MATEMÁTICA
DOCENTE: ALEJANDRO BERARDI ALUMNO:………………………………………………………………………………………….. AÑO: 5º ES
CapÄąĚ tulo 6
MATRICES Y DETERMINANTES 6.1.
IntroduccioĚ n
Las matrices y los determinantes son herramientas del aĚ lgebra que facilitan el ordenamiento de datos, asÄąĚ como su manejo. Los conceptos de matriz y todos los relacionados fueron desarrollados baĚ sicamente en el siglo XIX por matemaĚ ticos como los ingleses J.J. Sylvester y Arthur Cayley y el irlandeĚ s William Hamilton. Las matrices se encuentran en aquellos aĚ mbitos en los que se trabaja con datos regularmente ordenados y aparecen en situaciones propias de las Ciencias Sociales , EconoĚ micas y BioloĚ gicas.
6.2.
Matrices. DeďŹ nicioĚ n y primeros ejemplos
Una matriz es una tabla rectangular de  a11 a12 a13  a21 a22 a23  A= . .. .. ďŁ .. . .
am1 am2 am3
nuĚ meros reales dispuestos en ďŹ las y columnas del modo:   . . . a1n â†?   . . . a2n  â†? ..  â†? Filas de la matriz A .. . .     . . . amn â†?
Columnas de la matriz A
Abreviadamente se puede expresar A = (aij ). Cada elemento de la matriz lleva dos subÄąĚ ndices. El primero de ellos “iâ€?, indica la ďŹ la en la que se encuentra el elemento, y el segundo, “jâ€?, la columna. AsÄąĚ el elemento a23 estaĚ en la ďŹ la 2 y columna 3. Las matrices siempre se representaraĚ n con letras mayuĚ sculas. Ejemplos: Son ejemplos de matrices los siguientes:   3 1 0 √  2 −4 0  2 1 6 −4 0 √  C = A= B= ďŁâˆ’1 1 2 1 2 1 3 4 5 1 0 0 A tiene 2 ďŹ las y 2 columnas, diremos que su tamanĚƒo es 2 x 2.ÂżQueĚ elemento es a21 ?. B tiene 2 ďŹ las y 3 columnas, diremos que su tamanĚƒo es 2 x 3.ÂżQueĚ elemento es b23?. C tiene 4 ďŹ las y 3 columnas, diremos que su tamanĚƒo es 4 x 3.ÂżQueĚ elemento es c42 ?. En general, si una matriz A tiene m ďŹ las y n columnas, diremos que su tamanĚƒo o dimensioĚ n es m x n (se lee “m por nâ€?), siempre en primer lugar el nÂş de ďŹ las y en segundo lugar el de columnas.
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CAPIĚ TULO 6. MATRICES Y DETERMINANTES
6.3.
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Tipos de matrices
1. Se llama matriz nula a la que tiene todos los elementos cero. Por ejemplo,
0 0 0 0 0 A= 0 0 0 0 0
es una matriz nula de tamanĚƒo 2x5. 2. Se llama matriz ďŹ la a la que soĚ lo tiene una ďŹ la, es decir su dimensioĚ n es 1x n. Por ejemplo,
1 0 −4 9
es una matriz ďŹ la de tamanĚƒo 1 x 4. 3. Se llama matriz columna a la que soĚ lo consta de una columna, es decir su dimensioĚ n seraĚ m x 1, como por ejemplo:   1 0  C=ďŁ âˆš − 8 es una matriz columna de tamanĚƒo 3 x 1. 4. Una matriz es cuadrada cuando tiene el mismo nuĚ mero de ďŹ las que de columnas, es decir su dimensioĚ n es n x n. La matriz ( 23 14 ) del primer ejemplo anterior es cuadrada de tamanĚƒo 2 x 2 o simplemente de orden 2. Otro ejemplo de matriz cuadrada es:

 1 2 3 D=ďŁ 6 5 4 −3 −4 0
de orden 3. Dentro de las matrices cuadradas llamaremos diagonal principal a la formada por los elementos a11 , a22 , a33, . . . , ann , siendo la matriz:   a11 a12 a13 . . . a1n a21 a22 a23 . . . a2n    A= . .. .. ..  .. ďŁ .. . . . .  an1 an2 an3 . . .
ann
En la matriz D del ejemplo anterior, su diagonal principal estarÄąĚ a formada por 1, 5, 0. Se llama traza de la matriz a la suma de los elementos de la diagonal. Es decir, Traza (A)=a11 + a22 + a33 + . . . + ann , y en el caso de D, Traza (D)= 1+5+0 = 6. La diagonal secundaria es la formada por los elementos a1n , a2,n−1, a3,n−2, . . . , an1 . En la matriz D estarÄąĚ a formada por 3, 5, -3. Una clase especial de matrices cuadradas son las matrices triangulares. Una matriz es triangular superior si todos los elementos por debajo de la diagonal principal son nulos y triangular inferior si son nulos todos los elementos situados por encima de dicha diagonal. Son ejemplos de estas matrices:     1 0 0 0 1 4 13 0 −4 0  0  F = ďŁ0 9 −5 E= ďŁ3 4 5 0  0 0 Ď€ 1 3 16 −78 Triangular superior Triangular inferior
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Si una matriz es a la vez triangular superior e inferior, soĚ lo tiene elementos en la diagonal principal. Una matriz de este tipo se denomina matriz diagonal. Un ejemplo de matriz diagonal serÄąĚ a:   1 0 0 0 0 −45 0 0  G= ďŁ0 0 3 0 0 0 0 0 Por uĚ ltimo, si una matriz diagonal tiene en su diagonal principal soĚ lo unos, se denomina matriz unidad o identidad. Se suelen representar por In , donde n es el orden o tamanĚƒo de la matriz. Algunas matrices identidad son:     1 0 0 0 1 0 0 0 1 0 0 1 0  ďŁ I3 = 0 1 0 I4 =  I2 = ďŁ0 0 1 0 0 1 0 0 1 0 0 0 1
6.4.
Aplicaciones de las matrices
Las matrices se utilizan en el contexto de las ciencias como elementos que sirven para clasiďŹ car valores numeĚ ricos atendiendo a dos criterios o variables. Ejemplo: Un importador de globos los importa de dos colores, naranja (N) y fresa (F). Todos ellos se envasan en paquetes de 2, 5 y 10 unidades, que se venden al precio (en euros) indicado por la tabla siguiente: Color N Color F
2 unid. 0’04 0’03
5 unid. 0’08 0’05
10 unid. 0’12 0’08
Sabiendo que en un anĚƒo se venden el siguiente nuĚ mero de paquetes: 2 unid. 5 unid. 10 unid.
Color N 700000 600000 500000
Color F 50000 40000 500000
Resumir la informacioĚ n anterior en 2 matrices A y B, de tamanĚƒo respectivo 2x3 y 3x2 que recojan las ventas en un anĚƒo (A) y los precios (B). Nos piden que organicemos la informacioĚ n anterior en dos matrices de tamanĚƒo concreto. Si nos ďŹ jamos en las tablas, es sencillo obtener las matrices: N  2 ud 5 ud 10 ud 0 04 700000 600000 500000 N ďŁ A= B = 0 08 50000 40000 500000 F 0 12
F  0 03 2 ud 0 05 5 ud 0 08 10 ud
Estas matrices se denominan matrices de informacioĚ n, y simplemente recogen los datos numeĚ ricos del problema en cuestioĚ n. Otras matrices son las llamadas matrices de relacioĚ n, que indican si ciertos elementos estaĚ n o no relacionados entre sÄąĚ . En general, la existencia de relacioĚ n se expresa con un 1 en la matriz y la ausencia de dicha relacioĚ n de expresa con un 0. Estas matrices se utilizan cuando queremos trasladar la informacioĚ n dada por un grafo y expresarla numeĚ ricamente.
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En Matemáticas, un grafo es una colección cualquiera de puntos conectados por lineas. Existen muchos tipos de grafos. Entre ellos, podemos destacar: * Grafo simple: Es el grafo que no contiene ciclos, es decir, lineas que unan un punto consigo mismo, ni lineas paralelas, es decir, lineas que conectan el mismo par de puntos. * Grafo dirigido: Es el grafo que indica un sentido de recorrido de cada linea, mediante una flecha. Estos tipos de grafo pueden verse en la figura:
Figura 6.1: Grafo, Grafo simple y Grafo dirigido.
Relacionadas con los grafos se pueden definir algunas matrices. Entre todas ellas, nosotros nos fijaremos en la llamada matriz de adyacencia, que es aquella formada por ceros y unos exclusivamente, de tal forma que: * un 1 en el lugar (i,j) expresa la posibilidad de ir desde el punto de la fila i hasta el punto de la columna j mediante una linea que los una directamente. * un 0 en el lugar (i,j) expresa la imposibilidad de ir del primer punto al segundo mediante una linea que los una directamente. La matriz de adyacencia del grafo dirigido de la figura anterior será: A A 0 B 0 C 1 D 0
B 1 0 0 0
C 0 1 0 0
D 1 0 0 0
Ejercicio 1) Escribe las correspondientes matrices de adyacencia de los grafos:
2) Dibuja los grafos dirigidos que correspondan a las matrices de adyacencia: A A 0 B 1 C 0
B 1 0 0
C 0 1 0
A A 0 B 0 C 1 D 0
B 1 0 0 1
C 1 0 0 1
D 1 1 0 0
CAPIĚ TULO 6. MATRICES Y DETERMINANTES
6.5.
86
Operaciones con matrices
6.5.1.
Suma y diferencia
Dadas dos matrices A y B podemos realizar su suma o diferencia de acuerdo a la siguiente regla. Para sumar o restar dos matrices del mismo tamanĚƒo, se suman o restan los elementos que se encuentren en la misma posicioĚ n, resultando otra matriz de igual tamanĚƒo. Por ejemplo: 2 0 4 0 1 −1 2 1 3 − = 3 2 5 −7 0 −4 −4 2 1 2x3
2x3
2x3
Si las matrices tienen diferente tamanĚƒo, no se pueden sumar o restar entre sÄąĚ . Propiedades de la suma (y diferencia) de matrices: a) Conmutativa: A + B = B + A b) Asociativa: A + (B + C) = (A + B) + C c) Elemento neutro: La matriz nula del tamanĚƒo correspondiente. d) Elemento opuesto de A: La matriz -A, que resulta de cambiar de signo a los elementos de A. Ejemplo: Si     0 1 0 −1 A = ďŁâˆ’4 −2 =⇒ −A = ďŁ 4 2 −3 9 3 −9 3x2
porque:
3x2
     0 1 0 0 0 −1 ďŁâˆ’4 −2 + ďŁ 4 2 = ďŁ0 0 −3 9 0 0 3 −9 
3x2
3x2
3x2
Ejercicios: 1. Las exportaciones, en millones de euros, de 3 paÄąĚ ses A, B, C a otros tres X, Y, Z, en los anĚƒos 2000 y 2001 vienen dadas por las matrices: X Y Z  A 11 6 7 0 5 A2000 = B ďŁ14 5 10 1 2 C 20 9 3 2 2 3 
X Y Z  A 13 3 7 1 = B ďŁ15 7 11 1 3 2 C 21 0 2 4 3 
A2001
Calcula y expresa en forma de matriz el total de exportaciones para el conjunto de los dos anĚƒos. ÂżCuaĚ ntos millones ha exportado el paÄąĚ s B al Z en total? Calcula el incremento de las exportaciones del anĚƒo 2000 al 2001 con los datos del ejemplo anterior. 2. Calcula x, y, z en la suma:       x − y −1 2 y 0 z −1 −1 3 ďŁ 1 y −x + ďŁâˆ’z 2 3 = ďŁ 0 4 4 0 z 2 −2 3 x −2 4 1 3. Calcula a, b, c para que se cumpla la igualdad: 3−a b −2 2 a+b 4 −1 a 2 + = 4 −c + 1 6 1−c 2 0 2 0 6
CAPIĚ TULO 6. MATRICES Y DETERMINANTES
6.5.2.
87
Producto por un nuĚ mero real
Dada una matriz cualquiera A y un nuĚ mero real k, el producto k¡A se realiza multiplicando todos los elementos de A por k, resultando otra matriz de igual tamanĚƒo. (Evidentemente la misma regla sirve para dividir una matriz por un nuĚ mero real). Por ejemplo: −10 −5 −15 2 1 3 = −5 ¡ 20 −10 −5 −4 2 1 2x3
2x3
Propiedades: a) Distributiva respecto de la suma de matrices: k¡(A + B) = k¡A + k¡B b) Distributiva respecto de la suma de nuĚ meros: (k + d)¡A= k¡A + d¡A c) Asociativa: k¡(d¡A)=(k¡d)¡A d) Elemento neutro, el nuĚ mero 1: 1¡A=A Ejercicios: 1. Si A =
1 1 0 1
−1 0 yB= , halla una matriz X que veriďŹ que la ecuacioĚ n: 0 2 2¡X −4¡A = B
2. Determina las matrices X y Y sabiendo que:  1   3X − 5Y = 8 2    −X + 3Y = 3
6.5.3.
−2 1 4 0
TrasposicioĚ n de matrices
Dada una matriz cualquiera A, se llama matriz traspuesta de A, y se representa por At a la matriz que resulta de intercambiar las columnas de A. las ďŹ las y 2 1 0 7 Por ejemplo, si A = , entonces la matriz traspuesta de A es: −3 4 2 1   2 −3 1 4   At =  ďŁ0 2  7 1 Evidentemente, si A es una matriz de tamanĚƒo m x n, su traspuesta At tendraĚ tamanĚƒo n x m, pues el nuĚ mero de columnas pasa a ser el de ďŹ las y viceversa. Si la matriz A es cuadrada, su traspuesta tendraĚ el mismo tamanĚƒo. Propiedades: a) (At )t = A, es decir, la traspuesta de la traspuesta es la matriz inicial. b) (A + B)t = At + B t c) (k ¡ A)t = k ¡ At En base a esta nueva operacioĚ n, podemos deďŹ nir otras dos clases de matrices, que son: Matriz simeĚ trica, que es aquella para la que se cumple  2 1 ďŁ A= 1 0 3 −2
que At = A, por ejemplo la matriz:  3 −2 √ 7
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es simeĚ trica (comprueĚ balo). En una matriz simeĚ trica, los elementos son simeĚ tricos respecto a la diagonal principal. Ejercicio: ÂżPuede ser simeĚ trica una matriz que no sea cuadrada?ÂżPor queĚ ?. Matriz antisimeĚ trica, es aquella para la que se cumple que At = −A. Por ejemplo:   0 1 3 B = ďŁâˆ’1 0 −2 −3 2 0 es antisimeĚ trica (comprueba). En una matriz antisimeĚ trica, los elementos de la diagonal principal son siempre nulos (Âżpor queĚ ?), y los restantes son opuestos respecto a dicha diagonal. Ejercicios: 
   1 3 3 1 1 2 1. Dadas las matrices A = ďŁ1 4 3 y B = ďŁ 2 0 −1 calcula 3At − B t . 1 3 4 −6 −1 0 2. Obtener las matrices X e Y  1   2X − 3Y = 4 a) −1   X − Y = 3
6.5.4.
que veriďŹ quen los sistemas:  5 2 1   X + Y = 2 3 0 b) 0 6 2   X − Y = 6 0 1
 3 1   2X + Y = 0 −2 c) 1 0   X + 2Y = −2 4
Producto de matrices
Hay que dejar claro ya desde el principio que no todas las matrices pueden multiplicarse. Dos matrices se pueden multiplicar cuando se cumple la siguiente condicioĚ n: “Para multiplicar dos matrices A y B, en este orden, A¡B , es condicioĚ n indispensable que el nuĚ mero de columnas de A sea igual al nuĚ mero de ďŹ las de Bâ€? Si no se cumple esta condicioĚ n, el producto A¡B no puede realizarse, de modo que esta es una condicioĚ n que debemos comprobar previamente a la propia multiplicacioĚ n. Una vez comprobado que el producto A¡B se puede realizar, si A es una matriz m x n y B es una matriz n x p (observemos que el nÂş de columnas de A = n = nÂş de ďŹ las de B), entonces el producto A¡B da como resultado una matriz C de tamanĚƒo n x p del siguiente modo: “El elemento que se encuentra en la ďŹ la i y la columna j de la matriz C=A¡B, se obtiene multiplicando los elementos de la ďŹ la i de A por la columna j de B y sumando los resultadosâ€? VeaĚ moslo mediante un ejemplo: Para multiplicar las matrices: A=
−3 2 1 4 2 5 3 −2 2x4

y
 0 −4 1 1 −2 1  B= ďŁ2 0 2 3 2 1 4x3
primero comprobamos que se puede realizar el producto A¡B, pues el nÂş de columnas de A es 4 y el nÂş de ďŹ las de B tambieĚ n es 4, y el resultado, seguĚ n lo dicho seraĚ una matriz de tamanĚƒo 2 x 3, tiene 2
CAPIĚ TULO 6. MATRICES Y DETERMINANTES ďŹ las y 3 columnas:
89
  0 −4 1 1 −2 1 −3 2 1 4   ¡ = 2 5 3 −2 ďŁ2 0 2 2x4 2x3 3 2 1 4x3
SoĚ lo nos falta completar los elementos de la matriz producto. Para ello, seguimos la regla anterior: El elemento de la ďŹ la 1 y columna 1 de A¡B proviene de multiplicar elemento a elemento la ďŹ la 1 de A por la columna 1 de B y sumar, es decir: (−3) ¡ 0 + 2 ¡ 1 + 1 ¡ 2 + 4 ¡ 3 = 0 + 2 + 2 + 12 = 16 El elemento de la ďŹ la 1 y columna 2 de A¡B proviene de multiplicar elemento a elemento la ďŹ la 1 de A y la columna 2 de B y sumar: (−3) ¡ (−4) + 2 ¡ (−2) + 1 ¡ 0 + 4 ¡ 2 = 12 − 4 + 0 + 8 = 16 El elemento de la ďŹ la 1 y columna 3 de A¡B proviene de multiplicar elemento a elemento la ďŹ la 1 de A y la columna 3 de B y sumar: (−3) ¡ 1 + 2 ¡ 1 + 1 ¡ 2 + 4 ¡ 1 = −3 + 2 + 2 + 4 = 5 AsÄąĚ sucesivamente se obtienen (comprueba):
16 16 5 5 −22 11 2x3
Ejercicios: 1. Para las matrices A y B anteriores, calcula B¡A 1 −3 3 −5 2. Si A = ,B= , calcula si es posible A¡B y B¡A. ÂżCoinciden?. −2 6 2 1   1 −1 3 0 2 ďŁ ďŁ¸ 3. Lo mismo si A = 0 −2 , B = . 1 −1 5 4 1 4. Calcula todos los productos posibles entre las matrices:     1 2 3 1 A = ďŁ1 1 1  B = ďŁ2 0 2 −1 1
C=
2 1 0 3 4 5
AdemaĚ s, calcula A2 y A3 . 5. Para las matrices 1 −1 2 A= 4 0 −3
B=
0 3 4 −1 −2 3

 2 3 0 1 C = ďŁâˆ’5 1 4 −2 1 0 0 −3
  2 D = ďŁ1 3
calcula: A + B, 3A − 4B, A ¡ B, A ¡ D, B ¡ C, C ¡ D, At ¡ C, Dt ¡ At , B t ¡ A, Dt ¡ D, D ¡ Dt
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Propiedades del producto de matrices a) Asociativa: A¡(B¡C) = (A¡B)¡C b) Distributiva respecto de la suma: A ¡ (B + C) = A ¡ B + A ¡ C (B + C) ¡ A = B ¡ A + C ¡ A c) Elemento neutro, la matriz identidad correpondiente, si A es m x n: A ¡ In = A Im ¡ A = A d) En general el producto de matrices no es conmutativo A ¡ B = B ¡ A Pueden verse ejemplos en los ejercicios anteriores. Esta es una propiedad muy importante. e) El producto de dos matrices no nulas A y B puede dar lugar a una matriz nula:   5 0 2 1 3 ďŁ ďŁ¸ ¡ 2 = 0 0 2 1 −4 2x3 2x1 3x1
Se dice que el conjunto de las matrices con la operacioĚ n producto tiene divisores de cero, es decir, hay matrices no nulas cuyo producto es nulo. Ejercicios: 1. Si A y B son dos matrices cuadradas del mismo orden, Âżson ciertas las propiedades siguientes, que son ciertas para las operaciones con nuĚ meros reales?: a) (A + B)2 = A2 + B 2 + 2 ¡ A ¡ B b) (A − B)2 = A2 + B 2 − 2 ¡ A ¡ B c) (A + B) ¡ (A − B) = A2 − B 2
2. Determina los valores de a y b de la matriz A = 3. ÂżQueĚ matrices conmutan con la matriz
6.6.
2 −1 a b
para que A2 = A.
1 2 ?. 0 1
La matriz inversa
Sabemos ya multiplicar matrices y hemos visto algunas de las propiedades de esta operacioĚ n. Recordemos, en primer lugar, que no siempre es posible efectuĚ ar la multiplicacioĚ n de dos matrices, y en segundo lugar, que aunque sea posible hacer esta multiplicacioĚ n, en general no es conmutativo, es decir A¡B es distinto de B¡A. En el caso particular de que tratemos con matrices cuadradas del mismo orden A y B, es claro que podemos efectuar los productos A¡B y B¡A, que daraĚ n como resultado otra matriz del mismo orden, aunque, como ya se ha dicho, las matrices resultantes seraĚ n, en general, distintas. Sabemos tambieĚ n que el elemento neutro del producto de matrices es la matriz identidad In . Por analogÄąĚ a con el caso de los nuĚ meros reales, podemos plantearnos la siguiente cuestioĚ n:
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Si tenemos un nuĚ mero real, por ejemplo el 2, podemos interesarnos en buscar el inverso del 2 para el producto, es decir un nuĚ mero real x tal que 2¡x = 1, el producto de 2 por x sea igual al elemento neutro, el 1. Evidentemente, en el caso de los nuĚ meros reales es bien faĚ cil despejar x para obtener, en nuestro 1 caso, que x = , es decir, el inverso de un nuĚ mero real es otro nuĚ mero que multiplicado por eĚ l da el 2 elemento neutro, el 1. Todo nuĚ mero real, salvo el 0, tiene inverso. Trasladando esto a las matrices, nos podemos plantear si dada una matriz cuadrada A de orden n, cualquiera, existe su inversa X para el producto de matrices,tal que A ¡ X = In es decir, el producto de A por su inversa produce el elemento neutro matricial, la matriz identidad In . Sin embargo, hay algunas diferencias con respecto al caso de los nuĚ meros reales: In 1) No podemos “despejarâ€? la matriz X del modo X = , porque no hemos deďŹ nido la divisioĚ n de A matrices. 2) No todas las matrices cuadradas no nulas tienen matriz “inversaâ€? (sea lo que sea, por analogÄąĚ a con los nuĚ meros). DeďŹ namos, en primer lugar, el teĚ rmino de matriz inversa: Dada una matriz cuadrada de orden n , A, se dice que A es invertible (o que posee inversa o que es no singular o que es regular ), si existe otra matriz del mismo orden, denominada matriz inversa de A y representada por A−1 y tal que: A ¡ A−1 = In y
A−1 ¡ A = In
Si A no tiene inversa, se dice que es singular o no invertible. Si una matriz tiene inversa, dicha matriz inversa es uĚ nica (soĚ lo hay una). Para calcular dicha matriz inversa, podemos utilizar dos vÄąĚ as:
6.6.1.
MeĚ todo directo:
Consiste en determinar A−1 planteando sistema de ecuaciones, es decir, si por ejemplo queremos un 1 2 determinar la inversa de la matriz A = , lo que estoy buscando es otra matriz de igual tamanĚƒo −1 1 x y −1 −1 −1 , se tiene que cumplir que : (orden 2) tal que A ¡ A = I2 y A ¡ A = I2 , es decir, si A = z t 1 2 x y 1 0 x + 2z y + 2t 1 0 −1 ¡ = =⇒ = A ¡ A = I2 =⇒ −1 1 z t 0 1 −x + z −y + t 0 1  x + 2z = 1    y + 2t = 0 −x +z =0    −y + t = 1 Es decir, hemos de resolver un sistema de 4 ecuaciones con 4 incoĚ gnitas, aunque en realidad son 2 sistemas de dos ingoĚ nitas cada uno (uno con x y z y otro con y y t). Resolviendo el sistema se obtiene que x=
−2 1 1 1 ,y = ,z = ,t = 3 3 3 3
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por lo que la matriz inversa es: −1
A
1 =
3 1 3
−2 3 1 3
1 1 −2 = ¡ 1 1 3
Se puede comprobar que tambieĚ n se cumple que A−1 ¡ A = I2 , luego A es invertible, tiene inversa. Si el sistema no tiene solucioĚ n, la matriz no tiene inversa. 1 1 Por ejemplo, en el caso en que A = , del mismo modo : 2 2 −1
A¡A
= I2 =⇒
1 1 x y 1 0 x+z y+t 1 0 ¡ = =⇒ = 2 2 z t 0 1 2x + 2z 2y + 2t 0 1  x+z = 1    y+t = 0 2x + 2z = 0    2y + 2t = 1
Y por ejemplo de 2x+2z=0 se obtiene x = -z, si se sustituye en la primera ecuacioĚ n es -z+z=1, es decir 0 = 1 (imposible). El sistema no tiene solucioĚ n. Por tanto A no es invertible, es singular. Este meĚ todo directo soĚ lo se suele utilizar para matrices cuadradas de tamanĚƒo 2, puesto que para las de tamanĚƒo 3 obtenemos un sistemas de ÂĄ9 ecuaciones con 9 incoĚ gnitas! que realmente es difÄąĚ cil de resolver.
6.6.2.
MeĚ todo de Gauss-Jordan:
Consiste en hacer transformaciones elementales en las ďŹ las de la matriz para llegar a obtener la matriz identidad. Realizando estas mismas transformaciones con la matriz identidad llegamos a la matriz A−1 . Se llama transformacioĚ n elemental en una matriz a: T1) Multiplicar o dividir una ďŹ la por un nuĚ mero real distinto de cero. T2) Sumar o restar a una ďŹ la otra multiplicada por un nuĚ mero real no nulo. T3) Intercambiar el lugar de dos ďŹ las entre sÄąĚ . 1 2 Veamos como se realiza el meĚ todo de Gauss-Jordan, realizaĚ ndolo a la vez con la matriz . −1 1 i) Consideramos la matriz formada por A y la matriz identidad correspondiente . En nuestro caso: 1 2 1 0 (A|I2 ) = −1 1 0 1 ii) Se hace la matriz triangular superior (es decir, hacemos ceros por debajo de la diagonal principal) usando transformaciones elementales en ďŹ las. La mejor forma de realizar esto es hacer cero los elementos por debajo de la diagonal en la primera columna usando la ďŹ la 1. Luego, hacer cero los elementos por debajo de la diagonal en la segunda columna usando la ďŹ la 2, y asÄąĚ sucesivamente. En nuestro caso, basta sumar la ďŹ la 2 con la ďŹ la 1, y se obtiene: 1 2 1 0 F2 +F1 1 2 1 0 −−−−→ (A|I2 ) = −1 1 0 1 0 3 1 1 iii) Una vez hecha la matriz triangular superior, se hace la matriz triangular inferior, haciendo ceros a los elementos por encima de la diagonal. El proceso es parecido al anterior:
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Hacer cero los elementos por encima de la diagonal en la uĚ ltima columna usando la uĚ ltima ďŹ la. Luego, hacer cero los elementos por encima de la diagonal en la penuĚ ltima columna usando la penuĚ mtima ďŹ la, y asÄąĚ sucesivamente. En nuestro caso: 1 2 1 0 3¡F1 −2¡F2 3 0 1 −2 −−−−−−→ 0 3 1 1 0 3 1 1 iv) Ya tenemos una matriz diagonal. Lo uĚ nico que falta es dividir a cada ďŹ la entre el nuĚ mero adecuado para obtener unos en la diagonal principal, es decir, para obtener la matriz identidad en la parte izquierda: F1 F2 , 3 1 0 13 −2 3 0 1 −2 3 3 − − − − → 0 1 31 13 0 3 1 1 v) Una vez se tiene la matriz identidad en la parte de la izquierda, la parte derecha es la matriz inversa, es decir, llegamos a: 1 −2 1 1 −2 1 0 13 −2 −1 3 3 3 (I2 , A−1 ) = ¡ =⇒ A = 1 1 = 1 1 0 1 31 13 3 3 3 matriz que habÄąĚ amos obtenido antes por el meĚ todo directo. Si al realizar el meĚ todo de Gauss-Jordan en alguĚ n momento alguna ďŹ la es de ceros, la matriz no tiene inversa. Cuanto mayor sea el orden de la matriz, mejor es este meĚ todo frente al directo. Veamos otro ejemplo:   1 1 0 Calcular la inversa de la matriz B = ďŁâˆ’1 1 2 por el meĚ todo de Gauss-Jordan. 1 0 1 Siguiendo los pasos anteriores:       1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 0 F +F1 F3 −F1 ďŁ0 2 2 1 1 0 − −−−→ ďŁ0 2 2 1 1 0 (B|I3 ) = ďŁâˆ’1 1 2 0 1 0 −−2−−→ 1 0 1 0 0 1 1 0 1 0 0 1 0 −1 1 −1 0 1     1 1 0 1 0 0 1 1 0 1 0 0 2¡F +F2 2¡F2 −F3 4¡F −F2 ďŁ0 2 2 1 1 0 − −−−3−−→ −−−−→ ďŁ0 4 0 3 1 −2 −−−1−−→ 0 0 4 −1 1 2 0 0 4 −1 1 2     2 4 0 0 1 −1 2 1 0 0 14 −1 F1 F2 F3 4 4 , , 4¡F −F2 4 4 1 −2  ďŁ0 4 0 3 −−−1−−→ 1 −2 −− −− −−4→ ďŁ0 1 0 43 = (I3 |B −1 ) 4 4 2 0 0 4 −1 1 2 0 0 1 −1 14 4  1 −1 1  4 =⇒ B −1 = ďŁ
4 3 4 −1 4
4 1 4 1 4
2 −1  2 1 2
TambieĚ n se puede expresar, sacando factor comuĚ n:   1 −1 2 1 1 −2 B −1 = ¡ ďŁ 3 4 −1 1 2 1 1 Si calculamos por este meĚ todo la inversa de A = resulta: 2 2 1 1 1 0 F2 −2¡F1 1 1 1 0 −−−−−→ (A|I2 ) = 2 2 0 1 0 0 −2 1
es la inversa de B.
Como aparece una ďŹ la de ceros, la matriz A no tiene inversa. Ejercicios:
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1. Calcular por el meĚ todo de Gauss-Jordan la inversa de las matrices:     1 2 −3 −2 1 4 A = ďŁ3 2 −4 B=ďŁ 0 1 2  2 −1 0 1 0 −1 
3 0 2. Dada la matriz diagonal D = ďŁ0 −2 0 0 raĚ pida la inversa de una matriz diagonal
6.7.
 0 0 calcula su inversa. ÂżCoĚ mo calcularÄąĚ as de forma 5 cualquiera?.
Rango de una matriz
Un concepto muy importante relacionado con las matrices es el de rango. El concepto de rango se encuentra ligado al de “independencia linealâ€? de ďŹ las o columnas de una matriz, pero no se introduciraĚ de esta manera porque se requieren conceptos que no conocemos. Baste saber que se deďŹ ne el rango de una matriz como el nuĚ mero maĚ ximo de ďŹ las o columnas linealmente independientes. Sin embargo, el caĚ lculo del rango de una matriz lo abordaremos desde otra perspectiva, utilizando el meĚ todo de Gauss. Supongamos que tenemos una matriz cualquiera A a la que aplicamos el meĚ todo de Gauss con el ďŹ n de simpliďŹ carla lo maĚ s posible (es decir, consiguiendo que tenga el mayor nuĚ mero de ceros posible), realizando operaciones elementales en ďŹ las. Llamaremos rango de la matriz A y lo representaremos por Rg(A) al nuĚ mero de ďŹ las no nulas de la matriz tras aplicarle el meĚ todo de Gauss. Ejemplo: Calcular el rango de las siguientes matrices:   1 1 0 1 1 0 3 A= B= C=ďŁ 2 1 1  2 2 1 1 −1 1 −2
D=
2 4 6 −1 −2 −3
1 1 F2 −2¡F1 1 1 −−−−−→ , Rg(A)=1 ,soĚ lo una ďŹ la distinta de cero. a) 0 0 2 2 0 3 F2 F1 1 1 −−−−→ , Rg(B)=2 hay 2 ďŹ las no nulas. b) 0 3 1 1         1 1 0 1 1 0 1 1 0 1 1 0 F2 −2¡F1 ďŁ F +F1 ďŁ F +2¡F2 ďŁ âˆ’âˆ’âˆ’â†’ 0 −1 1  −−3−−→ 0 −1 1  −−3−−−→ 0 −1 1 c) ďŁ 2 1 1  −− −1 1 −2 0 2 −2 0 0 0 −1 1 −2 Rg(C)=2 hay 2 ďŹ las no nulas. 2 4 6 2 4 6 2¡F2 +F1 −−−−−→ , Rg(D)=1, soĚ lo una ďŹ la no nula. d) 0 0 0 −1 −2 −3 Los ejemplos anteriores ponen de maniďŹ esto que el rango de cualquier matriz siempre es menor o igual que el nuĚ mero de ďŹ las de la matriz. De hecho se veriďŹ ca que el rango de cualquier matriz siempre es menor o igual que su nuĚ mero de ďŹ las y de columnas, pues el proceso para hacer el meĚ todo de Gauss se puede hacer indistintamente mediante operaciones elementales en ďŹ las o en columnas. Esto permite, antes de calcular el rango de una matriz, saber entre queĚ valores va a estar ese rango. Por ejemplo, en el caso c) del ejemplo, como la matriz es 3x3 , el rango soĚ lo puede ser 0, 1, 2 oĚ 3, no hay otras posibilidades. En el caso del apartado d), como la matriz es 2 x 3, el rango soĚ lo puede ser 0,1 oĚ 2. (De hecho, podemos reducir esto algo maĚ s , pues una matriz soĚ lo tiene rango cero si es la matriz nula).Resumiendo:
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Propiedad: Si A es una matriz de tamanĚƒo m x n no nula se cumple que: 1 ≤ Rg(A) ≤ min{m, n} Ejemplo: Calcular en funcioĚ n de k el rango de la matriz: 1 1 2 A= 3 3 k Aplicando Gauss,
A=
1 1 2 3 3 k
1 1 2 −−−−−→ 0 0 k−6 F2 −3¡F1
Ahora es evidente que si k-6=0, la uĚ ltima ďŹ la es nula. Por tanto, si k=6, la uĚ ltima ďŹ la es nula y el rango de A es 1, Rg(A)=1, mientras que si k-6 es distinto de cero, es decir si k es distinto de 6, hay 2 ďŹ las no nulas y el rango de A es 2, Rg(A)=2. Resumiendo: Si k = 6, entonces Rg(A)=2 Si k=6, entonces Rg(A)=1 La siguiente propiedad permite relacionar el concepto de rango con el de matriz inversa visto anteriormente: Propiedad: Una matriz cuadrada A tiene inversa â‡?⇒ Rg(A) es maĚ ximo. Ejercicios: 
 1 −2 1 1. Calcula el rango de A seguĚ n los valores de k: A = ďŁ1 1 3.ÂżPara queĚ valores de k tiene A 5 −1 k inversa?. 2. Calcula el rango de las matrices:  0 1 0 1 A= B = ďŁ1 2 1 0 0    2 −1 1 1 2 0 0 1 0  ďŁâˆ’1 D = C = ďŁ2 1 1 1 1 0 0 0 1
6.8.
 2 1 0 −1 4 2  1 5 −1 8 2 3 4 5 3 10 11 13
Determinantes
Introduciremos a continuacioĚ n el concepto de determinante asociado a una matriz cuadrada. Este concepto permite simpliďŹ car operaciones matriciales tales como el caĚ lculo del rango o de la matriz inversa. DeďŹ nicioĚ n: Si es una matriz 2 x 2 se deďŹ ne el determinante de la matriz A, y se expresa como det(A) o bien |A|, como el nuĚ mero: a11 a12 = a11 ¡ a22 − a12 ¡ a21 det(A) = |A| = a21 a22 Ejemplos: El caĚ lculo de los determinantes de orden 2 es bien sencillo, por ejemplo:
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1 3 =1·4-(-1)·3=4+3=7. a) −1 4 −2 −3 =-10+6=-4. b) 2 5 Para definir determinantes de matrices de orden mayor que 2 es necesario introducir previamente algunos conceptos. Dada una matriz cuadrada A de orden n, definimos el menor complementario de un elemento de A,aij , como el determinante de la matriz que se obtiene al suprimir la fila i y la columna j en la que se encuentra dicho elemento aij . Se representa por Mij .
−2 4 5 Ejemplo: En la matriz A = 6 7 −3 , los menores complementarios de cada uno de los 3 0 2 elementos de la primera fila son: 7 −3 =14-0=14. Menor complementario de -2:M11 = 0 2 6 −3 =12+9=21. Menor complementario de 4:M12 = 3 2 6 7 =0-21=-21. Menor complementario de 5:M13 = 3 0 Y ası́ sucesivamente. Ejercicio: Obtener los restantes menores complementarios de los elementos de la matriz A. Estrechamente ligado al concepto de menor complementario se encuentra el de adjunto de una matriz. Dada una matriz cuadrada A de orden n, definimos el adjunto de un elemento aij de A como el número: Aij = (−1)i+j · Mij es decir, no es más que el menor complementario correspondiente acompañado de un signo más o menos dependiendo de la fila y la columna en la que se encuentre el elemento en cuestión. Por ejemplo, para la matriz anterior, los adjuntos de los elementos de la primera fila son: Adjunto de -2:A11 = (−1)1+1 · M11 = 1 · 14 = 14 (coincide con el menor complementario) Adjunto de 4:A12 = (−1)1+2 · M12 = (−1) · 21 = −21 (menor complementario con signo cambiado) Adjunto de 5:A13 = (−1)1+3 · M13 = 1 · −21 = −21 (coincide con el menor complementario). Ejercicio: Obtener los restantes adjuntos de los elementos de la matriz A. En general puede saberse si el signo del menor complementario y del adjunto coinciden o no utilizando una sencilla regla gráfica, por ejemplo, para matrices 3 x 3 y 4 x 4 basta fijarse en las matrices: + − + − + − + − + − + − + − + − + − + − + − + − + donde el + significa que el adjunto coincide con el menor complementario y el - indica que tienen signo contrario. Una vez vistos estos conceptos se puede definir ya: Definición: Dada una matriz cuadrada A de tamaño n se define su determinante como la suma
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del producto de los elementos de una linea cualquiera de la matriz (ďŹ la o columna) elegida, por sus correpondientes adjuntos. Se puede demostrar, aunque dicha demostracioĚ n excede los contenidos del curso, que el valor del determinante no depende de la ďŹ la o columna elegida para calcularlo. 
 −2 4 5 Ejemplo: Para la matriz A = ďŁ 6 7 −3 ,aplicando la deďŹ nicioĚ n, si elegimos la ďŹ la tercera queda: 3 0 2 4 5 −2 5 −2 4 = det(A) = 3 ¡ + 0 ¡ − +2¡ 7 −3 6 −3 6 7 = 3 ¡ (−12 − 35) + 0 ¡ (−(6 − 30)) + 2 ¡ (−14 − 24) = −141 + 0 − 76 = −217 Si hubieĚ semos elegido otra ďŹ la o columna, por ejemplo la columna 2, quedarÄąĚ a: 6 −3 −2 5 −2 5 +7¡ det(A) = 4 ¡ − 3 2 + 0 ¡ − 6 −3 = 3 2 = 4 ¡ (−(12 + 9)) + 7 ¡ (−4 − 15) + 0 ¡ (−(6 − 30)) = −84 − 133 + 0 = −217 Ejercicio: Calcula, desarrollando por la ďŹ la que tuĚ elijas los determinantes de las matrices:         1 8 1 3 4 −6 7 8 0 0 3 1 ďŁ1 7 0  ďŁ2 −1 1  ďŁ0 −7 3 ďŁâˆ’2 0 2 1 6 −1 5 3 −5 1 0 1 3 4 0  1 1  ďŁ1 0
6.9.
1 1 0 1
1 0 1 1
 0 1  1 1

1 2  ďŁ3 3
2 1 1 4
3 3 4 1
 4 1  3 2

1 2  ďŁ2 3
 0 −1 2 3 2 −2  4 2 1 1 5 −3
La regla de Sarrus
La deďŹ nicioĚ n de determinante es bastante engorrosa y se hace mucho maĚ s pesada a medida que aumenta el orden de la matriz A. En el caso de las matrices cuadradas de orden 3, esta regla facilita el caĚ lculo de dichos determinantes.   a11 a12 a13 Si la matriz es A = ďŁa21 a22 a23 , entonces el determinante de A se calcula mediante la resta a31 a32 a33 de dos expresiones obtenidas del siguiente modo: Llamaremos sumandos positivos a los obtenidos al multiplicar: - Los elementos de la diagonal principal,a11 ¡ a22 ¡ a33 . - Los elementos de la linea paralela superior a la diagonal principal por el elemento aislado de la esquina inferior izquierda:a12 ¡ a23 ¡ a31 . - Los elementos de la linea paralela inferior a la diagonal principal por el elemento aislado de la esquina superior derecha:a21 ¡ a32 ¡ a13 .
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Gráficamente:
Figura 6.2: Sumandos positivos
Llamaremos sumandos negativos a los obtenidos al multiplicar: - Los elementos de la diagonal secundaria,a13 · a22 · a31 . - Los elementos de la linea paralela superior a la diagonal secundaria por el elemento aislado de la esquina inferior derecha: a12 · a21 · a33 . - Los elementos de la linea paralela inferior a la diagonal secundaria por el elemento aislado de la esquina superior izquierda: a32 · a23 · a11 . Gráficamente:
Figura 6.3: Sumandos negativos
Y entonces det (A)= Sumandos positivos - Sumandos negativos. Por ejemplo, en el caso de la matriz anterior: −2 4 5 A = 6 7 −3 3 0 2 , se tiene que aplicando la regla de Sarrus: det(A)=(-2)·7·2+4·3·(-3)+6·5·0-(3·7·5+0·(-2)·(-3)+6·4·2)=-28-36-105-48=-217. Ejercicio: Comprobar, mediante la regla de Sarrus, los determinantes de orden 3 obtenidos en el ejercicio anterior.
6.10.
Propiedades de los determinantes
Algunas propiedades importantes que tienen los determinantes, y que se enuncian sin demostración, son: 1. Si una matriz tiene una linea (fila o columna) de ceros, el determinante vale cero. Esta propiedad es evidente, puesto que por definición de determinante, basta elegir dicha linea para desarrollar y el determinante será 0.
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2. Si una matriz tiene dos filas iguales o proporcionales, su determinante es nulo. 3. Si permutamos dos lineas paralelas de una matriz cuadrada, su determinante cambia de signo, por ejemplo: 0 1 0 1 2 −3 2 −3 1 3 1 3 2 −5 2 −5 = 91 =⇒ = −91 2 4 3 1 3 −2 −8 1 2 4 3 −2 −8 1 3 1 4. Si multiplicamos todos los elementos de una linea de un determinante por un número, el determinante queda multiplicado por ese número. Por ejemplo: 0 1 0 1 2 −3 2 −3 1 3 2 6 2 −5 4 −10 = 91 =⇒ = 182 2 4 3 1 3 1 2 4 3 −2 −8 1 3 −2 −8 1 0 2 4 −6 2 6 4 −10 pero = 16 · 91 = 1456 6 2 4 8 6 −4 −16 2 5. Si a una linea de una matriz se le suma otra linea multiplicada por un número, el determinante no cambia. Esta propiedad permite utilizar un método más sencillo para calcular determinantes de orden mayor que 3. 6. El determinante de una matriz es igual al de su traspuesta, |A| = |At| 7. Si A tiene matriz inversa, A−1 , se verifica que: det(A−1 ) =
1 det(A)
Una estrategia a tener en cuenta en este caso de determinantes de orden 4 o superior, o incluso de orden 3 si la matriz es compleja, es el método de “hacer ceros”, puesto que el valor del determinante no varı́a al realizar a la matriz ciertas transformaciones elementales en filas,como indica la propiedad 5 anterior, si bien hemos de ser cuidadosos al aplicar dicha propiedad. Ası́ pues la mejor forma de calcular un determinante es hacer ceros en una fila o columna y desarrollar por dicha fila o columna, porque entonces sólo tendremos que calcular un adjunto. Por ejemplo, si calculamos: 0 1 0 0 1 2 −3 1 2 −3 2 −3 1 3 2 −5 F4 −3·F2 1 3 2 −5 2 −5 F3 −2·F2 1 3 = = = 2 4 3 1 0 −2 −1 11 0 −2 −1 11 3 −2 −8 1 0 −11 −14 16 3 −2 −8 1 Desarrollando por la columna 1 1 2 −3 = 1 · − −2 −1 11 = −11 −14 16 = −(−16 − 242 − 84 − (−33 − 154 − 64)) = 91
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Como hemos dicho, hemos de tener especial cuidado al aplicar esta regla con determinantes, puesto que no podemos hacer las mismas operaciones que con las matrices, lo que puede confundir. Por ejemplo, si queremos calcular el determinante: 1 2 3 C = 0 1 2 4 1 5 mediante la regla de Sarrus es: det(C)=5+16+0-(12+2+0)=21-14=7. Si hiciésemos ceros en la primera columna, y desarrollásemos nos deberı́a dar lo mismo. Ahora bien,podemos hacer cero el 4 de la primera columna mediante: 1 2 1 2 3 3 1 2 1 0 1 2 F3 −4·F 2 = = −7 + 14 = 7. = 0 1 −7 −7 0 −7 −7 4 1 5 lo que es correcto. Sin embargo, si queremos hacer cero el 1 de la primera columna serı́a un error hacer: 1 2 3 4·F1 −F3 0 7 7 0 1 2 −→ 0 1 2 = 4 · 7 7 = 4 · (14 − 7) = 28. 1 2 4 1 5 4 1 5 no obtenemos lo mismo, porque hemos multiplicado la fila sustituida por un número y eso altera el valor del determinante. Luego la fila a sustituir conviene no multiplicarla, como en el primer ejemplo, puesto que si no nos damos cuenta, podemos variar el valor del determinante.
6.11.
Relación entre la inversa y los determinantes
Hay una estrecha relación entre la inversa de una matriz cuadrada y su determinante. De hecho se verifica que: Propiedad: Una matriz cuadrada A tiene inversa ⇐⇒ |A| = 0. −1 Además, en este caso, la matriz inversa de A, A se calcula de la manera: A−1 =
(Adj(A)t |A|
donde Adj(A) denota la matriz adjunta de A, es decir, aquella que se obtiene de sustituir cada elemento de A por su adjunto.
1 0 −1 Ejemplo: Calcular, si es posible, la inversa de la matriz A = 0 1 −3. −1 1 0 1 0 −1 En primer lugar,|A| = 0 1 −3 = 0 + 0 + 0 − (1 − 3 + 0) = 2 y por tanto A tiene inversa. −1 1 0 Calculando Adj(A), se obtiene: 1 −3 − 0 −3 0 1 −1 0 −1 1 1 0 3 3 1 0 −1 1 −1 1 0 Adj(A) = − 1 0 −1 0 − −1 1 = −1 −1 −1 1 3 1 0 −1 1 −1 1 0 − 1 −3 0 −3 0 1
CAPIĚ TULO 6. MATRICES Y DETERMINANTES Por tanto,
101

 3 −1 1 (Adj(A)t) = ďŁ3 −1 3 1 −1 1
Y entonces, se obtiene:
3 A−1 =
2 ďŁ3 2 1 2
−1 2 −1 2 −1 2

1 2 3 2 1 2
Ejercicio: Calcular la inversa anterior por el meĚ todo de Gauss.
6.12.
AplicacioĚ n de los determinantes al caĚ lculo del rango
Los determinantes tambieĚ n proporcionan una forma sencilla de calcular el rango de una matriz cualquiera. Un deďŹ nicioĚ n alternativa de rango de una matriz es: El Rango de una matriz A es el tamanĚƒo del mayor menor complementario no nulo que esteĚ incluido dentro de la matriz. Aplicando este criterio, calculemos el rango de las matrices siguientes:   1 1 0 1 1 0 3 2 4 6 ďŁ ďŁ¸ A= B= C= 2 1 1 D= 2 2 1 1 −1 −2 −3 −1 1 −2 a) SoĚ lo hay un menor de orden 2, que es: 1 1 2 2 = 0 Como es nulo, el rango de la matriz NO es 2. Menores de orden 1 hay 4, por ejemplo |1| = 1, que es no nulo,luego el rango de la matriz es Rg(A)=1 (el tamanĚƒo de dicho menor complementario). b) SoĚ lo hay un menor de orden 2, que es: 0 3 1 1 = 0 − 3 = −3 Como no es nulo, el rango de la matriz es Rg(B)=2 (el tamanĚƒo de dicho menor complementario). c) SoĚ lo hay un menor de orden 3, que es: 1 1 0 2 1 1 = −2 − 1 + 0 − (0 + 1 − 4) = −3 + 3 = 0 −1 1 −2 Como es nulo, podemos asegurar que el rango NO es 3. Menores de orden 2 hay 9. Calculando alguno: 1 0 1 1 = 1 resulta que es no nulo, luego el rango es Rg(C)=2 (el tamanĚƒo de dicho menor complementario). d) El menor maĚ s grande que podemos formar es de orden 2. Hay 3 de ellos: 2 2 4 4 6 6 −1 −2 = −4 + 4 = 0 −1 −3 = −6 + 6 = 0 −2 −3 = −12 + 12 = 0
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Son todos nulos, luego el rango NO es 2. Menores de orden 1 hay 6, y por ejemplo |6| = 6 = 0, es no nulo, luego el rango es Rg(D)=1. Ejercicio Calcula,utilizando los determinantes, el rango de  0 1 0 1 A= B = ďŁ1 2 1 0 0    2 −1 1 1 2 0 0 1 0   ďŁ C =ďŁ D = −1 2 1 1 1 1 0 0 0 1
las matrices:  2 1 0 −1 4 2  1 5 −1 8 2 3 4 5 3 10 11 13
Bolzano
Weierstrass H) Si una funciรณn f es continua en un intervalo cerrado [a,b] T) Entonces hay al menos dos puntos x1, x2 pertenecientes a [a,b] donde f alcanza valores extremos absolutos, es decir: , para cualquier
Valor Intermedio Sea f una funciรณn: Continua en un intervalo [a,b] y k un escalar real / f (a) < k < f (b). Entonces existe al menos un c dentro de (a,b) tal que f (c) = k
1) Demuestra que la ecuación x3 + 6x2 + 15x –23 = 0 tiene raíces reales. 2) Demuestra que la ecuación x cos (x/2) + 15 sen (x) = 0 tiene alguna raíz. 3) Prueba que la ecuación x2 =18 Ln(x) posee al menos una raíz real. 4) ¿Posee la ecuación x · ln (x) – 1 = 0 alguna solución en el intervalo [1,e]? 5) Dada la ecuación x3 + 5x2 – 10 = 0, encontrar tres intervalos cerrados de forma que en el interior de cada uno de ellos haya una raíz de la ecuación. 6) Demuestra que la función y = sen (2x) / x tiene al menos una raiz en [0;2Pi]. 7) Comprueba que la ecuación x3 – 3x + 1 = 0 tiene tres raíces reales. 8) Dada la función f(x) = x3 , estudiar si está acotada superiormente e inferiormente en el intervalo [1, 5] e indica si alcanza sus valores máximos y mínimos. 9) Probar que la función f(x) = x + sen x − 1 es continua para todo R y probar que existe al menos una raíz real de la ecuación x + sen x − 1 = 0. 10) Sean f y g dos funciones continuas en [a, b] y tales que f(a) > g(a) y f(b) < g(b). Demostrar que existe c en (a, b) tal que f(c) = g(c).
684
❙❙❙❙
45–48
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
||||
45. r 3 sin , 47. r 2, ■
55. (a) Use Formula 10.2.7 to show that the area of the surface
Find the exact length of the polar curve.
■
0 3
0 2 ■
■
■
2
48. r , ■
■
■
generated by rotating the polar curve
0 2
46. r e ,
r f
0 2 ■
■
■
■
(where f is continuous and 0 a b ) about the polar axis is
49–52 |||| Use a calculator to find the length of the curve correct to four decimal places.
49. r 3 sin 2 51. r sin 2 ■
■
b
S y 2 r sin a
50. r 4 sin 3
■
■
■
■
■
■
■
r2
dr d
2
d
(b) Use the formula in part (a) to find the surface area generated by rotating the lemniscate r 2 cos 2 about the polar axis.
52. r 1 cos 3 ■
a b
■
■
56. (a) Find a formula for the area of the surface generated by
; 53–54
||||
Graph the curve and find its length.
53. r cos 4 4 ■
■
■
||||10.5
54. r cos 2 2 ■
■
■
■
■
■
■
■
■
rotating the polar curve r f , a b (where f is continuous and 0 a b ), about the line 2. (b) Find the surface area generated by rotating the lemniscate r 2 cos 2 about the line 2.
Conic Sections In this section we give geometric definitions of parabolas, ellipses, and hyperbolas and derive their standard equations. They are called conic sections, or conics, because they result from intersecting a cone with a plane as shown in Figure 1.
ellipse
parabola
hyperbola
FIGURE 1
Conics
parabola
axis focus
vertex FIGURE 2
F
directrix
Parabolas A parabola is the set of points in a plane that are equidistant from a fixed point F (called the focus) and a fixed line (called the directrix). This definition is illustrated by Figure 2. Notice that the point halfway between the focus and the directrix lies on the parabola; it is called the vertex. The line through the focus perpendicular to the directrix is called the axis of the parabola. In the 16th century Galileo showed that the path of a projectile that is shot into the air at an angle to the ground is a parabola. Since then, parabolic shapes have been used in designing automobile headlights, reflecting telescopes, and suspension bridges.
SECTION 10.5 CONIC SECTIONS
y
P(x, y) F(0, p)
y p x
O
y=_p
â?&#x2122;â?&#x2122;â?&#x2122;â?&#x2122;
685
(See Problem 16 on page 276 for the reflection property of parabolas that makes them so useful.) We obtain a particularly simple equation for a parabola if we place its vertex at the origin O and its directrix parallel to the x-axis as in Figure 3. If the focus is the point 0, p , then the directrix has the equation y p. If P x, y is any point on the parabola, then the distance from P to the focus is
PF sx y p and the distance from P to the directrix is y p . (Figure 3 illustrates the case where 2
2
p 0.) The defining property of a parabola is that these distances are equal:
FIGURE 3
sx 2 y p 2 y p
We get an equivalent equation by squaring and simplifying:
x 2 y p 2 y p
2
y p 2
x 2 y 2 2py p 2 y 2 2py p 2 x 2 4py An equation of the parabola with focus 0, p and directrix y p is
1
x 2 4py If we write a 1 4p , then the standard equation of a parabola (1) becomes y ax 2. It opens upward if p 0 and downward if p 0 [see Figure 4, parts (a) and (b)]. The graph is symmetric with respect to the y-axis because (1) is unchanged when x is replaced by x. y
y
y
y=_p
(0, p)
x
(0, p)
y=_p
(a) â&#x2030;&#x2C6;=4py, p>0
( p, 0)
( p, 0)
0 x
0
y
(b) â&#x2030;&#x2C6;=4py, p<0
0
x
x
0
x=_p
x=_p
(c) ÂĽ=4px, p>0
(d) ÂĽ=4px, p<0
FIGURE 4
If we interchange x and y in (1), we obtain 2
y 2 4px
which is an equation of the parabola with focus p, 0 and directrix x p. (Interchanging x and y amounts to reflecting about the diagonal line y x.) The parabola opens to the right if p 0 and to the left if p 0 [see Figure 4, parts (c) and (d)]. In both cases the graph is symmetric with respect to the x-axis, which is the axis of the parabola.
686
â?&#x2122;â?&#x2122;â?&#x2122;â?&#x2122;
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
y
EXAMPLE 1 Find the focus and directrix of the parabola y 2 10x 0 and sketch
ÂĽ+10x=0
the graph. SOLUTION If we write the equation as y 2 10x and compare it with Equation 2, we see
â&#x20AC;?_
that 4p 10, so p 52 . Thus, the focus is p, 0 ( 52, 0) and the directrix is x 52 . The sketch is shown in Figure 5.
5 , 0â&#x20AC;&#x2122; 2
x
0
Ellipses
5 x= 2
An ellipse is the set of points in a plane the sum of whose distances from two fixed points F1 and F2 is a constant (see Figure 6). These two fixed points are called the foci (plural of focus). One of Keplerâ&#x20AC;&#x2122;s laws is that the orbits of the planets in the solar system are ellipses with the Sun at one focus. In order to obtain the simplest equation for an ellipse, we place the foci on the x-axis at the points c, 0 and c, 0 as in Figure 7 so that the origin is halfway between the foci. Let the sum of the distances from a point on the ellipse to the foci be 2a 0. Then P x, y is a point on the ellipse when PF1 PF2 2a
FIGURE 5 P
FÂĄ
Fâ&#x201E;˘
FIGURE 6 y
that is,
s x c 2 y 2 s x c 2 y 2 2a
or
s x c 2 y 2 2a s x c 2 y 2
P(x, y)
Squaring both sides, we have F¥(_c, 0)
0
Fâ&#x201E;˘(c, 0)
x 2 2cx c 2 y 2 4a 2 4as x c 2 y 2 x 2 2cx c 2 y 2
x
which simplifies to
as x c 2 y 2 a 2 cx
We square again:
FIGURE 7
a 2 x 2 2cx c 2 y 2 a 4 2a 2cx c 2x 2 which becomes
a 2 c 2 x 2 a 2 y 2 a 2 a 2 c 2
From triangle F1 F2 P in Figure 7 we see that 2c 2a, so c a and, therefore, a 2 c 2 0. For convenience, let b 2 a 2 c 2. Then the equation of the ellipse becomes b 2x 2 a 2 y 2 a 2b 2 or, if both sides are divided by a 2b 2, x2 y2 1 2 a b2
3
y (0, b)
(_a, 0)
a
b (_c, 0)
0
c (0, _b)
FIGURE 8
â&#x2030;&#x2C6; ÂĽ + Â =1 a@ b@
(a, 0)
(c, 0)
x
Since b 2 a 2 c 2 a 2, it follows that b a. The x-intercepts are found by setting y 0. Then x 2 a 2 1, or x 2 a 2, so x a. The corresponding points a, 0 and a, 0 are called the vertices of the ellipse and the line segment joining the vertices is called the major axis. To find the y-intercepts we set x 0 and obtain y 2 b 2, so y b. Equation 3 is unchanged if x is replaced by x or y is replaced by y, so the ellipse is symmetric about both axes. Notice that if the foci coincide, then c 0, so a b and the ellipse becomes a circle with radius r a b. We summarize this discussion as follows (see also Figure 8). 4
The ellipse
x2 y2 1 2 a b2
a b 0
has foci c, 0 , where c 2 a 2 b 2, and vertices a, 0 .
SECTION 10.5 CONIC SECTIONS
❙❙❙❙
687
If the foci of an ellipse are located on the y-axis at 0, c , then we can find its equation by interchanging x and y in (4). (See Figure 9.)
y
(0, a) (0, c)
5 (_b, 0)
The ellipse
(b, 0) 0
x2 y2 1 b2 a2
x
(0, _c)
a b 0
has foci 0, c , where c 2 a 2 b 2, and vertices 0, a . (0, _a)
EXAMPLE 2 Sketch the graph of 9x 2 16y 2 144 and locate the foci.
FIGURE 9
≈ ¥ + =1, a˘b b@ a@
SOLUTION Divide both sides of the equation by 144:
x2 y2 1 16 9 The equation is now in the standard form for an ellipse, so we have a 2 16, b 2 9, a 4, and b 3. The x-intercepts are 4 and the y-intercepts are 3. Also, c 2 a 2 b 2 7, so c s7 and the foci are ( s7, 0). The graph is sketched in Figure 10.
y (0, 3)
(_4, 0) {_œ„7, 0}
(4, 0)
0
{œ„7, 0}
x
EXAMPLE 3 Find an equation of the ellipse with foci 0, 2 and vertices 0, 3 . SOLUTION Using the notation of (5), we have c 2 and a 3. Then we obtain
b 2 a 2 c 2 9 4 5, so an equation of the ellipse is (0, _3)
x2 y2 1 5 9
FIGURE 10
9≈+16¥=144
Another way of writing the equation is 9x 2 5y 2 45. Like parabolas, ellipses have an interesting reflection property that has practical consequences. If a source of light or sound is placed at one focus of a surface with elliptical cross-sections, then all the light or sound is reflected off the surface to the other focus (see Exercise 59). This principle is used in lithotripsy, a treatment for kidney stones. A reflector with elliptical cross-section is placed in such a way that the kidney stone is at one focus. High-intensity sound waves generated at the other focus are reflected to the stone and destroy it without damaging surrounding tissue. The patient is spared the trauma of surgery and recovers within a few days.
y
Hyperbolas P(x, y)
F¡(_c, 0)
0
FIGURE 11
P is on the hyperbola when | PF¡|-| PF™ |= 2a
F™(c, 0) x
A hyperbola is the set of all points in a plane the difference of whose distances from two fixed points F1 and F2 (the foci) is a constant. This definition is illustrated in Figure 11. Hyperbolas occur frequently as graphs of equations in chemistry, physics, biology, and economics (Boyle’s Law, Ohm’s Law, supply and demand curves). A particularly significant application of hyperbolas is found in the navigation systems developed in World Wars I and II (see Exercise 51). Notice that the definition of a hyperbola is similar to that of an ellipse; the only change is that the sum of distances has become a difference of distances. In fact, the derivation of the equation of a hyperbola is also similar to the one given earlier for an ellipse. It is left
688
â?&#x2122;â?&#x2122;â?&#x2122;â?&#x2122;
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
as Exercise 52 to show that when the foci are on the x-axis at c, 0 and the difference of distances is PF1 PF2 2a, then the equation of the hyperbola is
x2 y2 1 a2 b2
6
b
y
where c 2 a 2 b 2. Notice that the x-intercepts are again a and the points a, 0 and a, 0 are the vertices of the hyperbola. But if we put x 0 in Equation 6 we get y 2 b 2, which is impossible, so there is no y-intercept. The hyperbola is symmetric with respect to both axes. To analyze the hyperbola further, we look at Equation 6 and obtain
b
y=_ a x
y= a x
(_a, 0)
(a, 0)
(_c, 0)
(c, 0)
0
x2 y2 1
1 a2 b2
x
This shows that x 2 a 2, so x sx 2 a. Therefore, we have x a or x a. This means that the hyperbola consists of two parts, called its branches. When we draw a hyperbola it is useful to first draw its asymptotes, which are the dashed lines y b a x and y b a x shown in Figure 12. Both branches of the hyperbola approach the asymptotes; that is, they come arbitrarily close to the asymptotes. [See Exercise 67 in Section 4.5, where y b a x is shown to be a slant asymptote.]
FIGURE 12 ÂĽ â&#x2030;&#x2C6; - Â =1 b@ a@
7
The hyperbola x2 y2 1 2 a b2
y
(0, c) a y=_ b x
a
has foci c, 0 , where c 2 a 2 b 2, vertices a, 0 , and asymptotes y b a x.
y= b x (0, a) (0, _a)
0
x
If the foci of a hyperbola are on the y-axis, then by reversing the roles of x and y we obtain the following information, which is illustrated in Figure 13.
(0, _c)
8
FIGURE 13 ÂĽ â&#x2030;&#x2C6; - Â =1 a@ b@
The hyperbola y2 x2 1 a2 b2
has foci 0, c , where c 2 a 2 b 2, vertices 0, a , and asymptotes y a b x. 3
y=_ 4 x
y
3
y= 4 x
EXAMPLE 4 Find the foci and asymptotes of the hyperbola 9x 2 16y 2 144 and sketch
its graph. (_4, 0) (_5, 0)
SOLUTION If we divide both sides of the equation by 144, it becomes
(4, 0) 0
FIGURE 14
9â&#x2030;&#x2C6;-16ÂĽ=144
(5, 0) x
x2 y2 1 16 9 which is of the form given in (7) with a 4 and b 3. Since c 2 16 9 25, the foci are 5, 0 . The asymptotes are the lines y 34 x and y 34 x. The graph is shown in Figure 14.
SECTION 10.5 CONIC SECTIONS
❙❙❙❙
689
EXAMPLE 5 Find the foci and equation of the hyperbola with vertices 0, 1 and asymptote y 2x. SOLUTION From (8) and the given information, we see that a 1 and a b 2. Thus,
b a 2 12 and c 2 a 2 b 2 54 . The foci are (0, s5 2) and the equation of the hyperbola is y 2 4x 2 1
Shifted Conics As discussed in Appendix C, we shift conics by taking the standard equations (1), (2), (4), (5), (7), and (8) and replacing x and y by x h and y k. EXAMPLE 6 Find an equation of the ellipse with foci 2, 2 , 4, 2 and vertices
1, 2 , 5, 2 .
SOLUTION The major axis is the line segment that joins the vertices 1, 2 , 5, 2 and has length 4, so a 2. The distance between the foci is 2, so c 1. Thus, b 2 a 2 c 2 3. Since the center of the ellipse is 3, 2 , we replace x and y in (4) by x 3 and y 2 to obtain
x 3 2 y 2 2 1 4 3 as the equation of the ellipse. EXAMPLE 7 Sketch the conic
9x 2 4y 2 72x 8y 176 0 y
and find its foci.
3
y-1=_ 2 (x-4)
SOLUTION We complete the squares as follows:
4 y 2 2y 9 x 2 8x 176 (4, 4)
4 y 2 2y 1 9 x 2 8x 16 176 4 144
(4, 1)
4 y 1 2 9 x 4 2 36 x
0 (4, _2)
3
y-1= 2 (x-4) FIGURE 15
9≈-4¥-72x+8y+176=0
y 1 2 x 4 2 1 9 4 This is in the form (8) except that x and y are replaced by x 4 and y 1. Thus, a 2 9, b 2 4, and c 2 13. The hyperbola is shifted four units to the right and one unit upward. The foci are (4, 1 s13 ) and (4, 1 s13 ) and the vertices are 4, 4 and 4, 2 . The asymptotes are y 1 32 x 4 . The hyperbola is sketched in Figure 15.
❙❙❙❙
690
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
|||| 10.5 1–8
Exercises 23. 2y 2 3x 2 4y 12x 8 0
Find the vertex, focus, and directrix of the parabola and sketch its graph. ||||
1. x 2y
2
■
3. 4x y
4. y 12x
5. x 2 8 y 3
6. x 1 y 5 2
7. y 2 2y 12x 25 0
8. y 12 x 2x 2 16
2
■
9–10
■
■
9.
■
■
■
■
10.
y
■
■
■
■
■
■
■
■
■
■
■
_2
y
■
■
■
■
27. x 2 4y 2y 2
28. y 2 8y 6x 16
29. y 2 2y 4x 2 3
30. 4x 2 4x y 2 0
■
■
■
■
■
■
■
■
■
■
■
|||| Find an equation for the conic that satisfies the given conditions.
x
2
0
■
26. x 2 y 2 1
31–48 1
x
25. x 2 y 1
■
1
■
■
|||| Identify the type of conic section whose equation is given and find the vertices and foci.
Find an equation of the parabola. Then find the focus and directrix.
■
■
25–30
||||
■
■
2
2
■
24. 16x 2 9y 2 64x 90y 305
2. 4y x 0
2
■
■
■
■
■
31. Parabola,
vertex 0, 0 , focus 0, 2
32. Parabola,
vertex 1, 0 , directrix x 5
33. Parabola,
focus 4, 0 , directrix x 2
34. Parabola,
focus 3, 6 , vertex 3, 2
vertex 0, 0 , axis the x-axis, passing through (1, 4)
35. Parabola, 11–16
||||
Find the vertices and foci of the ellipse and sketch
its graph.
vertical axis, passing through 2, 3 , 0, 3 ,
36. Parabola,
and 1, 9
x2 y2 1 11. 9 5
x2 y2 1 12. 64 100
37. Ellipse,
foci 2, 0 , vertices 5, 0
13. 4x y 16
14. 4x 25y 25
38. Ellipse,
foci 0, 5 , vertices 0, 13
39. Ellipse,
foci 0, 2 , 0, 6 vertices 0, 0 , 0, 8
40. Ellipse,
foci 0, 1 , 8, 1 , vertex 9, 1
41. Ellipse,
center 2, 2 , focus 0, 2 , vertex 5, 2
42. Ellipse,
foci 2, 0 , passing through 2, 1
2
2
2
2
15. 9x 18x 4y 27 2
2
16. x 2y 6x 4y 7 0 2
■
2
■
17–18
■
||||
■
■
■
■
■
■
■
■
■
Find an equation of the ellipse. Then find its foci.
17.
18.
y
y
1 0
1
1
x
x
2
43. Hyperbola,
foci 0, 3 , vertices 0, 1
44. Hyperbola,
foci 6, 0 , vertices 4, 0
45. Hyperbola,
foci 1, 3 and 7, 3 , vertices 2, 3 and 6, 3
46. Hyperbola,
foci 2, 2 and 2, 8 , vertices 2, 0 and 2, 6
47. Hyperbola,
vertices 3, 0 , asymptotes y 2x
foci 2, 2 and 6, 2 , asymptotes y x 2 and y 6 x
48. Hyperbola, ■ ■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
■
49. The point in a lunar orbit nearest the surface of the moon is 19–24
Find the vertices, foci, and asymptotes of the hyperbola and sketch its graph.
19.
||||
x2 y2 1 144 25
21. y 2 x 2 4
20.
y2 x2 1 16 36
22. 9x 2 4y 2 36
called perilune and the point farthest from the surface is called apolune. The Apollo 11 spacecraft was placed in an elliptical lunar orbit with perilune altitude 110 km and apolune altitude 314 km (above the moon). Find an equation of this ellipse if the radius of the moon is 1728 km and the center of the moon is at one focus.
SECTION 10.5 CONIC SECTIONS
50. A cross-section of a parabolic reflector is shown in the figure.
The bulb is located at the focus and the opening at the focus is 10 cm. (a) Find an equation of the parabola. (b) Find the diameter of the opening CD , 11 cm from the vertex.
691
56. (a) Show that the equation of the tangent line to the parabola
y 2 4px at the point x 0 , y 0 can be written as y 0 y 2p x x 0
(b) What is the x-intercept of this tangent line? Use this fact to draw the tangent line. 57. Use Simpson’s Rule with n 10 to estimate the length of the
C A
ellipse x 2 4y 2 4.
58. The planet Pluto travels in an elliptical orbit around the Sun
5 cm 11 cm F 5 cm
V
❙❙❙❙
(at one focus). The length of the major axis is 1.18 10 10 km and the length of the minor axis is 1.14 10 10 km. Use Simpson’s Rule with n 10 to estimate the distance traveled by the planet during one complete orbit around the Sun. 59. Let P x 1, y1 be a point on the ellipse x 2 a 2 y 2 b 2 1 with
B D
51. In the LORAN (LOng RAnge Navigation) radio navigation
system, two radio stations located at A and B transmit simultaneous signals to a ship or an aircraft located at P. The onboard computer converts the time difference in receiving these signals into a distance difference PA PB , and this, according to the definition of a hyperbola, locates the ship or aircraft on one branch of a hyperbola (see the figure). Suppose that station B is located 400 mi due east of station A on a coastline. A ship received the signal from B 1200 microseconds ( s) before it received the signal from A. (a) Assuming that radio signals travel at a speed of 980 ft s, find an equation of the hyperbola on which the ship lies. (b) If the ship is due north of B, how far off the coastline is the ship?
foci F1 and F2 and let and be the angles between the lines PF1, PF2 and the ellipse as in the figure. Prove that . This explains how whispering galleries and lithotripsy work. Sound coming from one focus is reflected and passes through the other focus. [Hint: Use the formula in Problem 15 on page 276 to show that tan tan .] y
P(⁄, ›) å ∫
F¡
0
x
F™
¥ ≈ + b@ =1 a@
60. Let P x 1, y1 be a point on the hyperbola x 2 a 2 y 2 b 2 1
P
A
coastline
B
with foci F1 and F2 and let and be the angles between the lines PF1 , PF2 and the hyperbola as shown in the figure. Prove that . (This is the reflection property of the hyperbola. It shows that light aimed at a focus F2 of a hyperbolic mirror is reflected toward the other focus F1 .) y
400 mi sending stations
P
å
52. Use the definition of a hyperbola to derive Equation 6 for a
∫
hyperbola with foci c, 0 and vertices a, 0 . 53. Show that the function defined by the upper branch of the
F¡
0
F™
hyperbola y a x b 1 is concave upward. 2
2
2
2
54. Find an equation for the ellipse with foci 1, 1 and 1, 1
and major axis of length 4. 55. Determine the type of curve represented by the equation
y2 x2 1 k k 16 in each of the following cases: (a) k 16, (b) 0 k 16, and (c) k 0. (d) Show that all the curves in parts (a) and (b) have the same foci, no matter what the value of k is.
P F¡
F™
x
Wind velocity is a vector because it has both magnitude and direction. Pictured are velocity vectors indicating the wind pattern over San Francisco Bay at 12:00 P.M. on June 11, 2002.
Vectors and the Geometry of Space
In this chapter we introduce vectors and coordinate systems for three-dimensional space. This will be the setting for our study of the calculus of functions of two variables in Chapter 14 because the graph of such a function is a surface in space. In this chapter we will see that vectors provide particularly simple descriptions of lines and planes in space.
|||| 12.1
Three-Dimensional Coordinate Systems
z
O y x
FIGURE 1
Coordinate axes z
y
To locate a point in a plane, two numbers are necessary. We know that any point in the plane can be represented as an ordered pair a, b of real numbers, where a is the x-coordinate and b is the y-coordinate. For this reason, a plane is called two-dimensional. To locate a point in space, three numbers are required. We represent any point in space by an ordered triple a, b, c of real numbers. In order to represent points in space, we first choose a fixed point O (the origin) and three directed lines through O that are perpendicular to each other, called the coordinate axes and labeled the x-axis, y-axis, and z-axis. Usually we think of the x- and y-axes as being horizontal and the z-axis as being vertical, and we draw the orientation of the axes as in Figure 1. The direction of the z-axis is determined by the right-hand rule as illustrated in Figure 2: If you curl the fingers of your right hand around the z-axis in the direction of a 90 counterclockwise rotation from the positive x-axis to the positive y-axis, then your thumb points in the positive direction of the z-axis. The three coordinate axes determine the three coordinate planes illustrated in Figure 3(a). The xy-plane is the plane that contains the x- and y-axes; the yz-plane contains the y- and z-axes; the xz-plane contains the x- and z-axes. These three coordinate planes divide space into eight parts, called octants. The first octant, in the foreground, is determined by the positive axes. z
x
z
FIGURE 2
Right-hand rule y z-plan
ane
l xz-p
x
FIGURE 3
e
left
O
xy-plane (a) Coordinate planes
y
x
right w
all
l wal O
floor
y
(b)
Because many people have some difficulty visualizing diagrams of three-dimensional figures, you may find it helpful to do the following [see Figure 3(b)]. Look at any bottom corner of a room and call the corner the origin. The wall on your left is in the xz-plane, the wall on your right is in the yz-plane, and the floor is in the xy-plane. The x-axis runs along the intersection of the floor and the left wall. The y-axis runs along the intersection of the floor and the right wall. The z-axis runs up from the floor toward the ceiling along the intersection of the two walls. You are situated in the first octant, and you can now imagine seven other rooms situated in the other seven octants (three on the same floor and four on the floor below), all connected by the common corner point O.
794
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
z
Now if P is any point in space, let a be the (directed) distance from the yz-plane to P, let b be the distance from the xz-plane to P, and let c be the distance from the xy-plane to P. We represent the point P by the ordered triple a, b, c of real numbers and we call a, b, and c the coordinates of P; a is the x-coordinate, b is the y-coordinate, and c is the z-coordinate. Thus, to locate the point a, b, c we can start at the origin O and move a units along the x-axis, then b units parallel to the y-axis, and then c units parallel to the z-axis as in Figure 4. The point P a, b, c determines a rectangular box as in Figure 5. If we drop a perpendicular from P to the xy-plane, we get a point Q with coordinates a, b, 0 called the projection of P on the xy-plane. Similarly, R 0, b, c and S a, 0, c are the projections of P on the yz-plane and xz-plane, respectively. As numerical illustrations, the points 4, 3, 5 and 3, 2, 6 are plotted in Figure 6.
P(a, b, c)
c
O
a
y x
b
FIGURE 4
z
z
z
3
(0, 0, c) S(a, 0, c)
0
_4
R(0, b, c)
0
P(a, b, c)
_5
y
y
x 0
x
(_4, 3, _5)
(0, b, 0) (a, 0, 0)
3
_2
_6
y
x
(3, _2, _6)
Q(a, b, 0)
FIGURE 6
FIGURE 5
The Cartesian product x, y, z x, y, z is the set of all ordered triples of real numbers and is denoted by 3. We have given a one-to-one correspondence between points P in space and ordered triples a, b, c in 3. It is called a threedimensional rectangular coordinate system. Notice that, in terms of coordinates, the first octant can be described as the set of points whose coordinates are all positive. In two-dimensional analytic geometry, the graph of an equation involving x and y is a curve in 2. In three-dimensional analytic geometry, an equation in x, y, and z represents a surface in 3. EXAMPLE 1 What surfaces in 3 are represented by the following equations?
(a) z 3
(b) y 5
SOLUTION
(a) The equation z 3 represents the set x, y, z z 3 , which is the set of all points in 3 whose z-coordinate is 3. This is the horizontal plane that is parallel to the xy-plane and three units above it as in Figure 7(a). z
z
y 5
3 0 x
FIGURE 7
0
(a) z=3, a plane in R#
y
x
5
(b) y=5, a plane in R#
0 y
(c) y=5, a line in R@
x
SECTION 12.1 THREE-DIMENSIONAL COORDINATE SYSTEMS
❙❙❙❙
795
(b) The equation y 5 represents the set of all points in 3 whose y-coordinate is 5. This is the vertical plane that is parallel to the xz-plane and five units to the right of it as in Figure 7(b). NOTE
When an equation is given, we must understand from the context whether it represents a curve in 2 or a surface in 3. In Example 1, y 5 represents a plane in 3, but of course y 5 can also represent a line in 2 if we are dealing with two-dimensional analytic geometry. See Figure 7(b) and (c). In general, if k is a constant, then x k represents a plane parallel to the yz-plane, y k is a plane parallel to the xz-plane, and z k is a plane parallel to the xy-plane. In Figure 5, the faces of the rectangular box are formed by the three coordinate planes x 0 (the yz-plane), y 0 (the xz-plane), and z 0 (the xy-plane), and the planes x a, y b, and z c.
z
y 0
■
EXAMPLE 2 Describe and sketch the surface in 3 represented by the equation y x. SOLUTION The equation represents the set of all points in 3 whose x- and y-coordinates
x
are equal, that is, x, x, z x , z . This is a vertical plane that intersects the xy-plane in the line y x, z 0. The portion of this plane that lies in the first octant is sketched in Figure 8.
FIGURE 8
The plane y=x
The familiar formula for the distance between two points in a plane is easily extended to the following three-dimensional formula.
Distance Formula in Three Dimensions The distance P1 P2 between the points
P1 x 1, y1, z1 and P2 x 2 , y2 , z2 is
P P s x 1
z P¡(⁄, ›, z¡)
x 1 2 y2 y1 2 z2 z1 2
2
To see why this formula is true, we construct a rectangular box as in Figure 9, where P1 and P2 are opposite vertices and the faces of the box are parallel to the coordinate planes. If A x 2 , y1, z1 and B x 2 , y2 , z1 are the vertices of the box indicated in the figure, then
P™(¤, fi, z™)
P A x 1
2
x1
AB y
y1
2
BP z 2
2
z1
Because triangles P1 BP2 and P1 AB are both right-angled, two applications of the Pythagorean Theorem give
0 x
2
P P P B
B(¤, fi, z¡)
1
A(¤, ›, z¡) y
and
2
2
1
FIGURE 9
2
P A P1 B
2
2
1
AB BP2
2
2
Combining these equations, we get
P P 1
2
2
AB BP x x y y z P1 A 2
2
2
1
2
2
2
1
2
2
2
z1
2
x 2 x 1 2 y2 y1 2 z2 z1 2 Therefore
P P s x 1
2
2
x 1 2 y2 y1 2 z2 z1 2
796
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
EXAMPLE 3 The distance from the point P 2, 1, 7 to the point Q 1, 3, 5 is
PQ s 1 2
2
3 1 2 5 7 2
s1 4 4 3 EXAMPLE 4 Find an equation of a sphere with radius r and center C h, k, l . z
SOLUTION By definition, a sphere is the set of all points P x, y, z whose distance from
C is r. (See Figure 10.) Thus, P is on the sphere if and only if PC r. Squaring both sides, we have PC 2 r 2 or
P(x, y, z)
r
x h 2 y k 2 z l 2 r 2
C(h, k, l)
The result of Example 4 is worth remembering.
0
Equation of a Sphere An equation of a sphere with center C h, k, l and radius r is
x
x h 2 y k 2 z l 2 r 2
y
FIGURE 10
In particular, if the center is the origin O, then an equation of the sphere is x 2 y 2 z2 r 2 EXAMPLE 5 Show that x 2 y 2 z 2 4x 6y 2z 6 0 is the equation of a
sphere, and find its center and radius. SOLUTION We can rewrite the given equation in the form of an equation of a sphere if we complete squares:
x 2 4x 4 y 2 6y 9 z 2 2z 1 6 4 9 1 x 2 2 y 3 2 z 1 2 8 Comparing this equation with the standard form, we see that it is the equation of a sphere with center 2, 3, 1 and radius s8 2s2. EXAMPLE 6 What region in 3 is represented by the following inequalities?
1 x 2 y 2 z2 4
z 0
SOLUTION The inequalities z
1 x 2 y 2 z2 4 can be rewritten as 1 sx 2 y 2 z 2 2
0 1 2 x
FIGURE 11
y
so they represent the points x, y, z whose distance from the origin is at least 1 and at most 2. But we are also given that z 0, so the points lie on or below the xy-plane. Thus, the given inequalities represent the region that lies between (or on) the spheres x 2 y 2 z 2 1 and x 2 y 2 z 2 4 and beneath (or on) the xy-plane. It is sketched in Figure 11.
SECTION 12.1 THREE-DIMENSIONAL COORDINATE SYSTEMS
|||| 12.1
❙❙❙❙
797
Exercises
1. Suppose you start at the origin, move along the x-axis a
distance of 4 units in the positive direction, and then move downward a distance of 3 units. What are the coordinates of your position? 2. Sketch the points 0, 5, 2 , 4, 0, 1 , 2, 4, 6 , and 1, 1, 2
on a single set of coordinate axes. 3. Which of the points P 6, 2, 3 , Q 5, 1, 4 , and R 0, 3, 8 is
closest to the xz-plane? Which point lies in the yz-plane? 4. What are the projections of the point (2, 3, 5) on the xy-, yz-,
15–18
|||| Show that the equation represents a sphere, and find its center and radius.
15. x 2 y 2 z 2 6x 4y 2z 11 16. x 2 y 2 z 2 4x 2y 17. x 2 y 2 z 2 x y z 18. 4x 2 4y 2 4z 2 8x 16y 1 ■
■
8. Find the lengths of the sides of the triangle with vertices
A 1, 2, 3 , B 3, 4, 2 , and C 3, 2, 1 . Is ABC a right triangle? Is it an isosceles triangle? 9. Determine whether the points lie on a straight line.
(a) A 5, 1, 3 , B 7, 9, 1 , C 1, 15, 11 (b) K 0, 3, 4 , L 1, 2, 2 , M 3, 0, 1 10. Find the distance from 3, 7, 5 to each of the following.
(a) The xy-plane (c) The xz-plane (e) The y-axis
(b) The yz-plane (d) The x-axis (f ) The z-axis
11. Find an equation of the sphere with center 1, 4, 3 and
radius 5. What is the intersection of this sphere with the xz-plane? 12. Find an equation of the sphere with center 6, 5, 2 and
radius s7. Describe its intersection with each of the coordinate planes.
■
■
■
■
■
■
x 1 x 2 y1 y2 z1 z2 , , 2 2 2
20. Find an equation of a sphere if one of its diameters has end-
points 2, 1, 4 and 4, 3, 10 .
3
and R 1, 1, 2 is an equilateral triangle.
■
(b) Find the lengths of the medians of the triangle with vertices A 1, 2, 3 , B 2, 0, 5 , and C 4, 1, 5 .
tion x y 2.
7. Show that the triangle with vertices P 2, 4, 0 , Q 1, 2, 1 ,
■
P1 x 1, y1, z1 to P2 x 2 , y2 , z2 is
5. Describe and sketch the surface in 3 represented by the equa-
it represent in ? Illustrate with sketches. (b) What does the equation y 3 represent in 3 ? What does z 5 represent? What does the pair of equations y 3, z 5 represent? In other words, describe the set of points x, y, z such that y 3 and z 5. Illustrate with a sketch.
■
19. (a) Prove that the midpoint of the line segment from
and xz-planes? Draw a rectangular box with the origin and 2, 3, 5 as opposite vertices and with its faces parallel to the coordinate planes. Label all vertices of the box. Find the length of the diagonal of the box.
6. (a) What does the equation x 4 represent in 2 ? What does
■
21. Find equations of the spheres with center 2, 3, 6 that touch
(a) the xy-plane, (b) the yz-plane, (c) the xz-plane. 22. Find an equation of the largest sphere with center (5, 4, 9) that
is contained in the first octant. Describe in words the region of 3 represented by the equation or inequality.
23–34
||||
23. y 4
24. x 10
25. x 3
26. y 0
27. 0 z 6
28. y z
29. x y z 1 2
2
2
30. 1 x 2 y 2 z 2 25 31. x 2 y 2 z 2 2z 3
32. x 2 y 2 1
33. x 2 z 2 9 ■
■
35–38
■
||||
34. xyz 0 ■
■
■
■
■
■
■
■
■
Write inequalities to describe the region.
35. The half-space consisting of all points to the left of the
xz-plane 36. The solid rectangular box in the first octant bounded by the
planes x 1, y 2, and z 3 37. The region consisting of all points between (but not on) the
spheres of radius r and R centered at the origin, where r R
13. Find an equation of the sphere that passes through the point
4, 3, 1 and has center 3, 8, 1 .
38. The solid upper hemisphere of the sphere of radius 2 centered
at the origin
14. Find an equation of the sphere that passes through the origin
and whose center is 1, 2, 3 .
■
■
■
■
■
■
■
■
■
■
■
■
798
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
39. The figure shows a line L 1 in space and a second line L 2 ,
which is the projection of L 1 on the xy-plane. (In other words, z
L¡
the points on L 2 are directly beneath, or above, the points on L 1.) (a) Find the coordinates of the point P on the line L 1. (b) Locate on the diagram the points A, B, and C, where the line L1 intersects the xy-plane, the yz-plane, and the xz-plane, respectively. 40. Consider the points P such that the distance from P to
P
A 1, 5, 3 is twice the distance from P to B 6, 2, 2 . Show that the set of all such points is a sphere, and find its center and radius.
1
0 1
41. Find an equation of the set of all points equidistant from the
L™
1
points A 1, 5, 3 and B 6, 2, 2 . Describe the set. y
x
42. Find the volume of the solid that lies inside both of the spheres
x 2 y 2 z 2 4x 2y 4z 5 0 and
|||| 12.2
x 2 y 2 z2 4
Vectors
D
B
u
v C
A
FIGURE 1
Equivalent vectors
The term vector is used by scientists to indicate a quantity (such as displacement or velocity or force) that has both magnitude and direction. A vector is often represented by an arrow or a directed line segment. The length of the arrow represents the magnitude of the vector and the arrow points in the direction of the vector. We denote a vector by printing a letter in boldface v or by putting an arrow above the letter vl . For instance, suppose a particle moves along a line segment from point A to point B. The corresponding displacement vector v, shown in Figure 1, has initial point A (the tail) l and terminal point B (the tip) and we indicate this by writing v AB. Notice that the vecl tor u CD has the same length and the same direction as v even though it is in a different position. We say that u and v are equivalent (or equal) and we write u v. The zero vector, denoted by 0, has length 0. It is the only vector with no specific direction.
Combining Vectors C B
A FIGURE 2
l Suppose a particle moves from A to B, so its displacement vector is AB. Then the particle l changes direction and moves from B to C, with displacement vector BC as in Figure 2. The combined effect of these displacements is that the particle has moved from A to C. The l l l resulting displacement vector AC is called the sum of AB and BC and we write l l l AC AB BC In general, if we start with vectors u and v, we first move v so that its tail coincides with the tip of u and define the sum of u and v as follows. Definition of Vector Addition If u and v are vectors positioned so the initial point of v
is at the terminal point of u, then the sum u v is the vector from the initial point of u to the terminal point of v.
SECTION 12.2 VECTORS
❙❙❙❙
799
The definition of vector addition is illustrated in Figure 3. You can see why this definition is sometimes called the Triangle Law. u u+v
v v
u v v+ u+
v
u
u
FIGURE 4 The Parallelogram Law
FIGURE 3 The Triangle Law
In Figure 4 we start with the same vectors u and v as in Figure 3 and draw another copy of v with the same initial point as u. Completing the parallelogram, we see that u v v u. This also gives another way to construct the sum: If we place u and v so they start at the same point, then u v lies along the diagonal of the parallelogram with u and v as sides. (This is called the Parallelogram Law.) EXAMPLE 1 Draw the sum of the vectors a and b shown in Figure 5. a
b
SOLUTION First we translate b and place its tail at the tip of a, being careful to draw a copy of b that has the same length and direction. Then we draw the vector a b [see Figure 6(a)] starting at the initial point of a and ending at the terminal point of the copy of b. Alternatively, we could place b so it starts where a starts and construct a b by the Parallelogram Law as in Figure 6(b).
FIGURE 5
a Visual 12.2 shows how the Triangle and Parallelogram Laws work for various vectors u and v.
FIGURE 6
a
b a+b
a+b b
(a)
(b)
It is possible to multiply a vector by a real number c. (In this context we call the real number c a scalar to distinguish it from a vector.) For instance, we want 2v to be the same vector as v v, which has the same direction as v but is twice as long. In general, we multiply a vector by a scalar as follows. Definition of Scalar Multiplication If c is a scalar and v is a vector, then the scalar mul-
2v
v
_v
_1.5v
FIGURE 7
Scalar multiples of v
1 2v
tiple cv is the vector whose length is c times the length of v and whose direction is the same as v if c 0 and is opposite to v if c 0. If c 0 or v 0, then cv 0. This definition is illustrated in Figure 7. We see that real numbers work like scaling factors here; that’s why we call them scalars. Notice that two nonzero vectors are parallel if they are scalar multiples of one another. In particular, the vector v 1 v has the same length as v but points in the opposite direction. We call it the negative of v. By the difference u v of two vectors we mean u v u v
800
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
So we can construct u v by first drawing the negative of v, v, and then adding it to u by the Parallelogram Law as in Figure 8(a). Alternatively, since v u v u, the vector u v, when added to v, gives u. So we could construct u v as in Figure 8(b) by means of the Triangle Law.
v
u u-v
u-v
_v
v u
FIGURE 8
Drawing u-v
(a)
(b)
EXAMPLE 2 If a and b are the vectors shown in Figure 9, draw a 2b. SOLUTION We first draw the vector 2b pointing in the direction opposite to b and twice as long. We place it with its tail at the tip of a and then use the Triangle Law to draw a 2b as in Figure 10. a
_2b a b
a-2b
FIGURE 9
FIGURE 10
Components y
For some purposes it’s best to introduce a coordinate system and treat vectors algebraically. If we place the initial point of a vector a at the origin of a rectangular coordinate system, then the terminal point of a has coordinates of the form a1, a2 or a1, a2, a3 , depending on whether our coordinate system is two- or three-dimensional (see Figure 11). These coordinates are called the components of a and we write
(a¡, a™)
a O
x
a a 1, a 2
a=ka¡, a™l z (a¡, a™, a£)
a O y
x
or
a a 1, a 2 , a 3
We use the notation a1, a2 for the ordered pair that refers to a vector so as not to confuse it with the ordered pair a1, a2 that refers to a point in the plane. For instance, the vectors shown in Figure 12 are all equivalent to the vector l OP 3, 2 whose terminal point is P 3, 2 . What they have in common is that the terminal point is reached from the initial point by a displacement of three units to the right and two upward. We can think of all these geometric vectors as representations of the y
a=ka¡, a™, a£l
z
(4, 5)
FIGURE 11 (1, 3)
position vector of P
P(3, 2)
P (a¡, a™, a£) 0
x
O
x
FIGURE 12
Representations of the vector a=k3, 2l
B(x+a¡, y+a™, z+a£)
A(x, y, z)
FIGURE 13 Representations of a=ka¡, a™, a£l
y
SECTION 12.2 VECTORS
❙❙❙❙
801
l algebraic vector a 3, 2 . The particular representation OP from the origin to the point P 3, 2 is called the position vector of the point P. l In three dimensions, the vector a OP a1, a2, a3 is the position vector of the l point P a1, a2, a3 . (See Figure 13.) Let’s consider any other representation AB of a, where the initial point is A x 1, y1, z1 and the terminal point is B x 2 , y2 , z2 . Then we must have x 1 a 1 x 2, y1 a 2 y2, and z1 a 3 z2 and so a 1 x 2 x 1, a 2 y2 y1, and a 3 z2 z1. Thus, we have the following result. 1 Given the points A x 1, y1, z1 and B x 2 , y2 , z2 , the vector a with represenl tation AB is a x 2 x 1, y2 y1, z2 z1
EXAMPLE 3 Find the vector represented by the directed line segment with initial point
A 2, 3, 4) and terminal point B 2, 1, 1 . l
SOLUTION By (1), the vector corresponding to AB is
a 2 2, 1 3 , 1 4 4, 4, 3
The magnitude or length of the vector v is the length of any of its representations and is denoted by the symbol v or v . By using the distance formula to compute the length of a segment OP, we obtain the following formulas.
The length of the two-dimensional vector a a 1, a 2 is
a sa
2 1
a 22
The length of the three-dimensional vector a a 1, a 2 , a 3 is y
(a¡+b¡, a™+b™)
a+b
b¡ a 0
a™
a™
a¡
2 1
a 22 a 23
How do we add vectors algebraically? Figure 14 shows that if a a 1, a 2 and b b 1, b 2 , then the sum is a b a1 b1, a2 b2 , at least for the case where the components are positive. In other words, to add algebraic vectors we add their components. Similarly, to subtract vectors we subtract components. From the similar triangles in Figure 15 we see that the components of ca are ca1 and ca2. So to multiply a vector by a scalar we multiply each component by that scalar.
b™
b
a sa
x
b¡
If a a 1, a 2 and b b1, b2 , then
FIGURE 14
a b a 1 b1, a 2 b2
a b a 1 b1, a 2 b2
ca ca1, ca2
Similarly, for three-dimensional vectors, ca a
ca™
a™
a 1, a 2 , a 3 b1, b2 , b3 a 1 b1, a 2 b2 , a 3 b3
a 1, a 2 , a 3 b1, b2 , b3 a 1 b1, a 2 b2 , a 3 b3
a¡ FIGURE 15
ca¡
c a 1, a 2 , a 3 ca1, ca2 , ca3
802
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
EXAMPLE 4 If a 4, 0, 3 and b 2, 1, 5 , find a and the vectors a b, a b,
3b, and 2a 5b.
a s4
SOLUTION
2
0 2 32 s25 5
a b 4, 0, 3 2, 1, 5 4 2, 0 1, 3 5 2, 1, 8 a b 4, 0, 3 2, 1, 5 4 2 , 0 1, 3 5 6, 1, 2 3b 3 2, 1, 5 3 2 , 3 1 , 3 5 6, 3, 15 2a 5b 2 4, 0, 3 5 2, 1, 5 8, 0, 6 10, 5, 25 2, 5, 31 We denote by V2 the set of all two-dimensional vectors and by V3 the set of all threedimensional vectors. More generally, we will later need to consider the set Vn of all n-dimensional vectors. An n-dimensional vector is an ordered n-tuple: |||| Vectors in n dimensions are used to list various quantities in an organized way. For instance, the components of a six-dimensional vector p p1 , p2 , p3 , p4 , p5 , p6 might represent the prices of six different ingredients required to make a particular product. Four-dimensional vectors x, y, z, t are used in relativity theory, where the first three components specify a position in space and the fourth represents time.
a a1, a 2, . . . , a n where a1, a 2, . . . , a n are real numbers that are called the components of a. Addition and scalar multiplication are defined in terms of components just as for the cases n 2 and n 3. Properties of Vectors If a, b, and c are vectors in Vn and c and d are scalars, then 1. a b b a
2. a b c a b c
3. a 0 a
4. a a 0
5. c a b ca cb
6. c d a ca da
7. cd a c da
8. 1a a
These eight properties of vectors can be readily verified either geometrically or algebraically. For instance, Property 1 can be seen from Figure 4 (it’s equivalent to the Parallelogram Law) or as follows for the case n 2: a b a 1, a 2 b1, b2 a 1 b1, a 2 b2 b1 a 1, b2 a 2 b1, b2 a 1, a 2
Q
b a
c
(a+b)+c =a+(b+c)
b
a+b b+c
P FIGURE 16
We can see why Property 2 (the associative law) is true by looking at Figure 16 and l applying the Triangle Law several times: The vector PQ is obtained either by first constructing a b and then adding c or by adding a to the vector b c. Three vectors in V3 play a special role. Let
a
i 1, 0, 0
j 0, 1, 0
k 0, 0, 1
SECTION 12.2 VECTORS
❙❙❙❙
803
Then i , j, and k are vectors that have length 1 and point in the directions of the positive x-, y-, and z-axes. Similarly, in two dimensions we define i 1, 0 and j 0, 1 . (See Figure 17.) y
z
(0, 1)
j 0
k x
i
j
i
(1, 0)
FIGURE 17
y
x
(a)
Standard basis vectors in V™ and V£
(b)
If a a 1, a 2 , a 3 , then we can write a a 1, a 2 , a 3 a 1, 0, 0 0, a 2 , 0 0, 0, a 3 a 1 1, 0, 0 a 2 0, 1, 0 a 3 0, 0, 1
y (a¡, a™)
a
a™ j
a¡i
0
2
a a1 i a2 j a3 k
Thus, any vector in V3 can be expressed in terms of the standard basis vectors i , j, and k. For instance,
x
1, 2, 6 i 2j 6k (a) a=a¡i+a™ j
Similarly, in two dimensions, we can write z
3
a a1, a2 a1 i a2 j
(a¡, a™, a£)
See Figure 18 for the geometric interpretation of Equations 3 and 2 and compare with Figure 17.
a
a£k
a¡i
y
x
a™ j (b) a=a¡i+a™ j+a£k
EXAMPLE 5 If a i 2j 3k and b 4i 7 k, express the vector 2a 3b in terms of i , j, and k. SOLUTION Using Properties 1, 2, 5, 6, and 7 of vectors, we have
FIGURE 18
2a 3b 2 i 2j 3k 3 4i 7k 2i 4j 6k 12i 21k 14i 4j 15k A unit vector is a vector whose length is 1. For instance, i , j, and k are all unit vectors. In general, if a 0, then the unit vector that has the same direction as a is 4
u
1 a a a a
In order to verify this, we let c 1 a . Then u ca and c is a positive scalar, so u has the same direction as a. Also 1
u ca c a a a 1
804
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
EXAMPLE 6 Find the unit vector in the direction of the vector 2i j 2k. SOLUTION The given vector has length
2i j 2k s2
2
1 2 2 2 s9 3
so, by Equation 4, the unit vector with the same direction is 1 3
2i j 2k 23 i 13 j 23 k
Applications Vectors are useful in many aspects of physics and engineering. In Chapter 13 we will see how they describe the velocity and acceleration of objects moving in space. Here we look at forces. A force is represented by a vector because it has both a magnitude (measured in pounds or newtons) and a direction. If several forces are acting on an object, the resultant force experienced by the object is the vector sum of these forces. 50°
32°
T¡
T™
EXAMPLE 7 A 100-lb weight hangs from two wires as shown in Figure 19. Find the tensions (forces) T1 and T2 in both wires and their magnitudes. SOLUTION We first express T1 and T2 in terms of their horizontal and vertical components. From Figure 20 we see that
100 FIGURE 19
T cos 32 i T sin 32 j
5
T1 T1 cos 50 i T1 sin 50 j
6
T2
2
2
The resultant T1 T2 of the tensions counterbalances the weight w and so we must have 50° T¡
T™
50°
32°
32°
T1 T2 w 100j Thus
( T1 cos 50 T2 cos 32 ) i ( T1 sin 50 T2 sin 32 ) j 100j
w
Equating components, we get FIGURE 20
T sin 50 T sin 32 100 Solving the first of these equations for T and substituting into the second, we get T cos 50 sin 32 100 T sin 50 T1 cos 50 T2 cos 32 0 1
2
2
1
1
cos 32
So the magnitudes of the tensions are
T 1
and
100 85.64 lb sin 50 tan 32 cos 50
T cos 50 T cos 32 64.91 lb 1
2
Substituting these values in (5) and (6), we obtain the tension vectors T1 55.05 i 65.60 j
T2 55.05 i 34.40 j
SECTION 12.2 VECTORS
|||| 12.2
805
Exercises 9. A 1, 1 ,
1. Are the following quantities vectors or scalars? Explain.
(a) (b) (c) (d)
❙❙❙❙
The cost of a theater ticket The current in a river The initial flight path from Houston to Dallas The population of the world
11. A 0, 3, 1 , ■
2. What is the relationship between the point (4, 7) and the
vector 4, 7 ? Illustrate with a sketch. 3. Name all the equal vectors in the parallelogram shown.
A
B
C
■
2, 4
15. 0, 1, 2 ,
0, 0, 3
■
■
||||
■
l l (b) RP PS l l l (d) RS SP PQ
■
■
B 4, 2, 1 ■
■
■
16. 1, 0, 2 , ■
17. a 4, 3 ,
b 6, 2
18. a 2 i 3 j,
b i 5j
■
23–25
■
||||
■
■
0, 4, 0 ■
■
■
■
■
■
b 1, 5, 2
20. a 3, 4, 1 ,
■
■
5, 7
22. a 3 i 2 k,
P
■
14. 2, 1 ,
■
21. a i 2 j k,
Q
12. A 4, 0, 2 , ■
Find a , a b, a b, 2a, and 3a 4b.
19. a 6, 2, 3 ,
4. Write each combination of vectors as a single vector.
■
13. 3, 1 ,
17–22
D
B 2, 3, 1
■
B 3, 0
13–16 |||| Find the sum of the given vectors and illustrate geometrically.
■
E
l l (a) PQ QR l l (c) QS PS
■
10. A 2, 2 ,
B 3, 4
b 6, 2, 3
b j 2k
b i j k ■
■
■
■
■
■
Find a unit vector that has the same direction as the given
vector. 23. 9, 5
S
R
25. 8 i j 4 k
5. Copy the vectors in the figure and use them to draw the
following vectors. (a) u v (c) v w
24. 12 i 5 j
■
■
■
■
■
■
■
■
■
■
■
26. Find a vector that has the same direction as 2, 4, 2 but has
(b) u v (d) w v u
length 6. 27. If v lies in the first quadrant and makes an angle 3 with the
u
v
positive x-axis and v 4, find v in component form.
w
28. If a child pulls a sled through the snow with a force of 50 N 6. Copy the vectors in the figure and use them to draw the follow-
ing vectors. (a) a b (c) 2a (e) 2a b
(b) a b (d) 12 b (f) b 3a
a
exerted at an angle of 38 above the horizontal, find the horizontal and vertical components of the force. 29. Two forces F1 and F2 with magnitudes 10 lb and 12 lb act
on an object at a point P as shown in the figure. Find the resultant force F acting at P as well as its magnitude and its direction. (Indicate the direction by finding the angle shown in the figure.) F
b
F¡ 7–12 |||| Find a vector a with representation given by the directed l l line segment AB. Draw AB and the equivalent representation starting at the origin.
7. A 2, 3 ,
B 2, 1
8. A 2, 2 ,
■
B 5, 3
F™ ¨ 45°
30° P
806
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
37. (a) Draw the vectors a 3, 2 , b 2, 1 , and c 7, 1 .
30. Velocities have both direction and magnitude and thus are
vectors. The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction N45 W at a speed of 50 km h. (This means that the direction from which the wind blows is 45 west of the northerly direction.) A pilot is steering a plane in the direction N60 E at an airspeed (speed in still air) of 250 km h. The true course, or track, of the plane is the direction of the resultant of the velocity vectors of the plane and the wind. The ground speed of the plane is the magnitude of the resultant. Find the true course and the ground speed of the plane. 31. A woman walks due west on the deck of a ship at 3 mi h. The
ship is moving north at a speed of 22 mi h. Find the speed and direction of the woman relative to the surface of the water. 32. Ropes 3 m and 5 m in length are fastened to a holiday decora-
tion that is suspended over a town square. The decoration has a mass of 5 kg. The ropes, fastened at different heights, make angles of 52 and 40 with the horizontal. Find the tension in each wire and the magnitude of each tension.
52° 3 m
40°
(b) Show, by means of a sketch, that there are scalars s and t such that c sa t b. (c) Use the sketch to estimate the values of s and t. (d) Find the exact values of s and t. 38. Suppose that a and b are nonzero vectors that are not parallel
and c is any vector in the plane determined by a and b. Give a geometric argument to show that c can be written as c sa t b for suitable scalars s and t. Then give an argument using components. 39. If r x, y, z and r0 x 0 , y0 , z0 , describe the set of all
points x, y, z such that r r0 1.
40. If r x, y , r1 x 1, y1 , and r2 x 2 , y2 , describe the
set of all points x, y such that r r1 r r2 k, where k r1 r2 .
41. Figure 16 gives a geometric demonstration of Property 2 of
vectors. Use components to give an algebraic proof of this fact for the case n 2. 42. Prove Property 5 of vectors algebraically for the case n 3.
Then use similar triangles to give a geometric proof.
5 m
43. Use vectors to prove that the line joining the midpoints of two
sides of a triangle is parallel to the third side and half its length. 44. Suppose the three coordinate planes are all mirrored and a 33. A clothesline is tied between two poles, 8 m apart. The line
is quite taut and has negligible sag. When a wet shirt with a mass of 0.8 kg is hung at the middle of the line, the midpoint is pulled down 8 cm. Find the tension in each half of the clothesline. 34. The tension T at each end of the chain has magnitude 25 N.
What is the weight of the chain?
37°
light ray given by the vector a a 1, a 2 , a 3 first strikes the xz-plane, as shown in the figure. Use the fact that the angle of incidence equals the angle of reflection to show that the direction of the reflected ray is given by b a 1, a 2 , a 3 . Deduce that, after being reflected by all three mutually perpendicular mirrors, the resulting ray is parallel to the initial ray. (American space scientists used this principle, together with laser beams and an array of corner mirrors on the Moon, to calculate very precisely the distance from the Earth to the Moon.) z
37°
35. If A, B, and C are the vertices of a triangle, find
l l l AB BC CA.
b
36. Let C be the point on the line segment AB that is twice as far
l l l from B as it is from A. If a OA, b OB, and c OC, show 2 1 that c 3 a 3 b.
a x
y
SECTION 12.3 THE DOT PRODUCT
|||| 12.3
❙❙❙❙
807
The Dot Product So far we have added two vectors and multiplied a vector by a scalar. The question arises: Is it possible to multiply two vectors so that their product is a useful quantity? One such product is the dot product, whose definition follows. Another is the cross product, which is discussed in the next section. 1 Definition If a a 1, a 2 , a 3 and b b1, b2 , b3 , then the dot product of a and b is the number a b given by
a b a 1 b1 a 2 b2 a 3 b3 Thus, to find the dot product of a and b we multiply corresponding components and add. The result is not a vector. It is a real number, that is, a scalar. For this reason, the dot product is sometimes called the scalar product (or inner product). Although Definition 1 is given for three-dimensional vectors, the dot product of two-dimensional vectors is defined in a similar fashion: a 1, a 2 b1, b2 a 1 b1 a 2 b2 EXAMPLE 1
2, 4 3, 1 2 3 4 1 2 1, 7, 4 6, 2, 12 1 6 7 2 4( 12 ) 6 i 2 j 3k 2 j k 1 0 2 2 3 1 7 The dot product obeys many of the laws that hold for ordinary products of real numbers. These are stated in the following theorem. 2
Properties of the Dot Product If a, b, and c are vectors in V3 and c is a scalar, then
1. a a a
2. a b b a
2
3. a b c a b a c
4. ca b c a b a cb
5. 0 a 0
These properties are easily proved using Definition 1. For instance, here are the proofs of Properties 1 and 3:
1. a a a 21 a 22 a 23 a
2
3. a b c a1, a2, a3 b1 c1, b2 c2, b3 c3
a 1 b1 c1 a 2 b2 c2 a 3 b3 c3 a 1 b1 a 1 c1 a 2 b2 a 2 c2 a 3 b3 a 3 c3 a 1 b1 a 2 b2 a 3 b3 a 1 c1 a 2 c2 a 3 c3 a b a c The proofs of the remaining properties are left as exercises. The dot product a b can be given a geometric interpretation in terms of the angle between a and b, which is defined to be the angle between the representations of a and b that start at the origin, where 0 . In other words, is the angle between the
808
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
l l line segments OA and OB in Figure 1. Note that if a and b are parallel vectors, then 0 or . The formula in the following theorem is used by physicists as the definition of the dot product.
z
B a-b
b 0 ¨ x
a
A
3
Theorem If is the angle between the vectors a and b, then
b cos
a b a
y
FIGURE 1
Proof If we apply the Law of Cosines to triangle OAB in Figure 1, we get
AB
2
4
OA
2
OB
2
2 OA
OB cos
(Observe that the Law of Cosines still applies in the limiting cases when 0 or , or a 0 or b 0.) But OA a , OB b , and AB a b , so Equation 4 becomes
a b
2
5
a
2
b
2
b cos
2 a
Using Properties 1, 2, and 3 of the dot product, we can rewrite the left side of this equation as follows: a b 2 a b a b
a a a b b a b b
a
2
2a b b
2
Therefore, Equation 5 gives
a
2
Thus or
2 a b cos 2a b 2 a b cos a b a b cos
2a b b
2
a
2
b
2
EXAMPLE 2 If the vectors a and b have lengths 4 and 6, and the angle between them is
3, find a b. SOLUTION Using Theorem 3, we have
b cos 3 4 6
a b a
1 2
12
The formula in Theorem 3 also enables us to find the angle between two vectors. 6
Corollary If is the angle between the nonzero vectors a and b, then
cos
a b a b
EXAMPLE 3 Find the angle between the vectors a 2, 2, 1 and b 5, 3, 2 . SOLUTION Since
a s2
2
2 2 1 2 3
and
b s5
2
3 2 2 2 s38
SECTION 12.3 THE DOT PRODUCT
❙❙❙❙
809
and since a b 2 5 2 3 1 2 2 we have, from Corollary 6, cos So the angle between a and b is
cos 1
a b 2 a b 3s38
2 3s38
1.46
or 84
Two nonzero vectors a and b are called perpendicular or orthogonal if the angle between them is 2. Then Theorem 3 gives
b cos 2 0
a b a
and conversely if a b 0, then cos 0, so 2. The zero vector 0 is considered to be perpendicular to all vectors. Therefore, we have the following method for determining whether two vectors are orthogonal. a and b are orthogonal if and only if a b 0.
7
EXAMPLE 4 Show that 2i 2j k is perpendicular to 5i 4j 2k. SOLUTION Since
2i 2j k 5i 4j 2k 2 5 2 4 1 2 0 a
¨
a
b
b
b
¨
a · b>0
a · b =0
a · b <0
a
these vectors are perpendicular by (7). Because cos 0 if 0 2 and cos 0 if 2 , we see that a b is positive for 2 and negative for 2. We can think of a b as measuring the extent to which a and b point in the same direction. The dot product a b is positive if a and b point in the same general direction, 0 if they are perpendicular, and negative if they point in generally opposite directions (see Figure 2). In the extreme case where a and b point in exactly the same direction, we have 0, so cos 1 and
b
FIGURE 2 Visual 12.3A shows an animation of Figure 2.
a b a
If a and b point in exactly opposite directions, then and so cos 1 and a b a b .
Direction Angles and Direction Cosines The direction angles of a nonzero vector a are the angles , , and (in the interval 0, that a makes with the positive x-, y-, and z-axes (see Figure 3 on page 810). The cosines of these direction angles, cos , cos , and cos , are called the direction cosines of the vector a. Using Corollary 6 with b replaced by i , we obtain 8
cos
a i a1 a i a
810
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
z
(This can also be seen directly from Figure 3.) Similarly, we also have cos
9
ç a¡
a ∫
a2 a
cos
a3 a
By squaring the expressions in Equations 8 and 9 and adding, we see that
å y
cos 2 cos 2 cos 2 1
10
x
We can also use Equations 8 and 9 to write
FIGURE 3
a a 1, a 2 , a 3 a cos , a cos , a cos
a cos , cos , cos Therefore 1 a cos , cos , cos a
11
which says that the direction cosines of a are the components of the unit vector in the direction of a. EXAMPLE 5 Find the direction angles of the vector a 1, 2, 3 .
SOLUTION Since a s1 2 2 2 3 2 s14, Equations 8 and 9 give
cos and so
cos 1
1 s14
1 s14
cos
2 s14
cos 1
74
2 s14
cos
3 s14
cos 1
58
3 s14
37
Projections l l Figure 4 shows representations PQ and PR of two vectors a and b with the same initial l point P. If S is the foot of the perpendicular from R to the line containing PQ, then the l vector with representation PS is called the vector projection of b onto a and is denoted by proja b. R Visual 12.3B shows how Figure 4 changes when we vary a and b.
R b
b
a
a
FIGURE 4
Vector projections
P
S
proj a b
Q
S
P
Q
proj a b
The scalar projection of b onto a (also called the component of b along a) is defined to be the magnitude of the vector projection, which is the number b cos , where is the
SECTION 12.3 THE DOT PRODUCT
b
P
811
angle between a and b. (See Figure 5; you can think of the scalar projection of b as being the length of a shadow of b.) This is denoted by compa b. Observe that it is negative if 2 . The equation
R
a
¨
â?&#x2122;â?&#x2122;â?&#x2122;â?&#x2122;
b cos a ( b cos )
a b a
Q
S
ĺ&#x2026;Š b ĺ&#x2026;Š cos ¨
shows that the dot product of a and b can be interpreted as the length of a times the scalar projection of b onto a. Since
FIGURE 5
Scalar projection
a b
b cos a
a b a
the component of b along a can be computed by taking the dot product of b with the unit vector in the direction of a. We summarize these ideas as follows.
Scalar projection of b onto a:
compa b
Vector projection of b onto a:
proja b
a b a
a b a
a a b a a a 2
Notice that the vector projection is the scalar projection times the unit vector in the direction of a. EXAMPLE 6 Find the scalar projection and vector projection of b 1, 1, 2 onto
a 2, 3, 1 .
SOLUTION Since a s 2 2 3 2 1 2 s14, the scalar projection of b onto a is
compa b
a b 2 1 3 1 1 2 3 a s14 s14
The vector projection is this scalar projection times the unit vector in the direction of a: proja b
F
Q D
FIGURE 6
W ( F cos ) D
S
P
One use of projections occurs in physics in calculating work. In Section 6.4 we defined the work done by a constant force F in moving an object through a distance d as W Fd, but this applies only when the force is directed along the line of motion of the object. l Suppose, however, that the constant force is a vector F PR pointing in some other direction as in Figure 6. If the force moves the object from P to Q, then the displacement l vector is D PQ. The work done by this force is defined to be the product of the component of the force along D and the distance moved:
R
¨
3 a 3 3 9 3 a , , 14 7 14 14 s14 a
But then, from Theorem 3, we have 12
D cos F D
W F
❙❙❙❙
812
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
Thus, the work done by a constant force F is the dot product F D, where D is the displacement vector. EXAMPLE 7 A crate is hauled 8 m up a ramp under a constant force of 200 N applied at an angle of 25 to the ramp. Find the work done. F 25°
SOLUTION If F and D are the force and displacement vectors, as pictured in Figure 7, then the work done is
D
D cos 25
W F D F
200 8 cos 25 1450 N m 1450 J
FIGURE 7
EXAMPLE 8 A force is given by a vector F 3i 4j 5k and moves a particle from the point P 2, 1, 0 to the point Q 4, 6, 2 . Find the work done. l SOLUTION The displacement vector is D PQ 2, 5, 2 , so by Equation 12, the work done is
W F D 3, 4, 5 2, 5, 2 6 20 10 36 If the unit of length is meters and the magnitude of the force is measured in newtons, then the work done is 36 joules.
|||| 12.3
Exercises
1. Which of the following expressions are meaningful? Which are
meaningless? Explain. (a) a b c (c) a b c (e) a b c
||||
If u is a unit vector, find u v and u w. 12.
11.
(b) a b c (d) a b c (f) a b c
11–12
u
u
v
v w
2. Find the dot product of two vectors if their lengths are 6
and 13 and the angle between them is 4. w 3–10
||||
Find a b.
3. a 4, 1 , 4. a
1 2
, 4 ,
b 3, 6
b 8, 3 b 3, 1, 10
6. a s, 2s, 3s ,
b t, t, 5t
8. a 4 j 3 k,
10.
■
■
■
■
■
■
■
■
15–20 |||| Find the angle between the vectors. (First find an exact expression and then approximate to the nearest degree.)
15. a 3, 4 ,
the angle between a and b is 120° ■
■
■
drinks on a given day. He charges $2 for a hamburger, $1.50 for a hot dog, and $1 for a soft drink. If A a, b, c and P 2, 1.5, 1 , what is the meaning of the dot product A P ?
b 2i 4 j 6k
■
■
14. A street vendor sells a hamburgers, b hot dogs, and c soft
b 15, the angle between a and b is 6
■
■
(b) Show that i i j j k k 1.
b 5i 9k
a 4, b 10,
9. a 12,
■
13. (a) Show that i j j k k i 0.
5. a 5, 0, 2 ,
7. a i 2 j 3 k ,
■
■
■
■
■
■
■
■
16. a s3, 1 ,
b 5, 12 b 0, 5
SECTION 12.3 THE DOT PRODUCT
17. a 1, 2, 3 ,
b 4, 0, 1
18. a 6, 3, 2 , 19. a j k,
37. a 4, 2, 0 ,
39. a i k ,
b i 2 j 3k
20. a 2 i j k,
■ ■
■
■
■
■
■
■
■
■
■
■
B 3, 6 ,
22. D 0, 1, 1 , ■
■
■
■
■
■
■
b i 6 j 2k ■
■
■
■
■
■
■
■
41. Show that the vector orth a b b proj a b is orthogonal to a.
(It is called an orthogonal projection of b.) drawing the vectors a, b, proj a b, and orth a b.
F 1, 2, 1 ■
■
42. For the vectors in Exercise 36, find orth a b and illustrate by
C 1, 4
E 2, 4, 3 ,
b i j
■
21–22 |||| Find, correct to the nearest degree, the three angles of the triangle with the given vertices.
21. A 1, 0 ,
b 3, 3, 4
40. a 2 i 3 j k ,
b 3i 2 j k
813
b 1, 1, 1
38. a 1, 2, 2 ,
b 2, 1, 2
❙❙❙❙
■
■
43. If a 3, 0, 1 , find a vector b such that comp a b 2. ■
■
■
■
44. Suppose that a and b are nonzero vectors. 23–24
(a) Under what circumstances is comp a b comp b a? (b) Under what circumstances is proj a b proj b a?
Determine whether the given vectors are orthogonal, parallel, or neither. ||||
23. (a) a 5, 3, 7 ,
b 6, 8, 2 (b) a 4, 6 , b 3, 2 (c) a i 2 j 5 k, b 3 i 4 j k (d) a 2 i 6 j 4 k, b 3 i 9 j 6 k
45. A constant force with vector representation
F 10 i 18 j 6 k moves an object along a straight line from the point 2, 3, 0 to the point 4, 9, 15 . Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.
24. (a) u 3, 9, 6 ,
v 4, 12, 8 (b) u i j 2 k, v 2 i j k (c) u a, b, c , v b, a, 0
■
■
■
■
■
■
■
■
46. Find the work done by a force of 20 lb acting in the direction
N50 W in moving an object 4 ft due west. ■
■
■
■
47. A woman exerts a horizontal force of 25 lb on a crate as she
pushes it up a ramp that is 10 ft long and inclined at an angle of 20 above the horizontal. Find the work done on the box.
25. Use vectors to decide whether the triangle with vertices
P 1, 3, 2 , Q 2, 0, 4 , and R 6, 2, 5 is right-angled.
48. A wagon is pulled a distance of 100 m along a horizontal path
26. For what values of b are the vectors 6, b, 2 and b, b 2, b
by a constant force of 50 N. The handle of the wagon is held at an angle of 30 above the horizontal. How much work is done?
orthogonal?
49. Use a scalar projection to show that the distance from a point
27. Find a unit vector that is orthogonal to both i j and i k.
P1 x 1, y1 to the line a x by c 0 is
28. Find two unit vectors that make an angle of 60 with
ax
1 by1 c sa 2 b 2
v 3, 4 . 29–33 |||| Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.)
Use this formula to find the distance from the point 2, 3 to the line 3x 4y 5 0.
29. 3, 4, 5
50. If r x, y, z , a a 1, a 2 , a 3 , and b b1, b2 , b3 , show
30. 1, 2, 1
that the vector equation r a r b 0 represents a sphere, and find its center and radius.
31. 2 i 3 j 6 k
51. Find the angle between a diagonal of a cube and one of its
32. 2 i j 2 k 33. c, c, c , ■
■
edges.
where c 0
■
■
■
52. Find the angle between a diagonal of a cube and a diagonal of ■
■
■
■
■
■
■
34. If a vector has direction angles 4 and 3, find the
third direction angle .
35–40
||||
Find the scalar and vector projections of b onto a.
35. a 3, 4 , 36. a 1, 2 ,
b 5, 0 b 4, 1
one of its faces. 53. A molecule of methane, CH 4, is structured with the four hydro-
gen atoms at the vertices of a regular tetrahedron and the carbon atom at the centroid. The bond angle is the angle formed by the H— C —H combination; it is the angle between the lines that join the carbon atom to two of the hydrogen atoms. Show that the bond angle is about 109.5 . [Hint: Take the vertices of the tetrahedron to be the points 1, 0, 0 , 0, 1, 0 ,
814
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
0, 0, 1 , and 1, 1, 1 as shown in the figure. Then the centroid is ( 12 , 12 , 12 ).]
57. Use Theorem 3 to prove the Cauchy-Schwarz Inequality:
a b a b
z
H
58. The Triangle Inequality for vectors is C
a b a b
H H
(a) Give a geometric interpretation of the Triangle Inequality. (b) Use the Cauchy-Schwarz Inequality from Exercise 57 to prove the Triangle Inequality. [Hint: Use the fact that a b 2 a b a b and use Property 3 of the dot product.]
y
H
x
54. If c a b b a, where a, b, and c are all nonzero
vectors, show that c bisects the angle between a and b.
59. The Parallelogram Law states that
55. Prove Properties 2, 4, and 5 of the dot product (Theorem 2).
a b
2
56. Suppose that all sides of a quadrilateral are equal in length and
opposite sides are parallel. Use vector methods to show that the diagonals are perpendicular.
|||| 12.4
a b
2
2 a
2
2 b
2
(a) Give a geometric interpretation of the Parallelogram Law. (b) Prove the Parallelogram Law. (See the hint in Exercise 58.)
The Cross Product The cross product a b of two vectors a and b, unlike the dot product, is a vector. For this reason it is also called the vector product. Note that a b is defined only when a and b are three-dimensional vectors. 1 Definition If a a 1, a 2 , a 3 and b b1, b2 , b3 , then the cross product of a and b is the vector
a b a 2 b3 a 3 b2 , a 3 b1 a 1 b3 , a 1 b2 a 2 b1 This may seem like a strange way of defining a product. The reason for the particular form of Definition 1 is that the cross product defined in this way has many useful properties, as we will soon see. In particular, we will show that the vector a b is perpendicular to both a and b. In order to make Definition 1 easier to remember, we use the notation of determinants. A determinant of order 2 is defined by
a c
2 6
For example,
b ad bc d
1 2 4 1 6 14 4
A determinant of order 3 can be defined in terms of second-order determinants as follows: 2
a1 b1 c1
a2 b2 c2
a3 b2 b3 a1 c2 c3
b3 b1 a2 c3 c1
b3 b1 a3 c3 c1
b2 c2
SECTION 12.4 THE CROSS PRODUCT
❙❙❙❙
815
Observe that each term on the right side of Equation 2 involves a number a i in the first row of the determinant, and a i is multiplied by the second-order determinant obtained from the left side by deleting the row and column in which a i appears. Notice also the minus sign in the second term. For example,
1 3 5
1 0 1 1 4 2
2 0 4
1 3 2 2 5
1 3 1 2 5
0 4
1 0 4 2 6 5 1 12 0 38
If we now rewrite Definition 1 using second-order determinants and the standard basis vectors i , j, and k, we see that the cross product of the vectors a a 1 i a 2 j a 3 k and b b 1 i b 2 j b 3 k is a b
3
a2 b2
a3 a1 i b3 b1
a3 a1 j b3 b1
a2 k b2
In view of the similarity between Equations 2 and 3, we often write
i a b a1 b1
4
j a2 b2
k a3 b3
Although the first row of the symbolic determinant in Equation 4 consists of vectors, if we expand it as if it were an ordinary determinant using the rule in Equation 2, we obtain Equation 3. The symbolic formula in Equation 4 is probably the easiest way of remembering and computing cross products. EXAMPLE 1 If a 1, 3, 4 and b 2, 7, 5 , then
i j k a b 1 3 4 2 7 5
3 7
4 1 i 5 2
4 1 j 5 2
3 k 7
15 28 i 5 8 j 7 6 k 43i 13j k EXAMPLE 2 Show that a a 0 for any vector a in V3. SOLUTION If a a 1, a 2 , a 3 , then
i a a a1 a1
j a2 a2
k a3 a3
a 2 a 3 a 3 a 2 i a 1 a 3 a 3 a 1 j a 1 a 2 a 2 a 1 k 0i 0j 0k 0
816
â?&#x2122;â?&#x2122;â?&#x2122;â?&#x2122;
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
One of the most important properties of the cross product is given by the following theorem. 5
Theorem The vector a b is orthogonal to both a and b.
Proof In order to show that a b is orthogonal to a, we compute their dot product as
follows: a b a
a2 b2
a3 a1 a1 b3 b1
a3 a1 a2 b3 b1
a2 a3 b2
a 1 a 2 b3 a 3 b2 a 2 a 1 b3 a 3 b1 a 3 a 1 b2 a 2 b1 a 1 a 2 b3 a 1 b2 a 3 a 1 a 2 b3 b1 a 2 a 3 a 1 b2 a 3 b1 a 2 a 3 0 A similar computation shows that a b b 0. Therefore, a b is orthogonal to both a and b.
axb
a
¨
b
If a and b are represented by directed line segments with the same initial point (as in Figure 1), then Theorem 5 says that the cross product a b points in a direction perpendicular to the plane through a and b. It turns out that the direction of a b is given by the right-hand rule: If the fingers of your right hand curl in the direction of a rotation (through an angle less than 180 ) from a to b, then your thumb points in the direction of a b. Now that we know the direction of the vector a b, the remaining thing we need to complete its geometric description is its length a b . This is given by the following theorem.
6
Theorem If is the angle between a and b (so 0 ), then
FIGURE 1
Visual 12.4 shows how a b changes as b changes.
a b a b sin Proof From the definitions of the cross product and length of a vector, we have
a b
2
a 2 b3 a 3 b2 2 a 3 b1 a 1 b3 2 a 1 b2 a 2 b1 2 a 22 b 23 2a 2 a 3 b2 b3 a 23 b 22 a 23 b 21 2a 1 a 3 b1 b3 a 21 b 23 a 21 b 22 2a 1 a 2 b1 b2 a 22 b 21 a 21 a 22 a 23 b 21 b 22 b 23 a 1 b1 a 2 b2 a 3 b3 2
b a b a b a b cos a b 1 cos a b sin a
2
2
2
2
2
2
2
2
2
2
2
2
(by Theorem 12.3.3)
2
2
Taking square roots and observing that ssin 2 sin because sin 0 when 0 , we have a b a b sin
Geometric characterization of a b
Since a vector is completely determined by its magnitude and direction, we can now say that a b is the vector that is perpendicular to both a and b, whose orientation is deter-
SECTION 12.4 THE CROSS PRODUCT
❙❙❙❙
817
b sin . In fact, that is exactly how
mined by the right-hand rule, and whose length is a physicists define a b. 7
Corollary Two nonzero vectors a and b are parallel if and only if
a b 0 Proof Two nonzero vectors a and b are parallel if and only if 0 or . In either case
sin 0, so a b 0 and therefore a b 0.
b
兩 b兩 sin ¨
¨ FIGURE 2
The geometric interpretation of Theorem 6 can be seen by looking at Figure 2. If a and b are represented by directed line segments with the same initial point, then they determine a parallelogram with base a , altitude b sin , and area
( b sin ) a b
A a
a
Thus, we have the following way of interpreting the magnitude of a cross product. The length of the cross product a b is equal to the area of the parallelogram determined by a and b. EXAMPLE 3 Find a vector perpendicular to the plane that passes through the points
P 1, 4, 6 , Q 2, 5, 1 , and R 1, 1, 1 . l l l l SOLUTION The vector PQ PR is perpendicular to both PQ and PR and is therefore perpendicular to the plane through P, Q, and R. We know from (12.2.1) that l PQ 2 1 i 5 4 j 1 6 k 3i j 7k l PR 1 1 i 1 4 j 1 6 k 5 j 5k We compute the cross product of these vectors:
i j k l l PQ PR 3 1 7 0 5 5
5 35 i 15 0 j 15 0 k 40 i 15 j 15k So the vector 40, 15, 15 is perpendicular to the given plane. Any nonzero scalar multiple of this vector, such as 8, 3, 3 , is also perpendicular to the plane. EXAMPLE 4 Find the area of the triangle with vertices P 1, 4, 6 , Q 2, 5, 1 , and R 1, 1, 1 . l l SOLUTION In Example 3 we computed that PQ PR 40, 15, 15 . The area of the parallelogram with adjacent sides PQ and PR is the length of this cross product:
l l PR s 40 PQ
2
15 2 15 2 5s82
The area A of the triangle PQR is half the area of this parallelogram, that is, 52 s82.
818
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
If we apply Theorems 5 and 6 to the standard basis vectors i , j, and k using 2, we obtain i j k
j k i
k i j
j i k
k j i
i k j
Observe that i j j i Thus, the cross product is not commutative. Also i i j i k j whereas i i j 0 j 0 So the associative law for multiplication does not usually hold; that is, in general, a b c a b c However, some of the usual laws of algebra do hold for cross products. The following theorem summarizes the properties of vector products. 8
Theorem If a, b, and c are vectors and c is a scalar, then
1. a b b a 2. (ca) b c(a b) a (cb) 3. a (b c) a b a c 4. (a b) c a c b c 5. a b c a b c 6. a b c a c b a b c
These properties can be proved by writing the vectors in terms of their components and using the definition of a cross product. We give the proof of Property 5 and leave the remaining proofs as exercises. Proof of Property 5 If a a 1, a 2 , a 3 , b b1, b2 , b3 , and c c1, c2 , c3 , then 9
a b c a 1 b2 c3 b3 c2 a 2 b3 c1 b1 c3 a 3 b1 c2 b2 c1 a 1 b2 c3 a 1 b3 c2 a 2 b3 c1 a 2 b1 c3 a 3 b1 c2 a 3 b2 c1 a 2 b3 a 3 b2 c1 a 3 b1 a 1 b3 c2 a 1 b2 a 2 b1 c3 a b c
The product a b c that occurs in Property 5 is called the scalar triple product of the vectors a, b, and c. Notice from Equation 9 that we can write the scalar triple product as a determinant: 10
a1 a b c b1 c1
a2 b2 c2
a3 b3 c3
SECTION 12.4 THE CROSS PRODUCT
❙❙❙❙
819
The geometric significance of the scalar triple product can be seen by considering the parallelepiped determined by the vectors a, b, and c (Figure 3). The area of the base parallelogram is A b c . If is the angle between a and b c, then the height h of the parallelepiped is h a cos . (We must use cos instead of cos in case 2.) Therefore, the volume of the parallelepiped is
bxc
h ¨ a c
b
V Ah b c
FIGURE 3
a cos a b c
Thus, we have proved the following formula. 11 The volume of the parallelepiped determined by the vectors a, b, and c is the magnitude of their scalar triple product:
V a b c
If we use the formula in (11) and discover that the volume of the parallelepiped determined by a, b, and c is 0, then the vectors must lie in the same plane; that is, they are coplanar. EXAMPLE 5 Use the scalar triple product to show that the vectors a 1, 4, 7 ,
b 2, 1, 4 , and c 0, 9, 18 are coplanar.
SOLUTION We use Equation 10 to compute their scalar triple product:
1 a b c 2 0 1
4 1 9
1 9
7 4 18
4 2 4 18 0
4 2 7 18 0
1 9
1 18 4 36 7 18 0 Therefore, by (11) the volume of the parallelepiped determined by a, b, and c is 0. This means that a, b, and c are coplanar.
r ¨
F
FIGURE 4
The idea of a cross product occurs often in physics. In particular, we consider a force F acting on a rigid body at a point given by a position vector r. (For instance, if we tighten a bolt by applying a force to a wrench as in Figure 4, we produce a turning effect.) The torque (relative to the origin) is defined to be the cross product of the position and force vectors
r F and measures the tendency of the body to rotate about the origin. The direction of the torque vector indicates the axis of rotation. According to Theorem 6, the magnitude of the torque vector is
r F r F sin where is the angle between the position and force vectors. Observe that the only component of F that can cause a rotation is the one perpendicular to r, that is, F sin . The magnitude of the torque is equal to the area of the parallelogram determined by r and F.
820
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
EXAMPLE 6 A bolt is tightened by applying a 40-N force to a 0.25-m wrench as shown in Figure 5. Find the magnitude of the torque about the center of the bolt. 75° 0.25 m
SOLUTION The magnitude of the torque vector is
r F r F sin 75 0.25 40 sin 75
40 N
10 sin 75 9.66 N m 9.66 J If the bolt is right-threaded, then the torque vector itself is
n 9.66 n
FIGURE 5
where n is a unit vector directed down into the page.
|||| 12.4
Exercises
|||| Find the cross product a b and verify that it is orthogonal to both a and b.
1–7
1. a 1, 2, 0 ,
b 0, 3, 1
2. a 5, 1, 4 ,
b 1, 0, 2
3. a 2 i j k,
z
b j 2k
4. a i j k,
b
b i j k a
5. a 3 i 2 j 4 k,
b i 2 j 3k
6. a i e t j e t k,
b 2 i e t j e t k
7. a t, t 2, t 3 , ■
■
■
x
■
■
■
■
■
■
■
■
■
16. Find two unit vectors orthogonal to both i j k and 2 i k.
why. If so, state whether it is a vector or a scalar. (a) a b c (b) a b c (c) a b c (d) a b c (e) a b c d (f) a b c d
17. Show that 0 a 0 a 0 for any vector a in V3. 18. Show that a b b 0 for all vectors a and b in V3. 19. Prove Property 1 of Theorem 8.
Find u v and determine whether u v is directed into the page or out of the page.
10.
20. Prove Property 2 of Theorem 8. 21. Prove Property 3 of Theorem 8.
11.
22. Prove Property 4 of Theorem 8.
| u |=6
| u |=5
23. Find the area of the parallelogram with vertices A 2, 1 ,
| v |=8
60°
that a b c a b c. 0, 4, 4 .
9. State whether each expression is meaningful. If not, explain
14. If a 3, 1, 2 , b 1, 1, 0 , and c 0, 0, 4 , show 15. Find two unit vectors orthogonal to both 1, 1, 1 and
a b as vectors starting at the origin.
||||
y
13. If a 1, 2, 1 and b 0, 1, 3 , find a b and b a.
b 1, 2t, 3t 2
8. If a i 2 k and b j k, find a b. Sketch a, b, and
10–11
(b) Use the right-hand rule to decide whether the components of a b are positive, negative, or 0.
B 0, 4 , C 4, 2 , and D 2, 1 .
150°
| v |=10
24. Find the area of the parallelogram with vertices K 1, 2, 3 , ■
■
■
■
■
■
■
■
■
■
■
12. The figure shows a vector a in the xy-plane and a vector b in
the direction of k. Their lengths are a 3 and b 2. (a) Find a b .
■
L 1, 3, 6 , M 3, 8, 6 , and N 3, 7, 3 . 25–28 |||| (a) Find a vector orthogonal to the plane through the points P, Q, and R, and (b) find the area of triangle PQR.
25. P 1, 0, 0 ,
Q 0, 2, 0 ,
R 0, 0, 3
SECTION 12.4 THE CROSS PRODUCT
26. P 2, 1, 5 ,
Q 1, 3, 4 , Q 4, 1, 2 ,
28. P 2, 0, 3 ,
Q 3, 1, 0 ,
■
■
■
■
■
R 5, 3, 1
39. (a) Let P be a point not on the line L that passes through the
R 5, 2, 2 ■
■
■
■
■
■
■
29. a 6, 3, 1 ,
b 0, 1, 2 ,
30. a i j k,
b i j k, c i j k
■
■
■
■
■
c 4, 2, 5
■
■
■
■
■
■
31–32
|||| Find the volume of the parallelepiped with adjacent edges PQ, PR, and PS.
31. P 2, 0, 1 , 32. P 0, 1, 2 , ■
■
■
R 3, 1, 1 , S 2, 2, 2
Q 4, 1, 0 , Q 2, 4, 5 , ■
■
■
■
■
■
■
■
33. Use the scalar triple product to verify that the vectors
a 2 i 3 j k, b i j, and c 7 i 3 j 2 k are coplanar. 34. Use the scalar triple product to determine whether the points
P 1, 0, 1 , Q 2, 4, 6 , R 3, 1, 2 , and S 6, 2, 8 lie in the same plane. 35. A bicycle pedal is pushed by a foot with a 60-N force as
shown. The shaft of the pedal is 18 cm long. Find the magnitude of the torque about P.
l l where a QR and b QP. (b) Use the formula in part (a) to find the distance from the point P 1, 1, 1 to the line through Q 0, 6, 8 and R 1, 4, 7 . 40. (a) Let P be a point not on the plane that passes through the
points Q, R, and S. Show that the distance d from P to the plane is a b c d a b
R 1, 0, 1 , S 6, 1, 4 ■
points Q and R. Show that the distance d from the point P to the line L is a b d a
29–30 |||| Find the volume of the parallelepiped determined by the vectors a, b, and c.
■
821
minimum values of the length of the vector u v. In what direction does u v point?
R 3, 0, 6
27. P 0, 2, 0 ,
❙❙❙❙
l l l where a QR, b QS, and c QP. (b) Use the formula in part (a) to find the distance from the point P 2, 1, 4 to the plane through the points Q 1, 0, 0 , R 0, 2, 0 , and S 0, 0, 3 . 41. Prove that a b a b 2 a b . 42. Prove part 6 of Theorem 8, that is,
a b c a c b a b c 43. Use Exercise 42 to prove that
a b c b c a c a b 0 60 N
44. Prove that
70°
a b c d 10°
P
a c b c a d b d
45. Suppose that a 0. 36. Find the magnitude of the torque about P if a 36-lb force is
applied as shown. 4 ft
(a) If a b a c, does it follow that b c? (b) If a b a c, does it follow that b c? (c) If a b a c and a b a c, does it follow that b c?
46. If v1, v2, and v3 are noncoplanar vectors, let
P 4 ft
30° 36 lb 37. A wrench 30 cm long lies along the positive y-axis and grips a
bolt at the origin. A force is applied in the direction 0, 3, 4 at the end of the wrench. Find the magnitude of the force needed to supply 100 J of torque to the bolt. 38. Let v 5j and let u be a vector with length 3 that starts at
the origin and rotates in the xy-plane. Find the maximum and
k1
v2 v3 v1 v2 v3 k3
k2
v3 v1 v1 v2 v3
v1 v2 v1 v2 v3
(These vectors occur in the study of crystallography. Vectors of the form n1 v1 n 2 v2 n3 v3 , where each ni is an integer, form a lattice for a crystal. Vectors written similarly in terms of k1, k2, and k3 form the reciprocal lattice.) (a) Show that k i is perpendicular to vj if i j. (b) Show that k i vi 1 for i 1, 2, 3. 1 (c) Show that k1 k2 k3 . v1 v2 v3
822
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
DISCOVERY PROJECT The Geometry of a Tetrahedron A tetrahedron is a solid with four vertices, P, Q, R, and S, and four triangular faces as shown in the figure.
P
1. Let v1 , v2 , v3 , and v4 be vectors with lengths equal to the areas of the faces opposite the
vertices P, Q, R, and S, respectively, and directions perpendicular to the respective faces and pointing outward. Show that v1 v2 v3 v4 0
S R
Q
2. The volume V of a tetrahedron is one-third the distance from a vertex to the opposite face,
times the area of that face. (a) Find a formula for the volume of a tetrahedron in terms of the coordinates of its vertices P, Q, R, and S. (b) Find the volume of the tetrahedron whose vertices are P 1, 1, 1 , Q 1, 2, 3 , R 1, 1, 2 , and S 3, 1, 2 . 3. Suppose the tetrahedron in the figure has a trirectangular vertex S. (This means that the three
angles at S are all right angles.) Let A, B, and C be the areas of the three faces that meet at S, and let D be the area of the opposite face PQR. Using the result of Problem 1, or otherwise, show that D 2 A2 B 2 C 2 (This is a three-dimensional version of the Pythagorean Theorem.)
|||| 12.5
Equations of Lines and Planes
z
P¸(x¸, y¸, z¸) a P(x, y, z)
L
r¸ O
r
v
x y
FIGURE 1
A line in the xy-plane is determined when a point on the line and the direction of the line (its slope or angle of inclination) are given. The equation of the line can then be written using the point-slope form. Likewise, a line L in three-dimensional space is determined when we know a point P0 x 0 , y0 , z0 on L and the direction of L. In three dimensions the direction of a line is conveniently described by a vector, so we let v be a vector parallel to L. Let P x, y, z be an arbitrary point on L and let r0 and r be the position vectors of P0 and P (that is, they have representations OP A0 and OP A). If a is the vector with representation P A, 0 P as in Figure 1, then the Triangle Law for vector addition gives r r0 a. But, since a and v are parallel vectors, there is a scalar t such that a tv. Thus 1
z
t=0
t>0 L
t<0 r¸
x
FIGURE 2
y
r r0 tv
which is a vector equation of L. Each value of the parameter t gives the position vector r of a point on L. In other words, as t varies, the line is traced out by the tip of the vector r. As Figure 2 indicates, positive values of t correspond to points on L that lie on one side of P0 , whereas negative values of t correspond to points that lie on the other side of P0 . If the vector v that gives the direction of the line L is written in component form as v a, b, c , then we have tv ta, tb, tc . We can also write r x, y, z and r0 x 0 , y0 , z0 , so the vector equation (1) becomes x, y, z x 0 ta, y0 tb, z0 tc
SECTION 12.5 EQUATIONS OF LINES AND PLANES
❙❙❙❙
823
Two vectors are equal if and only if corresponding components are equal. Therefore, we have the three scalar equations: 2
x x 0 at
y y0 bt
z z0 ct
where t . These equations are called parametric equations of the line L through the point P0 x 0 , y0 , z0 and parallel to the vector v a, b, c . Each value of the parameter t gives a point x, y, z on L. |||| Figure 3 shows the line L in Example 1 and its relation to the given point and to the vector that gives its direction.
z
(a) Here r0 5, 1, 3 5i j 3k and v i 4 j 2k, so the vector equation (1) becomes
r¸ v=i+4j-2k
x
(a) Find a vector equation and parametric equations for the line that passes through the point 5, 1, 3 and is parallel to the vector i 4 j 2k. (b) Find two other points on the line. SOLUTION
L (5, 1, 3)
EXAMPLE 1
r 5i j 3k t i 4 j 2k
y
r 5 t i 1 4t j 3 2t k
or
Parametric equations are FIGURE 3
x 5 t
y 1 4t
z 3 2t
(b) Choosing the parameter value t 1 gives x 6, y 5, and z 1, so 6, 5, 1 is a point on the line. Similarly, t 1 gives the point 4, 3, 5 . The vector equation and parametric equations of a line are not unique. If we change the point or the parameter or choose a different parallel vector, then the equations change. For instance, if, instead of 5, 1, 3 , we choose the point 6, 5, 1 in Example 1, then the parametric equations of the line become x 6 t
y 5 4t
z 1 2t
Or, if we stay with the point 5, 1, 3 but choose the parallel vector 2i 8j 4k, we arrive at the equations x 5 2t
y 1 8t
z 3 4t
In general, if a vector v a, b, c is used to describe the direction of a line L, then the numbers a, b, and c are called direction numbers of L. Since any vector parallel to v could also be used, we see that any three numbers proportional to a, b, and c could also be used as a set of direction numbers for L. Another way of describing a line L is to eliminate the parameter t from Equations 2. If none of a, b, or c is 0, we can solve each of these equations for t, equate the results, and obtain 3
x x0 y y0 z z0 a b c
❙❙❙❙
824
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
These equations are called symmetric equations of L . Notice that the numbers a, b, and c that appear in the denominators of Equations 3 are direction numbers of L, that is, components of a vector parallel to L. If one of a, b, or c is 0, we can still eliminate t. For instance, if a 0, we could write the equations of L as y y0 z z0 b c
x x0
This means that L lies in the vertical plane x x 0. |||| Figure 4 shows the line L in Example 2 and the point P where it intersects the xy-plane. z 1
B x
_1
4
y
L
(a) We are not explicitly given a vector parallel to the line, but observe that the vector v l with representation AB is parallel to the line and v 3 2, 1 4, 1 3 1, 5, 4
A
FIGURE 4
(a) Find parametric equations and symmetric equations of the line that passes through the points A 2, 4, 3 and B 3, 1, 1 . (b) At what point does this line intersect the xy-plane? SOLUTION
2
1
P
EXAMPLE 2
Thus, direction numbers are a 1, b 5, and c 4. Taking the point 2, 4, 3 as P0, we see that parametric equations (2) are x 2 t
y 4 5t
z 3 4t
and symmetric equations (3) are x 2 y 4 z 3 1 5 4 (b) The line intersects the xy-plane when z 0, so we put z 0 in the symmetric equations and obtain x 2 y 4 3 1 5 4 This gives x 114 and y 14 , so the line intersects the xy-plane at the point ( 114 , 14 , 0). In general, the procedure of Example 2 shows that direction numbers of the line L through the points P0 x 0 , y0 , z0 and P1 x 1, y1, z1 are x 1 x 0 , y1 y0 , and z1 z0 and so symmetric equations of L are x x0 y y0 z z0 x1 x0 y1 y0 z1 z0 Often, we need a description, not of an entire line, but of just a line segment. How, for instance, could we describe the line segment AB in Example 2? If we put t 0 in the parametric equations in Example 2(a), we get the point 2, 4, 3 and if we put t 1 we get 3, 1, 1 . So the line segment AB is described by the parametric equations x 2 t
y 4 5t
z 3 4t
0 t 1
or by the corresponding vector equation r t 2 t, 4 5t, 3 4t
0 t 1
SECTION 12.5 EQUATIONS OF LINES AND PLANES
❙❙❙❙
825
In general, we know from Equation 1 that the vector equation of a line through the (tip of the) vector r 0 in the direction of a vector v is r r 0 tv. If the line also passes through (the tip of) r1, then we can take v r1 r 0 and so its vector equation is r r 0 t r1 r 0 1 t r 0 tr1 The line segment from r 0 to r1 is given by the parameter interval 0 t 1. 4
The line segment from r 0 to r1 is given by the vector equation r t 1 t r 0 t r1
|||| The lines L 1 and L 2 in Example 3, shown in Figure 5, are skew lines.
EXAMPLE 3 Show that the lines L 1 and L 2 with parametric equations
z
L¡
5
L™
0 t 1
x 1 t
y 2 3t
z 4 t
x 2s
y 3 s
z 3 4s
are skew lines; that is, they do not intersect and are not parallel (and therefore do not lie in the same plane). SOLUTION The lines are not parallel because the corresponding vectors 1, 3, 1 and
5 10
5 x
y
2, 1, 4 are not parallel. (Their components are not proportional.) If L 1 and L 2 had a point of intersection, there would be values of t and s such that 1 t 2s
_5
2 3t 3 s 4 t 3 4s
FIGURE 5
But if we solve the first two equations, we get t 115 and s 85 , and these values don’t satisfy the third equation. Therefore, there are no values of t and s that satisfy the three equations. Thus, L 1 and L 2 do not intersect. Hence, L 1 and L 2 are skew lines.
Planes Although a line in space is determined by a point and a direction, a plane in space is more difficult to describe. A single vector parallel to a plane is not enough to convey the “direction” of the plane, but a vector perpendicular to the plane does completely specify its direction. Thus, a plane in space is determined by a point P0 x 0 , y0 , z0 in the plane and a vector n that is orthogonal to the plane. This orthogonal vector n is called a normal vector. Let P x, y, z be an arbitrary point in the plane, and let r0 and r be the position vectors of P0 and P. Then the vector r r0 is represented by P A. 0 P (See Figure 6.) The normal vector n is orthogonal to every vector in the given plane. In particular, n is orthogonal to r r0 and so we have
z
n P (x, y, z)
r 0
5
r-r¸
n r r0 0
which can be rewritten as
r¸ P¸(x¸, y¸, z¸)
x
6
n r n r0
y
FIGURE 6
Either Equation 5 or Equation 6 is called a vector equation of the plane.
826
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
To obtain a scalar equation for the plane, we write n a, b, c , r x, y, z , and r0 x 0 , y0 , z0 . Then the vector equation (5) becomes a, b, c x x 0 , y y0 , z z0 0 or 7
a x x 0 b y y0 c z z0 0
Equation 7 is the scalar equation of the plane through P0 x 0 , y0 , z0 with normal vector n a, b, c . EXAMPLE 4 Find an equation of the plane through the point 2, 4, 1 with normal vector
n 2, 3, 4 . Find the intercepts and sketch the plane.
SOLUTION Putting a 2, b 3, c 4, x 0 2, y0 4, and z0 1 in Equation 7, we see that an equation of the plane is
z (0, 0, 3)
2 x 2 3 y 4 4 z 1 0 (0, 4, 0) (6, 0, 0)
2x 3y 4z 12
or y
x
FIGURE 7
To find the x-intercept we set y z 0 in this equation and obtain x 6. Similarly, the y-intercept is 4 and the z-intercept is 3. This enables us to sketch the portion of the plane that lies in the first octant (see Figure 7). By collecting terms in Equation 7 as we did in Example 4, we can rewrite the equation of a plane as ax by cz d 0
8
where d ax 0 by0 cz0 . Equation 8 is called a linear equation in x, y, and z. Conversely, it can be shown that if a, b, and c are not all 0, then the linear equation (8) represents a plane with normal vector a, b, c . (See Exercise 73.) |||| Figure 8 shows the portion of the plane in Example 5 that is enclosed by triangle PQR.
EXAMPLE 5 Find an equation of the plane that passes through the points P 1, 3, 2 ,
Q 3, 1, 6 , and R 5, 2, 0 . l
Q(3, -1, 6)
a 2, 4, 4 P(1, 3, 2)
b 4, 1, 2
Since both a and b lie in the plane, their cross product a b is orthogonal to the plane and can be taken as the normal vector. Thus
y
i j k n a b 2 4 4 12 i 20 j 14 k 4 1 2
x
R(5, 2, 0) FIGURE 8
l
SOLUTION The vectors a and b corresponding to PQ and PR are
z
With the point P 1, 3, 2 and the normal vector n, an equation of the plane is 12 x 1 20 y 3 14 z 2 0 or
6x 10y 7z 50
SECTION 12.5 EQUATIONS OF LINES AND PLANES
❙❙❙❙
827
EXAMPLE 6 Find the point at which the line with parametric equations x 2 3t,
y 4t, z 5 t intersects the plane 4x 5y 2z 18.
SOLUTION We substitute the expressions for x, y, and z from the parametric equations into the equation of the plane:
4 2 3t 5 4t 2 5 t 18 This simplifies to 10t 20, so t 2. Therefore, the point of intersection occurs when the parameter value is t 2. Then x 2 3 2 4, y 4 2 8, z 5 2 3 and so the point of intersection is 4, 8, 3 . n™ ¨ n¡
Two planes are parallel if their normal vectors are parallel. For instance, the planes x 2y 3z 4 and 2x 4y 6z 3 are parallel because their normal vectors are n1 1, 2, 3 and n 2 2, 4, 6 and n 2 2n1 . If two planes are not parallel, then they intersect in a straight line and the angle between the two planes is defined as the acute angle between their normal vectors (see angle in Figure 9).
¨ FIGURE 9
EXAMPLE 7 |||| Figure 10 shows the planes in Example 7 and their line of intersection L.
x+y+z=1
x-2y+3z=1
(a) Find the angle between the planes x y z 1 and x 2y 3z 1. (b) Find symmetric equations for the line of intersection L of these two planes. SOLUTION
(a) The normal vectors of these planes are 6 4 2 z 0 _2 _4
n1 1, 1, 1
L
n 2 1, 2, 3
and so, if is the angle between the planes, Corollary 12.3.6 gives cos _2
0 y
2
2
0 x
n1 n 2 1 1 1 2 1 3 2 n1 n 2 s1 1 1 s1 4 9 s42
_2
cos 1
FIGURE 10
2 s42
72
(b) We first need to find a point on L. For instance, we can find the point where the line intersects the xy-plane by setting z 0 in the equations of both planes. This gives the equations x y 1 and x 2y 1, whose solution is x 1, y 0. So the point 1, 0, 0 lies on L. Now we observe that, since L lies in both planes, it is perpendicular to both of the normal vectors. Thus, a vector v parallel to L is given by the cross product |||| Another way to find the line of intersection is to solve the equations of the planes for two of the variables in terms of the third, which can be taken as the parameter.
v n1 n 2
i j k 1 1 1 5i 2 j 3 k 1 2 3
and so the symmetric equations of L can be written as x 1 y z 5 2 3 NOTE Since a linear equation in x, y, and z represents a plane and two nonparallel planes intersect in a line, it follows that two linear equations can represent a line. The points x, y, z that satisfy both a 1 x b1 y c1 z d1 0 and a 2 x b2 y c2 z d2 0 ■
828
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
L
y x-1 = _2 5
2 1 z 0 y _1
2
lie on both of these planes, and so the pair of linear equations represents the line of intersection of the planes (if they are not parallel). For instance, in Example 7 the line L was given as the line of intersection of the planes x y z 1 and x 2y 3z 1. The symmetric equations that we found for L could be written as
z
x 1 y 5 2
=3
and
y z 2 3
_2 _1 y
0
1
2
_2 0 _1 x
1
which is again a pair of linear equations. They exhibit L as the line of intersection of the planes x 1 5 y 2 and y 2 z 3 . (See Figure 11.) In general, when we write the equations of a line in the symmetric form
FIGURE 11 |||| Figure 11 shows how the line L in Example 7 can also be regarded as the line of intersection of planes derived from its symmetric equations.
x x0 y y0 z z0 a b c we can regard the line as the line of intersection of the two planes x x0 y y0 a b
and
y y0 z z0 b c
EXAMPLE 8 Find a formula for the distance D from a point P1 x 1, y1, z1 to the plane
ax by cz d 0.
SOLUTION Let P0 x 0 , y0 , z0 be any point in the given plane and let b be the vector corre-
sponding to PA. 0 P1 Then b x 1 x 0 , y1 y0 , z1 z0 From Figure 12 you can see that the distance D from P1 to the plane is equal to the absolute value of the scalar projection of b onto the normal vector n a, b, c . (See Section 12.3.) Thus
P¡ ¨ b
D
n b n
D compn b
n
P¸
a x
x0 b y1 y0 c z1 z0 sa 2 b 2 c 2
ax
by1 cz1 ax0 by0 cz0 sa 2 b 2 c 2
FIGURE 12
1
1
Since P0 lies in the plane, its coordinates satisfy the equation of the plane and so we have ax 0 by0 cz0 d 0. Thus, the formula for D can be written as
9
D
ax
by1 cz1 d sa 2 b 2 c 2
1
EXAMPLE 9 Find the distance between the parallel planes 10x 2y 2z 5 and 5x y z 1. SOLUTION First we note that the planes are parallel because their normal vectors
10, 2, 2 and 5, 1, 1 are parallel. To find the distance D between the planes, we choose any point on one plane and calculate its distance to the other plane. In particular, if we put y z 0 in the equation of the first plane, we get 10x 5 and so
❙❙❙❙
SECTION 12.5 EQUATIONS OF LINES AND PLANES
829
( 12 , 0, 0) is a point in this plane. By Formula 9, the distance between ( 12 , 0, 0) and the plane 5x y z 1 0 is D
5( ) 1 0 1 0 1 1 2
s5 1 1 2
2
2
3 2
3s3
s3 6
So the distance between the planes is s3 6. EXAMPLE 10 In Example 3 we showed that the lines
L1: x 1 t
y 2 3t
z 4 t
L 2 : x 2s
y 3 s
z 3 4s
are skew. Find the distance between them. SOLUTION Since the two lines L 1 and L 2 are skew, they can be viewed as lying on two parallel planes P1 and P2 . The distance between L 1 and L 2 is the same as the distance between P1 and P2 , which can be computed as in Example 9. The common normal vector to both planes must be orthogonal to both v1 1, 3, 1 (the direction of L 1 ) and v2 2, 1, 4 (the direction of L 2 ). So a normal vector is
n v1 v2
i j k 1 3 1 13i 6 j 5k 2 1 4
If we put s 0 in the equations of L 2 , we get the point 0, 3, 3 on L 2 and so an equation for P2 is 13 x 0 6 y 3 5 z 3 0
13x 6y 5z 3 0
or
If we now set t 0 in the equations for L 1 , we get the point 1, 2, 4 on P1 . So the distance between L 1 and L 2 is the same as the distance from 1, 2, 4 to 13x 6y 5z 3 0. By Formula 9, this distance is D
|||| 12.5
s13 2 6 2 5 2
8 0.53 s230
Exercises
1. Determine whether each statement is true or false.
(a) (b) (c) (d) (e) (f) (g) (h) (i) ( j) (k)
13 1 6 2 5 4 3
Two lines parallel to a third line are parallel. Two lines perpendicular to a third line are parallel. Two planes parallel to a third plane are parallel. Two planes perpendicular to a third plane are parallel. Two lines parallel to a plane are parallel. Two lines perpendicular to a plane are parallel. Two planes parallel to a line are parallel. Two planes perpendicular to a line are parallel. Two planes either intersect or are parallel. Two lines either intersect or are parallel. A plane and a line either intersect or are parallel.
2–5
|||| Find a vector equation and parametric equations for the line.
2. The line through the point 1, 0, 3 and parallel to the
vector 2 i 4 j 5 k
3. The line through the point 2, 4, 10 and parallel to the
vector 3, 1, 8
4. The line through the origin and parallel to the line x 2t,
y 1 t, z 4 3t
5. The line through the point (1, 0, 6) and perpendicular to the
plane x 3y z 5 ■
■
■
■
■
■
■
■
■
■
■
■
❙❙❙❙
830
6–12
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
||||
Find parametric equations and symmetric equations for the
line.
23–38
24. The plane through the point 4, 0, 3 and with normal
7. The line through the points 1, 3, 2 and 4, 3, 0
vector j 2 k
8. The line through the points 6, 1, 3 and 2, 4, 5
25. The plane through the point 1, 1, 1 and with normal vector
i j k
9. The line through the points (0, 2 , 1) and 2, 1, 3 1
26. The plane through the point 2, 8, 10 and perpendicular to
10. The line through 2, 1, 0 and perpendicular to both i j
the line x 1 t, y 2t, z 4 3t
and j k
27. The plane through the origin and parallel to the plane
11. The line through 1, 1, 1 and parallel to the line
2x y 3z 1
x 2 y z 3 1 2
28. The plane through the point 1, 6, 5 and parallel to the
12. The line of intersection of the planes x y z 1
plane x y z 2 0
and x z 0 ■
■
29. The plane through the point 4, 2, 3 and parallel to the plane
■
■
■
■
■
■
■
■
3x 7z 12
■
13. Is the line through 4, 6, 1 and 2, 0 3 parallel to the
30. The plane that contains the line x 3 2t, y t, z 8 t
line through 10, 18, 4 and 5, 3, 14 ?
and is parallel to the plane 2x 4y 8z 17
31. The plane through the points 0, 1, 1 , 1, 0, 1 , and 1, 1, 0
14. Is the line through 4, 1, 1 and 2, 5, 3 perpendicular to the
line through 3, 2, 0 and 5, 1, 4 ?
32. The plane through the origin and the points 2, 4, 6
and 5, 1, 3
15. (a) Find symmetric equations for the line that passes through
33. The plane through the points 3, 1, 2 , 8, 2, 4 , and
the point 0, 2, 1 and is parallel to the line with parametric equations x 1 2t, y 3t, z 5 7t. (b) Find the points in which the required line in part (a) intersects the coordinate planes.
1, 2, 3
34. The plane that passes through the point 1, 2, 3 and contains
the line x 3t, y 1 t, z 2 t
16. (a) Find parametric equations for the line through 5, 1, 0 that
is perpendicular to the plane 2x y z 1. (b) In what points does this line intersect the coordinate planes?
35. The plane that passes through the point 6, 0, 2 and contains
the line x 4 2t, y 3 5t, z 7 4 t
36. The plane that passes through the point 1, 1, 1 and
contains the line with symmetric equations x 2y 3z
17. Find a vector equation for the line segment from 2, 1, 4
37. The plane that passes through the point 1, 2, 1 and contains
to 4, 6, 1 .
the line of intersection of the planes x y z 2 and 2x y 3z 1
18. Find parametric equations for the line segment from 10, 3, 1
to 5, 6, 3 .
19–22
38. The plane that passes through the line of intersection of the
planes x z 1 and y 2z 3 and is perpendicular to the plane x y 2z 1
Determine whether the lines L 1 and L 2 are parallel, skew, or intersecting. If they intersect, find the point of intersection. ||||
■
19. L 1: x 6t,
y 1 9t,
L 2: x 1 2s, 20. L 1: x 1 2t,
L 2: x 1 s, 21. L 1:
z 3t
y 4 3s, y 3t,
x y 1 z 2 , 1 2 3
L 2: ■
■
■
■
■
■
■
■
■
■
■
■
■
■
Find the point at which the line intersects the given plane.
41. x y 1 2z ;
x 3 y 2 z 1 4 3 2
x y 2z 9
■
■
x 2y z 1 0
4x y 3z 8 ■
■
■
■
■
■
■
■
42. Where does the line through 1, 0, 1 and 4, 2, 2 intersect
the plane x y z 6 ?
43. Find direction numbers for the line of intersection of the planes
x y z 1 and x z 0.
x 2 y 6 z 2 1 1 3 ■
||||
■
40. x 1 2t, y 4t, z 2 3t ;
z 1 3s
L 2:
■
39. x 3 t, y 2 t, z 5t ;
z 2 t
y 4 s,
■
39–41
z s
x 1 y 3 z 2 22. L 1: 2 2 1
■
Find an equation of the plane.
vector 2, 1, 5
6. The line through the origin and the point 1, 2, 3
■
||||
23. The plane through the point 6, 3, 2 and perpendicular to the
44. Find the cosine of the angle between the planes x y z 0 ■
■
■
■
■
■
and x 2y 3z 1.
SECTION 12.5 EQUATIONS OF LINES AND PLANES
45–50 |||| Determine whether the planes are parallel, perpendicular, or neither. If neither, find the angle between them.
45. x 4y 3z 1, 46. 2z 4y x,
49. x 4y 2z,
■
51–52
■
L 4 : r 2, 1, 3 t 2, 2, 10
x 6y 4z 3
63–64 |||| Use the formula in Exercise 39 in Section 12.4 to find the distance from the point to the given line.
2x y 2z 1 ■
■
■
■
■
■
■
■
3x 4y 5z 6
52. x 2y z 1,
2x y z 1
■
■
■
■
■
■
■
63. 1, 2, 3 ; ■
■
65–66
■
■
■
■
■
53–54 |||| Find parametric equations for the line of intersection of the planes.
■
■
■
■
■
■
■
||||
■
■
■
■
■
■
55. Find an equation for the plane consisting of all points that are
■
■
■
■
and z-intercept c. 58. (a) Find the point at which the given lines intersect:
r 1, 1, 0 t 1, 1, 2 and
r 2, 0, 2 s 1, 1, 0
(b) Find an equation of the plane that contains these lines. 59. Find parametric equations for the line through the point
0, 1, 2 that is parallel to the plane x y z 2 and perpendicular to the line x 1 t, y 1 t, z 2t. 60. Find parametric equations for the line through the point
0, 1, 2 that is perpendicular to the line x 1 t, y 1 t, z 2t and intersects this line. 61. Which of the following four planes are parallel? Are any of
them identical? P1: 4 x 2y 6z 3
P2: 4 x 2y 2z 6
P3: 6 x 3y 9z 5
P4: z 2 x y 3
■
■
■
■
■
■
■
■
■
■
■
Find the distance between the given parallel planes. 3x 6y 3z 4
■
■
x 2y 3z 1 ■
■
■
■
■
■
■
69. Show that the distance between the parallel planes
ax by cz d1 0 and ax by cz d2 0 is D
56. Find an equation for the plane consisting of all points that are 57. Find an equation of the plane with x-intercept a, y-intercept b,
■
4x 6y z 5
equidistant from the points 1, 1, 0 and 0, 1, 1 . equidistant from the points 4, 2, 1 and 2, 4, 3 .
z 1 2t
■
x 2y 2z 1
68. 3x 6y 9z 4, ■
y 3t, ■
z 5t
Find the distance from the point to the given plane.
66. 3, 2, 7 , ■
y 2 3t,
x 5 t,
67. z x 2y 1,
x 3y z 2 0 ■
||||
67–68
2 x 5y z 1
54. 2 x 5z 3 0,
■
65. 2, 8, 5 , ■
53. z x y,
x 2 t,
64. 1, 0, 1 ;
(a) Find symmetric equations for the line of intersection of the planes and (b) find the angle between the planes. ||||
51. x y z 2,
z 2 5t
L 3 : x 1 t, y 4 t, z 1 t
8y 1 2 x 4z
■
y t,
L2: x 1 y 2 1 z
x y z 1
50. x 2y 2z 1, ■
identical? L 1 : x 1 t,
3x 6y 7z 0
48. 2x 3y 4z 5,
831
62. Which of the following four lines are parallel? Are any of them
3x 12y 6z 1
47. x y z 1,
❙❙❙❙
d1 d2 sa 2 b 2 c 2
70. Find equations of the planes that are parallel to the plane
x 2y 2z 1 and two units away from it. 71. Show that the lines with symmetric equations x y z and
x 1 y 2 z 3 are skew, and find the distance between these lines.
72. Find the distance between the skew lines with parametric
equations x 1 t, y 1 6t, z 2t, and x 1 2s, y 5 15s, z 2 6s. 73. If a, b, and c are not all 0, show that the equation
ax by cz d 0 represents a plane and a, b, c is a normal vector to the plane. Hint: Suppose a 0 and rewrite the equation in the form
a x
d a
b y 0 c z 0 0
74. Give a geometric description of each family of planes.
(a) x y z c (b) x y cz 1 (c) y cos z sin 1
■
832
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
LABORATORY PROJECT Putting 3D in Perspective Computer graphics programmers face the same challenge as the great painters of the past: how to represent a three-dimensional scene as a flat image on a two-dimensional plane (a screen or a canvas). To create the illusion of perspective, in which closer objects appear larger than those farther away, three-dimensional objects in the computer’s memory are projected onto a rectangular screen window from a viewpoint where the eye, or camera, is located. The viewing volume–t– he portion of space that will be visible–i– s the region contained by the four planes that pass through the viewpoint and edge of the screen window. If objects in the scene extend beyond these four planes, they must be truncated before pixel data are sent to the screen. These planes are therefore called clipping planes. 1. Suppose the screen is represented by a rectangle in the yz-plane with vertices 0, 400, 0
and 0, 400, 600 , and the camera is placed at 1000, 0, 0 . A line L in the scene passes through the points 230, 285, 102 and 860, 105, 264 . At what points should L be clipped by the clipping planes?
2. If the clipped line segment is projected on the screen window, identify the resulting line
segment. 3. Use parametric equations to plot the edges of the screen window, the clipped line segment,
and its projection on the screen window. Then add sight lines connecting the viewpoint to each end of the clipped segments to verify that the projection is correct. 4. A rectangle with vertices 621, 147, 206 , 563, 31, 242 , 657, 111, 86 , and
599, 67, 122 is added to the scene. The line L intersects this rectangle. To make the rectangle appear opaque, a programmer can use hidden line rendering which removes portions of objects that are behind other objects. Identify the portion of L that should be removed.
|||| 12.6
Cylinders and Quadric Surfaces We have already looked at two special types of surfaces — planes (in Section 12.5) and spheres (in Section 12.1). Here we investigate two other types of surfaces — cylinders and quadric surfaces. In order to sketch the graph of a surface, it is useful to determine the curves of intersection of the surface with planes parallel to the coordinate planes. These curves are called traces (or cross-sections) of the surface.
Cylinders A cylinder is a surface that consists of all lines (called rulings) that are parallel to a given line and pass through a given plane curve. EXAMPLE 1 Sketch the graph of the surface z x 2. SOLUTION Notice that the equation of the graph, z x 2, doesn’t involve y. This means that
any vertical plane with equation y k (parallel to the xz-plane) intersects the graph in a curve with equation z x 2. So these vertical traces are parabolas. Figure 1 shows how the graph is formed by taking the parabola z x 2 in the xz-plane and moving it in the
SECTION 12.6 CYLINDERS AND QUADRIC SURFACES
❙❙❙❙
833
direction of the y-axis. The graph is a surface, called a parabolic cylinder, made up of infinitely many shifted copies of the same parabola. Here the rulings of the cylinder are parallel to the y-axis. z
FIGURE 1
0
The surface z=≈ is a parabolic cylinder.
y
x
We noticed that the variable y is missing from the equation of the cylinder in Example 1. This is typical of a surface whose rulings are parallel to one of the coordinate axes. If one of the variables x, y, or z is missing from the equation of a surface, then the surface is a cylinder. EXAMPLE 2 Identify and sketch the surfaces. (a) x 2 y 2 1 (b) y 2 z 2 1 SOLUTION
(a) Since z is missing and the equations x 2 y 2 1, z k represent a circle with radius 1 in the plane z k, the surface x 2 y 2 1 is a circular cylinder whose axis is the z-axis (see Figure 2). Here the rulings are vertical lines. (b) In this case x is missing and the surface is a circular cylinder whose axis is the x-axis (see Figure 3). It is obtained by taking the circle y 2 z 2 1, x 0 in the yz-plane and moving it parallel to the x-axis. z
z
y 0
x y
x
FIGURE 2 ≈+¥=1
|
NOTE
FIGURE 3 ¥+z@=1
When you are dealing with surfaces, it is important to recognize that an equation like x y 2 1 represents a cylinder and not a circle. The trace of the cylinder x 2 y 2 1 in the xy-plane is the circle with equations x 2 y 2 1, z 0. ■
2
Quadric Surfaces A quadric surface is the graph of a second-degree equation in three variables x, y, and z. The most general such equation is Ax 2 By 2 Cz 2 Dxy Eyz Fxz Gx Hy Iz J 0
834
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
where A, B, C, . . . , J are constants, but by translation and rotation it can be brought into one of the two standard forms Ax 2 By 2 Cz 2 J 0
or
Ax 2 By 2 Iz 0
Quadric surfaces are the counterparts in three dimensions of the conic sections in the plane. (See Section 10.5 for a review of conic sections.) EXAMPLE 3 Use traces to sketch the quadric surface with equation
x2
y2 z2 1 9 4
SOLUTION By substituting z 0, we find that the trace in the xy-plane is x 2 y 2 9 1,
which we recognize as an equation of an ellipse. In general, the horizontal trace in the plane z k is x2
(0, 0, 2)
0
z k
which is an ellipse, provided that k 2 4, that is, 2 k 2. Similarly, the vertical traces are also ellipses:
z
(1, 0, 0)
y2 k2 1 9 4
(0, 3, 0) y
y2 z2 1 k2 9 4
x k
if 1 k 1
z2 k2 1 4 9
y k
if 3 k 3
x2
x
FIGURE 4
The ellipsoid ≈+
z@ y@ + =1 4 9
Figure 4 shows how drawing some traces indicates the shape of the surface. It’s called an ellipsoid because all of its traces are ellipses. Notice that it is symmetric with respect to each coordinate plane; this is a reflection of the fact that its equation involves only even powers of x, y, and z. EXAMPLE 4 Use traces to sketch the surface z 4x 2 y 2. SOLUTION If we put x 0, we get z y 2, so the yz-plane intersects the surface in a
parabola. If we put x k (a constant), we get z y 2 4k 2. This means that if we slice the graph with any plane parallel to the yz-plane, we obtain a parabola that opens upward. Similarly, if y k, the trace is z 4x 2 k 2, which is again a parabola that opens upward. If we put z k, we get the horizontal traces 4x 2 y 2 k, which we recognize as a family of ellipses. Knowing the shapes of the traces, we can sketch the graph in Figure 5. Because of the elliptical and parabolic traces, the quadric surface z 4x 2 y 2 is called an elliptic paraboloid. z
FIGURE 5 The surface z=4≈+¥ is an elliptic paraboloid. Horizontal traces are ellipses; vertical traces are parabolas.
0 x
y
❙❙❙❙
SECTION 12.6 CYLINDERS AND QUADRIC SURFACES
835
EXAMPLE 5 Sketch the surface z y 2 x 2. SOLUTION The traces in the vertical planes x k are the parabolas z y 2 k 2, which
open upward. The traces in y k are the parabolas z x 2 k 2, which open downward. The horizontal traces are y 2 x 2 k, a family of hyperbolas. We draw the families of traces in Figure 6, and we show how the traces appear when placed in their correct planes in Figure 7. z
z
y
1
2 0
_1
1
_1 0
1
FIGURE 6
Vertical traces are parabolas; horizontal traces are hyperbolas. All traces are labeled with the value of k.
x
y
x
0
2
1
Traces in x=k are z=¥-k @
Traces in y=k are z=_≈+k @
Traces in z=k are ¥-≈=k
z
z
z
1
0 y
0 FIGURE 7
x
_1
x
_1
1
Traces moved to their correct planes
Traces in x=k
In Module 12.6A you can investigate how traces determine the shape of a surface.
y
y x
_1 0
1
Traces in y=k
Traces in z=k
In Figure 8 we fit together the traces from Figure 7 to form the surface z y 2 x 2, a hyperbolic paraboloid. Notice that the shape of the surface near the origin resembles that of a saddle. This surface will be investigated further in Section 14.7 when we discuss saddle points. z
0 y x
FIGURE 8
The surface z=¥-≈ is a hyperbolic paraboloid. EXAMPLE 6 Sketch the surface
x2 z2 y2 1. 4 4
SOLUTION The trace in any horizontal plane z k is the ellipse
x2 k2 y2 1 4 4
z k
836
❙❙❙❙
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
z
but the traces in the xz- and yz-planes are the hyperbolas x2 z2 1 4 4
y2
and
z2 1 4
x 0
This surface is called a hyperboloid of one sheet and is sketched in Figure 9.
(0, 1, 0)
(2, 0, 0)
y 0
y
x
The idea of using traces to draw a surface is employed in three-dimensional graphing software for computers. In most such software, traces in the vertical planes x k and y k are drawn for equally spaced values of k, and parts of the graph are eliminated using hidden line removal. Table 1 shows computer-drawn graphs of the six basic types of quadric surfaces in standard form. All surfaces are symmetric with respect to the z-axis. If a quadric surface is symmetric about a different axis, its equation changes accordingly.
FIGURE 9
TABLE 1 Graphs of quadric surfaces
Surface
Equation x2 y2 z2 1 2 2 a b c2
Ellipsoid z
Surface
Equation x2 y2 z2 2 2 c a b2
Cone z
All traces are ellipses.
Horizontal traces are ellipses.
If a b c, the ellipsoid is a sphere. y
x
x
z x2 y2 2 2 c a b
Elliptic Paraboloid z
z
Horizontal traces are ellipses.
Vertical traces are parabolas.
Vertical traces are hyperbolas. x
y
The axis of symmetry corresponds to the variable whose coefficient is negative.
y
z x2 y2 2 2 c a b
Hyperbolic Paraboloid z
y x
y2 z2 x2 2 2 1 2 a b c
Hyperboloid of One Sheet
Horizontal traces are ellipses. The variable raised to the first power indicates the axis of the paraboloid. x
y
Vertical traces in the planes x k and y k are hyperbolas if k 0 but are pairs of lines if k 0.
Hyperboloid of Two Sheets z
y2 z2 x2 1 2 2 a b c2
Horizontal traces are hyperbolas.
Horizontal traces in z k are ellipses if k c or k c.
Vertical traces are parabolas.
Vertical traces are hyperbolas.
The case where c 0 is illustrated.
x
y
The two minus signs indicate two sheets.
SECTION 12.6 CYLINDERS AND QUADRIC SURFACES
In Module 12.6B you can see how changing a, b, and c in Table 1 affects the shape of the quadric surface.
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EXAMPLE 7 Identify and sketch the surface 4x 2 y 2 2z 2 4 0. SOLUTION Dividing by 4, we first put the equation in standard form:
x 2
y2 z2 1 4 2
Comparing this equation with Table 1, we see that it represents a hyperboloid of two sheets, the only difference being that in this case the axis of the hyperboloid is the y-axis. The traces in the xy- and yz-planes are the hyperbolas x 2 z
y2 1 4
y2 z2 1 4 2
and
x 0
The surface has no trace in the xz-plane, but traces in the vertical planes y k for k 2 are the ellipses z2 k2 x2 1 y k 2 4 which can be written as
(0, _2, 0) 0 y (0, 2, 0)
x
z 0
x2 k2 1 4
FIGURE 10
4≈-¥+2z@+4=0
z2
k2 2 1 4
1
y k
These traces are used to make the sketch in Figure 10.
z
EXAMPLE 8 Classify the quadric surface x 2 2z 2 6x y 10 0. SOLUTION By completing the square we rewrite the equation as
0
y 1 x 3 2 2z 2 y
Comparing this equation with Table 1, we see that it represents an elliptic paraboloid. Here, however, the axis of the paraboloid is parallel to the y-axis, and it has been shifted so that its vertex is the point 3, 1, 0 . The traces in the plane y k k 1 are the ellipses x 3 2 2z 2 k 1 y k
(3, 1, 0)
x
FIGURE 11
≈+2z@-6x-y+10=0
|||| 12.6
The trace in the xy-plane is the parabola with equation y 1 x 3 2, z 0. The paraboloid is sketched in Figure 11.
Exercises
1. (a) What does the equation y x 2 represent as a curve in 2 ?
(b) What does it represent as a surface in ? (c) What does the equation z y 2 represent? 3
2. (a) Sketch the graph of y e x as a curve in 2.
(b) Sketch the graph of y e x as a surface in 3. (c) Describe and sketch the surface z e y.
3–8
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Describe and sketch the surface.
3. y 4z 2 4 2
4. z 4 x 2
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5. x y 2 0
6. yz 4
7. z cos x
8. x 2 y 2 1
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9. (a) Find and identify the traces of the quadric surface
x 2 y 2 z 2 1 and explain why the graph looks like the graph of the hyperboloid of one sheet in Table 1. (b) If we change the equation in part (a) to x 2 y 2 z 2 1, how is the graph affected? (c) What if we change the equation in part (a) to x 2 y 2 2y z 2 0?
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CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
10. (a) Find and identify the traces of the quadric surface
29–36 |||| Reduce the equation to one of the standard forms, classify the surface, and sketch it.
x 2 y 2 z 2 1 and explain why the graph looks like the graph of the hyperboloid of two sheets in Table 1. (b) If the equation in part (a) is changed to x 2 y 2 z 2 1, what happens to the graph? Sketch the new graph. |||| Find the traces of the given surface in the planes x k, y k, z k. Then identify the surface and sketch it.
11–20
11. 4x 2 9y 2 36z 2 36
12. 4y x 2 z 2
13. y 2 x 2 z 2
14. z x 2 y 2
15. x 2 4y 2 z 2 4
16. 25y 2 z 2 100 4x 2
17. x 2 4z 2 y 0
18. x 2 4y 2 z 2 4
19. y z 2 x 2
20. 16x 2 y 2 4z 2
29. z 2 4x 2 9y 2 36
30. x 2 2y 2 3z 2
31. x 2y 2 3z 2
32. 4x y 2 4z 2 0
33. 4x 2 y 2 4 z 2 4y 24z 36 0 34. 4y 2 z 2 x 16y 4z 20 0 35. x 2 y 2 z 2 4x 2y 2z 4 0 36. x 2 y 2 z 2 2x 2y 4z 2 0 ■
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21–28 |||| Match the equation with its graph (labeled I–VIII). Give reasons for your choices.
21. x 2 4y 2 9z 2 1
22. 9x 2 4y 2 z 2 1
23. x 2 y 2 z 2 1
24. x 2 y 2 z 2 1
25. y 2x 2 z 2
26. y 2 x 2 2z 2
27. x 2 2z 2 1
28. y x 2 z 2
|||| Use a computer with three-dimensional graphing software to graph the surface. Experiment with viewpoints and with domains for the variables until you get a good view of the surface.
37. z 3x 2 5y 2
38. 8x 2 15y 2 5z 2 100
39. z 2 x 2 4y 2
40. z y 2 xy
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41. Sketch the region bounded by the surfaces z sx y 2
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2
and x 2 y 2 1 for 1 z 2.
42. Sketch the region bounded by the paraboloids z x 2 y 2
and z 2 x 2 y 2.
z
I
z
II
43. Find an equation for the surface obtained by rotating the
parabola y x 2 about the y-axis. y
x
44. Find an equation for the surface obtained by rotating the line
y
x
x 3y about the x-axis. 45. Find an equation for the surface consisting of all points that
z
III
are equidistant from the point 1, 0, 0 and the plane x 1. Identify the surface.
z
IV
46. Find an equation for the surface consisting of all points P for y
which the distance from P to the x-axis is twice the distance from P to the yz-plane. Identify the surface.
y
x x z
V
y
y
x
z
VII
z
VI
x
47. Show that if the point a, b, c lies on the hyperbolic parabo-
z
VIII
loid z y 2 x 2, then the lines with parametric equations x a t, y b t, z c 2 b a t and x a t, y b t, z c 2 b a t both lie entirely on this paraboloid. (This shows that the hyperbolic paraboloid is what is called a ruled surface; that is, it can be generated by the motion of a straight line. In fact, this exercise shows that through each point on the hyperbolic paraboloid there are two generating lines. The only other quadric surfaces that are ruled surfaces are cylinders, cones, and hyperboloids of one sheet.)
48. Show that the curve of intersection of the surfaces
x 2 2y 2 z 2 3x 1 and 2 x 2 4y 2 2z 2 5y 0 lies in a plane.
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x
x ■
2 2 2 ; 49. Graph the surfaces z x y and z 1 y on a common
y
y
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screen using the domain x 1.2, y 1.2 and observe the curve of intersection of these surfaces. Show that the projection of this curve onto the xy-plane is an ellipse.