Regla del cerdito para doble producto vectorial de tres vectores a , b y c . Según la fórmula,
A
×
(
B
×
C
)
{\displaystyle \mathbf {A} \times \left(\mathbf {B} \times \mathbf {C} \right)}
B y C .
La interpretación geométrica del vector
p
→
=
(
u
→
×
a
→
)
×
u
→
{\displaystyle {\vec {p}}=({\vec {u}}\times {\vec {a}})\times {\vec {u}}}
a
→
{\displaystyle {\vec {a}}}
u
→
{\displaystyle {\vec {u}}}
El producto vectorial no tiene la propiedad asociativa, ya que es antisimétrico (o anticonmutativo ). El vector
(
A
×
B
)
×
C
=
−
C
×
(
A
×
B
)
=
(
A
⋅
C
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B
−
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B
⋅
C
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A
{\displaystyle \,\left(\mathbf {A} \times \mathbf {B} \right)\times \mathbf {C} =-\mathbf {C} \times \left(\mathbf {A} \times \mathbf {B} \right)=\left(\mathbf {A} \cdot \mathbf {C} \right)\mathbf {B} -\left(\mathbf {B} \cdot \mathbf {C} \right)\mathbf {A} }
está contenido en el plano definido por los vectores A y B , por lo que, en general, será
A
×
(
B
×
C
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≠
(
A
×
B
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×
C
{\displaystyle \,\mathbf {A} \times \left(\mathbf {B} \times \mathbf {C} \right)\neq \left(\mathbf {A} \times \mathbf {B} \right)\times \mathbf {C} }
con lo cual resulta fundamental la colocación de los paréntesis.
A
×
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B
×
C
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+
C
×
(
A
×
B
)
+
B
×
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C
×
A
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=
0
{\displaystyle \mathbf {A} \times (\mathbf {B} \times \mathbf {C} )+\mathbf {C} \times (\mathbf {A} \times \mathbf {B} )+\mathbf {B} \times (\mathbf {C} \times \mathbf {A} )=0}
Gráfico tridimensional del doble producto vectorial A x (B x C) Sea
A
×
(
B
×
C
)
{\displaystyle \mathbf {A} \times \left(\mathbf {B} \times \mathbf {C} \right)}
B y C , cualquiera sea la dirección de A . Entonces, se puede descomponer al vector
A
×
(
B
×
C
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{\displaystyle \mathbf {A} \times (\mathbf {B} \times \mathbf {C} )}
B y otra paralela a C .
(1
A
×
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B
×
C
)
=
B
x
+
C
y
x
,
y
∈
R
{\displaystyle \mathbf {A} \times (\mathbf {B} \times \mathbf {C} )=\mathbf {B} x+\mathbf {C} y\quad x,y\in \mathbb {R} \quad }
Para facilitar la demostración primero se supondrá
B
⊥
C
{\displaystyle \mathbf {B} \bot \mathbf {C} }
B en (1
B
⋅
[
A
×
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B
×
C
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]
=
B
⋅
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B
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+
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y
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{\displaystyle \mathbf {B} \cdot [\mathbf {A} \times (\mathbf {B} \times \mathbf {C} )]=\mathbf {B} \cdot (\mathbf {B} x+\mathbf {C} y)}
Aplicamos propiedad distributiva en el segundo miembro (recordemos que B .C = 0 por ser perpendiculares ):
B
⋅
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B
x
+
C
y
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=
B
⋅
B
x
+
B
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C
y
=
|
B
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2
x
{\displaystyle \mathbf {B} \cdot (\mathbf {B} x+\mathbf {C} y)=\mathbf {B} \cdot \mathbf {B} x+\mathbf {B} \cdot \mathbf {C} y=\left|\mathbf {B} \right|^{2}x}
El primer miembro es un producto mixto y, como tal, puede intercambiar sus factores de esta manera:
B
⋅
[
A
×
(
B
×
C
)
]
=
A
⋅
[
(
B
×
C
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×
B
]
{\displaystyle \mathbf {B} \cdot [\mathbf {A} \times (\mathbf {B} \times \mathbf {C} )]=\mathbf {A} \cdot [(\mathbf {B} \times \mathbf {C} )\times \mathbf {B} ]}
Igualando las expresiones anteriores se tiene:
(2
A
⋅
[
(
B
×
C
)
×
B
]
=
|
B
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2
x
{\displaystyle \mathbf {A} \cdot [(\mathbf {B} \times \mathbf {C} )\times \mathbf {B} ]=\left|\mathbf {B} \right|^{2}x\quad }
El producto
(
B
×
C
)
×
B
{\displaystyle (\mathbf {B} \times \mathbf {C} )\times \mathbf {B} }
C (ver figura). Si averiguamos el módulo de este producto obtenemos:
|
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B
×
C
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×
B
|
=
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B
×
C
)
|
.
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B
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.
sen
π
2
=
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B
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.
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C
|
.
sen
π
2
.
|
B
|
.1
=
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B
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2
.
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C
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{\displaystyle \left|(\mathbf {B} \times \mathbf {C} )\times \mathbf {B} \right|=\left|(\mathbf {B} \times \mathbf {C} )\right|.\left|\mathbf {B} \right|.\operatorname {sen} {\pi \over 2}=\left|\mathbf {B} \right|.\left|\mathbf {C} \right|.\operatorname {sen} {\pi \over 2}.\left|\mathbf {B} \right|.1=\left|\mathbf {B} \right|^{2}.\left|\mathbf {C} \right|}
Como
(
B
×
C
)
×
B
{\displaystyle (\mathbf {B} \times \mathbf {C} )\times \mathbf {B} }
C , se puede expresar de la siguiente manera:
(
B
×
C
)
×
B
=
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B
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2
C
{\displaystyle (\mathbf {B} \times \mathbf {C} )\times \mathbf {B} =\quad \left|\mathbf {B} \right|^{2}\quad \mathbf {C} }
Identidad que, multiplicada escalarmente por el vector A , coincide con (2
∴
x
=
A
⋅
C
{\displaystyle \therefore \,x=\mathbf {A} \cdot \mathbf {C} }
Para averiguar y se sigue un proceso análogo, en el cual se efectúa en (1 C :
C
⋅
[
A
×
(
B
×
C
)
]
=
C
⋅
(
B
x
+
C
y
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{\displaystyle \mathbf {C} \cdot [\mathbf {A} \times (\mathbf {B} \times \mathbf {C} )]=\mathbf {C} \cdot (\mathbf {B} x+\mathbf {C} y)}
A
⋅
[
(
B
×
C
)
×
C
]
=
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C
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2
y
{\displaystyle \mathbf {A} \cdot [(\mathbf {B} \times \mathbf {C} )\times \mathbf {C} ]=\left|\mathbf {C} \right|^{2}y}
En este punto cabe destacar una diferencia importante, que se deduce de la imagen. Nótese que el vector
(
B
×
C
)
×
C
{\displaystyle (\mathbf {B} \times \mathbf {C} )\times \mathbf {C} }
B . Esto implica:
−
A
⋅
B
|
C
|
2
=
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C
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2
y
→
y
=
−
A
⋅
B
{\displaystyle -\mathbf {A} \cdot \mathbf {B} \left|\mathbf {C} \right|^{2}=\left|\mathbf {C} \right|^{2}y\quad \rightarrow \quad y=-\mathbf {A} \cdot \mathbf {B} }
Reemplazamos x e y en (1 B y C perpendiculares.
(*
A
×
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B
×
C
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=
B
(
A
⋅
C
)
−
C
(
A
⋅
B
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{\displaystyle \mathbf {A} \times (\mathbf {B} \times \mathbf {C} )=\mathbf {B} (\mathbf {A} \cdot \mathbf {C} )-\mathbf {C} (\mathbf {A} \cdot \mathbf {B} )\quad }
Fórmula general
editar
Considerando ahora un vector B , ya no necesariamente perpendicular a C , se puede descomponerlo en dos componentes diferentes, una perpendicular y otra paralela a C .
B
=
B
′
+
C
k
B
′
⊥
C
,
k
∈
R
{\displaystyle \mathbf {B} =\mathbf {B} '+\mathbf {C} k\quad \mathbf {B} '\bot \mathbf {C} ,k\in \mathbb {R} }
Se efectúa el doble producto vectorial y se lleva a la forma (*
A
×
(
B
×
C
)
=
A
×
[
(
B
′
+
C
k
)
×
C
]
=
A
×
(
B
′
×
C
+
C
k
×
C
)
=
A
×
(
B
′
×
C
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{\displaystyle \mathbf {A} \times (\mathbf {B} \times \mathbf {C} )=\mathbf {A} \times [(\mathbf {B} '+\mathbf {C} k)\times \mathbf {C} ]=\mathbf {A} \times (\mathbf {B} '\times \mathbf {C} +\mathbf {C} k\times \mathbf {C} )=\mathbf {A} \times (\mathbf {B} '\times \mathbf {C} )}
De modo que se puede desarrollar de esta manera:
A
×
(
B
′
×
C
)
=
B
′
(
A
⋅
C
)
−
C
(
A
⋅
B
′
)
{\displaystyle \mathbf {A} \times (\mathbf {B} '\times \mathbf {C} )=\mathbf {B} '(\mathbf {A} \cdot \mathbf {C} )-\mathbf {C} (\mathbf {A} \cdot \mathbf {B} ')}
Ahora, tenemos
B
=
B
′
+
C
k
⇒
B
′
=
B
−
C
k
{\displaystyle \mathbf {B} =\mathbf {B} '+\mathbf {C} k\Rightarrow \mathbf {B} '=\mathbf {B} -\mathbf {C} k}
B
′
(
A
⋅
C
)
−
C
(
A
⋅
B
′
)
=
(
B
−
C
k
)
(
A
⋅
C
)
−
C
[
A
⋅
(
B
−
C
k
)
]
=
{\displaystyle \mathbf {B} '(\mathbf {A} \cdot \mathbf {C} )-\mathbf {C} (\mathbf {A} \cdot \mathbf {B} ')=(\mathbf {B} -\mathbf {C} k)(\mathbf {A} \cdot \mathbf {C} )-\mathbf {C} [\mathbf {A} \cdot (\mathbf {B} -\mathbf {C} k)]=}
=
B
(
A
⋅
C
)
−
C
k
(
A
⋅
C
)
−
C
(
A
⋅
B
−
A
⋅
C
k
)
=
B
(
A
⋅
C
)
−
C
k
(
A
⋅
C
)
−
C
(
A
⋅
B
)
+
C
(
A
⋅
C
k
)
=
{\displaystyle =\mathbf {B} (\mathbf {A} \cdot \mathbf {C} )-\mathbf {C} k(\mathbf {A} \cdot \mathbf {C} )-\mathbf {C} (\mathbf {A} \cdot \mathbf {B} -\mathbf {A} \cdot \mathbf {C} k)=\mathbf {B} (\mathbf {A} \cdot \mathbf {C} )-\mathbf {C} k(\mathbf {A} \cdot \mathbf {C} )-\mathbf {C} (\mathbf {A} \cdot \mathbf {B} )+\mathbf {C} (\mathbf {A} \cdot \mathbf {C} k)=}
=
B
(
A
⋅
C
)
−
C
k
(
A
⋅
C
)
−
C
(
A
⋅
B
)
+
C
k
(
A
⋅
C
)
{\displaystyle =\mathbf {B} (\mathbf {A} \cdot \mathbf {C} ){\cancel {-\mathbf {C} k(\mathbf {A} \cdot \mathbf {C} )}}-\mathbf {C} (\mathbf {A} \cdot \mathbf {B} ){\cancel {+\mathbf {C} k(\mathbf {A} \cdot \mathbf {C} )}}}
∴
A
×
(
B
×
C
)
=
B
(
A
⋅
C
)
−
C
(
A
⋅
B
)
{\displaystyle \therefore \ \mathbf {A} \times (\mathbf {B} \times \mathbf {C} )=\mathbf {B} (\mathbf {A} \cdot \mathbf {C} )-\mathbf {C} (\mathbf {A} \cdot \mathbf {B} )}
Esta última identidad coincide con (* A , B y C .
![{\displaystyle \mathbf {B} \cdot [\mathbf {A} \times (\mathbf {B} \times \mathbf {C} )]=\mathbf {B} \cdot (\mathbf {B} x+\mathbf {C} y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/368bab4b0d44273ecd2de2f00d82a9172c9f4f81)
![{\displaystyle \mathbf {B} \cdot [\mathbf {A} \times (\mathbf {B} \times \mathbf {C} )]=\mathbf {A} \cdot [(\mathbf {B} \times \mathbf {C} )\times \mathbf {B} ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8fa7cc3ac570990af039dfa44d8434c1321b7f54)
![{\displaystyle \mathbf {A} \cdot [(\mathbf {B} \times \mathbf {C} )\times \mathbf {B} ]=\left|\mathbf {B} \right|^{2}x\quad }](https://wikimedia.org/api/rest_v1/media/math/render/svg/9cf29a77b8ca589a548cebbcdbb83c6e8a6dc2a8)
![{\displaystyle \mathbf {C} \cdot [\mathbf {A} \times (\mathbf {B} \times \mathbf {C} )]=\mathbf {C} \cdot (\mathbf {B} x+\mathbf {C} y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df4c9c66b0ad3befe018bfaef40aeef3c72041cc)
![{\displaystyle \mathbf {A} \cdot [(\mathbf {B} \times \mathbf {C} )\times \mathbf {C} ]=\left|\mathbf {C} \right|^{2}y}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a72771294665f9c18dfd7494d24f2a1e2e48abe2)
![{\displaystyle \mathbf {A} \times (\mathbf {B} \times \mathbf {C} )=\mathbf {A} \times [(\mathbf {B} '+\mathbf {C} k)\times \mathbf {C} ]=\mathbf {A} \times (\mathbf {B} '\times \mathbf {C} +\mathbf {C} k\times \mathbf {C} )=\mathbf {A} \times (\mathbf {B} '\times \mathbf {C} )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/76e3f343b730fed722fa6ef61ef887646383179a)
![{\displaystyle \mathbf {B} '(\mathbf {A} \cdot \mathbf {C} )-\mathbf {C} (\mathbf {A} \cdot \mathbf {B} ')=(\mathbf {B} -\mathbf {C} k)(\mathbf {A} \cdot \mathbf {C} )-\mathbf {C} [\mathbf {A} \cdot (\mathbf {B} -\mathbf {C} k)]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0b19e3868c03b098c2ac38ce26d6d874d687c646)
Datos: Q5812140