VA CIA KAIP NEREIKIA DARYTI, kad negauti klaidingo kampo:
Pavyzdžiui, duoti vektoriai a=(1; -2; 2), b=(3; 0; -4). Jų vektorinė sandauga lygi
a
×
b
=
|
i
j
k
1
−
2
2
3
0
−
4
|
=
|
−
2
2
0
−
4
|
i
−
|
1
2
3
−
4
|
j
+
|
1
−
2
3
0
|
k
=
8
i
+
10
j
+
6
k
=
(
8
;
10
;
6
)
.
{\displaystyle a\times b={\begin{vmatrix}i&j&k\\1&-2&2\\3&0&-4\end{vmatrix}}={\begin{vmatrix}-2&2\\0&-4\end{vmatrix}}i-{\begin{vmatrix}1&2\\3&-4\end{vmatrix}}j+{\begin{vmatrix}1&-2\\3&0\end{vmatrix}}k=8i+10j+6k=(8;10;6).}
Čia skaičiuodami vektorinę sandaugą panaudojome determinantą .
Vektorinės sandaugos modulis yra lygiagretainio plotas, kurį sudaro du vektoriai:
S
=
|
|
a
×
b
|
|
=
8
2
+
10
2
+
6
2
=
200
=
10
2
.
{\displaystyle S=||a\times b||={\sqrt {8^{2}+10^{2}+6^{2}}}={\sqrt {200}}=10{\sqrt {2}}.}
Trikampio plotas yra
S
=
1
2
|
|
a
×
b
|
|
=
5
2
.
{\displaystyle S={\frac {1}{2}}||a\times b||=5{\sqrt {2}}.}
Kampo tarp vektorių sinusas yra
sin
ϕ
=
|
|
a
×
b
|
|
|
|
a
|
|
⋅
|
|
b
|
|
=
10
2
3
⋅
5
=
2
2
3
,
{\displaystyle \sin \phi ={\frac {||a\times b||}{||a||\cdot ||b||}}={\frac {10{\sqrt {2}}}{3\cdot 5}}={\frac {2{\sqrt {2}}}{3}},}
ϕ
=
arcsin
2
2
3
=
1.230959417
{\displaystyle \phi =\arcsin {\frac {2{\sqrt {2}}}{3}}=1.230959417}
ϕ
=
70
,
52877937
{\displaystyle \phi =70,52877937}
|
|
a
|
|
=
1
2
+
(
−
2
)
2
+
2
2
=
9
=
3
,
{\displaystyle ||a||={\sqrt {1^{2}+(-2)^{2}+2^{2}}}={\sqrt {9}}=3,}
|
|
b
|
|
=
3
2
+
0
2
+
(
−
4
)
2
=
25
=
5.
{\displaystyle ||b||={\sqrt {3^{2}+0^{2}+(-4)^{2}}}={\sqrt {25}}=5.}
Taikydami kosinusų toeremą ir Herono formulę patikrinsime ar kampas
ϕ
{\displaystyle \phi }
S surasti teisingai. Atkarpos f ilgis iš taško a=(1; -2; 2) iki taško b=(3; 0; -4) yra lygus
f
=
(
1
−
3
)
2
+
(
−
2
−
0
)
2
+
(
2
−
(
−
4
)
)
2
=
4
+
4
+
36
=
44
=
2
11
=
6.633249581.
{\displaystyle f={\sqrt {(1-3)^{2}+(-2-0)^{2}+(2-(-4))^{2}}}={\sqrt {4+4+36}}={\sqrt {44}}=2{\sqrt {11}}=6.633249581.}
Pagal Herono formulę randame trikampio pusperimetrį
p
=
3
+
5
+
2
11
2
=
4
+
11
=
7.31662479.
{\displaystyle p={3+5+2{\sqrt {11}} \over 2}=4+{\sqrt {11}}=7.31662479.}
S
Δ
=
p
(
p
−
a
)
(
p
−
b
)
(
p
−
f
)
=
(
4
+
11
)
(
7.31662479
−
3
)
(
7.31662479
−
5
)
(
4
+
11
−
2
11
)
=
{\displaystyle S_{\Delta }={\sqrt {p(p-a)(p-b)(p-f)}}={\sqrt {(4+{\sqrt {11}})(7.31662479-3)(7.31662479-5)(4+{\sqrt {11}}-2{\sqrt {11}})}}=}
=
7.31662479
⋅
4.31662479
⋅
2.31662479
⋅
(
4
−
11
)
=
50
=
5
2
=
7.071067812.
{\displaystyle ={\sqrt {7.31662479\cdot 4.31662479\cdot 2.31662479\cdot (4-{\sqrt {11}})}}={\sqrt {50}}=5{\sqrt {2}}=7.071067812.}
Iš kosinusų teoremos žinome, kad
f
2
=
a
2
+
b
2
−
2
a
b
cos
(
ϕ
)
{\displaystyle f^{2}\ =a^{2}+b^{2}-2ab\cos(\phi )}
cos
ϕ
=
f
2
−
a
2
−
b
2
−
2
a
b
=
(
2
11
)
2
−
3
2
−
5
2
−
2
⋅
3
⋅
5
=
44
−
9
−
25
−
30
=
10
−
30
=
−
1
3
.
{\displaystyle \cos \phi ={f^{2}-a^{2}-b^{2} \over -2ab}={(2{\sqrt {11}})^{2}-3^{2}-5^{2} \over -2\cdot 3\cdot 5}={44-9-25 \over -30}={10 \over -30}=-{1 \over 3}.}
Nereikia bandyti duprasti spavloto paveikslelio,
sin
θ
{\displaystyle \sin \theta }
n prasmes, nes ten nieko teisingo ir konkretaus nera. Šis paveikslėlis labiau tiktų apibūdinti mišriąją vektorių sandauga, kai n yra statmenas lygiagretainiui, kuris gaunamas is vektorinės sandaugos 'modulyje'
|
|
a
×
b
|
|
{\displaystyle ||a\times b||}
V
=
|
|
n
|
|
⋅
|
|
a
×
b
|
|
{\displaystyle V=||n||\cdot ||a\times b||}
Vektorinė vektorių sandauga
Grafinis vektorinės sandaugos pavaizdavimas Vektorinės vektorių sandaugos rezultatas yra vektorius. Vektorinė vektorių sandauga turi prasmę tik didesnio nei dviejų matavimų erdvėse.
Vektorių a × b sandauga yra vektorius, statmenas a ir b ir yra aprašytas taip:
a
×
b
=
‖
a
‖
‖
b
‖
sin
(
θ
)
n
,
{\displaystyle \mathbf {a} \times \mathbf {b} =\left\|\mathbf {a} \right\|\left\|\mathbf {b} \right\|\sin(\theta )\,\mathbf {n} ,}
kur
θ
{\displaystyle \theta }
a ir b , o vektorius n yra statmenas a ir b vektoriams ir betkurioms tiesiems, kurios jungia bet kuriuos vektorių a ir b taškus.
Pavyzdžiui, duoti vektoriai a=(1; -2; 0), b=(3; 0; 0), n=(0; 0; 10). Kampas tarp vektoriaus a ir b yra lygus
θ
=
1.107148718
{\displaystyle \theta =1.107148718}
a
×
b
=
|
i
j
k
1
−
2
0
3
0
0
|
=
{\displaystyle \mathbf {a} \times \mathbf {b} ={\begin{vmatrix}i&j&k\\1&-2&0\\3&0&0\end{vmatrix}}=}
=
i
⋅
(
−
2
)
⋅
0
+
j
⋅
0
⋅
3
+
k
⋅
1
⋅
0
−
i
⋅
0
⋅
0
−
j
⋅
1
⋅
0
−
k
⋅
(
−
2
)
⋅
3
=
0
i
+
0
j
−
6
k
=
(
0
;
0
;
−
6
)
.
{\displaystyle =i\cdot (-2)\cdot 0+j\cdot 0\cdot 3+k\cdot 1\cdot 0-i\cdot 0\cdot 0-j\cdot 1\cdot 0-k\cdot (-2)\cdot 3=0i+0j-6k=(0;0;-6).}
‖
a
‖
=
1
2
+
(
−
2
)
2
+
0
2
=
5
≈
2.236067978.
{\displaystyle \left\|\mathbf {a} \right\|={\sqrt {1^{2}+(-2)^{2}+0^{2}}}={\sqrt {5}}\approx 2.236067978.}
‖
b
‖
=
3
2
+
0
2
+
0
2
=
9
=
3.
{\displaystyle \left\|\mathbf {b} \right\|={\sqrt {3^{2}+0^{2}+0^{2}}}={\sqrt {9}}=3.}
‖
a
‖
‖
b
‖
sin
(
θ
)
=
3
5
=
6.708203933.
{\displaystyle \left\|\mathbf {a} \right\|\left\|\mathbf {b} \right\|\sin(\theta )=3{\sqrt {5}}=6.708203933.}
‖
a
‖
‖
b
‖
sin
(
θ
)
=
3
5
⋅
sin
1.107148718
=
6.708203933
⋅
0.877913565
=
5.88922323.
{\displaystyle \left\|\mathbf {a} \right\|\left\|\mathbf {b} \right\|\sin(\theta )=3{\sqrt {5}}\cdot \sin 1.107148718=6.708203933\cdot 0.877913565=5.88922323.}
Kadangi vektorinė sandauga keičia ženklą esant veidrodiniam atspindžiui (P-simetrija ), jos rezultatas kartais vadinamas pseudo-vektoriumi .
Nei vienas is vektoriniu budu nedave teisingo atsakymo ir atsakymas pats keistas gavosi per Herono formule, tai greiciausiai blogas pavyzdis, t.y. vektorius c nestatmenas plokstumai ant kurios guli vektoriai a ir b. Problema yra tame, kad neaisku kuris vektorius yra statmenas kuriems.
Duoti vektoriai a=(1; 2; -2), b=(1; -2; 1), c=(1; -2; 3), kurių pradžios koordinatės yra (0; 0; 0). Rasime gretasienio tūrį :
V
=
(
a
×
b
)
⋅
c
=
|
a
x
a
y
a
z
b
z
b
y
b
z
c
x
c
y
c
z
|
=
c
x
⋅
(
−
1
)
3
+
1
|
a
y
a
z
b
y
b
z
|
+
c
y
⋅
(
−
1
)
3
+
2
|
a
x
a
z
b
x
b
z
|
+
c
z
⋅
(
−
1
)
3
+
3
|
a
x
a
y
b
x
b
y
|
=
{\displaystyle V=(a\times b)\cdot c={\begin{vmatrix}a_{x}&a_{y}&a_{z}\\b_{z}&b_{y}&b_{z}\\c_{x}&c_{y}&c_{z}\end{vmatrix}}=c_{x}\cdot (-1)^{3+1}{\begin{vmatrix}a_{y}&a_{z}\\b_{y}&b_{z}\end{vmatrix}}+c_{y}\cdot (-1)^{3+2}{\begin{vmatrix}a_{x}&a_{z}\\b_{x}&b_{z}\end{vmatrix}}+c_{z}\cdot (-1)^{3+3}{\begin{vmatrix}a_{x}&a_{y}\\b_{x}&b_{y}\end{vmatrix}}=}
=
|
1
2
−
2
1
−
2
1
1
−
2
3
|
=
|
1
2
−
2
2
0
−
1
2
0
1
|
=
2
⋅
(
−
1
)
1
+
2
|
2
−
1
2
1
|
=
−
8.
{\displaystyle ={\begin{vmatrix}1&2&-2\\1&-2&1\\1&-2&3\end{vmatrix}}={\begin{vmatrix}1&2&-2\\2&0&-1\\2&0&1\end{vmatrix}}=2\cdot (-1)^{1+2}{\begin{vmatrix}2&-1\\2&1\end{vmatrix}}=-8.}
Gretasienio tūris yra |-8|=8. Taip pat galima skaičiuot taip:
V
=
|
1
2
−
2
1
−
2
1
1
−
2
3
|
=
{\displaystyle V={\begin{vmatrix}1&2&-2\\1&-2&1\\1&-2&3\end{vmatrix}}=}
=
1
⋅
(
−
2
)
⋅
3
+
2
⋅
1
⋅
1
+
(
−
2
)
⋅
1
⋅
(
−
2
)
−
(
−
2
)
⋅
(
−
2
)
⋅
1
−
2
⋅
1
⋅
3
−
1
⋅
1
⋅
(
−
2
)
=
−
6
+
2
+
4
−
4
−
6
+
2
=
−
8.
{\displaystyle =1\cdot (-2)\cdot 3+2\cdot 1\cdot 1+(-2)\cdot 1\cdot (-2)-(-2)\cdot (-2)\cdot 1-2\cdot 1\cdot 3-1\cdot 1\cdot (-2)=-6+2+4-4-6+2=-8.}
Patikriname ar atsakymas bus toks pat naudojant vektorine sandauga sudaugina su statmeno vektoriaus ilgiu:
a
×
b
=
|
i
j
k
1
2
−
2
1
−
2
1
|
=
{\displaystyle a\times b={\begin{vmatrix}i&j&k\\1&2&-2\\1&-2&1\end{vmatrix}}=}
=
i
⋅
2
⋅
1
+
j
⋅
(
−
2
)
⋅
1
+
k
⋅
1
⋅
(
−
2
)
−
i
⋅
(
−
2
)
⋅
(
−
2
)
−
j
⋅
1
⋅
1
−
k
⋅
2
⋅
1
=
{\displaystyle =i\cdot 2\cdot 1+j\cdot (-2)\cdot 1+k\cdot 1\cdot (-2)-i\cdot (-2)\cdot (-2)-j\cdot 1\cdot 1-k\cdot 2\cdot 1=}
=
2
i
−
4
i
−
2
j
−
j
−
2
k
−
2
k
=
−
2
i
−
3
j
−
4
k
=
(
−
2
;
−
3
;
−
4
)
.
{\displaystyle =2i-4i-2j-j-2k-2k=-2i-3j-4k=(-2;-3;-4).}
|
|
a
×
b
|
|
=
(
−
2
)
2
+
(
−
3
)
2
+
(
−
4
)
2
=
4
+
9
+
16
=
29
=
5.385164807.
{\displaystyle ||a\times b||={\sqrt {(-2)^{2}+(-3)^{2}+(-4)^{2}}}={\sqrt {4+9+16}}={\sqrt {29}}=5.385164807.}
|
|
c
|
|
=
1
2
+
(
−
2
)
2
+
3
2
=
1
+
4
+
9
=
14
.
{\displaystyle ||c||={\sqrt {1^{2}+(-2)^{2}+3^{2}}}={\sqrt {1+4+9}}={\sqrt {14}}.}
V
=
|
|
a
×
b
|
|
⋅
|
|
c
|
|
=
29
⋅
14
=
406
=
20.14944168.
{\displaystyle V=||a\times b||\cdot ||c||={\sqrt {29}}\cdot {\sqrt {14}}={\sqrt {406}}=20.14944168.}
Patikriname taikydami Herono formulę.
|
|
a
|
|
=
1
2
+
2
2
+
(
−
2
)
2
=
1
+
4
+
4
=
9.
{\displaystyle ||a||={\sqrt {1^{2}+2^{2}+(-2)^{2}}}={\sqrt {1+4+4}}=9.}
|
|
b
|
|
=
1
2
+
(
−
2
)
2
+
1
2
=
1
+
4
+
1
=
6
=
2.449489743.
{\displaystyle ||b||={\sqrt {1^{2}+(-2)^{2}+1^{2}}}={\sqrt {1+4+1}}={\sqrt {6}}=2.449489743.}
p
=
a
+
b
+
c
2
=
9
+
6
+
14
2
=
7.595573565.
{\displaystyle p={a+b+c \over 2}={9+{\sqrt {6}}+{\sqrt {14}} \over 2}=7.595573565.}
S
=
p
(
p
−
a
)
(
p
−
b
)
(
p
−
c
)
=
7.595573565
(
7.595573565
−
9
)
(
7.595573565
−
6
)
(
7.595573565
−
14
)
=
{\displaystyle S={\sqrt {p(p-a)(p-b)(p-c)}}={\sqrt {7.595573565(7.595573565-9)(7.595573565-{\sqrt {6}})(7.595573565-{\sqrt {14}})}}=}
=
7.595573565
(
−
1.404426435
)
5.146083822
⋅
3.853916178
=
−
211.5625
=
211.5625
=
14.54518821.
{\displaystyle ={\sqrt {7.595573565(-1.404426435)5.146083822\cdot 3.853916178}}={\sqrt {-211.5625}}={\sqrt {211.5625}}=14.54518821.}
Rasime kampą tarp vektoriaus a=(1; 2; -2) ir vektoriaus b=(1; -2; 1).
cos
ϕ
=
a
⋅
b
|
|
a
|
|
⋅
|
|
b
|
|
=
1
⋅
1
+
2
⋅
(
−
2
)
+
(
−
2
)
⋅
1
1
2
+
2
2
+
(
−
2
)
2
⋅
1
2
+
(
−
2
)
2
+
1
2
=
1
−
4
−
2
1
+
4
+
4
⋅
1
+
4
+
1
=
{\displaystyle \cos \phi ={\frac {a\cdot b}{||a||\cdot ||b||}}={\frac {1\cdot 1+2\cdot (-2)+(-2)\cdot 1}{{\sqrt {1^{2}+2^{2}+(-2)^{2}}}\cdot {\sqrt {1^{2}+(-2)^{2}+1^{2}}}}}={\frac {1-4-2}{{\sqrt {1+4+4}}\cdot {\sqrt {1+4+1}}}}=}
=
−
5
9
⋅
6
=
−
5
3
⋅
6
=
−
0
,
680413817.
{\displaystyle ={\frac {-5}{{\sqrt {9}}\cdot {\sqrt {6}}}}={\frac {-5}{3\cdot {\sqrt {6}}}}=-0,680413817.}
ϕ
=
arccos
(
−
0
,
680413817
)
=
0
,
822469154
{\displaystyle \phi =\arccos(-0,680413817)=0,822469154}
Rasime kampą tarp vektoriaus a=(1; 2; -2) ir vektoriaus c=(1; -2; 3).
cos
ϕ
=
a
⋅
c
|
|
a
|
|
⋅
|
|
c
|
|
=
1
⋅
1
+
2
⋅
(
−
2
)
+
(
−
2
)
⋅
3
1
2
+
2
2
+
(
−
2
)
2
⋅
1
2
+
(
−
2
)
2
+
3
2
=
1
−
4
−
6
1
+
4
+
4
⋅
1
+
4
+
9
=
{\displaystyle \cos \phi ={\frac {a\cdot c}{||a||\cdot ||c||}}={\frac {1\cdot 1+2\cdot (-2)+(-2)\cdot 3}{{\sqrt {1^{2}+2^{2}+(-2)^{2}}}\cdot {\sqrt {1^{2}+(-2)^{2}+3^{2}}}}}={\frac {1-4-6}{{\sqrt {1+4+4}}\cdot {\sqrt {1+4+9}}}}=}
=
−
9
9
⋅
14
=
−
9
3
⋅
14
=
−
0
,
801783725.
{\displaystyle ={\frac {-9}{{\sqrt {9}}\cdot {\sqrt {14}}}}={\frac {-9}{3\cdot {\sqrt {14}}}}=-0,801783725.}
ϕ
=
arccos
(
−
0
,
801783725
)
=
0
,
640522312
{\displaystyle \phi =\arccos(-0,801783725)=0,640522312}
Rasime kampą tarp vektoriaus b=(1; -2; 1) ir vektoriaus c=(1; -2; 3).
cos
ϕ
=
b
⋅
c
|
|
b
|
|
⋅
|
|
c
|
|
=
1
⋅
1
+
(
−
2
)
⋅
(
−
2
)
+
1
⋅
3
1
2
+
(
−
2
)
2
+
1
2
⋅
1
2
+
(
−
2
)
2
+
3
2
=
1
+
4
+
3
1
+
4
+
1
⋅
1
+
4
+
9
=
{\displaystyle \cos \phi ={\frac {b\cdot c}{||b||\cdot ||c||}}={\frac {1\cdot 1+(-2)\cdot (-2)+1\cdot 3}{{\sqrt {1^{2}+(-2)^{2}+1^{2}}}\cdot {\sqrt {1^{2}+(-2)^{2}+3^{2}}}}}={\frac {1+4+3}{{\sqrt {1+4+1}}\cdot {\sqrt {1+4+9}}}}=}
=
8
6
⋅
14
=
8
84
=
0
,
095238095.
{\displaystyle ={\frac {8}{{\sqrt {6}}\cdot {\sqrt {14}}}}={\frac {8}{\sqrt {84}}}=0,095238095.}
ϕ
=
arccos
0
,
095238095
=
1
,
475413668
{\displaystyle \phi =\arccos 0,095238095=1,475413668}
