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
Δ
∇
=
|
|
a
×
b
|
|
=
8
2
+
10
2
+
6
2
=
200
=
10
2
.
{\displaystyle S_{\Delta \nabla }=||a\times b||={\sqrt {8^{2}+10^{2}+6^{2}}}={\sqrt {200}}=10{\sqrt {2}}.}
Trikampio plotas yra
S
Δ
=
1
2
|
|
a
×
b
|
|
=
5
2
.
{\displaystyle S_{\Delta }={\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}
radianų arba
ϕ
=
70
,
52877937
{\displaystyle \phi =70,52877937}
laipsnių, kur
|
|
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 }
ir trikampio plotas 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}.}
ϕ
=
arccos
(
−
1
3
)
=
1.910633236
{\displaystyle \phi =\arccos(-{1 \over 3})=1.910633236}
radiano arba 109.4712206 laipsnio.
Idomus faktas, jog
ϕ
=
arccos
1
3
=
1.230959417
{\displaystyle \phi =\arccos {1 \over 3}=1.230959417}
radiano arba 70.52877937 laipsnio.
Dar kitas būdas patikrinti:
cos
ϕ
=
a
⋅
b
‖
a
‖
⋅
‖
b
‖
=
1
⋅
3
+
(
−
2
)
⋅
0
+
2
⋅
(
−
4
)
1
2
+
(
−
2
)
2
+
2
2
⋅
3
2
+
0
2
+
(
−
4
)
2
=
3
+
0
−
8
1
+
4
+
4
⋅
9
+
0
+
16
=
{\displaystyle \cos \phi ={\frac {\mathbf {a} \cdot \mathbf {b} }{\|\mathbf {a} \|\cdot \|\mathbf {b} \|}}={\frac {1\cdot 3+(-2)\cdot 0+2\cdot (-4)}{{\sqrt {1^{2}+(-2)^{2}+2^{2}}}\cdot {\sqrt {3^{2}+0^{2}+(-4)^{2}}}}}={\frac {3+0-8}{{\sqrt {1+4+4}}\cdot {\sqrt {9+0+16}}}}=}
=
−
5
9
⋅
25
=
−
5
15
=
−
1
3
.
{\displaystyle ={\frac {-5}{{\sqrt {9}}\cdot {\sqrt {25}}}}={\frac {-5}{15}}=-{\frac {1}{3}}.}
Nereikia bandyti duprasti spavloto paveikslelio,
sin
θ
{\displaystyle \sin \theta }
ir 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||}
. Tuomet lygiagretainio gretasienio tūris yra
V
=
|
|
n
|
|
⋅
|
|
a
×
b
|
|
{\displaystyle V=||n||\cdot ||a\times b||}
. Zemiau pateiktas bandymas ka nors suprast:
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 }
yra kampas tarp vektorių 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}
arba 63,43494882 laipsnių.
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 .
Angliskoje wikipedijoje raso http://en.wikipedia.org/wiki/Cross_product#Alternative_formulation , kad
‖
a
×
b
‖
=
‖
a
‖
‖
b
‖
sin
θ
.
{\displaystyle \|\mathbf {a} \times \mathbf {b} \|=\|\mathbf {a} \|\|\mathbf {b} \|\sin \theta \ .}
Taciau pagal formule is skyriaus "kampas tarp vektoriu":
cos
θ
=
a
⋅
b
‖
a
‖
⋅
‖
b
‖
=
1
⋅
3
+
(
−
2
)
⋅
0
+
0
⋅
0
1
2
+
(
−
2
)
2
+
0
2
⋅
3
2
+
0
2
+
0
2
=
3
1
+
4
+
0
⋅
9
+
0
+
0
=
{\displaystyle \cos \theta ={\frac {\mathbf {a} \cdot \mathbf {b} }{\|\mathbf {a} \|\cdot \|\mathbf {b} \|}}={\frac {1\cdot 3+(-2)\cdot 0+0\cdot 0}{{\sqrt {1^{2}+(-2)^{2}+0^{2}}}\cdot {\sqrt {3^{2}+0^{2}+0^{2}}}}}={\frac {3}{{\sqrt {1+4+0}}\cdot {\sqrt {9+0+0}}}}=}
=
3
5
⋅
3
=
1
5
=
0.447213595.
{\displaystyle ={\frac {3}{{\sqrt {5}}\cdot 3}}={\frac {1}{\sqrt {5}}}=0.447213595.}
θ
=
arccos
1
5
=
arccos
0.447213595
=
1.107148718
{\displaystyle \theta =\arccos {\frac {1}{\sqrt {5}}}=\arccos 0.447213595=1.107148718}
radiano arba 63,43494882 laipsnių.
Taikydami kosinusų toeremą patikrinsime ar kampas
θ
{\displaystyle \theta }
surastas teisingai. Atkarpos t ilgis iš taško a=(1; -2; 0) iki taško b=(3; 0; 0) yra lygus
t
=
(
1
−
3
)
2
+
(
−
2
−
0
)
2
+
(
0
−
0
)
2
=
4
+
4
+
0
=
8
=
2
2
=
2.828427125.
{\displaystyle t={\sqrt {(1-3)^{2}+(-2-0)^{2}+(0-0)^{2}}}={\sqrt {4+4+0}}={\sqrt {8}}=2{\sqrt {2}}=2.828427125.}
Iš kosinusų teoremos žinome, kad
t
2
=
‖
a
‖
2
+
‖
b
‖
2
−
2
‖
a
‖
⋅
‖
b
‖
cos
θ
{\displaystyle t^{2}\ =\|\mathbf {a} \|^{2}+\|\mathbf {b} \|^{2}-2\|\mathbf {a} \|\cdot \|\mathbf {b} \|\cos \theta }
;
cos
θ
=
t
2
−
‖
a
‖
2
−
‖
b
‖
2
−
2
‖
a
‖
‖
b
‖
=
(
8
)
2
−
5
2
−
3
2
−
2
⋅
5
⋅
3
=
8
−
5
−
9
−
6
5
=
−
6
−
6
5
=
1
5
.
{\displaystyle \cos \theta ={t^{2}-\|\mathbf {a} \|^{2}-\|\mathbf {b} \|^{2} \over -2\|\mathbf {a} \|\|\mathbf {b} \|}={({\sqrt {8}})^{2}-{\sqrt {5}}^{2}-3^{2} \over -2\cdot {\sqrt {5}}\cdot 3}={8-5-9 \over -6{\sqrt {5}}}={-6 \over -6{\sqrt {5}}}={1 \over {\sqrt {5}}}.}
‖
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.}
Reiškia jei
‖
a
×
b
‖
=
‖
a
‖
‖
b
‖
sin
θ
{\displaystyle \|\mathbf {a} \times \mathbf {b} \|=\|\mathbf {a} \|\|\mathbf {b} \|\sin \theta \ }
tai
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
×
b
‖
=
0
2
+
0
2
+
(
−
6
)
2
=
36
=
6.
{\displaystyle \|\mathbf {a} \times \mathbf {b} \|={\sqrt {0^{2}+0^{2}+(-6)^{2}}}={\sqrt {36}}=6.}
sin
θ
=
‖
a
×
b
‖
‖
a
‖
‖
b
‖
=
6
5
⋅
3
=
2
5
=
0.894427191
;
{\displaystyle \sin \theta ={\frac {\|\mathbf {a} \times \mathbf {b} \|}{\|\mathbf {a} \|\|\mathbf {b} \|}}={\frac {6}{{\sqrt {5}}\cdot 3}}={\frac {2}{\sqrt {5}}}=0.894427191;}
θ
=
arcsin
2
5
=
arcsin
0.894427191
=
1.107148718
{\displaystyle \theta =\arcsin {\frac {2}{\sqrt {5}}}=\arcsin 0.894427191=1.107148718}
radiano arba 63.43494882 laipsnio.
Nei vienas is vektoriniu budu nedave teisingo atsakymo ir atsakymas pats keistas gavosi per Herono formule, tai greiciausiai blogas pavyzdys, 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}
arba 47,12401133 laipsnių.
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}
arba 36,6992252 laipsnių.
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}
arba 84,5349762 laipsnių.
Pavyzdis. Trikampės piramidės viršūnės yra taškai A (3; -1; 5), B (5; 2; 6), C (-1; 3; 4) ir D (7; 3; -1). Apskaičiuosime šios piramidės tūrį ir aukštinės, nuleistos iš taško D į sieną ABC , ilgį.
Sprendimas . Nubraižykime tris vektorius, išeinančius iš vieno taško, pavyzdžiui, iš taško A : AB , AC , AD . Žinome, kad trikampės piramidės tūris
V
p
i
r
.
=
1
6
|
(
A
B
×
A
C
)
⋅
A
D
|
.
{\displaystyle V_{pir.}={\frac {1}{6}}|(AB\times AC)\cdot AD|.}
Randame vektorių AB , AC ir AD koordinates:
AB =B-A=(5-3; 2-(-1); 6-5)={2; 3; 1},
AC =C-A=(-1-3; 3-(-1); 4-5)={-4; 4; -1},
AD =D-A=(7-3; 3-(-1); -1-5)={4; 4; -6}.
Apskaičiuojame mišriąją gautų vektorių sandaugą:
(
A
B
×
A
C
)
⋅
A
D
=
|
2
3
1
−
4
4
−
1
4
4
−
6
|
=
2
⋅
(
−
1
)
1
+
1
|
4
−
1
4
−
6
|
+
3
⋅
(
−
1
)
1
+
2
|
−
4
−
1
4
−
6
|
+
1
⋅
(
−
1
)
1
+
3
|
−
4
4
4
4
|
=
{\displaystyle (AB\times AC)\cdot AD={\begin{vmatrix}2&3&1\\-4&4&-1\\4&4&-6\end{vmatrix}}=2\cdot (-1)^{1+1}{\begin{vmatrix}4&-1\\4&-6\end{vmatrix}}+3\cdot (-1)^{1+2}{\begin{vmatrix}-4&-1\\4&-6\end{vmatrix}}+1\cdot (-1)^{1+3}{\begin{vmatrix}-4&4\\4&4\end{vmatrix}}=}
=
2
⋅
(
−
1
)
2
⋅
(
4
⋅
(
−
6
)
−
(
−
1
)
⋅
4
)
+
3
⋅
(
−
1
)
3
⋅
(
(
−
4
)
⋅
(
−
6
)
−
(
−
1
)
⋅
4
)
+
1
⋅
(
−
1
)
4
⋅
(
(
−
4
)
⋅
4
−
4
⋅
4
)
=
{\displaystyle =2\cdot (-1)^{2}\cdot (4\cdot (-6)-(-1)\cdot 4)+3\cdot (-1)^{3}\cdot ((-4)\cdot (-6)-(-1)\cdot 4)+1\cdot (-1)^{4}\cdot ((-4)\cdot 4-4\cdot 4)=}
=
2
⋅
(
−
24
+
4
)
−
3
⋅
(
24
+
4
)
+
1
⋅
(
−
16
−
16
)
=
2
⋅
(
−
20
)
−
3
⋅
28
−
32
=
−
40
−
84
−
32
=
−
156.
{\displaystyle =2\cdot (-24+4)-3\cdot (24+4)+1\cdot (-16-16)=2\cdot (-20)-3\cdot 28-32=-40-84-32=-156.}
Tada trikampės piramidės tūris
V
p
i
r
.
=
1
6
⋅
|
−
156
|
=
156
6
=
26.
{\displaystyle V_{pir.}={\frac {1}{6}}\cdot |-156|={\frac {156}{6}}=26.}
Norėdami rasti piramidės aukštinę h , pritaikykime kitą piramidės tūrio formulę:
V
p
i
r
.
=
1
3
⋅
S
Δ
A
B
C
⋅
h
.
{\displaystyle V_{pir.}={\frac {1}{3}}\cdot S_{\Delta ABC}\cdot h.}
Bet
S
Δ
A
B
C
=
1
2
⋅
‖
A
B
×
A
C
‖
,
{\displaystyle S_{\Delta ABC}={\frac {1}{2}}\cdot \|AB\times AC\|,}
todėl
V
p
i
r
.
=
1
6
⋅
‖
A
B
×
A
C
‖
⋅
h
.
{\displaystyle V_{pir.}={\frac {1}{6}}\cdot \|AB\times AC\|\cdot h.}
Sulygindami šią formulę su ankstesne piramidės formule, gauname:
1
6
|
(
A
B
×
A
C
)
⋅
A
D
|
=
1
6
⋅
‖
A
B
×
A
C
‖
⋅
h
;
{\displaystyle {\frac {1}{6}}|(AB\times AC)\cdot AD|={\frac {1}{6}}\cdot \|AB\times AC\|\cdot h;}
h
=
|
(
A
B
×
A
C
)
⋅
A
D
|
‖
A
B
×
A
C
‖
=
156
453
=
7.329519377
,
{\displaystyle h={\frac {|(AB\times AC)\cdot AD|}{\|AB\times AC\|}}={\frac {156}{\sqrt {453}}}=7.329519377,}
kur
A
B
×
A
C
=
|
i
j
k
2
3
1
−
4
4
−
1
|
=
i
⋅
(
−
1
)
1
+
1
|
3
1
4
−
1
|
+
j
⋅
(
−
1
)
1
+
2
|
2
1
−
4
−
1
|
+
k
⋅
(
−
1
)
1
+
3
|
2
3
−
4
4
|
=
{\displaystyle AB\times AC={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\2&3&1\\-4&4&-1\end{vmatrix}}=\mathbf {i} \cdot (-1)^{1+1}{\begin{vmatrix}3&1\\4&-1\end{vmatrix}}+\mathbf {j} \cdot (-1)^{1+2}{\begin{vmatrix}2&1\\-4&-1\end{vmatrix}}+\mathbf {k} \cdot (-1)^{1+3}{\begin{vmatrix}2&3\\-4&4\end{vmatrix}}=}
=
i
⋅
(
3
⋅
(
−
1
)
−
1
⋅
4
)
−
j
⋅
(
2
⋅
(
−
1
)
−
1
⋅
(
−
4
)
)
+
k
⋅
(
2
⋅
4
−
3
⋅
(
−
4
)
)
=
i
⋅
(
−
3
−
4
)
−
j
⋅
(
−
2
+
4
)
+
k
⋅
(
8
+
12
)
=
−
7
i
−
2
j
+
20
k
=
(
−
7
;
−
2
;
20
)
;
{\displaystyle =\mathbf {i} \cdot (3\cdot (-1)-1\cdot 4)-\mathbf {j} \cdot (2\cdot (-1)-1\cdot (-4))+\mathbf {k} \cdot (2\cdot 4-3\cdot (-4))=\mathbf {i} \cdot (-3-4)-\mathbf {j} \cdot (-2+4)+\mathbf {k} \cdot (8+12)=-7\mathbf {i} -2\mathbf {j} +20\mathbf {k} =(-7;-2;20);}
‖
A
B
×
A
C
‖
=
(
−
7
)
2
+
(
−
2
)
2
+
20
2
=
49
+
4
+
400
=
453
.
{\displaystyle \|AB\times AC\|={\sqrt {(-7)^{2}+(-2)^{2}+20^{2}}}={\sqrt {49+4+400}}={\sqrt {453}}.}
Toliau pabandysime įrodyti, kad piramidės tūris surastas teisingai. Mes jau turime vieno lygiagretainio plotą į kurį įeina trikampio ABC plotas:
S
Δ
∇
1
=
2
⋅
S
Δ
A
B
C
=
2
⋅
1
2
⋅
‖
A
B
×
A
C
‖
=
‖
A
B
×
A
C
‖
=
453
=
21.28379665.
{\displaystyle S_{\Delta \nabla 1}=2\cdot S_{\Delta ABC}=2\cdot {\frac {1}{2}}\cdot \|AB\times AC\|=\|AB\times AC\|={\sqrt {453}}=21.28379665.}
Antro lygiagretainio plotas yra
S
Δ
∇
2
=
2
⋅
S
Δ
A
C
D
=
‖
A
D
×
A
C
‖
.
{\displaystyle S_{\Delta \nabla 2}=2\cdot S_{\Delta ACD}=\|AD\times AC\|.}
A
D
×
A
C
=
|
i
j
k
4
4
−
6
−
4
4
−
1
|
=
i
⋅
(
−
1
)
1
+
1
|
4
−
6
4
−
1
|
+
j
⋅
(
−
1
)
1
+
2
|
4
−
6
−
4
−
1
|
+
k
⋅
(
−
1
)
1
+
3
|
4
4
−
4
4
|
=
{\displaystyle AD\times AC={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\4&4&-6\\-4&4&-1\end{vmatrix}}=\mathbf {i} \cdot (-1)^{1+1}{\begin{vmatrix}4&-6\\4&-1\end{vmatrix}}+\mathbf {j} \cdot (-1)^{1+2}{\begin{vmatrix}4&-6\\-4&-1\end{vmatrix}}+\mathbf {k} \cdot (-1)^{1+3}{\begin{vmatrix}4&4\\-4&4\end{vmatrix}}=}
=
i
⋅
(
4
⋅
(
−
1
)
−
(
−
6
)
⋅
4
)
−
j
⋅
(
4
⋅
(
−
1
)
−
(
−
6
)
⋅
(
−
4
)
)
+
k
⋅
(
4
⋅
4
−
4
⋅
(
−
4
)
)
=
i
⋅
(
−
4
+
24
)
−
j
⋅
(
−
4
−
24
)
+
k
⋅
(
16
+
16
)
=
20
i
+
28
j
+
32
k
=
(
20
;
28
;
32
)
;
{\displaystyle =\mathbf {i} \cdot (4\cdot (-1)-(-6)\cdot 4)-\mathbf {j} \cdot (4\cdot (-1)-(-6)\cdot (-4))+\mathbf {k} \cdot (4\cdot 4-4\cdot (-4))=\mathbf {i} \cdot (-4+24)-\mathbf {j} \cdot (-4-24)+\mathbf {k} \cdot (16+16)=20\mathbf {i} +28\mathbf {j} +32\mathbf {k} =(20;28;32);}
S
Δ
∇
2
=
‖
A
D
×
A
C
‖
=
20
2
+
28
2
+
32
2
=
400
+
784
+
1024
=
2208
=
46.9893605.
{\displaystyle S_{\Delta \nabla 2}=\|AD\times AC\|={\sqrt {20^{2}+28^{2}+32^{2}}}={\sqrt {400+784+1024}}={\sqrt {2208}}=46.9893605.}
Dabar, jeigu tiesė CD =t yra statmena plokštumai ACD , tai piramidės tūris yra lygus
V
p
i
r
.
=
1
3
⋅
1
2
⋅
‖
A
B
×
A
C
‖
⋅
t
=
1
6
⋅
453
⋅
89
=
33.46515601
,
{\displaystyle V_{pir.}={\frac {1}{3}}\cdot {\frac {1}{2}}\cdot \|AB\times AC\|\cdot t={\frac {1}{6}}\cdot {\sqrt {453}}\cdot {\sqrt {89}}=33.46515601,}
kur
C
D
=
t
=
(
7
−
(
−
1
)
)
2
+
(
3
−
3
)
2
+
(
−
1
−
4
)
2
=
8
2
+
0
2
+
(
−
5
)
2
=
64
+
0
+
25
=
89
=
9.433981132.
{\displaystyle CD=t={\sqrt {(7-(-1))^{2}+(3-3)^{2}+(-1-4)^{2}}}={\sqrt {8^{2}+0^{2}+(-5)^{2}}}={\sqrt {64+0+25}}={\sqrt {89}}=9.433981132.}
Piramidės Tūris gavosi neteisingas (turėjo būti 26). Vadinasi atkarpa CD nėra stati plokštumai ABC. Bet galbūt tiesė CB =f yra stati plokštumai ACD, tai mes ir pabandysime išsiaiškinti. Randame jos ilgį:
C
B
=
f
=
(
5
−
(
−
1
)
)
2
+
(
2
−
3
)
2
+
(
6
−
4
)
2
=
6
2
+
1
+
2
2
=
36
+
1
+
4
=
41
=
6.403124237.
{\displaystyle CB=f={\sqrt {(5-(-1))^{2}+(2-3)^{2}+(6-4)^{2}}}={\sqrt {6^{2}+1+2^{2}}}={\sqrt {36+1+4}}={\sqrt {41}}=6.403124237.}
V
p
i
r
.
=
1
3
⋅
1
2
⋅
‖
A
D
×
A
C
‖
⋅
f
=
1
6
⋅
453
⋅
41
=
22.71379904.
{\displaystyle V_{pir.}={\frac {1}{3}}\cdot {\frac {1}{2}}\cdot \|AD\times AC\|\cdot f={\frac {1}{6}}\cdot {\sqrt {453}}\cdot {\sqrt {41}}=22.71379904.}
Panašu, kad tai laiko gaišimas, kad patikrinti visų tiesių statumą ir, kad tokiu būdų nepavyks įrodyti mišrios vektorių sandaugos formulės teisingumo piramidei.
Duoti vektoriai a =(3; 4; 5), b =(4; 3; 5), c =(-3; -4; 5). Vektorius c su vektoriu b sudaro beveik 90 laipsnių kampą. Vektorius c su vektorium b sudaro kampą
cos
θ
=
b
⋅
c
|
|
b
|
|
⋅
|
|
c
|
|
=
4
⋅
(
−
3
)
+
3
⋅
(
−
4
)
+
5
⋅
5
4
2
+
3
2
+
5
2
⋅
(
−
3
)
2
+
(
−
4
)
2
+
5
2
=
1
16
+
9
+
25
⋅
9
+
16
+
25
=
{\displaystyle \cos \theta ={\frac {b\cdot c}{||b||\cdot ||c||}}={\frac {4\cdot (-3)+3\cdot (-4)+5\cdot 5}{{\sqrt {4^{2}+3^{2}+5^{2}}}\cdot {\sqrt {(-3)^{2}+(-4)^{2}}}+5^{2}}}={\frac {1}{{\sqrt {16+9+25}}\cdot {\sqrt {9+16+25}}}}=}
=
1
50
⋅
50
=
1
50
=
0.02
;
{\displaystyle ={\frac {1}{{\sqrt {50}}\cdot {\sqrt {50}}}}={\frac {1}{50}}=0.02;}
θ
=
arccos
1
50
=
1.550794993
{\displaystyle \theta =\arccos {\frac {1}{50}}=1.550794993}
arba 88.854008 laipsnių. Taigi vektorius c yra beveik status abiems vektorioms, ko ir reikia norint surasti apytikslu lygiagretainio gretasienio tūrį (vektoriu c parinkti taip, kad butu status abiems vektoriams yra begalo sunku). Galime patikrinti, kad vektorius c su vektorium a tikrai sudaro 90 laipsniu kampą:
cos
ϕ
=
a
⋅
c
|
|
a
|
|
⋅
|
|
c
|
|
=
3
⋅
(
−
3
)
+
4
⋅
(
−
4
)
+
5
⋅
5
3
2
+
4
2
+
5
2
⋅
(
−
3
)
2
+
(
−
4
)
2
+
5
2
=
0
9
+
16
+
25
⋅
9
+
16
+
25
=
{\displaystyle \cos \phi ={\frac {a\cdot c}{||a||\cdot ||c||}}={\frac {3\cdot (-3)+4\cdot (-4)+5\cdot 5}{{\sqrt {3^{2}+4^{2}+5^{2}}}\cdot {\sqrt {(-3)^{2}+(-4)^{2}}}+5^{2}}}={\frac {0}{{\sqrt {9+16+25}}\cdot {\sqrt {9+16+25}}}}=}
=
0
50
⋅
50
=
0.
{\displaystyle ={\frac {0}{{\sqrt {50}}\cdot {\sqrt {50}}}}=0.}
ϕ
=
arccos
0
=
π
2
=
1.570796327
{\displaystyle \phi =\arccos 0={\pi \over 2}=1.570796327}
arba 90 laipsnių.
Rasime lygiagretainio gretasienio tūrį :
V
=
|
3
4
5
4
3
5
−
3
−
4
5
|
=
3
⋅
3
⋅
5
+
4
⋅
5
⋅
(
−
3
)
+
5
⋅
4
⋅
(
−
4
)
−
3
⋅
5
⋅
(
−
4
)
−
4
⋅
4
⋅
5
−
5
⋅
2
⋅
(
−
3
)
=
{\displaystyle V={\begin{vmatrix}3&4&5\\4&3&5\\-3&-4&5\end{vmatrix}}=3\cdot 3\cdot 5+4\cdot 5\cdot (-3)+5\cdot 4\cdot (-4)-3\cdot 5\cdot (-4)-4\cdot 4\cdot 5-5\cdot 2\cdot (-3)=}
=
45
−
60
−
80
+
60
−
80
+
30
=
−
85.
{\displaystyle =45-60-80+60-80+30=-85.}
Patikriname:
a
×
b
=
|
i
j
k
3
4
5
4
3
5
|
=
{\displaystyle a\times b={\begin{vmatrix}i&j&k\\3&4&5\\4&3&5\end{vmatrix}}=}
=
i
⋅
4
⋅
5
+
j
⋅
5
⋅
4
+
k
⋅
3
⋅
3
−
i
⋅
5
⋅
3
−
j
⋅
3
⋅
5
−
k
⋅
4
⋅
4
=
{\displaystyle =i\cdot 4\cdot 5+j\cdot 5\cdot 4+k\cdot 3\cdot 3-i\cdot 5\cdot 3-j\cdot 3\cdot 5-k\cdot 4\cdot 4=}
=
20
i
−
15
i
+
20
j
−
15
j
+
9
k
−
16
k
=
0
i
+
0
j
−
4
k
=
(
5
;
5
;
−
7
)
.
{\displaystyle =20i-15i+20j-15j+9k-16k=0i+0j-4k=(5;5;-7).}
|
|
a
×
b
|
|
=
5
2
+
5
2
+
(
−
7
)
2
=
25
+
25
+
49
=
99
=
9.949874371.
{\displaystyle ||a\times b||={\sqrt {5^{2}+5^{2}+(-7)^{2}}}={\sqrt {25+25+49}}={\sqrt {99}}=9.949874371.}
|
|
c
|
|
=
(
−
3
)
2
+
(
−
4
)
2
+
5
2
=
50
=
7.071067812.
{\displaystyle ||c||={\sqrt {(-3)^{2}+(-4)^{2}+5^{2}}}={\sqrt {50}}=7.071067812.}
V
=
|
|
a
×
b
|
|
⋅
|
|
c
|
|
=
99
⋅
50
=
4950
=
70.3562364.
{\displaystyle V=||a\times b||\cdot ||c||={\sqrt {99}}\cdot {\sqrt {50}}={\sqrt {4950}}=70.3562364.}
Patikriname taikydami Herono formulę.
|
|
a
|
|
=
3
2
+
4
2
+
5
2
=
50
=
7.071067812.
{\displaystyle ||a||={\sqrt {3^{2}+4^{2}+5^{2}}}={\sqrt {50}}=7.071067812.}
|
|
b
|
|
=
4
2
+
3
2
+
5
2
=
50
=
7.071067812.
{\displaystyle ||b||={\sqrt {4^{2}+3^{2}+5^{2}}}={\sqrt {50}}=7.071067812.}
Atstumas tarp taškų a=(3; 4; 5) ir b=(4; 3; 5) yra lygus:
f
=
(
3
−
4
)
2
+
(
4
−
3
)
2
+
(
5
−
5
)
2
=
2
=
1.414213562.
{\displaystyle f={\sqrt {(3-4)^{2}+(4-3)^{2}+(5-5)^{2}}}={\sqrt {2}}=1.414213562.}
p
=
a
+
b
+
f
2
=
50
+
50
+
2
2
=
50
+
2
2
=
7.778174593.
{\displaystyle p={a+b+f \over 2}={{\sqrt {50}}+{\sqrt {50}}+{\sqrt {2}} \over 2}={\sqrt {50}}+{{\sqrt {2}} \over 2}=7.778174593.}
S
Δ
=
p
(
p
−
a
)
(
p
−
b
)
(
p
−
f
)
=
{\displaystyle S_{\Delta }={\sqrt {p(p-a)(p-b)(p-f)}}=}
=
7.778174593
(
50
+
2
2
−
50
)
(
50
+
2
2
−
50
)
(
50
+
2
2
−
2
)
=
{\displaystyle ={\sqrt {7.778174593({\sqrt {50}}+{{\sqrt {2}} \over 2}-{\sqrt {50}})({\sqrt {50}}+{{\sqrt {2}} \over 2}-{\sqrt {50}})({\sqrt {50}}+{{\sqrt {2}} \over 2}-{\sqrt {2}})}}=}
=
7.778174593
⋅
2
2
⋅
2
2
⋅
(
50
−
2
2
)
=
7.778174593
⋅
1
2
⋅
6.363961031
=
24.75
=
4.974937186.
{\displaystyle ={\sqrt {7.778174593\cdot {{\sqrt {2}} \over 2}\cdot {{\sqrt {2}} \over 2}\cdot ({\sqrt {50}}-{{\sqrt {2}} \over 2})}}={\sqrt {7.778174593\cdot {1 \over 2}\cdot 6.363961031}}={\sqrt {24.75}}=4.974937186.}
S
=
2
S
Δ
=
2
⋅
4.974937186
=
9.949874371.
{\displaystyle S=2S_{\Delta }=2\cdot 4.974937186=9.949874371.}
V
=
S
⋅
|
|
c
|
|
=
9.949874371
⋅
50
=
70.3562364.
{\displaystyle V=S\cdot ||c||=9.949874371\cdot {\sqrt {50}}=70.3562364.}
Pavyzdis . Duoti vektoriai a =(3; 4; 5), b =(4; 3; 5), c =(-3; -4; 5). Vektorius c su vektorium b sudaro kampą
cos
θ
=
b
⋅
c
‖
b
‖
⋅
‖
c
‖
=
4
⋅
(
−
3
)
+
3
⋅
(
−
4
)
+
5
⋅
5
4
2
+
3
2
+
5
2
⋅
(
−
3
)
2
+
(
−
4
)
2
+
5
2
=
1
16
+
9
+
25
⋅
9
+
16
+
25
=
{\displaystyle \cos \theta ={\frac {\mathbf {b} \cdot \mathbf {c} }{\|\mathbf {b} \|\cdot \|\mathbf {c} \|}}={\frac {4\cdot (-3)+3\cdot (-4)+5\cdot 5}{{\sqrt {4^{2}+3^{2}+5^{2}}}\cdot {\sqrt {(-3)^{2}+(-4)^{2}}}+5^{2}}}={\frac {1}{{\sqrt {16+9+25}}\cdot {\sqrt {9+16+25}}}}=}
=
1
50
⋅
50
=
1
50
=
0.02
;
{\displaystyle ={\frac {1}{{\sqrt {50}}\cdot {\sqrt {50}}}}={\frac {1}{50}}=0.02;}
θ
=
arccos
1
50
=
1.550794993
{\displaystyle \theta =\arccos {\frac {1}{50}}=1.550794993}
arba 88.854008 laipsnių.
Vektorius c su vektorium a sudaro kampą:
cos
ϕ
=
a
⋅
c
‖
a
‖
⋅
‖
c
‖
=
3
⋅
(
−
3
)
+
4
⋅
(
−
4
)
+
5
⋅
5
3
2
+
4
2
+
5
2
⋅
(
−
3
)
2
+
(
−
4
)
2
+
5
2
=
0
9
+
16
+
25
⋅
9
+
16
+
25
=
{\displaystyle \cos \phi ={\frac {\mathbf {a} \cdot \mathbf {c} }{\|\mathbf {a} \|\cdot \|\mathbf {c} \|}}={\frac {3\cdot (-3)+4\cdot (-4)+5\cdot 5}{{\sqrt {3^{2}+4^{2}+5^{2}}}\cdot {\sqrt {(-3)^{2}+(-4)^{2}}}+5^{2}}}={\frac {0}{{\sqrt {9+16+25}}\cdot {\sqrt {9+16+25}}}}=}
=
0
50
⋅
50
=
0.
{\displaystyle ={\frac {0}{{\sqrt {50}}\cdot {\sqrt {50}}}}=0.}
ϕ
=
arccos
(
0
)
=
π
2
=
1.570796327
{\displaystyle \phi =\arccos(0)={\pi \over 2}=1.570796327}
arba 90 laipsnių.
Kadangi vektorius a su vektoriumi c sudaro statų kampą, tai vektoriaus c ilgis yra aukštinė piramidės, kurios pagrindas yra trikampis sudarytas iš vektorių a ir b . Apskaičiuosime to trikampio plotą
a
×
b
=
|
i
j
k
3
4
5
4
3
5
|
=
i
⋅
4
⋅
5
+
j
⋅
5
⋅
4
+
k
⋅
3
⋅
3
−
i
⋅
5
⋅
3
−
j
⋅
3
⋅
5
−
k
⋅
4
⋅
4
=
{\displaystyle \mathbf {a} \times \mathbf {b} ={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\3&4&5\\4&3&5\end{vmatrix}}=i\cdot 4\cdot 5+j\cdot 5\cdot 4+k\cdot 3\cdot 3-i\cdot 5\cdot 3-j\cdot 3\cdot 5-k\cdot 4\cdot 4=}
=
20
i
−
15
i
+
20
j
−
15
j
+
9
k
−
16
k
=
0
i
+
0
j
−
4
k
=
(
5
;
5
;
−
7
)
,
{\displaystyle =20i-15i+20j-15j+9k-16k=0i+0j-4k=(5;5;-7),}
S
Δ
=
1
2
⋅
‖
a
×
b
‖
=
1
2
⋅
5
2
+
5
2
+
(
−
7
)
2
=
1
2
⋅
25
+
25
+
49
=
1
2
⋅
99
=
4.974937186.
{\displaystyle S_{\Delta }={\frac {1}{2}}\cdot \|\mathbf {a} \times \mathbf {b} \|={\frac {1}{2}}\cdot {\sqrt {5^{2}+5^{2}+(-7)^{2}}}={\frac {1}{2}}\cdot {\sqrt {25+25+49}}={\frac {1}{2}}\cdot {\sqrt {99}}=4.974937186.}
Randame piramidės, kurią sudaro vektoriai a , b ir c aukštinę h , taigi:
h
=
‖
c
‖
=
(
−
3
)
2
+
(
−
4
)
2
+
5
2
=
9
+
16
+
25
=
50
=
5
2
=
7.071067812.
{\displaystyle h=\|\mathbf {c} \|={\sqrt {(-3)^{2}+(-4)^{2}+5^{2}}}={\sqrt {9+16+25}}={\sqrt {50}}=5{\sqrt {2}}=7.071067812.}
Randame piramidės, kurią sudaro vektoriai a , b ir c tūrį:
V
p
i
r
.
=
1
3
⋅
S
Δ
⋅
h
=
1
3
⋅
1
2
99
⋅
5
2
=
5
198
6
=
11.7260394.
{\displaystyle V_{pir.}={\frac {1}{3}}\cdot S_{\Delta }\cdot h={\frac {1}{3}}\cdot {\frac {1}{2}}{\sqrt {99}}\cdot 5{\sqrt {2}}={\frac {5{\sqrt {198}}}{6}}=11.7260394.}
Bandydami gauti piramidės tūrį, kurią sudaro vektoriai a , b , c , naudodami mišriąją vektorių sandaugą, gausime neteisingą atsakymą:
(
a
×
b
)
⋅
c
=
|
3
4
5
4
3
5
−
3
−
4
5
|
=
3
⋅
3
⋅
5
+
4
⋅
5
⋅
(
−
3
)
+
5
⋅
4
⋅
(
−
4
)
−
3
⋅
5
⋅
(
−
4
)
−
4
⋅
4
⋅
5
−
5
⋅
2
⋅
(
−
3
)
=
{\displaystyle (\mathbf {a} \times \mathbf {b} )\cdot \mathbf {c} ={\begin{vmatrix}3&4&5\\4&3&5\\-3&-4&5\end{vmatrix}}=3\cdot 3\cdot 5+4\cdot 5\cdot (-3)+5\cdot 4\cdot (-4)-3\cdot 5\cdot (-4)-4\cdot 4\cdot 5-5\cdot 2\cdot (-3)=}
=
45
−
60
−
80
+
60
−
80
+
30
=
−
85
;
{\displaystyle =45-60-80+60-80+30=-85;}
V
p
i
r
.
=
1
6
⋅
|
(
a
×
b
)
⋅
c
|
=
1
6
⋅
|
−
85
|
=
85
6
=
14.16666667.
{\displaystyle V_{pir.}={\frac {1}{6}}\cdot |(\mathbf {a} \times \mathbf {b} )\cdot \mathbf {c} |={\frac {1}{6}}\cdot |-85|={\frac {85}{6}}=14.16666667.}
Pavyzdis. Trikampės piramidės viršūnės yra taškai A (3; -1; 5), B (5; 2; 6), C (-1; 3; 4) ir D (7; 3; -1). Apskaičiuosime šios piramidės tūrį ir aukštinės, nuleistos iš taško D į sieną ABC , ilgį.
Sprendimas . Nubraižykime tris vektorius, išeinančius iš vieno taško, pavyzdžiui, iš taško A : AB , AC , AD . Žinome, kad trikampės piramidės tūris
V
p
i
r
.
=
1
6
|
(
A
B
×
A
C
)
⋅
A
D
|
.
{\displaystyle V_{pir.}={\frac {1}{6}}|(AB\times AC)\cdot AD|.}
Randame vektorių AB , AC ir AD koordinates:
AB =B-A=(5-3; 2-(-1); 6-5)={2; 3; 1},
AC =C-A=(-1-3; 3-(-1); 4-5)={-4; 4; -1},
AD =D-A=(7-3; 3-(-1); -1-5)={4; 4; -6}.
Apskaičiuojame mišriąją gautų vektorių sandaugą:
(
A
B
×
A
C
)
⋅
A
D
=
|
2
3
1
−
4
4
−
1
4
4
−
6
|
=
2
⋅
(
−
1
)
1
+
1
|
4
−
1
4
−
6
|
+
3
⋅
(
−
1
)
1
+
2
|
−
4
−
1
4
−
6
|
+
1
⋅
(
−
1
)
1
+
3
|
−
4
4
4
4
|
=
{\displaystyle (AB\times AC)\cdot AD={\begin{vmatrix}2&3&1\\-4&4&-1\\4&4&-6\end{vmatrix}}=2\cdot (-1)^{1+1}{\begin{vmatrix}4&-1\\4&-6\end{vmatrix}}+3\cdot (-1)^{1+2}{\begin{vmatrix}-4&-1\\4&-6\end{vmatrix}}+1\cdot (-1)^{1+3}{\begin{vmatrix}-4&4\\4&4\end{vmatrix}}=}
=
2
⋅
(
−
1
)
2
⋅
(
4
⋅
(
−
6
)
−
(
−
1
)
⋅
4
)
+
3
⋅
(
−
1
)
3
⋅
(
(
−
4
)
⋅
(
−
6
)
−
(
−
1
)
⋅
4
)
+
1
⋅
(
−
1
)
4
⋅
(
(
−
4
)
⋅
4
−
4
⋅
4
)
=
{\displaystyle =2\cdot (-1)^{2}\cdot (4\cdot (-6)-(-1)\cdot 4)+3\cdot (-1)^{3}\cdot ((-4)\cdot (-6)-(-1)\cdot 4)+1\cdot (-1)^{4}\cdot ((-4)\cdot 4-4\cdot 4)=}
=
2
⋅
(
−
24
+
4
)
−
3
⋅
(
24
+
4
)
+
1
⋅
(
−
16
−
16
)
=
2
⋅
(
−
20
)
−
3
⋅
28
−
32
=
−
40
−
84
−
32
=
−
156.
{\displaystyle =2\cdot (-24+4)-3\cdot (24+4)+1\cdot (-16-16)=2\cdot (-20)-3\cdot 28-32=-40-84-32=-156.}
Tada trikampės piramidės tūris
V
p
i
r
.
=
1
6
⋅
|
−
156
|
=
156
6
=
26.
{\displaystyle V_{pir.}={\frac {1}{6}}\cdot |-156|={\frac {156}{6}}=26.}
Norėdami rasti piramidės aukštinę h , pritaikykime kitą piramidės tūrio formulę:
V
p
i
r
.
=
1
3
⋅
S
Δ
A
B
C
⋅
h
.
{\displaystyle V_{pir.}={\frac {1}{3}}\cdot S_{\Delta ABC}\cdot h.}
Bet
S
Δ
A
B
C
=
1
2
⋅
‖
A
B
×
A
C
‖
,
{\displaystyle S_{\Delta ABC}={\frac {1}{2}}\cdot \|AB\times AC\|,}
todėl
V
p
i
r
.
=
1
6
⋅
‖
A
B
×
A
C
‖
⋅
h
.
{\displaystyle V_{pir.}={\frac {1}{6}}\cdot \|AB\times AC\|\cdot h.}
Sulygindami šią formulę su ankstesne piramidės formule, gauname:
1
6
|
(
A
B
×
A
C
)
⋅
A
D
|
=
1
6
⋅
‖
A
B
×
A
C
‖
⋅
h
;
{\displaystyle {\frac {1}{6}}|(AB\times AC)\cdot AD|={\frac {1}{6}}\cdot \|AB\times AC\|\cdot h;}
h
=
|
(
A
B
×
A
C
)
⋅
A
D
|
‖
A
B
×
A
C
‖
=
156
453
=
7.329519377
,
{\displaystyle h={\frac {|(AB\times AC)\cdot AD|}{\|AB\times AC\|}}={\frac {156}{\sqrt {453}}}=7.329519377,}
kur
A
B
×
A
C
=
|
i
j
k
2
3
1
−
4
4
−
1
|
=
i
⋅
(
−
1
)
1
+
1
|
3
1
4
−
1
|
+
j
⋅
(
−
1
)
1
+
2
|
2
1
−
4
−
1
|
+
k
⋅
(
−
1
)
1
+
3
|
2
3
−
4
4
|
=
{\displaystyle AB\times AC={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\2&3&1\\-4&4&-1\end{vmatrix}}=\mathbf {i} \cdot (-1)^{1+1}{\begin{vmatrix}3&1\\4&-1\end{vmatrix}}+\mathbf {j} \cdot (-1)^{1+2}{\begin{vmatrix}2&1\\-4&-1\end{vmatrix}}+\mathbf {k} \cdot (-1)^{1+3}{\begin{vmatrix}2&3\\-4&4\end{vmatrix}}=}
=
i
⋅
(
3
⋅
(
−
1
)
−
1
⋅
4
)
−
j
⋅
(
2
⋅
(
−
1
)
−
1
⋅
(
−
4
)
)
+
k
⋅
(
2
⋅
4
−
3
⋅
(
−
4
)
)
=
i
⋅
(
−
3
−
4
)
−
j
⋅
(
−
2
+
4
)
+
k
⋅
(
8
+
12
)
=
−
7
i
−
2
j
+
20
k
=
(
−
7
;
−
2
;
20
)
;
{\displaystyle =\mathbf {i} \cdot (3\cdot (-1)-1\cdot 4)-\mathbf {j} \cdot (2\cdot (-1)-1\cdot (-4))+\mathbf {k} \cdot (2\cdot 4-3\cdot (-4))=\mathbf {i} \cdot (-3-4)-\mathbf {j} \cdot (-2+4)+\mathbf {k} \cdot (8+12)=-7\mathbf {i} -2\mathbf {j} +20\mathbf {k} =(-7;-2;20);}
‖
A
B
×
A
C
‖
=
(
−
7
)
2
+
(
−
2
)
2
+
20
2
=
49
+
4
+
400
=
453
=
21.28379665.
{\displaystyle \|AB\times AC\|={\sqrt {(-7)^{2}+(-2)^{2}+20^{2}}}={\sqrt {49+4+400}}={\sqrt {453}}=21.28379665.}
Toliau įrodysime, kad piramidės tūris surastas teisingai. Rasime pusiaukampinę kampo tarp vektorių AB ir AC . Sudėję šių vektorių ortus gausime naują vektorių AG , kurio koordinatės yra:
A
G
→
=
A
B
→
‖
A
B
→
‖
+
A
C
→
‖
A
C
→
‖
=
2
i
+
3
j
+
1
k
2
2
+
3
2
+
1
2
+
−
4
i
+
4
j
−
1
k
(
−
4
)
2
+
4
2
+
(
−
1
)
2
=
2
i
+
3
j
+
1
k
4
+
9
+
1
+
−
4
i
+
4
j
−
1
k
16
+
16
+
1
=
{\displaystyle {\vec {AG}}={\frac {\vec {AB}}{\|{\vec {AB}}\|}}+{\frac {\vec {AC}}{\|{\vec {AC}}\|}}={\frac {2i+3j+1k}{\sqrt {2^{2}+3^{2}+1^{2}}}}+{\frac {-4i+4j-1k}{\sqrt {(-4)^{2}+4^{2}+(-1)^{2}}}}={\frac {2i+3j+1k}{\sqrt {4+9+1}}}+{\frac {-4i+4j-1k}{\sqrt {16+16+1}}}=}
=
2
i
+
3
j
+
1
k
14
+
−
4
i
+
4
j
−
1
k
33
=
2
i
14
+
−
4
i
33
+
3
j
14
+
4
j
33
+
k
14
+
−
k
33
=
2
33
−
4
14
14
⋅
33
i
+
3
33
+
4
14
14
⋅
33
j
+
33
−
14
14
⋅
33
k
=
{\displaystyle ={\frac {2i+3j+1k}{\sqrt {14}}}+{\frac {-4i+4j-1k}{\sqrt {33}}}={\frac {2i}{\sqrt {14}}}+{\frac {-4i}{\sqrt {33}}}+{\frac {3j}{\sqrt {14}}}+{\frac {4j}{\sqrt {33}}}+{\frac {k}{\sqrt {14}}}+{\frac {-k}{\sqrt {33}}}={\frac {2{\sqrt {33}}-4{\sqrt {14}}}{{\sqrt {14}}\cdot {\sqrt {33}}}}i+{\frac {3{\sqrt {33}}+4{\sqrt {14}}}{{\sqrt {14}}\cdot {\sqrt {33}}}}j+{\frac {{\sqrt {33}}-{\sqrt {14}}}{{\sqrt {14}}\cdot {\sqrt {33}}}}k=}
=
2
33
−
4
14
462
i
+
3
33
+
4
14
462
j
+
33
−
14
462
k
=
−
0.16178814
i
+
1.49809435
j
+
0.093183585
k
.
{\displaystyle ={\frac {2{\sqrt {33}}-4{\sqrt {14}}}{\sqrt {462}}}i+{\frac {3{\sqrt {33}}+4{\sqrt {14}}}{\sqrt {462}}}j+{\frac {{\sqrt {33}}-{\sqrt {14}}}{\sqrt {462}}}k=-0.16178814i+1.49809435j+0.093183585k.}
Rasime kampą tarp vektoriaus AG ={-0.16178814; 1.49809435; 0.093183585} ir vektoriaus AB ={2; 3; 1}. Taigi,
cos
ϕ
1
=
−
0.16178814
⋅
2
+
1.49809435
⋅
3
+
0.093183585
⋅
1
(
−
0.16178814
)
2
+
1.49809435
2
+
0.093183585
2
⋅
2
2
+
3
2
+
1
2
=
{\displaystyle \cos \phi _{1}={\frac {-0.16178814\cdot 2+1.49809435\cdot 3+0.093183585\cdot 1}{{\sqrt {(-0.16178814)^{2}+1.49809435^{2}+0.093183585^{2}}}\cdot {\sqrt {2^{2}+3^{2}+1^{2}}}}}=}
=
−
0.32357628
+
4.49428305
+
0.093183585
0.026175402
+
2.244286682
+
0.00868318
⋅
4
+
9
+
1
=
4.263890355
2.279145264
⋅
14
=
4.263890355
31.9080337
=
4.263890355
5.648719651
=
0.754841914.
{\displaystyle ={\frac {-0.32357628+4.49428305+0.093183585}{{\sqrt {0.026175402+2.244286682+0.00868318}}\cdot {\sqrt {4+9+1}}}}={\frac {4.263890355}{{\sqrt {2.279145264}}\cdot {\sqrt {14}}}}={\frac {4.263890355}{\sqrt {31.9080337}}}={\frac {4.263890355}{5.648719651}}=0.754841914.}
ϕ
1
=
arccos
(
0.754841914
)
=
0.715383259
{\displaystyle \phi _{1}=\arccos(0.754841914)=0.715383259}
radiano arba 40.98844149 laipsnio.
Patikrinimui, rasime kampą tarp vektoriaus AC ={-4; 4; -1} ir AB ={2; 3; 1}, taigi
cos
ϕ
2
=
(
−
4
)
⋅
2
+
4
⋅
3
+
(
−
1
)
⋅
1
(
−
4
)
2
+
4
2
+
(
−
1
)
2
⋅
2
2
+
3
2
+
1
2
=
−
8
+
12
−
1
16
+
16
+
1
⋅
4
+
9
+
1
=
3
33
⋅
14
=
3
462
=
0.139572631.
{\displaystyle \cos \phi _{2}={\frac {(-4)\cdot 2+4\cdot 3+(-1)\cdot 1}{{\sqrt {(-4)^{2}+4^{2}+(-1)^{2}}}\cdot {\sqrt {2^{2}+3^{2}+1^{2}}}}}={\frac {-8+12-1}{{\sqrt {16+16+1}}\cdot {\sqrt {4+9+1}}}}={\frac {3}{{\sqrt {33}}\cdot {\sqrt {14}}}}={\frac {3}{\sqrt {462}}}=0.139572631.}
ϕ
2
=
arccos
3
462
=
1.430766518
{\displaystyle \phi _{2}=\arccos {\frac {3}{\sqrt {462}}}=1.430766518}
radiano arba 81,97688296 laipsnio. Patikriname, kad
ϕ
2
=
2
⋅
ϕ
1
=
2
⋅
40.98844149
=
81.97688298.
{\displaystyle \phi _{2}=2\cdot \phi _{1}=2\cdot 40.98844149=81.97688298.}
Surandame kampą tarp vektorių AG ={-0.16178814; 1.49809435; 0.093183585} ir AD ={4; 4; -6}, taigi:
cos
ϕ
3
=
−
0.16178814
⋅
4
+
1.49809435
⋅
4
+
0.093183585
⋅
(
−
6
)
(
−
0.16178814
)
2
+
1.49809435
2
+
0.093183585
2
⋅
4
2
+
4
2
+
(
−
6
)
2
=
{\displaystyle \cos \phi _{3}={\frac {-0.16178814\cdot 4+1.49809435\cdot 4+0.093183585\cdot (-6)}{{\sqrt {(-0.16178814)^{2}+1.49809435^{2}+0.093183585^{2}}}\cdot {\sqrt {4^{2}+4^{2}+(-6)^{2}}}}}=}
=
−
0.64715256
+
5.9923774
−
0.55910151
0.026175402
+
2.244286682
+
0.00868318
⋅
16
+
16
+
36
=
4.78612333
2.279145264
⋅
68
=
4.78612333
154.981878
=
4.78612333
12.44917178
=
0.384453152.
{\displaystyle ={\frac {-0.64715256+5.9923774-0.55910151}{{\sqrt {0.026175402+2.244286682+0.00868318}}\cdot {\sqrt {16+16+36}}}}={\frac {4.78612333}{{\sqrt {2.279145264}}\cdot {\sqrt {68}}}}={\frac {4.78612333}{\sqrt {154.981878}}}={\frac {4.78612333}{12.44917178}}=0.384453152.}
ϕ
3
=
arccos
(
0.384453152
)
=
1.176180953
{\displaystyle \phi _{3}=\arccos(0.384453152)=1.176180953}
radiano arba 67.39020453 laipsnio.
Piramidės aukštinės h (kuri nuleista iš taško D į piramidės pagrindą ABC ) ilgis yra tiesės AD ilgis padaugintas iš
sin
ϕ
3
,
{\displaystyle \sin \phi _{3},}
todėl gauname:
h
=
‖
A
D
→
‖
⋅
sin
ϕ
3
=
4
2
+
4
2
+
(
−
6
)
2
⋅
sin
(
1.176180953
)
=
16
+
16
+
36
⋅
0.923144503
=
68
⋅
0.923144503
=
8.246211251
⋅
0.923144503
=
7.61244459.
{\displaystyle h=\|{\vec {AD}}\|\cdot \sin \phi _{3}={\sqrt {4^{2}+4^{2}+(-6)^{2}}}\cdot \sin(1.176180953)={\sqrt {16+16+36}}\cdot 0.923144503={\sqrt {68}}\cdot 0.923144503=8.246211251\cdot 0.923144503=7.61244459.}
Randame ABCD piramidės tūrį:
V
=
1
3
⋅
h
⋅
S
Δ
A
B
C
=
1
3
⋅
h
⋅
1
2
⋅
‖
A
B
→
×
A
C
→
‖
=
7.61244459
6
⋅
453
=
1.268740765
⋅
21.28379665
=
27.00362045.
{\displaystyle V={\frac {1}{3}}\cdot h\cdot S_{\Delta ABC}={\frac {1}{3}}\cdot h\cdot {\frac {1}{2}}\cdot \|{\vec {AB}}\times {\vec {AC}}\|={\frac {7.61244459}{6}}\cdot {\sqrt {453}}=1.268740765\cdot 21.28379665=27.00362045.}
Tašką į kurį nuleista aukštinė iš taško D , pavadiname tašku E . Randame projekcija vektoriaus AD ={4; 4; -6} vektoriuje AG ={-0.16178814; 1.49809435; 0.093183585}, tos projekcijos ilgis yra AE . Gauname:
A
E
=
p
r
A
G
→
A
D
→
=
A
D
→
⋅
A
G
→
‖
A
G
→
‖
=
4
⋅
(
−
0.16178814
)
+
4
⋅
1.49809435
+
(
−
6
)
⋅
0.093183585
(
−
0.16178814
)
2
+
1.49809435
2
+
0.093183585
2
=
{\displaystyle AE=pr_{\vec {AG}}{\vec {AD}}={\frac {{\vec {AD}}\cdot {\vec {AG}}}{\|{\vec {AG}}\|}}={\frac {4\cdot (-0.16178814)+4\cdot 1.49809435+(-6)\cdot 0.093183585}{\sqrt {(-0.16178814)^{2}+1.49809435^{2}+0.093183585^{2}}}}=}
=
−
0.64715256
+
5.9923774
−
0.55910151
0.026175402
+
2.244286682
+
0.00868318
=
4.78612333
2.279145264
=
4.78612333
1.509683829
=
3.170281908.
{\displaystyle ={\frac {-0.64715256+5.9923774-0.55910151}{\sqrt {0.026175402+2.244286682+0.00868318}}}={\frac {4.78612333}{\sqrt {2.279145264}}}={\frac {4.78612333}{1.509683829}}=3.170281908.}
Iš Pitagoro teoremos randame aukštine h=DE, taigi:
D
E
=
h
=
A
D
2
−
A
E
2
=
(
68
)
2
−
3.170281908
2
=
68
−
10.05068737
=
57.94931263
=
7.612444589.
{\displaystyle DE=h={\sqrt {AD^{2}-AE^{2}}}={\sqrt {({\sqrt {68}})^{2}-3.170281908^{2}}}={\sqrt {68-10.05068737}}={\sqrt {57.94931263}}=7.612444589.}
Patikrinsime ar vektoriai AG ={-0.16178814; 1.49809435; 0.093183585}, AB ={2; 3; 1} ir AC ={-4; 4; -1} komplanarūs (ar vektoriai guli toje pačioje plokštumoje):
(
A
G
→
×
A
B
→
)
⋅
A
C
→
=
|
−
0.16178814
1.49809435
0.093183585
2
3
1
−
4
4
−
1
|
=
{\displaystyle ({\vec {AG}}\times {\vec {AB}})\cdot {\vec {AC}}={\begin{vmatrix}-0.16178814&1.49809435&0.093183585\\2&3&1\\-4&4&-1\end{vmatrix}}=}
=
−
0.16178814
⋅
(
−
1
)
1
+
1
|
3
1
4
−
1
|
+
1.49809435
⋅
(
−
1
)
1
+
2
|
2
1
−
4
−
1
|
+
0.093183585
⋅
(
−
1
)
1
+
3
|
2
3
−
4
4
|
=
{\displaystyle =-0.16178814\cdot (-1)^{1+1}{\begin{vmatrix}3&1\\4&-1\end{vmatrix}}+1.49809435\cdot (-1)^{1+2}{\begin{vmatrix}2&1\\-4&-1\end{vmatrix}}+0.093183585\cdot (-1)^{1+3}{\begin{vmatrix}2&3\\-4&4\end{vmatrix}}=}
=
−
0.16178814
⋅
(
−
3
−
4
)
−
1.49809435
⋅
(
−
2
−
(
−
4
)
)
+
0.093183585
⋅
(
8
−
(
−
12
)
)
=
{\displaystyle =-0.16178814\cdot (-3-4)-1.49809435\cdot (-2-(-4))+0.093183585\cdot (8-(-12))=}
=
−
0.16178814
⋅
(
−
7
)
−
1.49809435
⋅
2
+
0.093183585
⋅
20
=
1.13251698
−
2.9961887
+
1.8636717
=
−
0.00000002.
{\displaystyle =-0.16178814\cdot (-7)-1.49809435\cdot 2+0.093183585\cdot 20=1.13251698-2.9961887+1.8636717=-0.00000002.}
Mišrios vektorių sandaugos rezultatas yra 0, todėl vektoriai AG , AB ir AC priklauso tai pačiai plokštumai.