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Para evaluar la norma L2 de un vector,definimos la siguiente funci\[OAcute]n\
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Consideremos dos puntos P1(x1,y1,z1), P2(x2,y2,z2) que pertenecen al plano \
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El vector VU = P2 - P1 pertenece tambi\[EAcute]n al plano \[CapitalPi]1 y \
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Comparando este resultado con CD2 - CD1 que sabemos es igual a 0 por ser CD1 \
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Luego, PE = CD2 - CD1 = 0, lo que significa que el vector VV es perpendicular \
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Mientras que la normal al plano positiva hacia el lado del origen VN (Ver \
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Despejando z en funci\[OAcute]n de x e y en la ecuaci\[OAcute]n del plano, \
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Cell["\<\
Ejercicio 7: Encontrar la ubicaci\[OAcute]n de todos los puntos (x,y,z) tales \
que un vector VA desde el punto P0 = (2,-1,4) al punto P = (x,y,z) sea \
perpendicular al vector VB desde P1 = (2,-1,4) a P2 = (3,3,2)\
\>", "Exercise"],

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El vector gen\[EAcute]rico que representa todos los puntos del conjunto \
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Cell["\[LineSeparator]Y la condici\[OAcute]n de normalidad de VX a VB \
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Cell["\<\
Que es la ecuaci\[OAcute]n de un plano, tal como se muestra en la siguiente \
figura\
\>", "ExerciseText"],

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Cell["\<\
Ejercicio 8: Encuentra el \[AAcute]ngulo entre los dos vectores definidos por \
el origen del sistema de coordenadas hacia puntos PA (4,3,2) y PB (-2,4,3)\
\>", "Exercise"],

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Cell["Y el angulo en grados resulta", "ExerciseText"],

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Cell["\<\
Ejercicio 9: Encuentra el \[AAcute]rea del paralelogramo formado por los dos \
vectores del Ejercicio 8\
\>", "Exercise"],

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Cell["\<\
El \[AAcute]rea del paralelogramo formado por los vectores VA y VB es igual \
al m\[OAcute]dulo del producto vectorial entre ambos vectores, por lo que\
\>", "ExerciseText"],

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Cell["\<\
Ejercicio 10: Encontrar el vector unitario perpendicular a los vectores VA y \
VB del Ejercicio 8\
\>", "Exercise"],

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Cell["\<\
El vector VAB definido en el ejercicio anterior es perpendicular a VA y VB, \
solo resta normalizarlo\
\>", "ExerciseText"],

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Cell["\<\
Ejercicio 11: Encontrar el vector unitario paralelo a la linea definida por x \
- z = y = 2z. Pista: La l\[IAcute]nea esta definida por la \
intersecci\[OAcute]n de dos planos, x - y - z = 0 e y - 2z = 0\
\>", "Exercise"],

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Cell["\<\
Reemplazando por una expresi\[OAcute]n en la otra, resulta la \
ecuaci\[OAcute]n de la linea en funci\[OAcute]n de x\
\>", "ExerciseText"],

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Cell["\<\
La l\[IAcute]nea definida pasa por el origen ya que para x = 0 -> y = z = 0, \
luego todos los vectores posici\[OAcute]n de todos los puntos de la \
l\[IAcute]nea pertenecen a la misma l\[IAcute]nea (s\[OAcute]lo ocurre para l\
\[IAcute]neas que pasan por el origen). En consecuencia, la misma ecuaci\
\[OAcute]n de la linea define el conjunto de vectores contenidos por la \
linea, y puede expresarse como\
\>", "ExerciseText"],

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Cell["\<\
Y el vector unitario correspondiente se obtiene, para un valor arbitrario de \
x = 1, como\
\>", "ExerciseText"],

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Cell["\<\
A continuaci\[OAcute]n el gr\[AAcute]fico muestra los planos y la linea dados \
en el enunciado\
\>", "ExerciseText"],

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Ejercicio 12: Se escogen nuevos ejes Xp, Yp, Zp,de forma que el eje Xp forma \
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Cell[TextData[{
 "Ejercicio 18: Probar la identidad \[Epsilon]- \[Delta], ",
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      SubscriptBox["\[Delta]", "ks"]}], "-", 
     RowBox[{
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      SubscriptBox["\[Delta]", "kr"]}]}]}], TraditionalForm]]],
 "\nPista: Dos de los cuatro indices libres de la expresi\[OAcute]n de la \
izquierda, jkrs, deben ser iguales (por ser i,j,k,r,s = 1..3). Demostrar que \
para (j) = (k) \[OAcute] (r) = (s) los lados izquierdo y derecho son id\
\[EAcute]nticamente nulos (El par\[EAcute]ntesis indica que estamos \
considerando un valor particular de j \[OAcute] k y que su valor permanece \
fijo en la expresi\[OAcute]n). Probar luego que a\[UAcute]n para (j) \
\[NotEqual] (k) y (r) \[NotEqual] (s) ambos t\[EAcute]rminos desaparecen para \
(j) = (r) a menos que (k) = (s). Finalmente probar que \n",
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       RowBox[{"r", " ", "\[NotEqual]", " ", "k"}]}]}]}]}], TraditionalForm]]],
 "\n\[DownQuestion]Que casos falta considerar?"
}], "Exercise"],

Cell[TextData[{
 StyleBox["Prueba te\[OAcute]rica",
  FontVariations->{"Underline"->True}],
 "\nTal como dice el enunciado, en el RHS son cuatro sub\[IAcute]ndices \
((j),(k),(r),(s)) con un valor arbitrario entre 1 y 3, por lo que como m\
\[IAcute]nimo, dos de ellos deben ser iguales para cualquier elecci\[OAcute]n \
posible. Adem\[AAcute]s de estos cuatro sub\[IAcute]ndices, en el LHS = ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", "ijk"], 
    SubscriptBox["\[Epsilon]", "irs"]}], TraditionalForm]]],
 " aparece el sub\[IAcute]ndice i que est\[AAcute] repetido, por lo que para \
todo valor de (j),(k),(r),(s); el valor de LHS = ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", 
     RowBox[{"1", "jk"}]], 
    SubscriptBox["\[Epsilon]", 
     RowBox[{"1", "rs"}]]}], TraditionalForm]]],
 "+",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", 
     RowBox[{"2", "jk"}]], 
    SubscriptBox["\[Epsilon]", 
     RowBox[{"2", "rs"}]]}], TraditionalForm]]],
 "+",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", 
     RowBox[{"3", "jk"}]], 
    SubscriptBox["\[Epsilon]", 
     RowBox[{"3", "rs"}]]}], TraditionalForm]]],
 "(En las expresiones que siguen se omiten los par\[EAcute]ntesis () \
alrededor de los \[IAcute]ndices que indican que estos toman un valor \
particular y no var\[IAcute]an)\nAnalizaremos los posibles valores del LHS y \
en funci\[OAcute]n de los valores que deben tomar los sub\[IAcute]ndices no \
repetidos (j,k,r,s) para cada situaci\[OAcute]n, evaluaremos RHS y \
verificaremos la igualdad LHS == RHS \n\tCASO A)  LHS = 0 por repetici\
\[OAcute]n de sub\[IAcute]ndices: Para que LHS sea = 0, deben repetirse los \
sub\[IAcute]ndices en uno de los dos factores del mismo, es decir\n\t\tCASO \
A.1) LHS = ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", "ijk"], 
    SubscriptBox["\[Epsilon]", "irs"]}], TraditionalForm]]],
 " = ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", "ijj"], 
    SubscriptBox["\[Epsilon]", "irs"]}], TraditionalForm]]],
 "= 0; esto implica que j = k. En esta situaci\[OAcute]n, cada uno de los \
tres sumandos del LHS tiene un factor nulo, por lo que LHS = 0. En cuanto al \
RHS = ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    RowBox[{
     SubscriptBox["\[Delta]", "jr"], 
     SubscriptBox["\[Delta]", "js"]}], "-", 
    RowBox[{
     SubscriptBox["\[Delta]", "js"], 
     SubscriptBox["\[Delta]", "jr"]}]}], TraditionalForm]]],
 " al repetirse el sub\[IAcute]ndice j con un valor particular, resulta RHS = \
 ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    RowBox[{
     SubscriptBox["\[Delta]", "jr"], 
     SubscriptBox["\[Delta]", "js"]}], "-", 
    RowBox[{
     SubscriptBox["\[Delta]", "js"], 
     SubscriptBox["\[Delta]", "jr"]}]}], TraditionalForm]]],
 ", como el orden de los factores no altera el producto y los signos de cada \
uno de estos sumandos son opuestos, RHS = 0 \[ForAll] r, s. Luego, se \
verifica que LHS == RHS\n\t\tCASO A.2) LHS = ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", "ijk"], 
    SubscriptBox["\[Epsilon]", "irs"]}], TraditionalForm]]],
 " = ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", "ijk"], 
    SubscriptBox["\[Epsilon]", "irr"]}], TraditionalForm]]],
 "= 0; esto implica que r = s. En esta situaci\[OAcute]n, cada uno de los \
tres sumandos del LHS tiene un factor nulo, por lo que LHS = 0. \tEn cuanto \
al RHS = ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    RowBox[{
     SubscriptBox["\[Delta]", "jr"], 
     SubscriptBox["\[Delta]", "kr"]}], "-", 
    RowBox[{
     SubscriptBox["\[Delta]", "jr"], 
     SubscriptBox["\[Delta]", "kr"]}]}], TraditionalForm]]],
 " al repetirse el sub\[IAcute]ndice r con un valor particular, resulta RHS = \
",
 Cell[BoxData[
  FormBox[
   RowBox[{
    RowBox[{
     SubscriptBox["\[Delta]", "jr"], 
     SubscriptBox["\[Delta]", "kr"]}], "-", 
    RowBox[{
     SubscriptBox["\[Delta]", "jr"], 
     SubscriptBox["\[Delta]", "kr"]}]}], TraditionalForm]]],
 " ambos sumandos son iguales por lo que RHS = 0 \[ForAll] j, k. Luego, se \
verifica que LHS == RHS\n\tCASO B) LHS = ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", "ijk"], 
    SubscriptBox["\[Epsilon]", "irs"]}], TraditionalForm]]],
 " con sub\[IAcute]ndices repetidos en ambos factores, esto da lugar a los \
siguientes casos \n\t\tCASO B.1) Repetici\[OAcute]n de los sub\[IAcute]ndices \
centrales, LHS = ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", "ijk"], 
    SubscriptBox["\[Epsilon]", "ijs"]}], TraditionalForm]]],
 ", esto implica que r = j,  j \[NotEqual] k y r \[NotEqual] s, ya que los \
casos en los que j = k y r = s fueron analizados en el caso A. En esta \
situaci\[OAcute]n, los tres sumandos del LHS resultan ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", 
     RowBox[{"1", "jk"}]], 
    SubscriptBox["\[Epsilon]", 
     RowBox[{"1", "js"}]]}], TraditionalForm]]],
 "+",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", 
     RowBox[{"2", "jk"}]], 
    SubscriptBox["\[Epsilon]", 
     RowBox[{"2", "js"}]]}], TraditionalForm]]],
 "+",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", 
     RowBox[{"3", "jk"}]], 
    SubscriptBox["\[Epsilon]", 
     RowBox[{"3", "js"}]]}], TraditionalForm]]],
 ". Como j \[NotEqual] k \[NotEqual] s y cada sub\[IAcute]ndice puede tomar \
los valores 1, 2 o 3; se presentan las siguientes situaciones:\n\t\t\tCASO \
B.1.1) k = s, en este caso, en uno de los tres sumandos va a resultar ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", "ijk"], 
    SubscriptBox["\[Epsilon]", "ijk"]}], TraditionalForm]]],
 " con i \[NotEqual] j \[NotEqual] k, mientras que en los restantes sumados \
al menos un sub\[IAcute]ndice va a estar repetido, con lo que el resultado va \
a ser LHS = 1 .  1 o LHS = (-1) . (-1), en ambos casos LHS = 1. En cuanto al \
RHS, resulta RHS = ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    RowBox[{
     SubscriptBox["\[Delta]", "jj"], 
     SubscriptBox["\[Delta]", "kk"]}], "-", 
    RowBox[{
     SubscriptBox["\[Delta]", "jk"], 
     SubscriptBox["\[Delta]", "kj"]}]}], TraditionalForm]]],
 ", el primer sumando es (1) . (1), mientras que el segundo es (0) . (0) por \
ser j \[NotEqual] k, luego RHS = 1. Se verifica que LHS == RHS \n\t\t\tCASO \
B.1.2) k \[NotEqual] s, en este caso, en uno de los tres sumandos el primer \
factor va a resultar ",
 Cell[BoxData[
  FormBox[
   SubscriptBox["\[Epsilon]", "ijk"], TraditionalForm]]],
 " con i \[NotEqual] j \[NotEqual] k por lo que su valor va a ser 1 o -1 \
dependiendo de la permutaci\[OAcute]n, mientras que en los restantes sumandos \
el primer factor debe ser 0 al resultar k = i o k = j, por lo que esos \
sumandos van a ser 0. En el sumando para el cual el primer factor es no nulo, \
el segundo factor resulta ",
 Cell[BoxData[
  FormBox[
   SubscriptBox["\[Epsilon]", "ijs"], TraditionalForm]]],
 " al menos un sub\[IAcute]ndice va a estar necesariamente repetido debido a \
que k \[NotEqual] s, con lo su valor va a ser nulo, y la suma de LHS va a \
tener tres sumandos nulos por lo que LHS = 0. En cuanto al RHS, resulta RHS = \
",
 Cell[BoxData[
  FormBox[
   RowBox[{
    RowBox[{
     SubscriptBox["\[Delta]", "jj"], 
     SubscriptBox["\[Delta]", "ks"]}], "-", 
    RowBox[{
     SubscriptBox["\[Delta]", "js"], 
     SubscriptBox["\[Delta]", "kj"]}]}], TraditionalForm]]],
 ", el primer sumando es (1) . (0) = 0, mientras que el segundo es (x) . (0) \
= 0 por ser j \[NotEqual] k, luego RHS = 0. Se verifica que LHS == RHS \n\t\t\
CASO B.2) Repetici\[OAcute]n de los sub\[IAcute]ndices derechos o finales, \
LHS = ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", "ijk"], 
    SubscriptBox["\[Epsilon]", "irk"]}], TraditionalForm]]],
 ", esto implica que s = k,  nuevamente k \[NotEqual] j y s \[NotEqual] r, ya \
que los casos en los que j = k y r = s fueron analizados en el caso A. En \
esta situaci\[OAcute]n, los tres sumandos del LHS resultan ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", 
     RowBox[{"1", "jk"}]], 
    SubscriptBox["\[Epsilon]", 
     RowBox[{"1", "rk"}]]}], TraditionalForm]]],
 "+",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", 
     RowBox[{"2", "jk"}]], 
    SubscriptBox["\[Epsilon]", 
     RowBox[{"2", "rk"}]]}], TraditionalForm]]],
 "+",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", 
     RowBox[{"3", "jk"}]], 
    SubscriptBox["\[Epsilon]", 
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 ". Como j \[NotEqual] k \[NotEqual] s y cada sub\[IAcute]ndice puede tomar \
los valores 1, 2 o 3; se presentan las siguientes situaciones:\n\t\t\tCASO \
B.2.1) j = r, en este caso, en uno de los tres sumandos va a resultar ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", "ijk"], 
    SubscriptBox["\[Epsilon]", "ijk"]}], TraditionalForm]]],
 " con i \[NotEqual] j \[NotEqual] k, mientras que en los restantes sumados \
al menos un sub\[IAcute]ndice va a estar repetido, con lo que el resultado va \
a ser LHS = 1 .  1 o LHS = (-1) . (-1), en ambos casos LHS = 1. En cuanto al \
RHS, resulta RHS = ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    RowBox[{
     SubscriptBox["\[Delta]", "jj"], 
     SubscriptBox["\[Delta]", "kk"]}], "-", 
    RowBox[{
     SubscriptBox["\[Delta]", "jk"], 
     SubscriptBox["\[Delta]", "kj"]}]}], TraditionalForm]]],
 ", id\[EAcute]ntico al caso B.1.1), luego RHS = 1. Se verifica que LHS == \
RHS \n\t\t\tCASO B.2.2) j \[NotEqual] r, en este caso, en uno de los tres \
sumandos el primer factor va a resultar ",
 Cell[BoxData[
  FormBox[
   SubscriptBox["\[Epsilon]", "ijk"], TraditionalForm]]],
 " con i \[NotEqual] j \[NotEqual] k por lo que su valor va a ser 1 o -1 \
dependiendo de la permutaci\[OAcute]n, mientras que en los restantes sumandos \
el primer factor debe ser 0 al resultar j = i o j = k, por lo que esos \
sumandos van a ser 0. En el sumando para el cual el primer factor es no nulo, \
el segundo factor resulta ",
 Cell[BoxData[
  FormBox[
   SubscriptBox["\[Epsilon]", "irk"], TraditionalForm]]],
 " al menos un sub\[IAcute]ndice va a estar necesariamente repetido debido a \
que j \[NotEqual] r, con lo su valor va a ser nulo, y la suma de LHS va a \
tener tres sumandos nulos por lo que LHS = 0. En cuanto al RHS, resulta RHS = \
",
 Cell[BoxData[
  FormBox[
   RowBox[{
    RowBox[{
     SubscriptBox["\[Delta]", "jr"], 
     SubscriptBox["\[Delta]", "kk"]}], "-", 
    RowBox[{
     SubscriptBox["\[Delta]", "jk"], 
     SubscriptBox["\[Delta]", "kr"]}]}], TraditionalForm]]],
 ", el primer sumando es (0) . (1) = 0, mientras que el segundo es (0) . (0) \
= 0 por ser k \[NotEqual] j, luego RHS = 0. Se verifica que LHS == RHS\n\t\t\
CASO B.3) Repetici\[OAcute]n del sub\[IAcute]ndice central del primer factor \
y el sub\[IAcute]ndice final del segundo, LHS = ",
 Cell[BoxData[
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    SubscriptBox["\[Epsilon]", "ijk"], 
    SubscriptBox["\[Epsilon]", "irj"]}], TraditionalForm]]],
 ", esto implica que j = s,  nuevamente k \[NotEqual] j y s \[NotEqual] r, ya \
que los casos en los que j = k y r = s fueron analizados en el caso A. En \
esta situaci\[OAcute]n, los tres sumandos del LHS resultan ",
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  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", 
     RowBox[{"1", "jk"}]], 
    SubscriptBox["\[Epsilon]", 
     RowBox[{"1", "rj"}]]}], TraditionalForm]]],
 "+",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", 
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    SubscriptBox["\[Epsilon]", 
     RowBox[{"2", "rj"}]]}], TraditionalForm]]],
 "+",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", 
     RowBox[{"3", "jk"}]], 
    SubscriptBox["\[Epsilon]", 
     RowBox[{"3", "rj"}]]}], TraditionalForm]]],
 ". Como j \[NotEqual] k \[NotEqual] r y cada sub\[IAcute]ndice puede tomar \
los valores 1, 2 o 3; se presentan las siguientes situaciones:\n\t\t\tCASO \
B.3.1) k = r, en este caso, en uno de los tres sumandos va a resultar ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", "ijk"], 
    SubscriptBox["\[Epsilon]", "ikj"]}], TraditionalForm]]],
 " con i \[NotEqual] j \[NotEqual] k, mientras que en los restantes sumados \
al menos un sub\[IAcute]ndice va a estar repetido. Si suponemos que los sub\
\[IAcute]ndices ijk forman una permutaci\[OAcute]n c\[IAcute]clica, \
necesariamente ikj ser\[AAcute] una permutaci\[OAcute]n antic\[IAcute]clica, \
y viceversa, con lo que el resultado va a ser LHS = 1 . (-1) o LHS = (-1) . \
1, en ambos casos LHS = -1. En cuanto al RHS, resulta RHS = ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    RowBox[{
     SubscriptBox["\[Delta]", "jk"], 
     SubscriptBox["\[Delta]", "kj"]}], "-", 
    RowBox[{
     SubscriptBox["\[Delta]", "jj"], 
     SubscriptBox["\[Delta]", "kk"]}]}], TraditionalForm]]],
 ", el primer sumando es siempre 0 dado que j \[NotEqual] k, mientras que el \
segundo es 1 por sub\[IAcute]ndices repetidos, con lo que resulta  RHS = 0 - \
1 = -1. Por lo tanto, se verifica que LHS == RHS \n\t\t\tCASO B.3.2) k \
\[NotEqual] r, en este caso, en uno de los tres sumandos el primer factor va \
a resultar ",
 Cell[BoxData[
  FormBox[
   SubscriptBox["\[Epsilon]", "ijk"], TraditionalForm]]],
 " con i \[NotEqual] j \[NotEqual] k por lo que su valor va a ser 1 o -1 \
dependiendo de la permutaci\[OAcute]n, mientras que en los restantes sumandos \
el primer factor debe ser 0 al resultar j = i o j = k, por lo que esos \
sumandos van a ser 0. En el sumando para el cual el primer factor es no nulo, \
el segundo factor resulta ",
 Cell[BoxData[
  FormBox[
   SubscriptBox["\[Epsilon]", "irj"], TraditionalForm]]],
 " al menos un sub\[IAcute]ndice va a estar necesariamente repetido debido a \
que k \[NotEqual] r, con lo su valor va a ser nulo, y la suma de LHS va a \
tener tres sumandos nulos por lo que LHS = 0. En cuanto al RHS, resulta RHS = \
",
 Cell[BoxData[
  FormBox[
   RowBox[{
    RowBox[{
     SubscriptBox["\[Delta]", "jr"], 
     SubscriptBox["\[Delta]", "kj"]}], "-", 
    RowBox[{
     SubscriptBox["\[Delta]", "jj"], 
     SubscriptBox["\[Delta]", "kr"]}]}], TraditionalForm]]],
 ", el primer sumando es (x) . (0) = 0 ya que k \[NotEqual] j, mientras que \
el segundo es (1) . (0) = 0 por ser k \[NotEqual] r, luego RHS = 0. \
Nuevamente, se verifica que LHS == RHS\n\t\tCASO B.4) Repetici\[OAcute]n del \
sub\[IAcute]ndice final del primer factor y el sub\[IAcute]ndice central del \
segundo, LHS = ",
 Cell[BoxData[
  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", "ijk"], 
    SubscriptBox["\[Epsilon]", "iks"]}], TraditionalForm]]],
 ", esto implica que k = r,  nuevamente k \[NotEqual] j y s \[NotEqual] r, ya \
que los casos en los que j = k y r = s fueron analizados en el caso A. En \
esta situaci\[OAcute]n, los tres sumandos del LHS resultan ",
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  FormBox[
   RowBox[{
    SubscriptBox["\[Epsilon]", 
     RowBox[{"1", "jk"}]], 
    SubscriptBox["\[Epsilon]", 
     RowBox[{"1", "ks"}]]}], TraditionalForm]]],
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Ejercicio 2 : Una fuerza de magnitud F act\[UAcute]a en direcci\[OAcute]n \
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