1.1.8c-8e

Demuestre las propiedades vistas en clase de el producto punto.

**c)**

a
→

⋅
(

b
→

c
→

# )

a
→

⋅

b
→

a
→

⋅

c
→

{\displaystyle {\vec {a}}\cdot ({\vec {b}}+{\vec {c}})={\vec {a}}\cdot {\vec {b}}+{\vec {a}}\cdot {\vec {c}}}

Sea:

a
→

=
(

a
1

,

a
2

,

a
3

)

{\displaystyle \vec{a}=(a\_1,a\_2,a\_3)}

b
→

=
(

b

1

,

b

2

,

b

3

)

{\displaystyle {\vec {b}}=(b\_{1},b\_{2},b\_{3})}

c
→

=
(

c

1

,

c

2

,

c

3

)

{\displaystyle {\vec {c}}=(c\_{1},c\_{2},c\_{3})}

Encontremos a cuanto equivale

b
→

c
→

{\displaystyle {\vec {b}}+{\vec {c}}}

:

b
→

c
→

=
(

b

1

,

b

2

,

b

3

)
(

c

1

,

c

2

,

c

3

# )
(

b

1

c

1

,

b

2

c

2

,

b

3

c

3

)

{\displaystyle {\vec {b}}+{\vec {c}}=(b\_{1},b\_{2},b\_{3})+(c\_{1},c\_{2},c\_{3})=(b\_{1}+c\_{1},b\_{2}+c\_{2},b\_{3}+c\_{3})}

Con lo anterior veamos a cuanto equivale

a
→

⋅
(

b
→

c
→

)

{\displaystyle {\vec {a}}\cdot ({\vec {b}}+{\vec {c}})}

:

a
→

⋅
(

b
→

c
→

# )
(

a

1

,

a

2

,

a

3

)
⋅
(

b

1

c

1

,

b

2

c

2

,

b

3

c

3

)

{\displaystyle {\vec {a}}\cdot ({\vec {b}}+{\vec {c}})=(a\_{1},a\_{2},a\_{3})\cdot (b\_{1}+c\_{1},b\_{2}+c\_{2},b\_{3}+c\_{3})}

=

a

1

(

b

1

c

1

)

a

2

(

b

2

c

2

)

a

3

(

b

3

c

3

)

{\displaystyle =a\_{1}(b\_{1}+c\_{1})+a\_{2}(b\_{2}+c\_{2})+a\_{3}(b\_{3}+c\_{3})}

=

a

1

b

1

a

1

c

1

a

2

b

2

a

2

c

2

a

3

b

3

a

3

c

3

{\displaystyle =a\_{1}b\_{1}+a\_{1}c\_{1}+a\_{2}b\_{2}+a\_{2}c\_{2}+a\_{3}b\_{3}+a\_{3}c\_{3}}

Encontremos a cuanto equivale

a
→

⋅

b
→

{\displaystyle \vec{a} \cdot \vec{b}}

 y

a
→

⋅

c
→

{\displaystyle {\vec {a}}\cdot {\vec {c}}}

:

a
→

⋅

b
→

=
(

a

1

,

a

2

,

a

3

)
⋅
(

b

1

,

b

2

,

b

3

# )

a

1

b

1

a

2

b

2

a

3

b

3

{\displaystyle {\vec {a}}\cdot {\vec {b}}=(a\_{1},a\_{2},a\_{3})\cdot (b\_{1},b\_{2},b\_{3})=a\_{1}b\_{1}+a\_{2}b\_{2}+a\_{3}b\_{3}}

a
→

⋅

c
→

=
(

a

1

,

a

2

,

a

3

)
⋅
(

c

1

,

c

2

,

c

3

# )

a

1

c

1

a

2

c

2

a

3

c

3

{\displaystyle {\vec {a}}\cdot {\vec {c}}=(a\_{1},a\_{2},a\_{3})\cdot (c\_{1},c\_{2},c\_{3})=a\_{1}c\_{1}+a\_{2}c\_{2}+a\_{3}c\_{3}}

Con lo anterior podemos ver a cuanto equivale

a
→

⋅

b
→

a
→

⋅

c
→

{\displaystyle {\vec {a}}\cdot {\vec {b}}+{\vec {a}}\cdot {\vec {c}}}

:

a
→

⋅

b
→

a
→

⋅

c
→

=
(

a

1

b

1

a

2

b

2

a

3

b

3

)
(

a

1

c

1

a

2

c

2

a

3

c

3

)

{\displaystyle {\vec {a}}\cdot {\vec {b}}+{\vec {a}}\cdot {\vec {c}}=(a\_{1}b\_{1}+a\_{2}b\_{2}+a\_{3}b\_{3})+(a\_{1}c\_{1}+a\_{2}c\_{2}+a\_{3}c\_{3})}

Si desarrollamos y conmutamos términos:

=

a

1

b

1

a

2

b

2

a

3

b

3

a

1

c

1

a

2

c

2

a

3

c

3

{\displaystyle =a\_{1}b\_{1}+a\_{2}b\_{2}+a\_{3}b\_{3}+a\_{1}c\_{1}+a\_{2}c\_{2}+a\_{3}c\_{3}}

=

a

1

b

1

a

1

c

1

a

2

b

2

a

2

c

2

a

3

b

3

a

3

c

3

{\displaystyle =a\_{1}b\_{1}+a\_{1}c\_{1}+a\_{2}b\_{2}+a\_{2}c\_{2}+a\_{3}b\_{3}+a\_{3}c\_{3}}

Notemos que obtenemos lo mismo que en

a
→

⋅
(

b
→

c
→

)

{\displaystyle {\vec {a}}\cdot ({\vec {b}}+{\vec {c}})}

.

Por lo tanto, queda demostrado que

a
→

⋅
(

b
→

c
→

# )

a
→

⋅

b
→

a
→

⋅

c
→

{\displaystyle {\vec {a}}\cdot ({\vec {b}}+{\vec {c}})={\vec {a}}\cdot {\vec {b}}+{\vec {a}}\cdot {\vec {c}}}

.

**d)**

(
r

a
→

)
⋅

b
→

=
r
(

a
→

⋅

b
→

# )

a
→

⋅
(
r

b
→

)

{\displaystyle (r{\vec {a}})\cdot {\vec {b}}=r({\vec {a}}\cdot {\vec {b}})={\vec {a}}\cdot (r{\vec {b}})}

Sea:

a
→

=
(

a
1

,

a
2

,

a
3

)

{\displaystyle \vec{a}=(a\_1,a\_2,a\_3)}

b
→

=
(

b

1

,

b

2

,

b

3

)

{\displaystyle {\vec {b}}=(b\_{1},b\_{2},b\_{3})}

Y

r

{\displaystyle r}

un escalar.

Demostremos que los tres términos mostrados en las igualdades son equivalentes.

*I)*

(
r

a
→

)
⋅

b
→

{\displaystyle (r{\vec {a}})\cdot {\vec {b}}}

(
r

a
→

)
⋅

b
→

=
(
r
(

a

1

,

a

2

,

a

3

)
)
⋅
(

b

1

,

b

2

,

b

3

)

{\displaystyle (r{\vec {a}})\cdot {\vec {b}}=(r(a\_{1},a\_{2},a\_{3}))\cdot (b\_{1},b\_{2},b\_{3})}

=
(
r
⋅

a

1

,
r
⋅

a

2

,
r
⋅

a

3

)
⋅
(

b

1

,

b

2

,

b

3

)

{\displaystyle =(r\cdot a\_{1},r\cdot a\_{2},r\cdot a\_{3})\cdot (b\_{1},b\_{2},b\_{3})}

=
r
⋅

a

1

⋅

b

1

r
⋅

a

2

⋅

b

2

r
⋅

a

3

⋅

b

3

{\displaystyle =r\cdot a\_{1}\cdot b\_{1}+r\cdot a\_{2}\cdot b\_{2}+r\cdot a\_{3}\cdot b\_{3}}

=
r
(

a

1

⋅

b

1

a

2

⋅

b

2

a

3

⋅

b

3

)

{\displaystyle =r(a\_{1}\cdot b\_{1}+a\_{2}\cdot b\_{2}+a\_{3}\cdot b\_{3})}

*II)*

r
(

a
→

⋅

b
→

)

{\displaystyle r({\vec {a}}\cdot {\vec {b}})}

r
(

a
→

⋅

b
→

# )
r
(
(

a

1

,

a

2

,

a

3

)
⋅
(

b

1

,

b

2

,

b

3

)
)

{\displaystyle r({\vec {a}}\cdot {\vec {b}})=r((a\_{1},a\_{2},a\_{3})\cdot (b\_{1},b\_{2},b\_{3}))}

=
r
(

a

1

⋅

b

1

a

2

⋅

b

2

a

3

⋅

b

3

)

{\displaystyle =r(a\_{1}\cdot b\_{1}+a\_{2}\cdot b\_{2}+a\_{3}\cdot b\_{3})}

*III)*

a
→

⋅
(
r

b
→

)

{\displaystyle {\vec {a}}\cdot (r{\vec {b}})}

r

b
→

=
r
(

b

1

,

b

2

,

b

3

# )
(
r
⋅

b

1

,
r
⋅

b

2

,
r
⋅

b

3

)

{\displaystyle r{\vec {b}}=r(b\_{1},b\_{2},b\_{3})=(r\cdot b\_{1},r\cdot b\_{2},r\cdot b\_{3})}

a
→

⋅
(
r

b
→

# )
(

a

1

,

a

2

,

a

3

)
⋅
(
r
⋅

b

1

,
r
⋅

b

2

,
r
⋅

b

3

)

{\displaystyle {\vec {a}}\cdot (r{\vec {b}})=(a\_{1},a\_{2},a\_{3})\cdot (r\cdot b\_{1},r\cdot b\_{2},r\cdot b\_{3})}

=
r
⋅

a

1

⋅

b

1

r
⋅

a

2

⋅

b

2

r
⋅

a

3

⋅

b

3

{\displaystyle =r\cdot a\_{1}\cdot b\_{1}+r\cdot a\_{2}\cdot b\_{2}+r\cdot a\_{3}\cdot b\_{3}}

=
r
(

a

1

⋅

b

1

a

2

⋅

b

2

a

3

⋅

b

3

)

{\displaystyle =r(a\_{1}\cdot b\_{1}+a\_{2}\cdot b\_{2}+a\_{3}\cdot b\_{3})}

Notemos que los tres términos son iguales,

Por lo tanto, queda demostrado que es cierto que

(
r

a
→

)
⋅

b
→

=
r
(

a
→

⋅

b
→

# )

a
→

⋅
(
r

b
→

)

{\displaystyle (r{\vec {a}})\cdot {\vec {b}}=r({\vec {a}}\cdot {\vec {b}})={\vec {a}}\cdot (r{\vec {b}})}

.

**e)**

0
→

⋅

a
→

=
0

{\displaystyle {\vec {0}}\cdot {\vec {a}}=0}

Sea:

a
→

=
(
x
,
y
,
z
)

{\displaystyle {\vec {a}}=(x,y,z)}

0
→

=
(
0
,
0
,
0
)

{\displaystyle {\vec {0}}=(0,0,0)}

Como

(
x
,
y
,
z
)
⋅
(

x
′

,

y
′

,

z
′

# )
x
⋅

x
′

y
⋅

y
′

z
⋅

z
′

{\displaystyle (x,y,z)\cdot (x',y',z')=x\cdot x'+y\cdot y'+z\cdot z'}

Entonces

0
→

⋅

a
→

=
(
0
,
0
,
0
)
⋅
(
x
,
y
,
z
)

{\displaystyle {\vec {0}}\cdot {\vec {a}}=(0,0,0)\cdot (x,y,z)}

=
x
⋅
0
y
⋅
0
z
⋅
0

{\displaystyle =x\cdot 0+y\cdot 0+z\cdot 0}

=
0
0
0

{\displaystyle =0+0+0}

=
0

{\displaystyle = 0}

Dicho lo anterior, concluimos que es cierto que

0
→

⋅

a
→

=
0

{\displaystyle {\vec {0}}\cdot {\vec {a}}=0}

.