Векторното произведение на два вектора
a
{\displaystyle \mathbf {a} }
b
{\displaystyle \mathbf {b} }
вектор , перпендикулярен на равнината, определена от векторите
a
{\displaystyle \mathbf {a} }
b
{\displaystyle \mathbf {b} }
дясна тройка с тях и има дължина, равна на произведението от големините на двата вектора и синуса на ъгъла между тях.
Ъгълът между два вектора приема стойности от
0
∘
{\displaystyle 0^{\circ }}
180
∘
{\displaystyle 180^{\circ }}
‖
a
×
b
‖
=
‖
a
‖
‖
b
‖
sin
∢
(
a
;
b
)
{\displaystyle \Vert \mathbf {a} \times \mathbf {b} \Vert =\Vert \mathbf {a} \Vert \Vert \mathbf {b} \Vert \sin \sphericalangle (\mathbf {a} ;\ \mathbf {b} )}
Самото векторно произведение на два вектора се дефинира така:
a
×
b
=
‖
a
‖
‖
b
‖
sin
∢
(
a
;
b
)
n
^
{\displaystyle \mathbf {a} \times \mathbf {b} =\Vert \mathbf {a} \Vert \Vert \mathbf {b} \Vert \sin \sphericalangle (\mathbf {a} ;\ \mathbf {b} )\ \mathbf {\hat {n}} }
като тук
n
^
=
a
×
b
‖
a
×
b
‖
{\displaystyle \mathbf {\hat {n}} ={\frac {\mathbf {a} \times \mathbf {b} }{\Vert \mathbf {a} \times \mathbf {b} \Vert }}}
Векторното произведение на
a
{\displaystyle \mathbf {a} }
b
{\displaystyle \mathbf {b} }
Ако са нанесени векторите
a
{\displaystyle \mathbf {a} }
b
{\displaystyle \mathbf {b} }
(
a
×
b
)
{\displaystyle (\mathbf {a} \times \mathbf {b} )}
a
{\displaystyle \mathbf {a} }
b
{\displaystyle \mathbf {b} }
(
a
,
b
,
a
×
b
)
{\displaystyle (\mathbf {a} ,\mathbf {b} ,\mathbf {a} \times \mathbf {b} )}
Ако векторите
a
{\displaystyle \mathbf {a} }
b
{\displaystyle \mathbf {b} }
a
=
(
a
1
,
a
2
,
a
3
)
{\displaystyle \mathbf {a} =(a_{1},a_{2},a_{3})}
b
=
(
b
1
,
b
2
,
b
3
)
{\displaystyle \mathbf {b} =(b_{1},b_{2},b_{3})}
i
,
j
,
k
{\displaystyle \mathbf {i} ,\mathbf {j} ,\mathbf {k} }
a
×
b
=
d
e
t
(
i
j
k
a
1
a
2
a
3
b
1
b
2
b
3
)
{\displaystyle \mathbf {a} \times \mathbf {b} =\mathrm {det} \left({\begin{matrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\end{matrix}}\right)}
По-подробно горната формула изглежда така:
a
×
b
=
i
d
e
t
(
a
2
a
3
b
2
b
3
)
−
j
d
e
t
(
a
1
a
3
b
1
b
3
)
+
k
d
e
t
(
a
1
a
2
b
1
b
2
)
=
i
(
a
2
b
3
−
a
3
b
2
)
−
j
(
a
1
b
3
−
a
3
b
1
)
+
k
(
a
1
b
2
−
a
2
b
1
)
{\displaystyle {\begin{aligned}\mathbf {a} \times \mathbf {b} &=\mathbf {i} \ \mathrm {det} \left({\begin{matrix}a_{2}&a_{3}\\b_{2}&b_{3}\end{matrix}}\right)-\mathbf {j} \ \mathrm {det} \left({\begin{matrix}a_{1}&a_{3}\\b_{1}&b_{3}\end{matrix}}\right)+\mathbf {k} \ \mathrm {det} \left({\begin{matrix}a_{1}&a_{2}\\b_{1}&b_{2}\end{matrix}}\right)\\&=\mathbf {i} (a_{2}b_{3}-a_{3}b_{2})-\mathbf {j} (a_{1}b_{3}-a_{3}b_{1})+\mathbf {k} (a_{1}b_{2}-a_{2}b_{1})\end{aligned}}}
Антикомутативност:
a
×
b
=
−
b
×
a
{\displaystyle \mathbf {a} \times \mathbf {b} =-\mathbf {b} \times \mathbf {a} }
Доказателство:
a
×
b
=
d
e
t
(
i
j
k
a
1
a
2
a
3
b
1
b
2
b
3
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=
−
d
e
t
(
i
j
k
b
1
b
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b
3
a
1
a
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a
3
)
=
−
b
×
a
{\displaystyle {\begin{aligned}\mathbf {a} \times \mathbf {b} &=\mathrm {det} \left({\begin{matrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\end{matrix}}\right)\\&=-\mathrm {det} \left({\begin{matrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\b_{1}&b_{2}&b_{3}\\a_{1}&a_{2}&a_{3}\end{matrix}}\right)\\&=-\mathbf {b} \times \mathbf {a} \end{aligned}}}
Дистрибутивност:
(
a
+
b
)
×
c
=
a
×
c
+
b
×
c
{\displaystyle (\mathbf {a} +\mathbf {b} )\times \mathbf {c} =\mathbf {a} \times \mathbf {c} +\mathbf {b} \times \mathbf {c} }
Доказателство:
Тъй като
a
+
b
=
(
a
1
+
b
1
,
a
2
+
b
2
,
a
3
+
b
3
)
{\displaystyle \mathbf {a} +\mathbf {b} =(a_{1}+b_{1},\ a_{2}+b_{2},\ a_{3}+b_{3})}
(
a
+
b
)
×
c
=
d
e
t
(
i
j
k
a
1
+
b
1
a
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b
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b
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=
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i
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a
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a
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+
d
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(
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b
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=
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×
c
+
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×
c
{\displaystyle {\begin{aligned}(\mathbf {a} +\mathbf {b} )\times \mathbf {c} &=\mathrm {det} \left({\begin{matrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\{a_{1}+b_{1}}&{a_{2}+b_{2}}&{a_{3}+b_{3}}\\c_{1}&c_{2}&c_{3}\end{matrix}}\right)\\&=\mathrm {det} \left({\begin{matrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\a_{1}&a_{2}&a_{3}\\c_{1}&c_{2}&c_{3}\end{matrix}}\right)+\mathrm {det} \left({\begin{matrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\b_{1}&b_{2}&b_{3}\\c_{1}&c_{2}&c_{3}\end{matrix}}\right)\\&=\mathbf {a} \times \mathbf {c} +\mathbf {b} \times \mathbf {c} \end{aligned}}}
Линейност:
(
λ
a
)
×
(
μ
b
)
=
λ
μ
(
a
×
b
)
{\displaystyle (\lambda \mathbf {a} )\times (\mu \mathbf {b} )=\lambda \mu (\mathbf {a} \times \mathbf {b} )}
λ
{\displaystyle \lambda }
μ
{\displaystyle \mu }
Доказателство:
Понеже
λ
a
=
(
λ
a
1
,
λ
a
2
,
λ
a
3
)
{\displaystyle \lambda \mathbf {a} =(\lambda a_{1},\ \lambda a_{2},\ \lambda a_{3})}
μ
b
=
(
μ
b
1
,
μ
b
2
,
μ
b
3
)
{\displaystyle \mu \mathbf {b} =(\mu b_{1},\ \mu b_{2},\ \mu b_{3})}
(
λ
a
)
×
(
μ
b
)
=
d
e
t
(
i
j
k
λ
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1
λ
a
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μ
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μ
d
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i
j
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a
1
a
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a
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b
1
b
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=
λ
μ
(
a
×
b
)
{\displaystyle {\begin{aligned}(\lambda \mathbf {a} )\times (\mu \mathbf {b} )&=\mathrm {det} \left({\begin{matrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\\lambda a_{1}&\lambda a_{2}&\lambda a_{3}\\\mu b_{1}&\mu b_{2}&\mu b_{3}\end{matrix}}\right)\\&=\lambda \mu \ \mathrm {det} \left({\begin{matrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\end{matrix}}\right)\\&=\lambda \mu (\mathbf {a} \times \mathbf {b} )\end{aligned}}}
Ако
a
∥
b
{\displaystyle \mathbf {a} \parallel \mathbf {b} }
a
×
b
=
0
{\displaystyle \mathbf {a} \times \mathbf {b} =\mathbf {0} }
Доказателство:
Щом
a
∥
b
{\displaystyle \mathbf {a} \parallel \mathbf {b} }
∢
(
a
,
b
)
=
0
∘
{\displaystyle \sphericalangle (\mathbf {a} ,\ \mathbf {b} )=0^{\circ }}
a
×
b
=
‖
a
‖
‖
b
‖
sin
0
∘
n
^
=
0
{\displaystyle \mathbf {a} \times \mathbf {b} =\Vert \mathbf {a} \Vert \Vert \mathbf {b} \Vert \sin 0^{\circ }\ \mathbf {\hat {n}} =\mathbf {0} }
Нека
i
,
j
,
k
{\displaystyle \mathbf {i} ,\mathbf {j} ,\mathbf {k} }
i
×
j
=
k
j
×
k
=
i
k
×
i
=
j
{\displaystyle {\begin{aligned}{\boldsymbol {\mathrm {i} }}\times {\boldsymbol {\mathrm {j} }}={\boldsymbol {\mathrm {k} }}\\{\boldsymbol {\mathrm {j} }}\times {\boldsymbol {\mathrm {k} }}={\boldsymbol {\mathrm {i} }}\\{\boldsymbol {\mathrm {k} }}\times {\boldsymbol {\mathrm {i} }}={\boldsymbol {\mathrm {j} }}\end{aligned}}}
Понеже векторното произведение е антикомутативно, то:
j
×
i
=
−
k
k
×
j
=
−
i
i
×
k
=
−
j
{\displaystyle {\begin{aligned}{\boldsymbol {\mathrm {j} }}\times {\boldsymbol {\mathrm {i} }}=-{\boldsymbol {\mathrm {k} }}\\{\boldsymbol {\mathrm {k} }}\times {\boldsymbol {\mathrm {j} }}=-{\boldsymbol {\mathrm {i} }}\\{\boldsymbol {\mathrm {i} }}\times {\boldsymbol {\mathrm {k} }}=-{\boldsymbol {\mathrm {j} }}\end{aligned}}}
Освен това лесно може да се покаже, че
i
×
i
=
j
×
j
=
k
×
k
=
0
{\displaystyle \mathbf {i} \times \mathbf {i} =\mathbf {j} \times \mathbf {j} =\mathbf {k} \times \mathbf {k} =\mathbf {0} }
С помощта на тези равенства можем да изразим векторното произведение на векторите
a
{\displaystyle \mathbf {a} }
b
{\displaystyle \mathbf {b} }
Понеже
a
=
a
1
i
+
a
2
j
+
a
3
k
b
=
b
1
i
+
b
2
j
+
b
3
k
{\displaystyle {\begin{alignedat}{3}\mathbf {a} &=a_{1}\mathbf {i} &&+a_{2}\mathbf {j} &&+a_{3}\mathbf {k} \\\mathbf {b} &=b_{1}\mathbf {i} &&+b_{2}\mathbf {j} &&+b_{3}\mathbf {k} \end{alignedat}}}
то векторното произведение
a
×
b
{\displaystyle \mathbf {a} \times \mathbf {b} }
a
×
b
=
(
a
1
i
+
a
2
j
+
a
3
k
)
×
(
b
1
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b
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b
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k
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=
a
1
b
1
(
i
×
i
)
+
a
1
b
2
(
i
×
j
)
+
a
1
b
3
(
i
×
k
)
+
a
2
b
1
(
j
×
i
)
+
a
2
b
2
(
j
×
j
)
+
a
2
b
3
(
j
×
k
)
+
a
3
b
1
(
k
×
i
)
+
a
3
b
2
(
k
×
j
)
+
a
3
b
3
(
k
×
k
)
=
a
1
b
1
0
+
a
1
b
2
k
−
a
1
b
3
j
−
a
2
b
1
k
+
a
2
b
2
0
+
a
2
b
3
i
+
a
3
b
1
j
−
a
3
b
2
i
+
a
3
b
3
0
=
(
a
2
b
3
−
a
3
b
2
)
i
+
(
a
3
b
1
−
a
1
b
3
)
j
+
(
a
1
b
2
−
a
2
b
1
)
k
{\displaystyle {\begin{aligned}\mathbf {a} \times \mathbf {b} ={}&(a_{1}\mathbf {i} +a_{2}\mathbf {j} +a_{3}\mathbf {k} )\times (b_{1}\mathbf {i} +b_{2}\mathbf {j} +b_{3}\mathbf {k} )\\={}&a_{1}b_{1}(\mathbf {i} \times \mathbf {i} )+a_{1}b_{2}(\mathbf {i} \times \mathbf {j} )+a_{1}b_{3}(\mathbf {i} \times \mathbf {k} )+{}\\&a_{2}b_{1}(\mathbf {j} \times \mathbf {i} )+a_{2}b_{2}(\mathbf {j} \times \mathbf {j} )+a_{2}b_{3}(\mathbf {j} \times \mathbf {k} )+{}\\&a_{3}b_{1}(\mathbf {k} \times \mathbf {i} )+a_{3}b_{2}(\mathbf {k} \times \mathbf {j} )+a_{3}b_{3}(\mathbf {k} \times \mathbf {k} )\\={}&\ a_{1}b_{1}\mathbf {0} +a_{1}b_{2}\mathbf {k} -a_{1}b_{3}\mathbf {j} \ -\\&a_{2}b_{1}\mathbf {k} +a_{2}b_{2}\mathbf {0} +a_{2}b_{3}\mathbf {i} \ +\\&a_{3}b_{1}\mathbf {j} \ -a_{3}b_{2}\mathbf {i} \ +a_{3}b_{3}\mathbf {0} \\={}&(a_{2}b_{3}-a_{3}b_{2})\mathbf {i} +(a_{3}b_{1}-a_{1}b_{3})\mathbf {j} +(a_{1}b_{2}-a_{2}b_{1})\mathbf {k} \\\end{aligned}}}
Нека с
S
{\displaystyle S}
θ
{\displaystyle \theta }
a
{\displaystyle \mathbf {a} }
b
{\displaystyle \mathbf {b} }
S
=
‖
a
‖
‖
b
‖
sin
θ
=
‖
a
×
b
‖
{\displaystyle S=\Vert \mathbf {a} \Vert \Vert \mathbf {b} \Vert \sin \theta =\Vert \mathbf {a} \times \mathbf {b} \Vert }
