править
В простейшем случае известны координаты векторов в ортонормированной системе координат.
Тогда скалярное произведение вычисляется как сумма произведения одноименных координат.
Смотри также Скалярное произведение векторов
В равностороннем треугольнике
△
A
B
C
{\displaystyle \triangle ABC}
A
B
→
⋅
B
C
→
+
B
C
→
⋅
C
A
→
+
C
A
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⋅
A
B
→
{\displaystyle {\overrightarrow {AB}}\cdot {\overrightarrow {BC}}+{\overrightarrow {BC}}\cdot {\overrightarrow {CA}}+{\overrightarrow {CA}}\cdot {\overrightarrow {AB}}}
A
B
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⋅
B
C
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+
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C
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⋅
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+
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A
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=
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cos
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^
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+
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cos
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^
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+
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cos
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C
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B
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^
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=
=
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⋅
1
⋅
cos
π
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cos
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cos
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−
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{\displaystyle {\begin{aligned}{\overrightarrow {AB}}\cdot {\overrightarrow {BC}}+{\overrightarrow {BC}}\cdot {\overrightarrow {CA}}+{\overrightarrow {CA}}\cdot {\overrightarrow {AB}}&=|{\overrightarrow {AB}}||{\overrightarrow {BC}}|\cos({\widehat {{\overrightarrow {AB}},{\overrightarrow {BC}}}})+|{\overrightarrow {BC}}||{\overrightarrow {CA}}|\cos({\widehat {{\overrightarrow {BC}},{\overrightarrow {CA}}}})+|{\overrightarrow {CA}}||{\overrightarrow {AB}}|\cos({\widehat {{\overrightarrow {CA}},{\overrightarrow {AB}}}})=\\&=1\cdot 1\cdot \cos {\frac {\pi }{3}}2+1\cdot 1\cdot \cos {\frac {\pi }{3}}2+1\cdot 1\cdot \cos {\frac {\pi }{3}}2=-{\frac {3}{2}}\end{aligned}}}
Даны два неколлинеарных вектора
a
{\displaystyle \mathbf {a} }
b
{\displaystyle \mathbf {b} }
x
{\displaystyle \mathbf {x} }
a
{\displaystyle \mathbf {a} }
b
{\displaystyle \mathbf {b} }
{
a
⋅
x
=
1
b
⋅
x
=
0
{\displaystyle {\begin{cases}\mathbf {a} \cdot \mathbf {x} =1\\\mathbf {b} \cdot \mathbf {x} =0\end{cases}}}
Поскольку векторы
a
{\displaystyle \mathbf {a} }
b
{\displaystyle \mathbf {b} }
x
=
λ
a
+
μ
b
{\displaystyle \mathbf {x} =\lambda \mathbf {a} +\mu \mathbf {b} }
Поэтому исходную систему можно переписать в виде
{
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⋅
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+
μ
b
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=
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{
λ
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{\displaystyle {\begin{cases}\mathbf {a} \cdot (\lambda \mathbf {a} +\mu \mathbf {b} )=1\\\mathbf {b} \cdot (\lambda \mathbf {a} +\mu \mathbf {b} )=0\end{cases}}\quad {\begin{cases}\lambda (\mathbf {a} \cdot \mathbf {a} )+\mu (\mathbf {a} \cdot \mathbf {b} )=1\\\lambda (\mathbf {b} \cdot \mathbf {a} )+\mu (\mathbf {b} \cdot \mathbf {b} )=0\end{cases}}}
Решение этой системы
{
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b
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b
(
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⋅
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(
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−
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{\displaystyle {\begin{cases}\lambda ={\frac {\mathbf {b} \cdot \mathbf {b} }{(\mathbf {a} \cdot \mathbf {a} )(\mathbf {b} \cdot \mathbf {b} )-(\mathbf {a} \cdot \mathbf {b} )^{2}}}\\\mu ={\frac {-(\mathbf {a} \cdot \mathbf {b} )}{(\mathbf {a} \cdot \mathbf {a} )(\mathbf {b} \cdot \mathbf {b} )-(\mathbf {a} \cdot \mathbf {b} )^{2}}}\\\end{cases}}}
Таким образом искомый вектор
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{\displaystyle \mathbf {x} ={\frac {(\mathbf {b} \cdot \mathbf {b} )\mathbf {a} +(\mathbf {a} \cdot \mathbf {b} )\mathbf {b} }{(\mathbf {a} \cdot \mathbf {a} )(\mathbf {b} \cdot \mathbf {b} )-(\mathbf {a} \cdot \mathbf {b} )^{2}}}}
править
Длина вектора связана со скалярным произведением формулой
|
a
|
=
a
⋅
a
{\displaystyle |\mathbf {a} |={\sqrt {\mathbf {a} \cdot \mathbf {a} }}}
a
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{
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1
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a
2
,
a
3
}
{\displaystyle \mathbf {a} =\{a_{1},a_{2},a_{3}\}}
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2
+
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+
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{\displaystyle \mathbf {a} \cdot \mathbf {a} =a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}
Таким образом
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+
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+
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{\displaystyle |\mathbf {a} |={\sqrt {a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}}}
Смотри также Длина вектора
Во многих случаях необходимо получить единичный вектор
x
{\displaystyle \mathbf {x} }
a
{\displaystyle \mathbf {a} }
x
=
k
a
{\displaystyle \mathbf {x} =k\mathbf {a} }
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x
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1
{\displaystyle |\mathbf {x} |=1}
k
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{\displaystyle k|\mathbf {a} |=1}
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1
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{\displaystyle k={\tfrac {1}{|\mathbf {a} |}}}
Смотри также Нормализация вектора
Угол между ненулевыми векторами связан со скалярным произведением формулой
cos
(
a
,
b
^
)
=
a
⋅
b
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{\displaystyle \cos({\widehat {\mathbf {a} ,\mathbf {b} }})={\frac {\mathbf {a} \cdot \mathbf {b} }{|\mathbf {a} ||\mathbf {b} |}}}
Если векторы заданы своими координатами
a
=
{
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1
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a
2
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3
}
{\displaystyle \mathbf {a} =\{a_{1},a_{2},a_{3}\}}
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{
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,
b
3
}
{\displaystyle \mathbf {b} =\{b_{1},b_{2},b_{3}\}}
cos
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^
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=
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b
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+
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+
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{\displaystyle \cos({\widehat {\mathbf {a} ,\mathbf {b} }})={\frac {a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}}{{\sqrt {a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}}{\sqrt {b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}}}}}
Смотри также Найти угол между векторами
Дан параллелограмм
O
A
C
B
{\displaystyle OACB}
|
O
A
|
=
a
,
|
O
B
|
=
b
{\displaystyle |OA|=a,|OB|=b}
∠
A
O
B
=
α
{\displaystyle \angle AOB=\alpha }
d
{\displaystyle d}
O
C
{\displaystyle OC}
Очевидно
O
C
→
=
O
A
→
+
O
B
→
{\displaystyle {\overrightarrow {OC}}={\overrightarrow {OA}}+{\overrightarrow {OB}}}
d
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=
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=
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⋅
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cos
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O
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=
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+
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+
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a
b
cos
α
{\displaystyle {\begin{aligned}d&=|{\overrightarrow {OC}}|=|{\overrightarrow {OA}}+{\overrightarrow {OB}}|={\sqrt {({\overrightarrow {OA}}+{\overrightarrow {OB}})\cdot ({\overrightarrow {OA}}+{\overrightarrow {OB}})}}=\\&={\sqrt {{\overrightarrow {OA}}\cdot {\overrightarrow {OA}}+{\overrightarrow {OB}}\cdot {\overrightarrow {OA}}+{\overrightarrow {OA}}\cdot {\overrightarrow {OB}}+{\overrightarrow {OB}}\cdot {\overrightarrow {OB}}}}=\\&={\sqrt {|{\overrightarrow {OA}}|+|{\overrightarrow {OB}}|+2|{\overrightarrow {OA}}||{\overrightarrow {OB}}|\cos({\widehat {{\overrightarrow {OA}},{\overrightarrow {OB}}}})}}={\sqrt {a^{2}+b^{2}+2ab\cos \alpha }}\end{aligned}}}
Углы между диагональю и сторонами
cos
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O
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O
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→
^
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=
O
A
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⋅
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|
O
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|
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=
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O
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+
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→
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|
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|
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|
=
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O
A
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2
+
O
A
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⋅
O
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|
O
A
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|
|
O
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|
=
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2
+
a
b
cos
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+
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b
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{\displaystyle {\begin{aligned}\cos({\widehat {{\overrightarrow {OA}},{\overrightarrow {OC}}}})&={\frac {{\overrightarrow {OA}}\cdot {\overrightarrow {OC}}}{|{\overrightarrow {OA}}||{\overrightarrow {OC}}|}}={\frac {{\overrightarrow {OA}}\cdot ({\overrightarrow {OA}}+{\overrightarrow {OB}})}{|{\overrightarrow {OA}}||{\overrightarrow {OC}}|}}={\frac {|{\overrightarrow {OA}}|^{2}+{\overrightarrow {OA}}\cdot {\overrightarrow {OB}}}{|{\overrightarrow {OA}}||{\overrightarrow {OC}}|}}={\frac {a^{2}+ab\cos \alpha }{a{\sqrt {a^{2}+b^{2}+2ab\cos \alpha }}}}=\\&={\frac {a+b\cos \alpha }{\sqrt {a^{2}+b^{2}+2ab\cos \alpha }}}\end{aligned}}}
cos
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B
→
,
O
C
→
^
)
=
O
B
→
⋅
O
C
→
|
O
B
→
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|
O
C
→
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=
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B
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⋅
(
O
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+
O
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→
)
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B
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O
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=
O
B
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⋅
O
A
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+
|
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B
→
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2
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O
B
→
|
|
O
C
→
|
=
a
b
cos
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+
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2
b
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+
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+
2
a
b
cos
α
=
=
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cos
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+
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+
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b
cos
α
{\displaystyle {\begin{aligned}\cos({\widehat {{\overrightarrow {OB}},{\overrightarrow {OC}}}})&={\frac {{\overrightarrow {OB}}\cdot {\overrightarrow {OC}}}{|{\overrightarrow {OB}}||{\overrightarrow {OC}}|}}={\frac {{\overrightarrow {OB}}\cdot ({\overrightarrow {OA}}+{\overrightarrow {OB}})}{|{\overrightarrow {OB}}||{\overrightarrow {OC}}|}}={\frac {{\overrightarrow {OB}}\cdot {\overrightarrow {OA}}+|{\overrightarrow {OB}}|^{2}}{|{\overrightarrow {OB}}||{\overrightarrow {OC}}|}}={\frac {ab\cos \alpha +b^{2}}{b{\sqrt {a^{2}+b^{2}+2ab\cos \alpha }}}}=\\&={\frac {a\cos \alpha +b}{\sqrt {a^{2}+b^{2}+2ab\cos \alpha }}}\end{aligned}}}
