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