1 − P0 ) −
Calculo II
MAT 102
(P1 − P0 ) · V
(−2)
−1
· V |=| (0, −1, 2) −
(3, 4, 1) | =| (−1, 0, 0) −
(1, 2, 3) |
2
26
14
|V |
1 1 3
=| (−1, 0, 0) + ( , , ) |
14 7 14
−13 1 3
, , )|
=| (
r14 7 14
13
1
3
= (− )2 + ( )2 + ( )2
14
7
14
r
182
=
196
r
13
=
14
(c) P1 = (5, 5, 0);x = t, y = 1 − t, z = 3 + t
Solución:
P0 =(0, 1, 3); P1 = (5, 5, 0), V = (1, −1, 1)
P1 − P0 = (5, 5, 0) − (0, 1, 3) = (1, −1, 1)
(P1 − P0 ) · V = (5, 4, −3) · (1, −1, 1) = 5 − 4 − 3 = −2
q
√
2
| V | = ( 12 + (−1)2 + 12 )2 = ( 3)2 = 3
d =| (P1 − P0 ) −
(P1 − P0 ) · V
(−2)
(1, −1, 1) |
· V | =| (5, 4, −3) −
2
3
|V |
2 2 2
=| (5, 4, −3) + ( , − , ) |
3 3 3
17 10 7
=| ( , , − ) |
r3 3 3
10
7
17
= ( )2 + ( )2 + (− )2
3
3
3
r
438
=
9
r
146
=
3
(d) P1 = (4, −1, 2) x = 1 + 3t, y = 2y, z = −t
Solución:
P0 =(1, 0, 0); P1 = (4, −1, 2), V = (3, 2, −1)
P1 − P0 = (4, −1, 2) − (1, 0, 0) = (3, −1, 2)
(P1 − P0 ) · V = (3, −1, 2) · (3, 2, −1) = 9 − 2 − 2 = 5
q
2
| V | = ( 32 + 22 + (−1)2 )2 = 14
4
Cecilia Tola Pacheco
Calculo II
d =| (P1 − P0 ) −
MAT 102
(P1 − P0 ) · V
(5)
· V | =| (3, −1, 2) −
(3, 2, −1) |
2
14
|V |
15 5 5
=| (3, −1, 2) − ( , , − ) |
14 7 14
27 12 33
=| ( , − , ) |
r14 7 14
27
12
33
= ( )2 + (− )2 + ( )2
14
7
14
r
2394
=
196
r
171
=
14
r
19
=3
14
4. Calcular la distancia del punto P0 a la recta que pasa por los puntos P1 y P2
(a) P0 = (5, 3, −1), P1 = (4, 0, 2), P2 = (5, 0, 0)
Solución:
P2 − P1 = (5, 0, 0) − (4, 0, 2) = (1, 0, −2) vector direccional
P1 − P0 = (4, 0, 2) − (5, 3, −1) = (−1, −3, 3)
(P1 − P0 ) · V = (−1, −3, 3) · (1, 0, −2) = −1 − 0 − 6 = −7
q
| V |2 = ( 12 + 02 + (−2)2 )2 = 5
d =| (P1 − P0 ) −
(P1 − P0 ) · V
(−7)
(1, 0, −2) |
·
V
|
=|
(−1,
−3,
3)
−
5
| V |2
7
14
=| (−1, −3, 3) + ( , 0, − ) |
5
5
2
1
=| ( , −3, ) |
5
r5
1
2
= ( )2 + (−3)2 + ( )2
5
5
r
230
=
25
r
46
=
5
5.Calcular el ángulo entre las rectas
(a) X = (1, 2, 1) + t(4, 1, 1) y X = (2, 0, 0) + t(1, 1, 0)
Solución:
U = (4, 1, 1) y V = (1, 1, 0) Vectores direccionales
(4, 1, 1) · (1, 1, 0)
U ·V
5
5
5
cosθ =
=√
=√ √ =√ =
√
| U || V |
18 2
36 6
42 + 12 + 12 12 + 12 + 02
5
Cecilia Tola Pacheco
Calculo II
θ =cos
−1
5
√
6
MAT 102
!
θ =33◦ 33′
(b) X = (1, 0, 5) + t(1, −1, 2) y X = (1, 1, 3) + t(4, −1, 2)
Solución:
U = (1, −1, 2) y V = (4, −1, 2) Vectores direccionales
(1, −1, 2) · (4, −1, 2)
9
U ·V
9
=√ √ =√
cosθ =
=p
p
| U || V |
6 21
126
12 + (−1)2 + 22 42 + (−1)2 + 22
!
9
θ =cos−1 √
126
◦
′
θ =36 41
(c) X = 4 − 2t, y = 3 + 2t, z = −t y x = 1 + t, y = 1 + 3t, z = 1 − 3t
Solución:
X =(4, 3, 0) + t(−2, 2, −1) y X = (1, 1, 1) + t(1, 3, −3)
U =(−2, 2, −1) y V = (1, 3, −3) Vectores direccionales
(−2, 2, −1) · (1, 3, −3)
7
7
U ·V
=p
=√ √ = √
cosθ =
p
| U || V |
9 19 3 19
(−2)2 + 22 + (−1)2 12 + 32 + (−3)2
!
7
θ =cos−1 √
3 19
◦
′
θ =57 38
(d) x = 1, y = 1, z = t y x = t, y = 1, z = 1
Solución:
X =(1, 1, 0) + t(0, 0, 1) y X = (0, 1, 1) + t(1, 0, 0)
U =(0, 0, 1) y V = (1, 0, 0) Vectores direccionales
(0, 0, , 1) · (1, 0, 0)
0
0
U ·V
=√
= √ √ = √ =0
cosθ =
√
| U || V |
1 1
1
02 + 02 + 12 12 + 02 + 02
θ =cos−1 (0)
θ =90◦
6. Hallar los ángulos que forma la recta x = t, y = 2t, z = t con los ejes coordenados
Solución:
x = t, y = 2t, z = t
x = (0, 0, 0) + t(1, 2, 1)
6
Cecilia Tola Pacheco
Calculo II
Con el eje X
P1 =(a, 0, 0)y P0 = (0, 0, 0)
V =P1 − P0 = (a, 0, 0) − (0, 0, 0) = (a, 0, 0)
U =(1, 2, 1)
(1, 2, 1)(a, 0, 0)
U ·V
a
1
cosθ =
=
=√
=√
√ √ 2
| U || V |
6·a
6
6 a
!
1
θ =cos−1 √
6
◦
′
θ =65 54
(a)
Con el eje Y
P1 =(0, b, 0)y P0 = (0, 0, 0)
V =P1 − P0 = (0, b, 0) − (0, 0, 0) = (0, b, 0)
U =(1, 2, 1)
(1, 2, 1)(0, b, 0)
U ·V
2b
2
cosα =
=
=√
=√
√
√
2
| U || V |
6·b
6
6 b
!
2
α =cos−1 √
6
◦
′
α =35 15
(b)
Con el eje Z
P1 =(0, 0, c)y P0 = (0, 0, 0)
V =P1 − P0 = (0, 0, c) − (0, 0, 0) = (0, 0, c)
U =(1, 2, 1)
(1, 2, 1)(0, 0, c)
U ·V
c
1
cosβ =
=
= √ =√
√ √ 2
| U || V |
c 6
6
6 c
!
1
β =cos−1 √
6
◦
′
β =65 54
(c)
7. Mostrar que las siguientes rectas forman un triangulo rectángulo en el
espacio:X = (1, 5, −3) + t(1, −1, 5); X = (3, 6, 1) + t(1, 2, −1); X = (1, 2, 3) + t(2, 1, 4)
Solución:
v1 = (1, −1, 5)
v2 = (1, 2, −1)
v3 = (2, 1, 4)
7
MAT 102
Cecilia Tola Pacheco
Calculo II
MAT 102
Sea el angulo entre v1 y v2
cosθ =
Se tiene
(1, −1, 5)(1, 2, −1)
−6
v1 · v2
6
6
=
= √ √ = −√
=− √
| v1 || v2 | | (1, −1, 5) || (1, 2, −1) |
27 6
162
9 2
!
6
θ =cos−1 − √
2
◦ ′
′′
θ =118 7 31
◦
180 − θ =φ
180◦ − 118,1255 = φ
φ =61◦ 52′ 28′′
Seaα el angulo entre v1 y v3
(1, −1, 5)(2, 1, 4)
v ·v
21
21
cosα = 1 3 =
=√ √ =√
| v1 || v3 | | (1, −1, 5) || (2, 1, 4) |
27 21
567
!
21
α =cos−1 √
567
◦ ′
′′
α =28 7 31
Seaβ el angulo entre v2 y v3
(1, 2, −1)(2, 1, 4)
0
v ·v
0
=√ √ =√
cosβ = 1 3 =
=0
| v1 || v3 | | (1, 2, −1) || (2, 1, 4) |
6 21
126
β =cos−1 (0)
α =90◦
Luego φ + α = 61◦ 52′ 28′′ + 28◦ 7′ 31′′ = 90◦ , es un triangulo rectángulo.
8. Calcular la distancia entre las siguientes rectas
(a) X = (1, 3, 2) + t(1, −1, 0) y X = (0,1,2) + t(4, 1, 1)
Solución:
U = (1, −1, 0) y V = (4, 1, 1)
P1 − P0 =(0, 1, 2) − (1, 3, 2) = (−1, −2, 0)
i j k
U × V = 1 −1 0 = −i − j + (1 + 4)k = −i − j + 5k = (−1, −1, 5)
4 1 1
(−1, −1, 5)
U ×V
= (−1, −2, 0) p
d = (P1 − P0 ) ·
| U ×V |
(−1)2 + (−1)2 + 52
= (−1, −2, 0) ·
(−1, −1, 5)
√
27
1+2+0
√
27
3
= √
3 3
1
=√
3
=
8
Cecilia Tola Pacheco
Calculo II
MAT 102
(b) X = (4, 2, 5) + t(1, 0, −1) y X = (1, 1, 2) + t(2, 0, −1)
Solución:
U = (1, 0, −1) y V = (2, 0, −1)
P1 − P0 =(1, 1, 2) − (4, 2, 5) = (−3, −1, −3)
i j k
U × V = 1 0 −1 = 0i − (−1 + 2)j + 0k = 0i − j + 0k = (0, −1, 0)
2 0 −1
(0, −1, 0)
U ×V
d = (P1 − P0 ) ·
= (−3, −1, −3) p
=| (−3, −1, −3) · (0, −1, 0) |= 1
| U ×V |
02 + (−1)2 + 02
(c) x = t, y = 2t, z = 3t y x = 1 − t, y = 2 + 2t, z = 1 − 3t
Solución:
X = (0, 0, 0)U = (1, 2, 3) y V = (−1, 2, −3)
P1 − P0 =(1, 2, 1) − (0, 0, 0) = (1, 2, 1)
i j k
U × V = 1 2 3 = (−6 − 6)i − (−3 + 3)j + (2 + 2)k = −12i − 0j + 4k = (−12, 0, 4)
−1 2 −3
d = (P1 − P0 ) ·
(−12, 0, 4)
(−12, 0, 4)
U ×V
= (1, 2, 1) p
= (1, 2, 1) √
| U ×V |
160
(−12)2 + 02 + 42
−12 + 0 + 4
=
√
160
−8
= √
160
8
=√
160
8
= √
4 10
2
=√
10
(d) X = 2 + t, y = −t, z = 1 + 3t y X = 2t, y = 1 − 2t, z = 2 + 6t
Solución:
X =(2, 0, 1) + t(1, −1, 3) y X = (0, 1, 2) + t(2, −2, 6)
U =(1, −1, 3) y V = (2, −2, 6)
P1 − P0 =(0, 1, 2) − (2, 0, 1) = (−2, 1, 1)
i j k
U × V = 1 −1 3 = (−6 + 6)i − (6 − 6)j + (−2 + 2)k = 0i − 0j + 0k = (0, 0, 0)
2 −2 6
d = (P1 − P0 ) ·
(0, 0, 0)
U ×V
= (−2, 1, 1) √
=0
| U ×V |
0
9
Cecilia Tola Pacheco
1.1.
Calculo II
EL PLANO
9. Hallar la ecuación vectorial del plano que pasa por los puntos:
(a) P0 = (7, 2, 3), P1 = (4, 5, 6), P2 = (−1, , 0, 1)
X =P0 + sU + tV
P0 − P0 = (4, 5, 6) − (7, 2, 3) = (−3, 3, 3)
P2 − P0 = (−1, 0, 1) − (7, 2, 3) = (−8, −2, −2)
X =P0 + sU + tV
X = (7, 2, 3) + s(−3, 3, 3) + t(−8, −2, −2)
(b) P0 = (0, 0, 0), P1 = (1, 0, 0), P2 = (0, 1, 0)
Solución:
P0 − P0 = (1, 0, 0) − (0, 0, 0) = (1, 0, 0)
P2 − P0 = (0, 1, 0) − (0, 0, 0) = (0, 1, 0)
X =P0 + sU + tV
X = (0, 0, 0) + s(1, 0, 0) + t(0, 1, 0)X = s(1, 0, 0) + t(0, 1, 0)
(c) P0 = (0, 0, 5), P1 = (4, 3, 0), P2 = (1, 5, 7)
Solución:
X =P0 + sU + tV
P0 − P0 = (4, 3, 0) − (0, 0, 5) = (4, 3, −5)
P2 − P0 = (1, 5, 7) − (0, 0, 5) = (1, 5, 2)
X =P0 + sU + tV
X = (0, 0, 5) + s(4, 3, −5) + t(1, 5, 2)
(10.) Hallar la ecuación cartesiana del plano que pasa por los puntos
(a)P0 = (1, 2, 1), P1 = (2, −1, 0), P2 = (4, 1, −2)
P1 − P0 = (2, −1, 0) − (1, 2, 1) = (1, −3, −1)
P2 − P0 = (4, 1, −2) − (1, 2, 1) = (3, −1, −3)
X = (x, y, z) = (1, 2, 1) + s(1, −3, −1) + t(3, −1, −3)
x = − + s + 3t
x = 1 + s + 3t 3y = 6 − 9s − 3t
y = 2 − 3s − t ⇒
...(1)
x + 3y = 7 − 8s
z = 1 − s − 3t
−3y = −6 + 9s + 3t
x = 1 + s + 3t
z = 1 − s − 3t
y = 2 − 3s − t ⇒
...(2)
z − 3y = −5 + 8s
z = 1 − s − 3t
De (1) y (2), se tiene:
10
MAT 102
Cecilia Tola Pacheco
Calculo II
x + 3y = 7 − 8s
z − 3y = −5 + 8s
x+z = 2
x+z−2 = 0
(b) P0 = (1, −1, 0), P1 = (4, 2, −3), P2 = (1, 2, 3)
Solución:
P1 − P0 = (4, 2, −3) − (1, −1, 0) = (3, 3, −3)
P2 − P0 = (1, 2, 3) − (1, −1, 0) = (0, 3, 3)
X = (x, y, z) = (1, 2, 3) + s(3, 3, −3) + t(0, 3, 3)
−y = 1 + 3s − 3t
x = 1 + 3s
z = −3s + 3t
y = −1 + 3s + 3t ⇒
...(1)
z − y = 1 − 6s
z = −3s + 3t
x = 1 + 3s
x = 1 + 3s/(2)
2x = 2 + 6s
y = −1 + 3s + 3t ⇒
⇒
z − y = 1 − 6s
z − y = 1 − 6s
z = −3s + 3t
2x = 2 + 6s
z − y = 1 − 6s
2x − y + z = 3
2x − y + z − 3 = 0
(c)P0 = (2, −1, 5), P0 = (0, −1, 0), P2 = (0, 0, 1)
Solución:
P1 − P0 = (0, −1, 0) − (2, −1, 5) = (−2, 0, −5)
P2 − P0 = (0, 0, 1) − (2, −1, 5) = (−2, 1, −4)
X = (x, y, z) = (2, −1, 5) + s(−2, 0, −5) + t(−2, 1, −4)
U = (−1, 0, −5) y V = (−2, 1, −4)
i j k
N = U × V = −2 0 −5 = 5i − (8 − 10)j − 2k = 5i + 2j − 2k = (5, 2, −2)
−2 1 −4
Sea X = (x, y, z)
N · (X − P0 ) = 0
(5, 2, −2) · ((x, y, z) − (2, −1, 5)) =0
(5, 2, −2) · (x − 2, y + 1, z − 5) =0
5(x − 2) + 2(y + 1) − 2(z − 5) =0
5x − 10 + 2y + 2 − 2z + 10 =0
5x + 2y − 2z + 2 =0
11
MAT 102
Cecilia Tola Pacheco
Calculo II
MAT 102
11. Hallar la distancia del punto P al plano dado
(a)P = (0, 0, 0), plano X = (1, 2, 3) + s(4, −1, 0) + t(1, 3, −1)
Solución:
i j K
N = U × V = 4 −1 0 = i + 4j + (12 + 18)k = i + 4 = i + 4j + 13k = (1, 4, 13)
1 3 −1
P − P0 =(0, 0, 0) − (1, 2, 3) = (−1, −2, −3)
√
√
| n |=| (1, 4, 13) |= 12 + 42 + 132 = 186
#
"
(1, 4, 13)
n
= (−1, −2, −3) · √
d = (P1 − P0 ) ·
|n|
186
"
#
−1 − 8 − 39
=
√
186
#
"
−48
= √
186
48
=√
186
(b) P = (−1, 2, 1), plano X = (4, 0, −1) + s(0, 1, 0) + t(0, 0, 1)
Solución:
i j K
N = U × V = 0 1 0 = i + 0j + 0k = (1, 0, 0)
0 0 1
P − P0 =(−1, 2, 1) − (4, 0, −1) = (−5, 2, 2)
√
| n |=| (1, 0, 0) |= 1 = 1
#
"
(1, 0, 0)
n
d = (P − P0 ) ·
= (−5, 2, 2) ·
|n|
1
= [−5 + 0 + 0]
= [−5]
=5
(c) P = (4, 1, 3); plano x + y + z = 6
Solución:
n = (1, 1, 1) es normal al plano x + y + z = 6, si x = 0, y = 0 entonces x + y + z = 6 ⇒ z = 6
Por tanto P0 = (0, 0, 6)
√
√
| n |=| (1, 1, 1) |= 12 + 12 + 12 = 3
P − P0 =(4, 1, 3) − (0, 0, 6) = (4, 1, −3)
12
Cecilia Tola Pacheco
Calculo II
#
"
(1, 1, 1)
n
d = (P − P0 ) ·
= (4, 1, −3) · √
|n|
3
" #
2
= √
3
" #
2
= √
3
(d) P = (1, 2, 3); plano x − y = 5
Solución:
n = (1, −1, 0); si y = 0, z = 0 ⇒ x − y = 5 ⇒ x = 5
por tantoP0 = (5, 0, 0)
q
√
| n |=| (1, −1, 0) |= 12 + (−1)2 + 02 = 2
P − P0 =(1, 2, 3) − (5, 0, 0) = (−4, 2, 3)
#
"
(1, 1, 1)
n
d = (P − P0 ) ·
= (4, 1, −3) · √
|n|
2
"
#
−4 − 2 + 0
=
√
2
" #
−6
= √
2
6
=√
2
√
6 2
= √
2
√
=3 2
1.1.1.
PROBLEMAS VARIOS
12. Hallar la intersección de la recta X = (3; 1; 3) + t(1; 1; 1) con cada uno de los planos
coordenados.
13. Determinar el punto donde la recta que pasa por (1; 3; 1) y es ortogonal al plano
3x − 2y + 5z = 15, intersecta a dicho plano.
Solución:
Sea
X = P0 + tV
P0 = (1, 3, 1)
N = V = (3, −2, 5)
Luego la recta es
X = (1, 3, 1) + t(3, −2, 5)
x = 1 + 3t
y = 3 − 2t
z = 1 + 5t
13
MAT 102
Cecilia Tola Pacheco
Calculo II
MAT 102
Reemplazando en la ecuacion del plano tenemos:
3x − 2y + 5z = 15
3(1 + 3t) − 2(3 − 2t) + 5(1 + 5t) = 15
3 + 9t − 6 + 4t + 5 + 25t = 15
38r + 2 = 15
38t = 15 − 2
35t = 13
13
t=
38
Como t =
13
tenemos el punto que intersecta a dicho plano
38
39 38 + 39 77
77
13
) = 1+
=
=
⇒x=
38
38
38
38
38
13
26 114 − 26 88
88
y = 3 − 2( ) = 3 −
=
=
⇒y=
38
38
38
38
38
13
65 38 + 65 103
103
z = 1 + 5( ) = 1 +
=
=
⇒z=
38
38
38
38
38
x = 1 + 3(
Por tanto tenemos el punto X = (
77 88 103
, ,
)
38 38 38
14. Mostrar que los planos X = (2; 0; 4) + s(1; 7; 3) + t(3; 8; 0) y X = (3; 2; 3) + s(4; 1; 3) + t(9; 5; 9)
son paralelos y encuentre la distancia entre ellos.
Solución:
De X = (2; 0; 4) + s(1; 7; 3) + t(3; 8; 0) tenemos U = (1, 7, 3) y V = (−3, 8, 0)
i j k
N1 = U × V = 1 7 3 = i (0 − 24) − j (0 + 9) + k (8 + 21) = (−24, −9, 29)
−3 8 0
∴ N1 = (−24, −9, 29)
De X = (3; 2; 3) + s(4; 1; 3) + t(9; 5; 9) tenemos U = (4, −1, 3) y V = (9, 5, 9)
i j k
N2 = U × V = 4 −1 3 = i (−9 − 15) − j (36 − 27) + k (20 + 9) = (−24, −9, 29)
9 5 9
∴ N2 = (−24, −9, 29)
Ahora recién podemos probar que:
N1 = kN2
(−24, −9, 29) = k(−24, −9, 29)
−24 = −24k ⇒ k = 1
−9 = −9k ⇒ k = 1
29 = 29k ⇒ k = 1
14
Cecilia Tola Pacheco
Calculo II
MAT 102
Luego, los planos paralelos
Debemos encontrar la ecuación del plano en su forma cartesiana
N1 · (X − P0 ) = 0
(−24, −9, 29) · ((x, y, z) − (2, 0, 4)) = 0
−24(x − 2) − 9y + 29(z − 4) = 0
−24x + 48 − 9y + 29z − 116 = 0
−24x − 9y + 29z − 68 = 0
−24x − 9y + 29z = 68
y
y
N2 · (X − P0 ) = 0
(−24, −9, 29) · ((x, y, z) − (3, 2, 3)) = 0
− 24(x − 3) − 9(y − 2) + 29(z − 3) = 0
− 24x + 72 − 9y + 18 + 29z − 87 = 0
− 24 − 9y + 29z + 3 = 0
− 24x − 9y + 29z = −3
y
y
y
y
Por ultimo nos falta encontrar la distancia entre los dos planos paralelos
La distancia d entre los planos paralelos ax + by + cz = d1 , ax + by + cz = d2 esta dada por
d =| √
−71
−3 − 68
71
|=| √
|=| √
=√
1498
1498
242 + 92 + 292
a2 + b2 + c2
d2 − d1
15. Encontrar la ecuación del plano que pasa por (1; 2; 3) y contiene a la recta
X = (1, 1, 1) + t(5, −2, 3).
Solución:
SeaP1 = (1, 2, −3)
P1 − P0 = (1, 2, 3) − (1, 1, 1) = (0, 1, −4) = V , entonces
X = P0 + sV + tU
X = (1, 1, 1) + s(0, 1, −4) + t(5, −2, 3)
x = 1 + 5t
y = 1 + s − 2t
z = 1 − 4s + 3t
Formando un sistema de ecuaciones se tiene:
x
= 1 + 5t
4y = 4 + 4s − 8t
4y + z = 5 − 5t
z = 1 − 4s + 3t
y = 1 + s − 2t ∗ (4)
⇒
⇒
z = 1 − 4s + 3t ∗ (1)
4y + z = 5 − 5t
x + 4y + z = 6
∴ x + 4y + z − 6 = 0
17. Hallar la recta que pasa por (1; 2; 3) y es perpendicular al plano x − y + 2z = 0
Solución:
X = P0 + tV
donde P0 = (1, 2, 3) y V = (a, b, c)
luego N = kV donde k = 1
Por tanto N = V = (1, −1, 2)
Por tanto la recta perdida es:
X = P0 + tV = (1, 2, 3) + t(1, −1, 2)
15
Cecilia Tola Pacheco
Calculo II
MAT 102
∴ X = (1, 2, 3) + t(1, −1, 2)
19. Mostrar que la distancia entre los planos paralelos X = P0 + sU + tV yX = P1 + sU + tV esta
dada por
d =| (P1 − P0 ) ·
U ×V
|
| U ×V |
Solución:
P1
P 1-P 0
UxV
Por una parte el vector U × V es perpendicular
aU yV
Por otra parte el vector P1 − −P0 va del vector P0
a P1
Entonces, la distancia entre los dos vectores es
la proyección del vector P1 − P0 sobre el vector
U × V es decir:
d =| (P1 − P0 ) ·
P2
Donde se va tomando el vector absoluto para
evitar distancias negativas.
P0
1.2.
U ×V
|
| U ×V |
CILINDROS Y SUPERFICIES CUADRÁTICAS
Graficar las siguientes superficies
34.x2 + y 2 = 9
Solución:
16
Cecilia Tola Pacheco
Calculo II
35. xy = 4
Solución:
36. y = ex
Solución:
17
MAT 102
Cecilia Tola Pacheco
Calculo II
37. x2 + y 2 − 4y = 0
Solución:
x2 + y 2 − 4y = 0
x2 y 2 − 4y + 4 − 4 = 0
x2 + (y − 2)2 = 4
Centro(0, 2); r = 2
38. y = cos x
−π < x < π
Solución:
18
MAT 102
Cecilia Tola Pacheco
Calculo II
40. y 2 + z2 = 16
Solución:
41. 9x2 + 4y 2 = 36
Solución:
9x2 + 4y 2 = 36
x2 y 2
+
=1
22 32
19
MAT 102
Cecilia Tola Pacheco
42. y = tan x
−
Calculo II
π
π
<x<
2
2
Solución:
43. y 2 + z = 2
Solución:
20
MAT 102