Vektorius – matematinis dydis, apibūdinamas reikšme ir kryptimi erdvėje. Grafiškai vektoriai vaizduojami tiesių atkarpomis su rodyklėmis.
Bendriausias vektoriaus pavyzdys fizikoje būtų jėga .
Skaitinių dydžių grupė abibūdinanti pasirinktą objektą gali būti užrašyta sugrupuotų skaičių sąrašu arba kitaip -- vektoriumi:
v = ( v 1 , v 2 , . . . , v n ) {\displaystyle v=(v_{1},v_{2},...,v_{n})} .
kur v yra d skaičių vektorius. Išraiškos su vektoriais yra naudojamos siekiant kompaktiškai užrašyti bei patogiai manipuliuoti ilgomis skaičių grupėmis. Kitas vektorinio užrašymo privalumas yra jo geometrinė interpretacija -- kiekvieną v galima įsivaizduoti kaip vektorių jungiantį n -matės erdvės koordinačių pradžią su tašku, kurio koordinatės nustatytos nariais sudarančiais v .
Vektoriaus daugyba iš skaliaro
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Vienas realaus dydžio skaičius yra vadinamas skaliaru . Vektoriaus daugyba iš skaliaro yra kiekvieno vektoriaus nario daugyba iš skaliaro ir gauta sandauga yra vektorius:
c v = ( c v 1 , c v 2 , . . . , c v n ) {\displaystyle cv=(cv_{1},cv_{2},...,cv_{n})} . Dviejų vektorių suma
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Du vektoriai sudedami sudedant kiekvieno iš jų atitinkamus narius:
v + w = ( v 1 + w 1 , v 2 + w 2 , . . . , v n + w n ) {\displaystyle v+w=(v_{1}+w_{1},v_{2}+w_{2},...,v_{n}+w_{n})} .
Atkreipkite dėmesį, jog vektorinė sudėtis yra komutatyvi , t. y., v+w=w+v.
Skaliarinė vektorių sandauga
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Išsamesnis straipsnis: Skaliarinė sandauga .
Skaliarinės sandaugos savoka yra glaudžiai susijusi su vektoriaus ilgio bei vektoriaus projekcijos sampratomis.
Norint vektorius sudauginti skaliariškai, abu vektoriai turi atitikti , t. y., abiejų vektorių narių skaičius turi būti vienodas. Skaliarinė dviejų vektorių sandauga yra suma visų kiekvieno iš vektoriaus atitinkamų narių sandaugų:
v ⋅ w = ∑ i = 1 n v i ⋅ w i = v 1 w 1 + v 2 w 2 + v 3 w 3 + . . . + v n w n . {\displaystyle v\cdot w=\sum _{i=1}^{n}v_{i}\cdot w_{i}=v_{1}w_{1}+v_{2}w_{2}+v_{3}w_{3}+...+v_{n}w_{n}.}
Skaliarinės vektorių sandaugos rezultatas yra ne vektorius, o skaliaras. Pavyzdžiui, yra vektoriai a(3; 5; 6) ir b(4; 0; 1), tai jų skaliarinė sandauga bus lygi:
a ⋅ b = 3 ⋅ 4 + 5 ⋅ 0 + 6 ⋅ 1 = 12 + 0 + 6 = 18. {\displaystyle a\cdot b=3\cdot 4+5\cdot 0+6\cdot 1=12+0+6=18.} Vektoriaus ilgis
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Išnagrinėkime atvejį, kai atliekama vektoriaus skaliarinė sandauga su juo pačiu. Plokštumos (2 -matės erdvės) bei įprastos koordinačių sistemos atveju turėsime:
v ⋅ v = ( v 1 ) 2 + ( v 2 ) 2 {\displaystyle v\cdot v=(v_{1})^{2}+(v_{2})^{2}} .Prisiminus Pitagoro teoremą , teigiančią, jog stataus trikampio įstrižainės ilgio kvadratas yra lygus trikampio kraštinių ilgių kvadratų sumai, tampa natūralus toks vektoriaus ilgio apibūdinimas:
| | v | | = v ⋅ v {\displaystyle ||v||={\sqrt {v\cdot v}}} .
Atkreipkite dėmesį, jog jei nors vienas iš vektoriaus narių bus didesnis nei kiti, tai jo pakėlimas kvadratu lems viso vektoriaus ilgį. Pavyzdžiui, vektoriaus a(3; -2; 4) ilgis (tai yra ilgis nuo taško (0; 0; 0) iki taško (3; -2; 4)):
| | a | | = a x 2 + a y 2 + a z 2 = 3 2 + ( − 2 ) 2 + 4 2 = 9 + 4 + 16 = 29 ≈ 5.385. {\displaystyle ||a||={\sqrt {a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}}={\sqrt {3^{2}+(-2)^{2}+4^{2}}}={\sqrt {9+4+16}}={\sqrt {29}}\approx 5.385.} Pavyzdžiui, žinomos vektoriaus pradžios A(3; 2; -4) ir galo B(6; -5; -2) koordinatės. Tada vektoriaus ilgis bus
A B = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 + ( z 2 − z 1 ) 2 = ( 6 − 3 ) 2 + ( − 5 − 2 ) 2 + ( − 2 − 4 ) 2 = {\displaystyle AB={\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}}}={\sqrt {(6-3)^{2}+(-5-2)^{2}+(-2-4)^{2}}}=} = 9 + 49 + 4 = 62 ≈ 7.874. {\displaystyle ={\sqrt {9+49+4}}={\sqrt {62}}\approx 7.874.}
Jeigu vektoriaus pradžios koordinatės A(0; 0; 0), o galo koordinatės B(6; -5; -2), tai vektoriaus AB ilgis bus:
A B = ( 6 − 0 ) 2 + ( − 5 − 0 ) 2 + ( − 2 − 0 ) 2 = 36 + 25 + 4 = 65 ≈ 8.062. {\displaystyle AB={\sqrt {(6-0)^{2}+(-5-0)^{2}+(-2-0)^{2}}}={\sqrt {36+25+4}}={\sqrt {65}}\approx 8.062.}
Vektoriaus sandaugos su skaliaru duos:
||cv ||=c ||v ||. Pavyzdžiui, vektorius a(3; -2; 4) ir skaliaras c=5. Tada ca =(15; -10; 20). c | | a | | = c a x 2 + a y 2 + a z 2 = 5 3 2 + ( − 2 ) 2 + 4 2 = 5 9 + 4 + 16 = 5 29 . {\displaystyle c||a||=c{\sqrt {a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}}=5{\sqrt {3^{2}+(-2)^{2}+4^{2}}}=5{\sqrt {9+4+16}}=5{\sqrt {29}}.}
| | c a | | = a x 2 + a y 2 + a z 2 = 15 2 + ( − 10 ) 2 + 20 2 = 225 + 100 + 400 = 725 = 5 29 . {\displaystyle ||ca||={\sqrt {a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}}={\sqrt {15^{2}+(-10)^{2}+20^{2}}}={\sqrt {225+100+400}}={\sqrt {725}}=5{\sqrt {29}}.} Trikampio nelygybė naudojama apibūdinti dviejų vektorių sumos ilgį:
||v +w ||<=||v ||+||w ||. Pavyzdžiui, yra vektoriai v=(3; -2; 4) ir w=(1; 5; 8). z =v +w =(3+1; -2+5; 4+8)=(4; 3; 12). | | z | | = 4 2 + 3 2 + 12 2 = 16 + 9 + 144 = 169 = 13. {\displaystyle ||z||={\sqrt {4^{2}+3^{2}+12^{2}}}={\sqrt {16+9+144}}={\sqrt {169}}=13.}
| | v | | = 3 2 + ( − 2 ) 2 + 4 2 = 29 ≈ 5.385164807. {\displaystyle ||v||={\sqrt {3^{2}+(-2)^{2}+4^{2}}}={\sqrt {29}}\approx 5.385164807.}
| | w | | = 1 2 + 5 2 + 8 2 = 1 + 25 + 64 = 90 = 3 10 ≈ 9.486832981. {\displaystyle ||w||={\sqrt {1^{2}+5^{2}+8^{2}}}={\sqrt {1+25+64}}={\sqrt {90}}=3{\sqrt {10}}\approx 9.486832981.}
||v||+||w||=5.385164807+9.486832981=14.87199779.
| | z | | = | | v + w | | = 13 ≤ 14.87199779 = | | v | | + | | w | | . {\displaystyle ||z||=||v+w||=13\leq 14.87199779=||v||+||w||.} Atstumas tarp vektorių
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Atstumas tarp vieno vektoriaus galo ir kito vektoriaus galo (atstumas tarp dviejų taškų n-matėje koordinačių sistemoje ) matuojamas pagal formulę:
‖ v − w ‖ = ∑ i = 1 n ( v i − w i ) 2 = ( v 1 − w 1 ) 2 + ( v 2 − w 2 ) 2 + . . . + ( v n − w n ) 2 . {\displaystyle \|v-w\|={\sqrt {\sum _{i=1}^{n}(v_{i}-w_{i})^{2}}}={\sqrt {(v_{1}-w_{1})^{2}+(v_{2}-w_{2})^{2}+...+(v_{n}-w_{n})^{2}}}.} Pavyzdžiai
Turime vektorius v=[3, 6], w=[7, 4]. Atstumas tarp jų galų: ( 3 − 7 ) 2 + ( 6 − 4 ) 2 = 20 ≈ 4 , 47. {\displaystyle {\sqrt {(3-7)^{2}+(6-4)^{2}}}={\sqrt {20}}\approx 4,47.}
Rasime trikampio, esančio trimatėje erdvėje, plotą. Trikampio viršunių koordinates (x; y; z) yra tokios: A(8; 3; -3); B(3; 2; -1); C(4; 0; -3). Dabar reikia surasti tiesių ilgius AB, AC ir BC: a = A B = ( 8 − 3 ) 2 + ( 3 − 2 ) 2 + ( − 3 − ( − 1 ) ) 2 = 30 , {\displaystyle a=AB={\sqrt {(8-3)^{2}+(3-2)^{2}+(-3-(-1))^{2}}}={\sqrt {30}},}
b = A C = ( 8 − 4 ) 2 + ( 3 − 0 ) 2 + ( − 3 − ( − 3 ) ) 2 = 5 , {\displaystyle b=AC={\sqrt {(8-4)^{2}+(3-0)^{2}+(-3-(-3))^{2}}}=5,}
c = B C = ( 3 − 4 ) 2 + ( 2 − 0 ) 2 + ( − 1 − ( − 3 ) ) 2 = 3. {\displaystyle c=BC={\sqrt {(3-4)^{2}+(2-0)^{2}+(-1-(-3))^{2}}}=3.} Taikydami Herono formule apskaičiuojame trikampio pusperimetrį p :
p = a + b + c 2 = 30 + 5 + 3 2 ≈ 13.477. {\displaystyle p={\frac {a+b+c}{2}}={\frac {{\sqrt {30}}+5+3}{2}}\approx 13.477.} Ir trikampio plotą S :
S = p ( p − a ) ( p − b ) ( p − c ) ≈ 13.477 ( 13.477 − 5.477 ) ( 13.477 − 5 ) ( 13.477 − 3 ) ≈ 97.855 {\displaystyle S={\sqrt {p(p-a)(p-b)(p-c)}}\approx {\sqrt {13.477(13.477-5.477)(13.477-5)(13.477-3)}}\approx 97.855} Rasime trikampio plotą, kurio višunės yra taškuose A(1; 3; -2), B(2; -1; 3), C(0; 2; 4). a = A B = ( 1 − 2 ) 2 + ( 3 + 1 ) 2 + ( − 2 − 3 ) 2 = 1 + 16 + 25 = 42 ≈ 6.48 , {\displaystyle a=AB={\sqrt {(1-2)^{2}+(3+1)^{2}+(-2-3)^{2}}}={\sqrt {1+16+25}}={\sqrt {42}}\approx 6.48,}
b = A C = ( 1 − 0 ) 2 + ( 3 − 2 ) 2 + ( − 2 − 4 ) 2 = 1 + 1 + 36 = 38 ≈ 6.16 , {\displaystyle b=AC={\sqrt {(1-0)^{2}+(3-2)^{2}+(-2-4)^{2}}}={\sqrt {1+1+36}}={\sqrt {38}}\approx 6.16,}
c = B C = ( 2 − 0 ) 2 + ( − 1 − 2 ) 2 + ( 3 − 4 ) 2 = 4 + 9 + 1 = 14 ≈ 3.74. {\displaystyle c=BC={\sqrt {(2-0)^{2}+(-1-2)^{2}+(3-4)^{2}}}={\sqrt {4+9+1}}={\sqrt {14}}\approx 3.74.}
p = 42 + 38 + 14 2 ≈ 8.193406044. {\displaystyle p={\frac {{\sqrt {42}}+{\sqrt {38}}+{\sqrt {14}}}{2}}\approx 8.193406044.} S ≈ 8.193406044 ( 8.193406044 − 42 ) ( 8.193406044 − 38 ) ( 8.193406044 − 14 ) ≈ {\displaystyle S\approx {\sqrt {8.193406044(8.193406044-{\sqrt {42}})(8.193406044-{\sqrt {38}})(8.193406044-{\sqrt {14}})}}\approx }
≈ 8.193406044 ⋅ 1 , 712665346 ⋅ 2 , 028992041 ⋅ 4 , 451748657 = 126 , 750000 = 11.25833025. {\displaystyle \approx {\sqrt {8.193406044\cdot 1,712665346\cdot 2,028992041\cdot 4,451748657}}={\sqrt {126,750000}}=11.25833025.}
Šio trikampio plotą galima apskaičiuoti naudojantis vektorine sandauga. AB=(2-1; -1-3; 3+2)=(1; -4; 5), AC=(0-1; 2-3; 4+2)=(-1; -1; 6).
A B × A C = | i j k 1 − 4 5 − 1 − 1 6 | = | − 4 5 − 1 6 | i − | 1 5 − 1 6 | j + | 1 − 4 − 1 − 1 | k = − 19 i − 11 j − 5 k = ( − 19 ; − 11 ; − 5 ) . {\displaystyle AB\times AC={\begin{vmatrix}i&j&k\\1&-4&5\\-1&-1&6\end{vmatrix}}={\begin{vmatrix}-4&5\\-1&6\end{vmatrix}}i-{\begin{vmatrix}1&5\\-1&6\end{vmatrix}}j+{\begin{vmatrix}1&-4\\-1&-1\end{vmatrix}}k=-19i-11j-5k=(-19;-11;-5).}
‖ A B × A C ‖ = ( − 19 ) 2 + ( − 11 ) 2 + ( − 5 ) 2 = 507 . {\displaystyle \|AB\times AC\|={\sqrt {(-19)^{2}+(-11)^{2}+(-5)^{2}}}={\sqrt {507}}.}
S Δ A B C = 1 2 ‖ A B × A C ‖ = 1 2 507 ≈ 11.25833025. {\displaystyle S_{\Delta ABC}={\frac {1}{2}}\|AB\times AC\|={\frac {1}{2}}{\sqrt {507}}\approx 11.25833025.} Vektoriaus projekcijos į koordinačių ašis
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Projekcija vieno vektoriaus į kitą vektorių
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Pusiaukampinė tarp vektorių
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Jei duoti vektoriai a =OA ir b =OB , tai pusiaukampinės ON =c (arba tiesiog, taško N koordinatės, kai taškas O (0; 0; 0)) koordinatės yra:
c → = a → ‖ a → ‖ + b → ‖ b → ‖ . {\displaystyle {\vec {c}}={\frac {\vec {a}}{\|{\vec {a}}\|}}+{\frac {\vec {b}}{\|{\vec {b}}\|}}.}
Vektoriaus a ortas yra a → ‖ a → ‖ . {\displaystyle {\frac {\vec {a}}{\|{\vec {a}}\|}}.}
Vektoriaus ON ortas yra:
a → ‖ a → ‖ + b → ‖ b → ‖ ‖ a → ‖ a → ‖ + b → ‖ b → ‖ ‖ . {\displaystyle {\frac {{\frac {\vec {a}}{\|{\vec {a}}\|}}+{\frac {\vec {b}}{\|{\vec {b}}\|}}}{\|{\frac {\vec {a}}{\|{\vec {a}}\|}}+{\frac {\vec {b}}{\|{\vec {b}}\|}}\|}}.}
Vektoriaus ortas ir vektorius yra vienakrypčiai, tačiau vektoriaus orto ilgis lygus 1.
‖ a → ‖ a → ‖ + b → ‖ b → ‖ ‖ a → ‖ a → ‖ + b → ‖ b → ‖ ‖ ‖ = 1. {\displaystyle \|{\frac {{\frac {\vec {a}}{\|{\vec {a}}\|}}+{\frac {\vec {b}}{\|{\vec {b}}\|}}}{\|{\frac {\vec {a}}{\|{\vec {a}}\|}}+{\frac {\vec {b}}{\|{\vec {b}}\|}}\|}}\|=1.}
Vektorių a ir b ortų ilgiai lygus vienam, ‖ a → ‖ a → ‖ ‖ = 1 ; ‖ b → ‖ b → ‖ ‖ = 1. {\displaystyle \|{\frac {\vec {a}}{\|{\vec {a}}\|}}\|=1;\,\,\,\|{\frac {\vec {b}}{\|{\vec {b}}\|}}\|=1.}
Vektorių padauginus iš skaliaro, vektroriaus ilgis pasikeičia, o kryptis išlieka ta pati. Pavyzdis . Duoti vektoriai a =(5; 3), b =(4; 20). Pusiaukampinė yra:c → = a → ‖ a → ‖ + b → ‖ b → ‖ = 5 i + 3 j 5 2 + 3 2 + 4 i + 20 j 4 2 + 20 2 = 5 i + 3 j 25 + 9 + 4 i + 20 j 16 + 400 = 5 i 34 + 3 j 34 + 4 i 416 + 20 j 416 = {\displaystyle {\vec {c}}={\frac {\vec {a}}{\|{\vec {a}}\|}}+{\frac {\vec {b}}{\|{\vec {b}}\|}}={\frac {5i+3j}{\sqrt {5^{2}+3^{2}}}}+{\frac {4i+20j}{\sqrt {4^{2}+20^{2}}}}={\frac {5i+3j}{\sqrt {25+9}}}+{\frac {4i+20j}{\sqrt {16+400}}}={\frac {5i}{\sqrt {34}}}+{\frac {3j}{\sqrt {34}}}+{\frac {4i}{\sqrt {416}}}+{\frac {20j}{\sqrt {416}}}=}
= 5 416 + 4 34 34 ⋅ 416 i + 3 416 + 20 34 34 ⋅ 416 j = 5 ⋅ 4 26 + 4 34 34 ⋅ 4 26 i + 3 ⋅ 4 26 + 20 34 34 ⋅ 4 26 j = {\displaystyle ={\frac {5{\sqrt {416}}+4{\sqrt {34}}}{{\sqrt {34}}\cdot {\sqrt {416}}}}i+{\frac {3{\sqrt {416}}+20{\sqrt {34}}}{{\sqrt {34}}\cdot {\sqrt {416}}}}j={\frac {5\cdot 4{\sqrt {26}}+4{\sqrt {34}}}{{\sqrt {34}}\cdot 4{\sqrt {26}}}}i+{\frac {3\cdot 4{\sqrt {26}}+20{\sqrt {34}}}{{\sqrt {34}}\cdot 4{\sqrt {26}}}}j=}
= 5 26 + 34 34 ⋅ 26 i + 3 26 + 5 34 34 ⋅ 26 j = 5 26 + 34 884 i + 3 26 + 5 34 884 j = {\displaystyle ={\frac {5{\sqrt {26}}+{\sqrt {34}}}{{\sqrt {34}}\cdot {\sqrt {26}}}}i+{\frac {3{\sqrt {26}}+5{\sqrt {34}}}{{\sqrt {34}}\cdot {\sqrt {26}}}}j={\frac {5{\sqrt {26}}+{\sqrt {34}}}{\sqrt {884}}}i+{\frac {3{\sqrt {26}}+5{\sqrt {34}}}{\sqrt {884}}}j=}
= 0.857492925 i + 0.514495755 j + 0.196116135 i + 0.980580675 j = 1.053609061 i + 1.495076431 j = ( 1.053609061 ; 1.495076431 ) . {\displaystyle =0.857492925i+0.514495755j+0.196116135i+0.980580675j=1.053609061i+1.495076431j=(1.053609061;1.495076431).}
Jei yra taškai O (0; 0), A (5; 3), B (4; 20), tai taškas N (1.053609061; 1.495076431) su tašku O (0; 0) sudaro tiesę ON , kuri yra pusiaukampinė tarp tiesių OA ir OB .
Randame kampą tarp tiesių ON ir OB .
cos ϕ 1 = c → ⋅ b → ‖ c → ‖ ⋅ ‖ b → ‖ = 1.053609061 ⋅ 4 + 1.495076431 ⋅ 20 1.05360906 2 + 1.495076431 2 ⋅ 4 2 + 20 2 = 4.214436243 + 29.90152862 3.345345588 ⋅ 416 = {\displaystyle \cos \phi _{1}={\frac {{\vec {c}}\cdot {\vec {b}}}{\|{\vec {c}}\|\cdot \|{\vec {b}}\|}}={\frac {1.053609061\cdot 4+1.495076431\cdot 20}{{\sqrt {1.05360906^{2}+1.495076431^{2}}}\cdot {\sqrt {4^{2}+20^{2}}}}}={\frac {4.214436243+29.90152862}{{\sqrt {3.345345588}}\cdot {\sqrt {416}}}}=}
= 34.11596487 37.30500991 = 0.914514295. {\displaystyle ={\frac {34.11596487}{37.30500991}}=0.914514295.}
= 83 155 ⋅ 117 = 83 18135 = 83 134.6662541 = 0.616338521. {\displaystyle ={\frac {83}{{\sqrt {155}}\cdot {\sqrt {117}}}}={\frac {83}{\sqrt {18135}}}={\frac {83}{134.6662541}}=0.616338521.}
ϕ 1 = arccos ( 0.912101751 ) = 0.416490632 {\displaystyle \phi _{1}=\arccos(0.912101751)=0.416490632} arba 23.86315547 laipsnio.
Randame kampą tarp tiesių OA ir OB , gauname:
cos ϕ 2 = a → ⋅ b → ‖ a → ‖ ⋅ ‖ b → ‖ = 5 ⋅ 4 + 3 ⋅ 20 5 2 + 3 2 ⋅ 4 2 + 20 2 = 20 + 60 34 ⋅ 416 = 80 14144 = 0.672672794. {\displaystyle \cos \phi _{2}={\frac {{\vec {a}}\cdot {\vec {b}}}{\|{\vec {a}}\|\cdot \|{\vec {b}}\|}}={\frac {5\cdot 4+3\cdot 20}{{\sqrt {5^{2}+3^{2}}}\cdot {\sqrt {4^{2}+20^{2}}}}}={\frac {20+60}{{\sqrt {34}}\cdot {\sqrt {416}}}}={\frac {80}{\sqrt {14144}}}=0.672672794.}
ϕ 2 = arccos 80 14144 = arccos ( 0.672672794 ) = 0.832981266 {\displaystyle \phi _{2}=\arccos {\frac {80}{\sqrt {14144}}}=\arccos(0.672672794)=0.832981266} arba 47.72631099 laipsnių. Palyginimui, 2 ϕ 1 = 2 ⋅ 0.416490632 = 0.832981265. {\displaystyle 2\phi _{1}=2\cdot 0.416490632=0.832981265.}
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.} Kampo tarp vektorių radimas su kosinusu
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Kampas tarp dviejų vektorių yra išreiškiamas per jų skaliarinę sandaugą:
cos ϕ = a ⋅ b | | a | | ⋅ | | b | | {\displaystyle \cos \phi ={\frac {\mathbf {a} \cdot \mathbf {b} }{||\mathbf {a} ||\cdot ||\mathbf {b} ||}}} .
ϕ = arccos a ⋅ b | | a | | ⋅ | | b | | . {\displaystyle \phi =\arccos {\frac {\mathbf {a} \cdot \mathbf {b} }{||\mathbf {a} ||\cdot ||\mathbf {b} ||}}.} Remiantis šia formule tampa akivaizdu kodėl yra sakoma, jog skaliarinė vektorių sandauga parodo vektorių atitikimą (panašumą) vienas kitam.
a ⋅ b = | | a | | ⋅ | | b | | ⋅ cos ϕ . {\displaystyle \mathbf {a} \cdot \mathbf {b} =||\mathbf {a} ||\cdot ||\mathbf {b} ||\cdot \cos \phi .} Pavyzdžiui, duoti vektoriai a=(1; -2; 2), b=(3; 0; -4). 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 − 8 1 + 4 + 4 ⋅ 9 + 0 + 16 = {\displaystyle \cos \phi ={\frac {a\cdot b}{||a||\cdot ||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 {3-8}{{\sqrt {1+4+4}}\cdot {\sqrt {9+0+16}}}}=}
= − 5 9 ⋅ 25 = − 5 3 ⋅ 5 = − 1 3 . {\displaystyle ={\frac {-5}{{\sqrt {9}}\cdot {\sqrt {25}}}}={\frac {-5}{3\cdot 5}}=-{\frac {1}{3}}.}
ϕ = arccos − 1 3 = 1.910633236 {\displaystyle \phi =\arccos {\frac {-1}{3}}=1.910633236} arba 109,4712206 laipsnių.
Taikydami kosinusų toeremą patikrinsime ar kampas ϕ {\displaystyle \phi } surastas 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.} 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}.} Pavyzdžiui, duoti vektoriai a=(1; -2; 0), b=(3; 0; 0). cos ϕ = a ⋅ b | | a | | ⋅ | | b | | = 1 ⋅ 3 + ( − 2 ) ⋅ 0 + 0 ⋅ 0 1 2 + ( − 2 ) 2 ⋅ 3 2 + 0 2 = 3 1 + 4 ⋅ 9 + 0 = {\displaystyle \cos \phi ={\frac {a\cdot b}{||a||\cdot ||b||}}={\frac {1\cdot 3+(-2)\cdot 0+0\cdot 0}{{\sqrt {1^{2}+(-2)^{2}}}\cdot {\sqrt {3^{2}+0^{2}}}}}={\frac {3}{{\sqrt {1+4}}\cdot {\sqrt {9+0}}}}=}
= 3 5 ⋅ 3 = 1 5 = 0 , 447213595. {\displaystyle ={\frac {3}{{\sqrt {5}}\cdot 3}}={1 \over {\sqrt {5}}}=0,447213595.}
ϕ = arccos 1 5 = 1.107148718 {\displaystyle \phi =\arccos {1 \over {\sqrt {5}}}=1.107148718} arba 63,43494882 laipsnių.Pavyzdžiui, duoti vektoriai a =(3; 5; 11), b =(7; 8; 2). cos ϕ = a ⋅ b | | a | | ⋅ | | b | | = 3 ⋅ 7 + 5 ⋅ 8 + 11 ⋅ 2 3 2 + 5 2 + 11 2 ⋅ 7 2 + 8 2 + 2 2 = 21 + 40 + 22 9 + 25 + 121 ⋅ 49 + 64 + 4 = {\displaystyle \cos \phi ={\frac {a\cdot b}{||a||\cdot ||b||}}={\frac {3\cdot 7+5\cdot 8+11\cdot 2}{{\sqrt {3^{2}+5^{2}+11^{2}}}\cdot {\sqrt {7^{2}+8^{2}+2^{2}}}}}={\frac {21+40+22}{{\sqrt {9+25+121}}\cdot {\sqrt {49+64+4}}}}=}
= 83 155 ⋅ 117 = 83 18135 = 83 134.6662541 = 0.616338521. {\displaystyle ={\frac {83}{{\sqrt {155}}\cdot {\sqrt {117}}}}={\frac {83}{\sqrt {18135}}}={\frac {83}{134.6662541}}=0.616338521.}
ϕ = arccos ( 0.616338521 ) = 0.906711738 {\displaystyle \phi =\arccos(0.616338521)=0.906711738} arba 51.95075583 laipsnių.
Taikydami kosinusų toeremą patikrinsime ar kampas ϕ {\displaystyle \phi } surastas teisingai. Atkarpos f ilgis iš taško a=(3; 5; 11) iki taško b=(7; 8; 2) yra lygus
f = ( 3 − 7 ) 2 + ( 5 − 8 ) 2 + ( 11 − 2 ) 2 = 16 + 9 + 81 = 106 = 10.29563014. {\displaystyle f={\sqrt {(3-7)^{2}+(5-8)^{2}+(11-2)^{2}}}={\sqrt {16+9+81}}={\sqrt {106}}=10.29563014.} 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 = ( 106 ) 2 − ( 155 ) 2 − ( 117 ) 2 − 2 ⋅ 155 ⋅ 117 = 106 − 155 − 117 − 2 18135 = − 166 − 2 18135 = {\displaystyle \cos \phi ={f^{2}-a^{2}-b^{2} \over -2ab}={({\sqrt {106}})^{2}-({\sqrt {155}})^{2}-({\sqrt {117}})^{2} \over -2\cdot {\sqrt {155}}\cdot {\sqrt {117}}}={106-155-117 \over -2{\sqrt {18135}}}={-166 \over -2{\sqrt {18135}}}=} = 83 18135 = 0.616338521. {\displaystyle ={83 \over {\sqrt {18135}}}=0.616338521.}
Rasti, kampą tarp vektoriaus a → = { 1 ; 1 ; 1 } {\displaystyle {\vec {a}}=\{1;1;1\}} ir vektoriaus a → ′ = { 2 ; 3 ; − 4 } . {\displaystyle {\vec {a}}'=\{2;3;-4\}.} cos α = a → ⋅ a → ′ ‖ a → ‖ ⋅ ‖ a → ′ ‖ = 1 ⋅ 2 + 1 ⋅ 3 + 1 ⋅ ( − 4 ) 1 2 + 1 2 + 1 2 ⋅ 2 2 + 3 2 + ( − 4 ) 2 = 2 + 3 − 4 1 + 1 + 1 ⋅ 4 + 9 + 16 = {\displaystyle \cos \alpha ={\frac {{\vec {a}}\cdot {\vec {a}}'}{\|{\vec {a}}\|\cdot \|{\vec {a}}'\|}}={\frac {1\cdot 2+1\cdot 3+1\cdot (-4)}{{\sqrt {1^{2}+1^{2}+1^{2}}}\cdot {\sqrt {2^{2}+3^{2}+(-4)^{2}}}}}={\frac {2+3-4}{{\sqrt {1+1+1}}\cdot {\sqrt {4+9+16}}}}=}
= 1 3 ⋅ 29 = 1 87 = 1 9.327379053 = 0.107211253. {\displaystyle ={\frac {1}{{\sqrt {3}}\cdot {\sqrt {29}}}}={\frac {1}{\sqrt {87}}}={\frac {1}{9.327379053}}=0.107211253.}
α = arccos 1 87 = arccos ( 0.107211253 ) = 1.463378618 {\displaystyle \alpha =\arccos {\frac {1}{\sqrt {87}}}=\arccos(0.107211253)=1.463378618} arba 83.84541865 laipsniai.
Įrodyti, kad kampas tarp vektoriaus a → = { 2 ; 2 t ; − t 2 } {\displaystyle {\vec {a}}=\{2;2t;-t^{2}\}} orto a → ∘ = { 2 2 + t 2 ; 2 t 2 + t 2 ; − t 2 2 + t 2 } {\displaystyle {\vec {a}}^{\circ }=\{{\frac {2}{2+t^{2}}};{\frac {2t}{2+t^{2}}};{\frac {-t^{2}}{2+t^{2}}}\}} ir vektoriaus a → {\displaystyle {\vec {a}}} orto išvestinės lygus 90 laipsnių, kai parametras t = 1 {\displaystyle t=1} . ‖ a → ‖ = 2 2 + ( 2 t ) 2 + ( − t 2 ) 2 = 4 + 4 t 2 + t 4 = ( t 2 + 2 ) 2 = t 2 + 2. {\displaystyle \|{\vec {a}}\|={\sqrt {2^{2}+(2t)^{2}+(-t^{2})^{2}}}={\sqrt {4+4t^{2}+t^{4}}}={\sqrt {(t^{2}+2)^{2}}}=t^{2}+2.}
Sprendimas .
Vektoriaus a → ∘ {\displaystyle {\vec {a}}^{\circ }} išvestinė yra:
( a → ∘ ) ′ = { ( 2 2 + t 2 ) ′ ; ( 2 t 2 + t 2 ) ′ ; ( − t 2 2 + t 2 ) ′ } = { − 2 ⋅ 2 t ( 2 + t 2 ) 2 ; 2 ( 2 + t 2 ) − 2 t ⋅ 2 t ( 2 + t 2 ) 2 ; − 2 t ( 2 + t 2 ) − ( − t 2 ) ⋅ 2 t ( 2 + t 2 ) 2 } = {\displaystyle ({\vec {a}}^{\circ })'=\{({\frac {2}{2+t^{2}}})';({\frac {2t}{2+t^{2}}})';({\frac {-t^{2}}{2+t^{2}}})'\}=\{-{\frac {2\cdot 2t}{(2+t^{2})^{2}}};{\frac {2(2+t^{2})-2t\cdot 2t}{(2+t^{2})^{2}}};{\frac {-2t(2+t^{2})-(-t^{2})\cdot 2t}{(2+t^{2})^{2}}}\}=}
= { − 4 t ( 2 + t 2 ) 2 ; 4 − 2 t 2 ( 2 + t 2 ) 2 ; − 4 t ( 2 + t 2 ) 2 } . {\displaystyle =\{{\frac {-4t}{(2+t^{2})^{2}}};{\frac {4-2t^{2}}{(2+t^{2})^{2}}};{\frac {-4t}{(2+t^{2})^{2}}}\}.}
Randame vektorių reikšmes taške M 1 ( 1 ; 1 ; 1 ) {\displaystyle M_{1}(1;1;1)} :
a → ∘ | t = 1 = { 2 2 + 1 2 ; 2 ⋅ 1 2 + 1 2 ; − 1 2 2 + 1 2 } = { 2 3 ; 2 3 ; − 1 3 } ; {\displaystyle {\vec {a}}^{\circ }|_{t=1}=\{{\frac {2}{2+1^{2}}};{\frac {2\cdot 1}{2+1^{2}}};{\frac {-1^{2}}{2+1^{2}}}\}=\{{\frac {2}{3}};{\frac {2}{3}};{\frac {-1}{3}}\};}
( a → ∘ ) ′ | t = 1 = { − 4 ⋅ 1 ( 2 + 1 2 ) 2 ; 4 − 2 ⋅ 1 2 ( 2 + 1 2 ) 2 ; − 4 ⋅ 1 ( 2 + 1 2 ) 2 } = { − 4 9 ; 2 9 ; − 4 9 } . {\displaystyle ({\vec {a}}^{\circ })'|_{t=1}=\{{\frac {-4\cdot 1}{(2+1^{2})^{2}}};{\frac {4-2\cdot 1^{2}}{(2+1^{2})^{2}}};{\frac {-4\cdot 1}{(2+1^{2})^{2}}}\}=\{{\frac {-4}{9}};{\frac {2}{9}};{\frac {-4}{9}}\}.}
Randame kampą tarp vektoriaus a → ∘ {\displaystyle {\vec {a}}^{\circ }} ir vektoriaus ( a → ∘ ) ′ {\displaystyle ({\vec {a}}^{\circ })'} taške M 1 ( 1 ; 1 ; 1 ) {\displaystyle M_{1}(1;1;1)} :
cos α = a → ∘ | t = 1 ⋅ ( a → ∘ ) ′ | t = 1 ‖ a → ∘ | t = 1 ‖ ⋅ ‖ ( a → ∘ ) ′ | t = 1 ‖ = 2 3 ⋅ − 4 9 + 2 3 ⋅ 2 9 + − 1 3 ⋅ − 4 9 ( 2 3 ) 2 + ( 2 3 ) 2 + ( − 1 3 ) 2 ⋅ ( − 4 9 ) 2 + ( 2 9 ) 2 + ( − 4 9 ) 2 = − 8 27 + 4 27 + 4 27 4 9 + 4 9 + 1 9 ⋅ 16 81 + 4 81 + 16 81 = {\displaystyle \cos \alpha ={\frac {{\vec {a}}^{\circ }|_{t=1}\cdot ({\vec {a}}^{\circ })'|_{t=1}}{\|{\vec {a}}^{\circ }|_{t=1}\|\cdot \|({\vec {a}}^{\circ })'|_{t=1}\|}}={\frac {{\frac {2}{3}}\cdot {\frac {-4}{9}}+{\frac {2}{3}}\cdot {\frac {2}{9}}+{\frac {-1}{3}}\cdot {\frac {-4}{9}}}{{\sqrt {({\frac {2}{3}})^{2}+({\frac {2}{3}})^{2}+({\frac {-1}{3}})^{2}}}\cdot {\sqrt {({\frac {-4}{9}})^{2}+({\frac {2}{9}})^{2}+({\frac {-4}{9}})^{2}}}}}={\frac {{\frac {-8}{27}}+{\frac {4}{27}}+{\frac {4}{27}}}{{\sqrt {{\frac {4}{9}}+{\frac {4}{9}}+{\frac {1}{9}}}}\cdot {\sqrt {{\frac {16}{81}}+{\frac {4}{81}}+{\frac {16}{81}}}}}}=}
= 0 1 ⋅ 36 81 = 0 1 ⋅ 6 9 = 0 2 3 = 0. {\displaystyle ={\frac {0}{{\sqrt {1}}\cdot {\sqrt {\frac {36}{81}}}}}={\frac {0}{1\cdot {\frac {6}{9}}}}={\frac {0}{\frac {2}{3}}}=0.}
α = arccos ( 0 ) = π 2 = 1.570796327 {\displaystyle \alpha =\arccos(0)={\frac {\pi }{2}}=1.570796327} arba 90 laipsnių.
Duota kreivė užrašyta parametrinėmis lygtimis x = ϕ ( t ) = t , y = ψ ( t ) = t 2 , z = ω ( t ) = t 3 . {\displaystyle x=\phi (t)=t,\quad y=\psi (t)=t^{2},\quad z=\omega (t)=t^{3}.}
Rasti:
a) kreivės liestinės vektorių;
b) normalizuotą kreivės liestinės vektorių (ortą);
c) kreivės normalės vektorių iš normalizuoto liestinės vektoriaus;
d) kampą tarp liestinės orto (vektoriaus) ir normalės vektoriaus, kai parametro t reikšmė lygi 1 (kai t = 1 {\displaystyle t=1} );
e) kampą tarp liestinės vektoriaus ir normalės vektoriaus, kai t = 1 {\displaystyle t=1} , naudojantis normalės vektororiaus formule n = { 1 ± ϕ ′ ( t ) ; 1 ± ψ ′ ( t ) ; 1 ± ω ′ ( t ) } , {\displaystyle \mathbf {n} =\{{\frac {1}{\pm \phi '(t)}};{\frac {1}{\pm \psi '(t)}};{\frac {1}{\pm \omega '(t)}}\},} kuri šiai kreivei yra n = { 2 − 3 ϕ ′ ( t ) ; 1 3 ψ ′ ( t ) ; 1 3 ω ′ ( t ) } ; {\displaystyle \mathbf {n} =\{{\frac {2}{-3\phi '(t)}};{\frac {1}{3\psi '(t)}};{\frac {1}{3\omega '(t)}}\};} padauginti iš 2 3 {\displaystyle {\frac {2}{3}}} reikia tą funkciją, kurios rodiklis ant t mažiausias; kadangi x = ϕ ( t ) {\displaystyle x=\phi (t)} reikia prilyginti t , jei ϕ ( t ) {\displaystyle \phi (t)} turi mažiausią rodiklį virš t , tai gaunasi, kad parametro t ir ϕ ( t ) {\displaystyle \phi (t)} reikšmė nekinta (funkcijos ϕ ( t ) = t {\displaystyle \phi (t)=t} kitimo greitis visose taškuose lygus nuliui, nes ϕ ′ ( t ) = t ′ = 1 {\displaystyle \phi '(t)=t'=1} ), bet linija vis tiek kyla aukštyn, todėl tą linijos kilimą reikia kompensuoti 1 3 {\displaystyle {\frac {1}{3}}} (nes trys koordinačių ašys), kad tikrai būtų liestinės normalė, bet taip padaryti galima tik padarius 1 / ϕ ′ ( t ) {\displaystyle 1/\phi '(t)} reikšmę viena trečiąja didesne.
f) kampą tarp liestinės vektoriaus ir normalės vektoriaus, kai t = 5 {\displaystyle t=5} , naudojantis normalės vektororiaus formule n = { 2 − 3 ϕ ′ ( t ) ; 1 3 ψ ′ ( t ) ; 1 3 ω ′ ( t ) } . {\displaystyle \mathbf {n} =\{{\frac {2}{-3\phi '(t)}};{\frac {1}{3\psi '(t)}};{\frac {1}{3\omega '(t)}}\}.}
Sprendimas .
a) Kreivės liestinės vektorius yra
a = { t ′ ; ( t 2 ) ′ ; ( t 3 ) ′ } = { 1 ; 2 t ; 3 t 2 } . {\displaystyle \mathbf {a} =\{t';(t^{2})';(t^{3})'\}=\{1;2t;3t^{2}\}.}
b) Vektoriaus a {\displaystyle \mathbf {a} } ilgis yra:
‖ a ‖ = 1 2 + ( 2 t ) 2 + ( 3 t 2 ) 2 = 1 + 4 t 2 + 9 t 4 . {\displaystyle \|\mathbf {a} \|={\sqrt {1^{2}+(2t)^{2}+(3t^{2})^{2}}}={\sqrt {1+4t^{2}+9t^{4}}}.}
Kreivės liestinės normalizuotas vektorius yra šis:
a ∘ = a ‖ a ‖ = { 1 1 + 4 t 2 + 9 t 4 ; 2 t 1 + 4 t 2 + 9 t 4 ; 3 t 2 1 + 4 t 2 + 9 t 4 } ; {\displaystyle \mathbf {a} ^{\circ }={\frac {\mathbf {a} }{\|\mathbf {a} \|}}=\{{\frac {1}{\sqrt {1+4t^{2}+9t^{4}}}};{\frac {2t}{\sqrt {1+4t^{2}+9t^{4}}}};{\frac {3t^{2}}{\sqrt {1+4t^{2}+9t^{4}}}}\};}
jo reikšmė, kai t = 1 {\displaystyle t=1} yra
a ∘ | t = 1 = { 1 1 + 4 + 9 ; 2 1 + 4 + 9 ; 3 1 + 4 + 9 } , {\displaystyle \mathbf {a} ^{\circ }|_{t=1}=\{{\frac {1}{\sqrt {1+4+9}}};{\frac {2}{\sqrt {1+4+9}}};{\frac {3}{\sqrt {1+4+9}}}\},}
a ∘ | t = 1 = { 1 14 ; 2 14 ; 3 14 } = { 0.267261241 ; 0.534522483 ; 0.801783725 } . {\displaystyle \mathbf {a} ^{\circ }|_{t=1}=\{{\frac {1}{\sqrt {14}}};{\frac {2}{\sqrt {14}}};{\frac {3}{\sqrt {14}}}\}=\{0.267261241;0.534522483;0.801783725\}.}
c) Kreivės normalės vektorius yra liestinės vektoriaus orto išvestinė:
n = ( a ∘ ) ′ = { ( 1 1 + 4 t 2 + 9 t 4 ) ′ ; ( 2 t 1 + 4 t 2 + 9 t 4 ) ′ ; ( 3 t 2 1 + 4 t 2 + 9 t 4 ) ′ } , {\displaystyle \mathbf {n} =(\mathbf {a} ^{\circ })'=\{({\frac {1}{\sqrt {1+4t^{2}+9t^{4}}}})';({\frac {2t}{\sqrt {1+4t^{2}+9t^{4}}}})';({\frac {3t^{2}}{\sqrt {1+4t^{2}+9t^{4}}}})'\},}
n = { − ( 1 + 4 t 2 + 9 t 4 ) ′ 2 ( 1 + 4 t 2 + 9 t 4 ) 3 ; ( 2 t ) ′ 1 + 4 t 2 + 9 t 4 − 2 t ( 1 + 4 t 2 + 9 t 4 ) ′ ( 1 + 4 t 2 + 9 t 4 ) 2 ; ( 3 t 2 ) ′ 1 + 4 t 2 + 9 t 4 − 3 t 2 ( 1 + 4 t 2 + 9 t 4 ) ′ ( 1 + 4 t 2 + 9 t 4 ) 2 } , {\displaystyle \mathbf {n} =\{-{\frac {(1+4t^{2}+9t^{4})'}{2{\sqrt {(1+4t^{2}+9t^{4})^{3}}}}};{\frac {(2t)'{\sqrt {1+4t^{2}+9t^{4}}}-2t({\sqrt {1+4t^{2}+9t^{4}}})'}{({\sqrt {1+4t^{2}+9t^{4}}})^{2}}};{\frac {(3t^{2})'{\sqrt {1+4t^{2}+9t^{4}}}-3t^{2}({\sqrt {1+4t^{2}+9t^{4}}})'}{({\sqrt {1+4t^{2}+9t^{4}}})^{2}}}\},}
n = { − 8 t + 36 t 3 2 ( 1 + 4 t 2 + 9 t 4 ) 3 ; 2 1 + 4 t 2 + 9 t 4 − 2 t ⋅ ( 1 + 4 t 2 + 9 t 4 ) ′ 2 1 + 4 t 2 + 9 t 4 1 + 4 t 2 + 9 t 4 ; 6 t 1 + 4 t 2 + 9 t 4 − 3 t 2 ⋅ ( 1 + 4 t 2 + 9 t 4 ) ′ 2 1 + 4 t 2 + 9 t 4 1 + 4 t 2 + 9 t 4 } , {\displaystyle \mathbf {n} =\{-{\frac {8t+36t^{3}}{2{\sqrt {(1+4t^{2}+9t^{4})^{3}}}}};{\frac {2{\sqrt {1+4t^{2}+9t^{4}}}-2t\cdot {\frac {(1+4t^{2}+9t^{4})'}{2{\sqrt {1+4t^{2}+9t^{4}}}}}}{1+4t^{2}+9t^{4}}};{\frac {6t{\sqrt {1+4t^{2}+9t^{4}}}-3t^{2}\cdot {\frac {(1+4t^{2}+9t^{4})'}{2{\sqrt {1+4t^{2}+9t^{4}}}}}}{1+4t^{2}+9t^{4}}}\},}
n = { − 8 t + 36 t 3 2 ( 1 + 4 t 2 + 9 t 4 ) 3 ; 2 1 + 4 t 2 + 9 t 4 − 2 t ⋅ 8 t + 36 t 3 2 1 + 4 t 2 + 9 t 4 1 + 4 t 2 + 9 t 4 ; 6 t 1 + 4 t 2 + 9 t 4 − 3 t 2 ⋅ 8 t + 36 t 3 2 1 + 4 t 2 + 9 t 4 1 + 4 t 2 + 9 t 4 } , {\displaystyle \mathbf {n} =\{-{\frac {8t+36t^{3}}{2{\sqrt {(1+4t^{2}+9t^{4})^{3}}}}};{\frac {2{\sqrt {1+4t^{2}+9t^{4}}}-2t\cdot {\frac {8t+36t^{3}}{2{\sqrt {1+4t^{2}+9t^{4}}}}}}{1+4t^{2}+9t^{4}}};{\frac {6t{\sqrt {1+4t^{2}+9t^{4}}}-3t^{2}\cdot {\frac {8t+36t^{3}}{2{\sqrt {1+4t^{2}+9t^{4}}}}}}{1+4t^{2}+9t^{4}}}\},}
n = { − 8 t + 36 t 3 2 ( 1 + 4 t 2 + 9 t 4 ) 3 ; 2 1 + 4 t 2 + 9 t 4 − 2 t ⋅ 8 t + 36 t 3 2 ( 1 + 4 t 2 + 9 t 4 ) 3 ; 6 t 1 + 4 t 2 + 9 t 4 − 3 t 2 ⋅ 8 t + 36 t 3 2 ( 1 + 4 t 2 + 9 t 4 ) 3 } , {\displaystyle \mathbf {n} =\{-{\frac {8t+36t^{3}}{2{\sqrt {(1+4t^{2}+9t^{4})^{3}}}}};{\frac {2}{\sqrt {1+4t^{2}+9t^{4}}}}-2t\cdot {\frac {8t+36t^{3}}{2{\sqrt {(1+4t^{2}+9t^{4})^{3}}}}};{\frac {6t}{\sqrt {1+4t^{2}+9t^{4}}}}-3t^{2}\cdot {\frac {8t+36t^{3}}{2{\sqrt {(1+4t^{2}+9t^{4})^{3}}}}}\},}
n = { − 4 t + 18 t 3 ( 1 + 4 t 2 + 9 t 4 ) 3 ; 2 1 + 4 t 2 + 9 t 4 − 2 t ( 4 t + 18 t 3 ) ( 1 + 4 t 2 + 9 t 4 ) 3 ; 6 t 1 + 4 t 2 + 9 t 4 − 3 t 2 ( 4 t + 18 t 3 ) ( 1 + 4 t 2 + 9 t 4 ) 3 } ; {\displaystyle \mathbf {n} =\{-{\frac {4t+18t^{3}}{\sqrt {(1+4t^{2}+9t^{4})^{3}}}};{\frac {2}{\sqrt {1+4t^{2}+9t^{4}}}}-{\frac {2t(4t+18t^{3})}{\sqrt {(1+4t^{2}+9t^{4})^{3}}}};{\frac {6t}{\sqrt {1+4t^{2}+9t^{4}}}}-{\frac {3t^{2}(4t+18t^{3})}{\sqrt {(1+4t^{2}+9t^{4})^{3}}}}\};}
su reikšme t = 1 {\displaystyle t=1} kreivės normalės vektorius statmenas liestienei yra:
n | t = 1 = { − 4 + 18 ( 1 + 4 + 9 ) 3 ; 2 1 + 4 + 9 − 2 ( 4 + 18 ) ( 1 + 4 + 9 ) 3 ; 6 1 + 4 + 9 − 3 ( 4 + 18 ) ( 1 + 4 + 9 ) 3 } , {\displaystyle \mathbf {n} |_{t=1}=\{-{\frac {4+18}{\sqrt {(1+4+9)^{3}}}};{\frac {2}{\sqrt {1+4+9}}}-{\frac {2(4+18)}{\sqrt {(1+4+9)^{3}}}};{\frac {6}{\sqrt {1+4+9}}}-{\frac {3(4+18)}{\sqrt {(1+4+9)^{3}}}}\},}
n | t = 1 = { − 22 14 3 ; 2 14 − 2 ⋅ 22 14 3 ; 6 14 − 3 ⋅ 22 14 3 } , {\displaystyle \mathbf {n} |_{t=1}=\{-{\frac {22}{\sqrt {14^{3}}}};{\frac {2}{\sqrt {14}}}-{\frac {2\cdot 22}{\sqrt {14^{3}}}};{\frac {6}{\sqrt {14}}}-{\frac {3\cdot 22}{\sqrt {14^{3}}}}\},}
n | t = 1 = { − 22 2744 ; 2 14 − 44 2744 ; 6 14 − 66 2744 } , {\displaystyle \mathbf {n} |_{t=1}=\{-{\frac {22}{\sqrt {2744}}};{\frac {2}{\sqrt {14}}}-{\frac {44}{\sqrt {2744}}};{\frac {6}{\sqrt {14}}}-{\frac {66}{\sqrt {2744}}}\},}
n | t = 1 = { − 22 52.38320341 ; 2 3.741657387 − 44 52.38320341 ; 6 3.741657387 − 66 52.38320341 } , {\displaystyle \mathbf {n} |_{t=1}=\{-{\frac {22}{52.38320341}};{\frac {2}{3.741657387}}-{\frac {44}{52.38320341}};{\frac {6}{3.741657387}}-{\frac {66}{52.38320341}}\},}
n | t = 1 = { − 0.419981951 ; 0.534522483 − 0.839963903 ; 1.603567451 − 1.259945855 } , {\displaystyle \mathbf {n} |_{t=1}=\{-0.419981951;0.534522483-0.839963903;1.603567451-1.259945855\},}
n | t = 1 = { − 0.419981951 ; − 0.305441419 ; 0.343621596 } ; {\displaystyle \mathbf {n} |_{t=1}=\{-0.419981951;-0.305441419;0.343621596\};}
normalizuotas kreivės normalės vektorius yra šis:
n ∘ | t = 1 = n | t = 1 ( − 0.419981951 ) 2 + ( − 0.305441419 ) 2 + ( 0.343621596 ) 2 , {\displaystyle \mathbf {n} ^{\circ }|_{t=1}={\frac {\mathbf {n} |_{t=1}}{\sqrt {(-0.419981951)^{2}+(-0.305441419)^{2}+(0.343621596)^{2}}}},}
n ∘ | t = 1 = n | t = 1 0.17638484 + 0.093294461 + 0.118075802 = n | t = 1 0.387755103 = n | t = 1 0.62269985 , {\displaystyle \mathbf {n} ^{\circ }|_{t=1}={\frac {\mathbf {n} |_{t=1}}{\sqrt {0.17638484+0.093294461+0.118075802}}}={\frac {\mathbf {n} |_{t=1}}{\sqrt {0.387755103}}}={\frac {\mathbf {n} |_{t=1}}{0.62269985}},}
n ∘ | t = 1 = { − 0.419981951 0.62269985 ; − 0.305441419 0.62269985 ; 0.343621596 0.62269985 } = { − 0.674453271 ; − 0.49051147 ; 0.551825403 } . {\displaystyle \mathbf {n} ^{\circ }|_{t=1}=\{{\frac {-0.419981951}{0.62269985}};{\frac {-0.305441419}{0.62269985}};{\frac {0.343621596}{0.62269985}}\}=\{-0.674453271;-0.49051147;0.551825403\}.}
d) Kampas α {\displaystyle \alpha } tarp kreivės liestinės normalizuoto vektoriaus ir kreivės normalės vektoriaus yra toks:
cos α = a ∘ | t = 1 ⋅ n ∘ | t = 1 ‖ a ∘ | t = 1 ‖ ⋅ ‖ n ∘ | t = 1 ‖ = {\displaystyle \cos \alpha ={\frac {\mathbf {a} ^{\circ }|_{t=1}\cdot \mathbf {n} ^{\circ }|_{t=1}}{\|\mathbf {a} ^{\circ }|_{t=1}\|\cdot \|\mathbf {n} ^{\circ }|_{t=1}\|}}=}
= 0.267261241 ⋅ ( − 0.674453271 ) + 0.534522483 ⋅ ( − 0.49051147 ) + 0.801783725 ⋅ 0.551825403 1 ⋅ 1 = {\displaystyle ={\frac {0.267261241\cdot (-0.674453271)+0.534522483\cdot (-0.49051147)+0.801783725\cdot 0.551825403}{{\sqrt {1}}\cdot {\sqrt {1}}}}=}
= − 0.180255218 − 0.262189408 + 0.442444627 = − 0.442444626 + 0.442444627 = 0.000000001 = 0 ; {\displaystyle =-0.180255218-0.262189408+0.442444627=-0.442444626+0.442444627=0.000000001=0;}
α = arccos ( 0 ) = π 2 = 1.570796327 {\displaystyle \alpha =\arccos(0)={\frac {\pi }{2}}=1.570796327} arba 90 laipsnių.
e) Kreivės liestinės vektorius yra
a = { 1 ; 2 t ; 3 t 2 } ; {\displaystyle \mathbf {a} =\{1;2t;3t^{2}\};}
kreivės liestinės vektorius, kai t = 1 {\displaystyle t=1} yra:
a | t = 1 = { 1 ; 2 ; 3 } ; {\displaystyle \mathbf {a} |_{t=1}=\{1;2;3\};}
kreivės liestinės vektoriaus ortas, kai t = 1 {\displaystyle t=1} , yra:
a ∘ | t = 1 = { 1 1 + 4 + 9 ; 2 1 + 4 + 9 ; 3 1 + 4 + 9 } , {\displaystyle \mathbf {a} ^{\circ }|_{t=1}=\{{\frac {1}{\sqrt {1+4+9}}};{\frac {2}{\sqrt {1+4+9}}};{\frac {3}{\sqrt {1+4+9}}}\},}
a ∘ | t = 1 = { 1 14 ; 2 14 ; 3 14 } = { 0.267261241 ; 0.534522483 ; 0.801783725 } . {\displaystyle \mathbf {a} ^{\circ }|_{t=1}=\{{\frac {1}{\sqrt {14}}};{\frac {2}{\sqrt {14}}};{\frac {3}{\sqrt {14}}}\}=\{0.267261241;0.534522483;0.801783725\}.}
kreivės pseudonormalės vektorius yra:
p = { 1 − ϕ ′ ( t ) ; 1 ψ ′ ( t ) ; 1 ω ′ ( t ) = { − 1 ; 1 2 t ; 1 3 t } ; {\displaystyle \mathbf {p} =\{{\frac {1}{-\phi '(t)}};{\frac {1}{\psi '(t)}};{\frac {1}{\omega '(t)}}=\{-1;{\frac {1}{2t}};{\frac {1}{3t}}\};}
kreivės normalės vektorius yra:
n = { 2 − 3 ϕ ′ ( t ) ; 1 3 ψ ′ ( t ) ; 1 3 ω ′ ( t ) = { − 2 3 ; 1 6 t ; 1 9 t 2 } ; {\displaystyle \mathbf {n} =\{{\frac {2}{-3\phi '(t)}};{\frac {1}{3\psi '(t)}};{\frac {1}{3\omega '(t)}}=\{-{\frac {2}{3}};{\frac {1}{6t}};{\frac {1}{9t^{2}}}\};}
kreivės pseudonormalės vektorius, kai t = 1 {\displaystyle t=1} , yra:
p | t = 1 = { − 1 ; 1 2 ; 1 3 } ; {\displaystyle \mathbf {p} |_{t=1}=\{-1;{\frac {1}{2}};{\frac {1}{3}}\};}
kreivės normalės vektorius, kai t = 1 {\displaystyle t=1} , yra:
n | t = 1 = { − 2 3 ; 1 6 ; 1 9 } ; {\displaystyle \mathbf {n} |_{t=1}=\{-{\frac {2}{3}};{\frac {1}{6}};{\frac {1}{9}}\};}
kreivės pseudonormalės normalizuotas vektorius, kai t = 1 {\displaystyle t=1} yra:
p ∘ | t = 1 = { − 1 ( − 1 ) 2 + ( 1 2 ) 2 + ( 1 3 ) 2 ; 1 2 ( − 1 ) 2 + ( 1 2 ) 2 + ( 1 3 ) 2 ; 1 3 ( − 1 ) 2 + ( 1 2 ) 2 + ( 1 3 ) 2 } , {\displaystyle \mathbf {p} ^{\circ }|_{t=1}=\{{\frac {-1}{\sqrt {(-1)^{2}+({1 \over 2})^{2}+({1 \over 3})^{2}}}};{\frac {\frac {1}{2}}{\sqrt {(-1)^{2}+({1 \over 2})^{2}+({1 \over 3})^{2}}}};{\frac {\frac {1}{3}}{\sqrt {(-1)^{2}+({1 \over 2})^{2}+({1 \over 3})^{2}}}}\},}
p ∘ | t = 1 = { − 1 1 + 1 4 + 1 9 ; 1 2 1 + 1 4 + 1 9 ; 1 3 1 + 1 4 + 1 9 } = { − 1 36 + 9 + 4 36 ; 1 2 36 + 9 + 4 36 ; 1 3 36 + 9 + 4 36 } , {\displaystyle \mathbf {p} ^{\circ }|_{t=1}=\{{\frac {-1}{\sqrt {1+{1 \over 4}+{1 \over 9}}}};{\frac {\frac {1}{2}}{\sqrt {1+{1 \over 4}+{1 \over 9}}}};{\frac {\frac {1}{3}}{\sqrt {1+{1 \over 4}+{1 \over 9}}}}\}=\{{\frac {-1}{\sqrt {36+9+4 \over 36}}};{\frac {\frac {1}{2}}{\sqrt {36+9+4 \over 36}}};{\frac {\frac {1}{3}}{\sqrt {36+9+4 \over 36}}}\},}
p ∘ | t = 1 = { − 1 49 36 ; 1 2 49 36 ; 1 3 49 36 } = { − 1 7 6 ; 1 2 7 6 ; 1 3 7 6 } , {\displaystyle \mathbf {p} ^{\circ }|_{t=1}=\{{\frac {-1}{\sqrt {49 \over 36}}};{\frac {\frac {1}{2}}{\sqrt {49 \over 36}}};{\frac {\frac {1}{3}}{\sqrt {49 \over 36}}}\}=\{{\frac {-1}{7 \over 6}};{\frac {\frac {1}{2}}{7 \over 6}};{\frac {\frac {1}{3}}{7 \over 6}}\},}
p ∘ | t = 1 = { − 6 7 ; 6 14 ; 6 21 } = { − 0.857142857 ; 0.428571428 ; 0.285714285 } ; {\displaystyle \mathbf {p} ^{\circ }|_{t=1}=\{{\frac {-6}{7}};{\frac {6}{14}};{\frac {6}{21}}\}=\{-0.857142857;0.428571428;0.285714285\};}
kreivės normalės normalizuotas vektorius, kai t = 1 {\displaystyle t=1} yra:
n ∘ | t = 1 = { − 2 3 ( − 2 3 ) 2 + ( 1 6 ) 2 + ( 1 9 ) 2 ; 1 6 ( − 2 3 ) 2 + ( 1 6 ) 2 + ( 1 9 ) 2 ; 1 9 ( − 2 3 ) 2 + ( 1 6 ) 2 + ( 1 9 ) 2 } , {\displaystyle \mathbf {n} ^{\circ }|_{t=1}=\{{\frac {-{\frac {2}{3}}}{\sqrt {(-{\frac {2}{3}})^{2}+({1 \over 6})^{2}+({1 \over 9})^{2}}}};{\frac {\frac {1}{6}}{\sqrt {(-{\frac {2}{3}})^{2}+({1 \over 6})^{2}+({1 \over 9})^{2}}}};{\frac {\frac {1}{9}}{\sqrt {(-{\frac {2}{3}})^{2}+({1 \over 6})^{2}+({1 \over 9})^{2}}}}\},}
n ∘ | t = 1 = { − 2 3 4 9 + 1 36 + 1 81 ; 1 6 4 9 + 1 36 + 1 81 ; 1 9 4 9 + 1 36 + 1 81 } = { − 2 3 4 ⋅ 36 + 9 + 4 324 ; 1 2 144 + 9 + 4 324 ; 1 3 144 + 9 + 4 324 } , {\displaystyle \mathbf {n} ^{\circ }|_{t=1}=\{{\frac {-{\frac {2}{3}}}{\sqrt {{4 \over 9}+{1 \over 36}+{1 \over 81}}}};{\frac {\frac {1}{6}}{\sqrt {{4 \over 9}+{1 \over 36}+{1 \over 81}}}};{\frac {\frac {1}{9}}{\sqrt {{4 \over 9}+{1 \over 36}+{1 \over 81}}}}\}=\{{\frac {-{\frac {2}{3}}}{\sqrt {4\cdot 36+9+4 \over 324}}};{\frac {\frac {1}{2}}{\sqrt {144+9+4 \over 324}}};{\frac {\frac {1}{3}}{\sqrt {144+9+4 \over 324}}}\},}
n ∘ | t = 1 = { − 2 3 157 324 ; 1 6 157 324 ; 1 9 157 324 } = { − 2 ⋅ 18 3 ⋅ 157 ; 18 6 ⋅ 157 ; 18 9 ⋅ 157 } , {\displaystyle \mathbf {n} ^{\circ }|_{t=1}=\{{\frac {-{\frac {2}{3}}}{\sqrt {157 \over 324}}};{\frac {\frac {1}{6}}{\sqrt {157 \over 324}}};{\frac {\frac {1}{9}}{\sqrt {157 \over 324}}}\}=\{-{2\cdot 18 \over 3\cdot {\sqrt {157}}};{\frac {18}{6\cdot {\sqrt {157}}}};{\frac {18}{9\cdot {\sqrt {157}}}}\},}
n ∘ | t = 1 = { − 12 157 ; 3 157 ; 2 157 } = { − 0.957704261 ; 0.239426065 ; 0.159617376 } ; {\displaystyle \mathbf {n} ^{\circ }|_{t=1}=\{{\frac {-12}{\sqrt {157}}};{\frac {3}{\sqrt {157}}};{\frac {2}{\sqrt {157}}}\}=\{-0.957704261;0.239426065;0.159617376\};}
kampas tarp pseudonormalės ir liestinės vektorių yra:
cos α = 0.267261241 ⋅ ( − 0.857142857 ) + 0.534522483 ⋅ 0.428571428 + 0.801783725 ⋅ 0.285714285 = {\displaystyle \cos \alpha =0.267261241\cdot (-0.857142857)+0.534522483\cdot 0.428571428+0.801783725\cdot 0.285714285=}
= − 0.229081063 + 0.229081063 + 0.229081063 = 0.229081063 ; {\displaystyle =-0.229081063+0.229081063+0.229081063=0.229081063;}
α = arccos ( 0.229081063 ) = 1.33966279 {\displaystyle \alpha =\arccos(0.229081063)=1.33966279} arba 76.75702383 laipsniai;
kampas tarp liestinės ir normalės vektoriaus yra:
cos θ = 0.267261241 ⋅ ( − 0.957704261 ) + 0.534522483 ⋅ 0.239426065 + 0.801783725 ⋅ 0.159617376 = {\displaystyle \cos \theta =0.267261241\cdot (-0.957704261)+0.534522483\cdot 0.239426065+0.801783725\cdot 0.159617376=}
= − 0.255957229 + 0.127978614 + 0.127978614 = − 0.255957229 + 0.255957229 = 0 ; {\displaystyle =-0.255957229+0.127978614+0.127978614=-0.255957229+0.255957229=0;}
θ = arccos ( 0 ) = π 2 {\displaystyle \theta =\arccos(0)={\frac {\pi }{2}}} arba 90 laipsnių.
f) Kampas tarp pseudonormalės ir liestinės yra α = 84.44409162 {\displaystyle \alpha =84.44409162} laipsniai;
kreivės normalės vektorius yra:
n = { 2 − 3 ϕ ′ ( t ) ; 1 3 ψ ′ ( t ) ; 1 3 ω ′ ( t ) = { − 2 3 ; 1 6 t ; 1 9 t 2 } ; {\displaystyle \mathbf {n} =\{{\frac {2}{-3\phi '(t)}};{\frac {1}{3\psi '(t)}};{\frac {1}{3\omega '(t)}}=\{-{\frac {2}{3}};{\frac {1}{6t}};{\frac {1}{9t^{2}}}\};}
kreivės normalės vektorius, kai t = 5 {\displaystyle t=5} yra:
n = { − 2 3 ; 1 6 ⋅ 5 ; 1 9 ⋅ 5 2 } = { − 2 3 ; 1 30 ; 1 225 } ; {\displaystyle \mathbf {n} =\{-{\frac {2}{3}};{\frac {1}{6\cdot 5}};{\frac {1}{9\cdot 5^{2}}}\}=\{-{\frac {2}{3}};{\frac {1}{30}};{\frac {1}{225}}\};}
kreivės liestinės vektorius yra
a = { 1 ; 2 t ; 3 t 2 } ; {\displaystyle \mathbf {a} =\{1;2t;3t^{2}\};}
kreivės liestinės vektorius, kai t = 5 {\displaystyle t=5} yra
a = { 1 ; 2 ⋅ 5 ; 3 ⋅ 5 2 } = { 1 ; 10 ; 75 } ; {\displaystyle \mathbf {a} =\{1;2\cdot 5;3\cdot 5^{2}\}=\{1;10;75\};}
Jeigu vektorių skaliarinė sandauga lygi nuliui, tai vektoriai yra statmeni vienas kitam:
a ⋅ n = 1 ⋅ ( − 2 3 ) + 10 ⋅ 1 30 + 75 ⋅ 1 225 = − 2 3 + 1 3 + 1 3 = 0. {\displaystyle \mathbf {a} \cdot \mathbf {n} =1\cdot (-{\frac {2}{3}})+10\cdot {\frac {1}{30}}+75\cdot {\frac {1}{225}}=-{\frac {2}{3}}+{\frac {1}{3}}+{\frac {1}{3}}=0.}
Duota kreivė užrašyta parametrinėmis lygtimis x = ϕ ( t ) = t 2 , y = ψ ( t ) = t 3 , z = ω ( t ) = t 4 . {\displaystyle x=\phi (t)=t^{2},\quad y=\psi (t)=t^{3},\quad z=\omega (t)=t^{4}.}
Rasti:
a) kreivės liestinės vektorių;
b) normalizuotą kreivės liestinės vektorių (ortą);
c) kreivės normalės vektorių iš normalizuoto liestinės vektoriaus;
d) kampą tarp liestinės orto (vektoriaus) ir normalės vektoriaus, kai parametro t reikšmė lygi 1;
e) normalizuotą liestinės vektorių naudojantis formule b = { 1 ; ψ ′ ( t ) ϕ ′ ( t ) ; ω ′ ( t ) ϕ ′ ( t ) } {\displaystyle \mathbf {b} =\{1;{\frac {\psi '(t)}{\phi '(t)}};{\frac {\omega '(t)}{\phi '(t)}}\}} ir palyginti su normalizuotu liestinės vektoriu a = { ϕ ′ ( t ) ; ψ ′ ( t ) ; ω ′ ( t ) } , {\displaystyle \mathbf {a} =\{\phi '(t);\psi '(t);\omega '(t)\},} kai t = 1 {\displaystyle t=1} ir kai t = 5 {\displaystyle t=5} ;
f) tikrąjį kreivės normalės vektorių naudojantis formule n = { 2 − 3 ϕ ′ ( t ) ; 1 3 ψ ′ ( t ) ; 1 3 ω ′ ( t ) } , {\displaystyle \mathbf {n} =\{{\frac {2}{-3\phi '(t)}};{\frac {1}{3\psi '(t)}};{\frac {1}{3\omega '(t)}}\},} kai t = 1 {\displaystyle t=1} ir kai t = 5 {\displaystyle t=5} ; apskaičiuoti kampą tarp šio normalės vektoriaus ir tarp liestinės vektoriaus, kai t = 1 {\displaystyle t=1} ir kai t = 5 {\displaystyle t=5} ;
g) tikrąjį kreivės normalės vektorių naudojantis formule n = { 2 − 3 ϕ ′ ( t ) ϕ ′ ( t ) ; 1 3 ψ ′ ( t ) ϕ ′ ( t ) ; 1 3 ω ′ ( t ) ϕ ′ ( t ) } = { − 2 3 ; ϕ ′ ( t ) 3 ψ ′ ( t ) ; ϕ ′ ( t ) 3 ω ′ ( t ) } , {\displaystyle \mathbf {n} =\{{\frac {2}{-3{\phi '(t) \over \phi '(t)}}};{\frac {1}{3{\psi '(t) \over \phi '(t)}}};{\frac {1}{3{\omega '(t) \over \phi '(t)}}}\}=\{-{\frac {2}{3}};{\frac {\phi '(t)}{3\psi '(t)}};{\frac {\phi '(t)}{3\omega '(t)}}\},} kai t = 1 {\displaystyle t=1} ir kai t = 5 {\displaystyle t=5} ; apskaičiuoti kampą tarp šio normalės vektoriaus ir tarp liestinės vektoriaus, kai t = 1 {\displaystyle t=1} ir kai t = 5 {\displaystyle t=5} .
Sprendimas .
a) Kreivės liestinės vektorius yra
a = { ( t 2 ) ′ ; ( t 3 ) ′ ; ( t 4 ) ′ } = { 2 t ; 3 t 2 ; 4 t 3 } . {\displaystyle \mathbf {a} =\{(t^{2})';(t^{3})';(t^{4})'\}=\{2t;3t^{2};4t^{3}\}.}
b) Vektoriaus a {\displaystyle \mathbf {a} } ilgis yra:
‖ a ‖ = ( 2 t ) 2 + ( 3 t 2 ) 2 + ( 4 t 3 ) 2 = 4 t 2 + 9 t 4 + 16 t 6 . {\displaystyle \|\mathbf {a} \|={\sqrt {(2t)^{2}+(3t^{2})^{2}+(4t^{3})^{2}}}={\sqrt {4t^{2}+9t^{4}+16t^{6}}}.}
Kreivės liestinės normalizuotas vektorius yra šis:
a ∘ = a ‖ a ‖ = { 2 t 4 t 2 + 9 t 4 + 16 t 6 ; 3 t 2 4 t 2 + 9 t 4 + 16 t 6 ; 4 t 3 4 t 2 + 9 t 4 + 16 t 6 } ; {\displaystyle \mathbf {a} ^{\circ }={\frac {\mathbf {a} }{\|\mathbf {a} \|}}=\{{\frac {2t}{\sqrt {4t^{2}+9t^{4}+16t^{6}}}};{\frac {3t^{2}}{\sqrt {4t^{2}+9t^{4}+16t^{6}}}};{\frac {4t^{3}}{\sqrt {4t^{2}+9t^{4}+16t^{6}}}}\};}
jo reikšmė, kai t = 1 {\displaystyle t=1} yra
a ∘ | t = 1 = { 2 4 + 9 + 16 ; 3 4 + 9 + 16 ; 4 4 + 9 + 16 } , {\displaystyle \mathbf {a} ^{\circ }|_{t=1}=\{{\frac {2}{\sqrt {4+9+16}}};{\frac {3}{\sqrt {4+9+16}}};{\frac {4}{\sqrt {4+9+16}}}\},}
a ∘ | t = 1 = { 2 29 ; 3 29 ; 4 29 } = { 0.371390676 ; 0.557086014 ; 0.742781352 } . {\displaystyle \mathbf {a} ^{\circ }|_{t=1}=\{{\frac {2}{\sqrt {29}}};{\frac {3}{\sqrt {29}}};{\frac {4}{\sqrt {29}}}\}=\{0.371390676;0.557086014;0.742781352\}.}
c) Kreivės normalės vektorius yra liestinės vektoriaus orto išvestinė:
n = ( a ∘ ) ′ = { ( 2 t 4 t 2 + 9 t 4 + 16 t 6 ) ′ ; ( 3 t 2 4 t 2 + 9 t 4 + 16 t 6 ) ′ ; ( 4 t 3 4 t 2 + 9 t 4 + 16 t 6 ) ′ } , {\displaystyle \mathbf {n} =(\mathbf {a} ^{\circ })'=\{({\frac {2t}{\sqrt {4t^{2}+9t^{4}+16t^{6}}}})';({\frac {3t^{2}}{\sqrt {4t^{2}+9t^{4}+16t^{6}}}})';({\frac {4t^{3}}{\sqrt {4t^{2}+9t^{4}+16t^{6}}}})'\},}
n = { ( 2 t ) ′ 4 t 2 + 9 t 4 + 16 t 6 − 2 t ( 4 t 2 + 9 t 4 + 16 t 6 ) ′ ( 4 t 2 + 9 t 4 + 16 t 6 ) 2 ; ( 3 t 2 ) ′ 4 t 2 + 9 t 4 + 16 t 6 − 3 t 2 ( 4 t 2 + 9 t 4 + 16 t 6 ) ′ ( 4 t 2 + 9 t 4 + 16 t 6 ) 2 ; ( 4 t 3 ) ′ 4 t 2 + 9 t 4 + 16 t 6 − 4 t 3 ( 4 t 2 + 9 t 4 + 16 t 6 ) ′ ( 4 t 2 + 9 t 4 + 16 t 6 ) 2 } , {\displaystyle \mathbf {n} =\{{\frac {(2t)'{\sqrt {4t^{2}+9t^{4}+16t^{6}}}-2t({\sqrt {4t^{2}+9t^{4}+16t^{6}}})'}{({\sqrt {4t^{2}+9t^{4}+16t^{6}}})^{2}}};{\frac {(3t^{2})'{\sqrt {4t^{2}+9t^{4}+16t^{6}}}-3t^{2}({\sqrt {4t^{2}+9t^{4}+16t^{6}}})'}{({\sqrt {4t^{2}+9t^{4}+16t^{6}}})^{2}}};{\frac {(4t^{3})'{\sqrt {4t^{2}+9t^{4}+16t^{6}}}-4t^{3}({\sqrt {4t^{2}+9t^{4}+16t^{6}}})'}{({\sqrt {4t^{2}+9t^{4}+16t^{6}}})^{2}}}\},}
n = { 2 4 t 2 + 9 t 4 + 16 t 6 − 2 t ⋅ ( 4 t 2 + 9 t 4 + 16 t 6 ) ′ 2 4 t 2 + 9 t 4 + 16 t 6 4 t 2 + 9 t 4 + 16 t 6 ; 6 t 4 t 2 + 9 t 4 + 16 t 6 − 3 t 2 ⋅ ( 4 t 2 + 9 t 4 + 16 t 6 ) ′ 2 4 t 2 + 9 t 4 + 16 t 6 4 t 2 + 9 t 4 + 16 t 6 ; 12 t 2 4 t 2 + 9 t 4 + 16 t 6 − 4 t 3 ⋅ ( 4 t 2 + 9 t 4 + 16 t 6 ) ′ 2 4 t 2 + 9 t 4 + 16 t 6 4 t 2 + 9 t 4 + 16 t 6 } , {\displaystyle \mathbf {n} =\{{\frac {2{\sqrt {4t^{2}+9t^{4}+16t^{6}}}-2t\cdot {\frac {(4t^{2}+9t^{4}+16t^{6})'}{2{\sqrt {4t^{2}+9t^{4}+16t^{6}}}}}}{4t^{2}+9t^{4}+16t^{6}}};{\frac {6t{\sqrt {4t^{2}+9t^{4}+16t^{6}}}-3t^{2}\cdot {\frac {(4t^{2}+9t^{4}+16t^{6})'}{2{\sqrt {4t^{2}+9t^{4}+16t^{6}}}}}}{4t^{2}+9t^{4}+16t^{6}}};{\frac {12t^{2}{\sqrt {4t^{2}+9t^{4}+16t^{6}}}-4t^{3}\cdot {\frac {(4t^{2}+9t^{4}+16t^{6})'}{2{\sqrt {4t^{2}+9t^{4}+16t^{6}}}}}}{4t^{2}+9t^{4}+16t^{6}}}\},}
n = { 2 4 t 2 + 9 t 4 + 16 t 6 − 2 t ⋅ 8 t + 36 t 3 + 96 t 5 2 4 t 2 + 9 t 4 + 16 t 6 4 t 2 + 9 t 4 + 16 t 6 ; 6 t 4 t 2 + 9 t 4 + 16 t 6 − 3 t 2 ⋅ 8 t + 36 t 3 + 96 t 5 2 4 t 2 + 9 t 4 + 16 t 6 4 t 2 + 9 t 4 + 16 t 6 ; 12 t 2 4 t 2 + 9 t 4 + 16 t 6 − 4 t 3 ⋅ 8 t + 36 t 3 + 96 t 5 2 4 t 2 + 9 t 4 + 16 t 6 4 t 2 + 9 t 4 + 16 t 6 } , {\displaystyle \mathbf {n} =\{{\frac {2{\sqrt {4t^{2}+9t^{4}+16t^{6}}}-2t\cdot {\frac {8t+36t^{3}+96t^{5}}{2{\sqrt {4t^{2}+9t^{4}+16t^{6}}}}}}{4t^{2}+9t^{4}+16t^{6}}};{\frac {6t{\sqrt {4t^{2}+9t^{4}+16t^{6}}}-3t^{2}\cdot {\frac {8t+36t^{3}+96t^{5}}{2{\sqrt {4t^{2}+9t^{4}+16t^{6}}}}}}{4t^{2}+9t^{4}+16t^{6}}};{\frac {12t^{2}{\sqrt {4t^{2}+9t^{4}+16t^{6}}}-4t^{3}\cdot {\frac {8t+36t^{3}+96t^{5}}{2{\sqrt {4t^{2}+9t^{4}+16t^{6}}}}}}{4t^{2}+9t^{4}+16t^{6}}}\},}
n = { 2 4 t 2 + 9 t 4 + 16 t 6 − 2 t ⋅ 8 t + 36 t 3 + 96 t 5 2 ( 4 t 2 + 9 t 4 + 16 t 6 ) 3 ; 6 t 4 t 2 + 9 t 4 + 16 t 6 − 3 t 2 ⋅ 8 t + 36 t 3 + 96 t 5 2 ( 4 t 2 + 9 t 4 + 16 t 6 ) 3 ; 12 t 2 4 t 2 + 9 t 4 + 16 t 6 − 4 t 3 ⋅ 8 t + 36 t 3 + 96 t 5 2 ( 4 t 2 + 9 t 4 + 16 t 6 ) 3 } , {\displaystyle \mathbf {n} =\{{\frac {2}{\sqrt {4t^{2}+9t^{4}+16t^{6}}}}-2t\cdot {\frac {8t+36t^{3}+96t^{5}}{2{\sqrt {(4t^{2}+9t^{4}+16t^{6})^{3}}}}};{\frac {6t}{\sqrt {4t^{2}+9t^{4}+16t^{6}}}}-3t^{2}\cdot {\frac {8t+36t^{3}+96t^{5}}{2{\sqrt {(4t^{2}+9t^{4}+16t^{6})^{3}}}}};{\frac {12t^{2}}{\sqrt {4t^{2}+9t^{4}+16t^{6}}}}-4t^{3}\cdot {\frac {8t+36t^{3}+96t^{5}}{2{\sqrt {(4t^{2}+9t^{4}+16t^{6})^{3}}}}}\},}
n = { 2 4 t 2 + 9 t 4 + 16 t 6 − 2 t ( 4 t + 18 t 3 + 48 t 5 ) ( 4 t 2 + 9 t 4 + 16 t 6 ) 3 ; 6 t 4 t 2 + 9 t 4 + 16 t 6 − 3 t 2 ( 4 t + 18 t 3 + 48 t 5 ) ( 4 t 2 + 9 t 4 + 16 t 6 ) 3 ; 12 t 2 4 t 2 + 9 t 4 + 16 t 6 − 4 t 3 ( 4 t + 18 t 3 + 48 t 5 ) ( 4 t 2 + 9 t 4 + 16 t 6 ) 3 } ; {\displaystyle \mathbf {n} =\{{\frac {2}{\sqrt {4t^{2}+9t^{4}+16t^{6}}}}-{\frac {2t(4t+18t^{3}+48t^{5})}{\sqrt {(4t^{2}+9t^{4}+16t^{6})^{3}}}};{\frac {6t}{\sqrt {4t^{2}+9t^{4}+16t^{6}}}}-{\frac {3t^{2}(4t+18t^{3}+48t^{5})}{\sqrt {(4t^{2}+9t^{4}+16t^{6})^{3}}}};{\frac {12t^{2}}{\sqrt {4t^{2}+9t^{4}+16t^{6}}}}-{\frac {4t^{3}(4t+18t^{3}+48t^{5})}{\sqrt {(4t^{2}+9t^{4}+16t^{6})^{3}}}}\};}
su reikšme t = 1 {\displaystyle t=1} kreivės normalės vektorius statmenas liestienei yra:
n | t = 1 = { 2 4 + 9 + 16 − 2 ( 4 + 18 + 48 ) ( 4 + 9 + 16 ) 3 ; 6 4 + 9 + 16 − 3 ( 4 + 18 + 48 ) ( 4 + 9 + 16 ) 3 ; 12 4 + 9 + 16 − 4 ( 4 + 18 + 48 ) ( 4 + 9 + 16 ) 3 } ; {\displaystyle \mathbf {n} |_{t=1}=\{{\frac {2}{\sqrt {4+9+16}}}-{\frac {2(4+18+48)}{\sqrt {(4+9+16)^{3}}}};{\frac {6}{\sqrt {4+9+16}}}-{\frac {3(4+18+48)}{\sqrt {(4+9+16)^{3}}}};{\frac {12}{\sqrt {4+9+16}}}-{\frac {4(4+18+48)}{\sqrt {(4+9+16)^{3}}}}\};}
n | t = 1 = { 2 29 − 2 ⋅ 70 29 3 ; 6 29 − 3 ⋅ 70 29 3 ; 12 29 − 4 ⋅ 70 29 3 } ; {\displaystyle \mathbf {n} |_{t=1}=\{{\frac {2}{\sqrt {29}}}-{\frac {2\cdot 70}{\sqrt {29^{3}}}};{\frac {6}{\sqrt {29}}}-{\frac {3\cdot 70}{\sqrt {29^{3}}}};{\frac {12}{\sqrt {29}}}-{\frac {4\cdot 70}{\sqrt {29^{3}}}}\};}
n | t = 1 = { 2 29 − 140 24389 ; 6 29 − 210 24389 ; 12 29 − 280 24389 } ; {\displaystyle \mathbf {n} |_{t=1}=\{{\frac {2}{\sqrt {29}}}-{\frac {140}{\sqrt {24389}}};{\frac {6}{\sqrt {29}}}-{\frac {210}{\sqrt {24389}}};{\frac {12}{\sqrt {29}}}-{\frac {280}{\sqrt {24389}}}\};}
n | t = 1 = { − 0.525069576 ; − 0.23051835 ; 0.435423551 } ; {\displaystyle \mathbf {n} |_{t=1}=\{-0.525069576;-0.23051835;0.435423551\};}
normalizuotas kreivės normalės vektorius yra šis:
n ∘ | t = 1 = n | t = 1 ( − 0.525069576 ) 2 + ( − 0.23051835 ) 2 + ( 0.435423551 ) 2 = n | t = 1 0.518430438 = n | t = 1 0.720021137 , {\displaystyle \mathbf {n} ^{\circ }|_{t=1}={\frac {\mathbf {n} |_{t=1}}{\sqrt {(-0.525069576)^{2}+(-0.23051835)^{2}+(0.435423551)^{2}}}}={\frac {\mathbf {n} |_{t=1}}{\sqrt {0.518430438}}}={\frac {\mathbf {n} |_{t=1}}{0.720021137}},}
n ∘ | t = 1 = { − 0.525069576 0.720021137 ; − 0.23051835 0.720021137 ; 0.435423551 0.720021137 } = { − 0.729241891 ; − 0.320154976 ; 0.604737178 } . {\displaystyle \mathbf {n} ^{\circ }|_{t=1}=\{{\frac {-0.525069576}{0.720021137}};{\frac {-0.23051835}{0.720021137}};{\frac {0.435423551}{0.720021137}}\}=\{-0.729241891;-0.320154976;0.604737178\}.}
d) Kampas α {\displaystyle \alpha } tarp kreivės liestinės normalizuoto vektoriaus ir kreivės normalės vektoriaus yra toks:
cos α = a ∘ | t = 1 ⋅ n ∘ | t = 1 ‖ a ∘ | t = 1 ‖ ⋅ ‖ n ∘ | t = 1 ‖ = {\displaystyle \cos \alpha ={\frac {\mathbf {a} ^{\circ }|_{t=1}\cdot \mathbf {n} ^{\circ }|_{t=1}}{\|\mathbf {a} ^{\circ }|_{t=1}\|\cdot \|\mathbf {n} ^{\circ }|_{t=1}\|}}=}
= 0.371390676 ⋅ ( − 0.729241891 ) + 0.557086014 ⋅ ( − 0.320154976 ) + 0.742781352 ⋅ 0.604737178 1 ⋅ 1 = {\displaystyle ={\frac {0.371390676\cdot (-0.729241891)+0.557086014\cdot (-0.320154976)+0.742781352\cdot 0.604737178}{{\sqrt {1}}\cdot {\sqrt {1}}}}=}
= − 0.270833638 − 0.178353859 + 0.449187498 = − 0.449187497 + 0.449187498 = 0.000000001 = 0 ; {\displaystyle =-0.270833638-0.178353859+0.449187498=-0.449187497+0.449187498=0.000000001=0;}
α = arccos ( 0 ) = π 2 = 1.570796327 {\displaystyle \alpha =\arccos(0)={\frac {\pi }{2}}=1.570796327} arba 90 laipsnių.
e) Kai t = 1 {\displaystyle t=1} normalizuotas liestinės vektorius yra:
a ∘ | t = 1 = { 2 29 ; 3 29 ; 4 29 } = { 0.371390676 ; 0.557086014 ; 0.742781352 } ; {\displaystyle \mathbf {a} ^{\circ }|_{t=1}=\{{\frac {2}{\sqrt {29}}};{\frac {3}{\sqrt {29}}};{\frac {4}{\sqrt {29}}}\}=\{0.371390676;0.557086014;0.742781352\};}
kai t = 5 {\displaystyle t=5} liestinės vektorius yra:
a | t = 5 = { 2 ⋅ 5 ; 3 ⋅ 5 2 ; 4 ⋅ 5 3 } = { 10 ; 75 ; 500 } ; {\displaystyle \mathbf {a} |_{t=5}=\{2\cdot 5;3\cdot 5^{2};4\cdot 5^{3}\}=\{10;75;500\};}
kai t = 5 {\displaystyle t=5} normalizuotas liestinės vektorius yra:
a ∘ | t = 5 = { 10 10 2 + 75 2 + 500 2 ; 75 10 2 + 75 2 + 500 2 ; 500 10 2 + 75 2 + 500 2 } , {\displaystyle \mathbf {a} ^{\circ }|_{t=5}=\{{\frac {10}{\sqrt {10^{2}+75^{2}+500^{2}}}};{\frac {75}{\sqrt {10^{2}+75^{2}+500^{2}}}};{\frac {500}{\sqrt {10^{2}+75^{2}+500^{2}}}}\},}
a ∘ | t = 5 = { 10 100 + 5625 + 250000 ; 75 255725 ; 500 255725 } , {\displaystyle \mathbf {a} ^{\circ }|_{t=5}=\{{\frac {10}{\sqrt {100+5625+250000}}};{\frac {75}{\sqrt {255725}}};{\frac {500}{\sqrt {255725}}}\},}
a ∘ | t = 5 = { 10 505.6925944 ; 75 505.6925944 ; 500 505.6925944 } , {\displaystyle \mathbf {a} ^{\circ }|_{t=5}=\{{\frac {10}{505.6925944}};{\frac {75}{505.6925944}};{\frac {500}{505.6925944}}\},}
a ∘ | t = 5 = { 0.019774859 ; 0.148311446 ; 0.988742974 } ; {\displaystyle \mathbf {a} ^{\circ }|_{t=5}=\{0.019774859;0.148311446;0.988742974\};}
b = { 1 ; ψ ′ ( t ) ϕ ′ ( t ) ; ω ′ ( t ) ϕ ′ ( t ) } = { 1 ; 3 t 2 2 t ; 4 t 3 2 t } , {\displaystyle \mathbf {b} =\{1;{\frac {\psi '(t)}{\phi '(t)}};{\frac {\omega '(t)}{\phi '(t)}}\}=\{1;{\frac {3t^{2}}{2t}};{\frac {4t^{3}}{2t}}\},}
b = { 1 ; 3 t 2 ; 2 t 2 } ; {\displaystyle \mathbf {b} =\{1;{\frac {3t}{2}};2t^{2}\};}
liestinės vektoriaus b reikšmė, kai t = 1 {\displaystyle t=1} yra:
b | t = 1 = { 1 ; 3 2 ; 2 } ; {\displaystyle \mathbf {b} |_{t=1}=\{1;{\frac {3}{2}};2\};}
normalizuoto liestinės vektoriaus b reikšmė, kai t = 1 {\displaystyle t=1} yra:
b ∘ | t = 1 = { 1 1 2 + 1.5 2 + 2 2 ; 1.5 1 2 + 1.5 2 + 2 2 ; 2 1 2 + 1.5 2 + 2 2 } , {\displaystyle \mathbf {b} ^{\circ }|_{t=1}=\{{\frac {1}{\sqrt {1^{2}+1.5^{2}+2^{2}}}};{\frac {1.5}{\sqrt {1^{2}+1.5^{2}+2^{2}}}};{\frac {2}{\sqrt {1^{2}+1.5^{2}+2^{2}}}}\},}
b ∘ | t = 1 = { 1 1 + 2.25 + 4 ; 1.5 7.25 ; 2 7.25 } = { 1 2.692582404 ; 1.5 2.692582404 ; 2 2.692582404 } , {\displaystyle \mathbf {b} ^{\circ }|_{t=1}=\{{\frac {1}{\sqrt {1+2.25+4}}};{\frac {1.5}{\sqrt {7.25}}};{\frac {2}{\sqrt {7.25}}}\}=\{{\frac {1}{2.692582404}};{\frac {1.5}{2.692582404}};{\frac {2}{2.692582404}}\},}
b ∘ | t = 1 = { 0.371390676 ; 0.557086014 ; 0.742781352 } ; {\displaystyle \mathbf {b} ^{\circ }|_{t=1}=\{0.371390676;0.557086014;0.742781352\};}
liestinės vektoriaus b reikšmė, kai t = 5 {\displaystyle t=5} yra:
b | t = 5 = { 1 ; 3 ⋅ 5 2 ; 2 ⋅ 5 2 } = { 1 ; 15 2 ; 50 } ; {\displaystyle \mathbf {b} |_{t=5}=\{1;{\frac {3\cdot 5}{2}};2\cdot 5^{2}\}=\{1;{\frac {15}{2}};50\};}
normalizuoto vektoriaus b reikšmė, kai t = 5 {\displaystyle t=5} yra tokia pati kaip normalizuoto a vektoriaus:
b ∘ | t = 5 = { 1 1 2 + 7.5 2 + 50 2 ; 15 2 1 + 56.25 + 2500 ; 50 2557.25 } ; {\displaystyle \mathbf {b} ^{\circ }|_{t=5}=\{{\frac {1}{\sqrt {1^{2}+7.5^{2}+50^{2}}}};{\frac {\frac {15}{2}}{\sqrt {1+56.25+2500}}};{\frac {50}{\sqrt {2557.25}}}\};}
b ∘ | t = 5 = { 1 50.56925944 ; 7.5 50.56925944 ; 50 50.56925944 } ; {\displaystyle \mathbf {b} ^{\circ }|_{t=5}=\{{\frac {1}{50.56925944}};{\frac {7.5}{50.56925944}};{\frac {50}{50.56925944}}\};}
b ∘ | t = 5 = { 0.019774859 ; 0.148311446 ; 0.988742974 } ; {\displaystyle \mathbf {b} ^{\circ }|_{t=5}=\{0.019774859;0.148311446;0.988742974\};}
f) 'Tikrasis' kreivės normalės vektorius yra:
m = { 2 − 3 ϕ ′ ( t ) ; 1 3 ψ ′ ( t ) ; 1 3 ω ′ ( t ) } = { 2 − 3 ⋅ 2 t ; 1 3 ⋅ 3 t 2 ; 1 3 ⋅ 4 t 3 } , {\displaystyle \mathbf {m} =\{{\frac {2}{-3\phi '(t)}};{\frac {1}{3\psi '(t)}};{\frac {1}{3\omega '(t)}}\}=\{{\frac {2}{-3\cdot 2t}};{\frac {1}{3\cdot 3t^{2}}};{\frac {1}{3\cdot 4t^{3}}}\},}
m = { − 1 3 t ; 1 9 t 2 ; 1 12 t 3 } ; {\displaystyle \mathbf {m} =\{-{\frac {1}{3t}};{\frac {1}{9t^{2}}};{\frac {1}{12t^{3}}}\};}
'tikrasis' kreivės normalės vektorius, kai t = 1 {\displaystyle t=1} yra:
m | t = 1 = { − 1 3 ; 1 9 ; 1 12 } ; {\displaystyle \mathbf {m} |_{t=1}=\{-{\frac {1}{3}};{\frac {1}{9}};{\frac {1}{12}}\};}
normalizuotas 'tikrasis' kreivės normalės vektorius, kai t = 1 {\displaystyle t=1} yra:
m ∘ | t = 1 = { − 1 3 ( − 1 3 ) 2 + ( 1 9 ) 2 + ( 1 12 ) 2 ; 1 9 ( − 1 3 ) 2 + ( 1 9 ) 2 + ( 1 12 ) 2 ; 1 12 ( − 1 3 ) 2 + ( 1 9 ) 2 + ( 1 12 ) 2 } , {\displaystyle \mathbf {m} ^{\circ }|_{t=1}=\{-{\frac {\frac {1}{3}}{\sqrt {(-{\frac {1}{3}})^{2}+({\frac {1}{9}})^{2}+({\frac {1}{12}})^{2}}}};{\frac {\frac {1}{9}}{\sqrt {(-{\frac {1}{3}})^{2}+({\frac {1}{9}})^{2}+({\frac {1}{12}})^{2}}}};{\frac {\frac {1}{12}}{\sqrt {(-{\frac {1}{3}})^{2}+({\frac {1}{9}})^{2}+({\frac {1}{12}})^{2}}}}\},}
m ∘ | t = 1 = { − 1 3 1 9 + 1 81 + 1 144 ; 1 9 1296 + 144 + 81 81 ⋅ 144 ; 1 12 1521 11664 } , {\displaystyle \mathbf {m} ^{\circ }|_{t=1}=\{-{\frac {\frac {1}{3}}{\sqrt {{\frac {1}{9}}+{\frac {1}{81}}+{\frac {1}{144}}}}};{\frac {\frac {1}{9}}{\sqrt {\frac {1296+144+81}{81\cdot 144}}}};{\frac {\frac {1}{12}}{\sqrt {\frac {1521}{11664}}}}\},}
m ∘ | t = 1 = { − 1 3 0.130401234 ; 1 9 0.130401234 ; 1 12 0.130401234 } = { − 1 3 0.361111111 ; 1 9 0.361111111 ; 1 12 0.361111111 } , {\displaystyle \mathbf {m} ^{\circ }|_{t=1}=\{-{\frac {\frac {1}{3}}{\sqrt {0.130401234}}};{\frac {\frac {1}{9}}{\sqrt {0.130401234}}};{\frac {\frac {1}{12}}{\sqrt {0.130401234}}}\}=\{-{\frac {\frac {1}{3}}{0.361111111}};{\frac {\frac {1}{9}}{0.361111111}};{\frac {\frac {1}{12}}{0.361111111}}\},}
m ∘ | t = 1 = { − 0.923076923 ; 0.307692307 ; 0.23076923 } ; {\displaystyle \mathbf {m} ^{\circ }|_{t=1}=\{-0.923076923;0.307692307;0.23076923\};}
'tikrasis' kreivės normalės vektorius, kai t = 5 {\displaystyle t=5} yra:
m | t = 5 = { − 1 3 ⋅ 5 ; 1 9 ⋅ 5 2 ; 1 12 ⋅ 5 3 } = { − 1 15 ; 1 9 ⋅ 25 ; 1 12 ⋅ 125 } , {\displaystyle \mathbf {m} |_{t=5}=\{-{\frac {1}{3\cdot 5}};{\frac {1}{9\cdot 5^{2}}};{\frac {1}{12\cdot 5^{3}}}\}=\{-{\frac {1}{15}};{\frac {1}{9\cdot 25}};{\frac {1}{12\cdot 125}}\},}
m | t = 5 = { − 1 15 ; 1 225 ; 1 1500 } ; {\displaystyle \mathbf {m} |_{t=5}=\{-{\frac {1}{15}};{\frac {1}{225}};{\frac {1}{1500}}\};}
normalizuotas 'tikrasis' kreivės normalės vektorius, kai t = 5 {\displaystyle t=5} yra:
m ∘ | t = 5 = { − 1 15 ( − 1 15 ) 2 + ( 1 225 ) 2 + ( 1 1500 ) 2 ; 1 225 0.004464641 ; 1 500 0.066817976 } , {\displaystyle \mathbf {m} ^{\circ }|_{t=5}=\{{\frac {-{\frac {1}{15}}}{\sqrt {(-{\frac {1}{15}})^{2}+({\frac {1}{225}})^{2}+({\frac {1}{1500}})^{2}}}};{\frac {\frac {1}{225}}{\sqrt {0.004464641}}};{\frac {\frac {1}{500}}{0.066817976}}\},}
m ∘ | t = 5 = { − 0.99735493 ; 0.066515699 ; 0.009977354 } ; {\displaystyle \mathbf {m} ^{\circ }|_{t=5}=\{-0.99735493;0.066515699;0.009977354\};}
kampas tarp liestinės vektoriaus a ir 'tikrojo' normalės vektoriaus m yra lygus 90 laipsnių, nes jų skaliarinė sandauga lygi nuliui:
a | t = 1 ⋅ m | t = 1 = 2 ⋅ ( − 1 3 ) + 3 ⋅ 1 9 + 4 ⋅ 1 12 = − 2 3 + 1 3 + 1 3 = 0 ; {\displaystyle \mathbf {a} |_{t=1}\cdot \mathbf {m} |_{t=1}=2\cdot (-{\frac {1}{3}})+3\cdot {\frac {1}{9}}+4\cdot {\frac {1}{12}}=-{\frac {2}{3}}+{\frac {1}{3}}+{\frac {1}{3}}=0;}
liestinės vektorius, kai t = 5 {\displaystyle t=5} yra:
a | t = 5 = { 2 ⋅ 5 ; 3 ⋅ 5 2 ; 4 ⋅ 5 3 } = { 10 ; 75 ; 500 } ; {\displaystyle \mathbf {a} |_{t=5}=\{2\cdot 5;3\cdot 5^{2};4\cdot 5^{3}\}=\{10;75;500\};}
kampas tarp liestinės vektoriaus ir 'tikrojo' normalės vektoriaus m yra 90 laipsnių, nes šių vektorių skaliarinė sandauga lygi nuliui:
a | t = 5 ⋅ m | t = 5 = 10 ⋅ ( − 1 15 ) + 75 ⋅ 1 225 + 500 ⋅ 1 1500 = − 2 3 + 1 3 + 1 3 = 0. {\displaystyle \mathbf {a} |_{t=5}\cdot \mathbf {m} |_{t=5}=10\cdot (-{\frac {1}{15}})+75\cdot {\frac {1}{225}}+500\cdot {\frac {1}{1500}}=-{\frac {2}{3}}+{\frac {1}{3}}+{\frac {1}{3}}=0.}
g) 'Tikrasis' kreivės normalės vektorius yra:
m = { − 2 3 ; ϕ ′ ( t ) 3 ψ ′ ( t ) ; ϕ ′ ( t ) 3 ω ′ ( t ) } = { − 2 3 ; 2 t 3 ⋅ 3 t 2 ; 2 t 3 ⋅ 4 t 3 } , {\displaystyle \mathbf {m} =\{-{\frac {2}{3}};{\frac {\phi '(t)}{3\psi '(t)}};{\frac {\phi '(t)}{3\omega '(t)}}\}=\{-{\frac {2}{3}};{\frac {2t}{3\cdot 3t^{2}}};{\frac {2t}{3\cdot 4t^{3}}}\},}
m = { − 2 3 ; 2 t 9 t 2 ; 2 t 12 t 3 } = { − 2 3 ; 2 t 9 t 2 ; t 6 t 3 } ; {\displaystyle \mathbf {m} =\{-{\frac {2}{3}};{\frac {2t}{9t^{2}}};{\frac {2t}{12t^{3}}}\}=\{-{\frac {2}{3}};{\frac {2t}{9t^{2}}};{\frac {t}{6t^{3}}}\};}
'tikrasis' kreivės normalės vektorius, kai t = 1 {\displaystyle t=1} yra:
m | t = 1 = { − 2 3 ; 2 9 ; 1 6 } ; {\displaystyle \mathbf {m} |_{t=1}=\{-{\frac {2}{3}};{\frac {2}{9}};{\frac {1}{6}}\};}
'tikrasis' kreivės normalės vektorius, kai t = 5 {\displaystyle t=5} yra:
m | t = 5 = { − 2 3 ; 2 ⋅ 5 9 ⋅ 5 2 ; 5 6 ⋅ 5 3 } = { − 2 3 ; 2 45 ; 1 150 } ; {\displaystyle \mathbf {m} |_{t=5}=\{-{\frac {2}{3}};{\frac {2\cdot 5}{9\cdot 5^{2}}};{\frac {5}{6\cdot 5^{3}}}\}=\{-{\frac {2}{3}};{\frac {2}{45}};{\frac {1}{150}}\};}
kampas tarp liestinės vektoriaus a ir 'tikrojo' normalės vektoriaus m yra lygus 90 laipsnių, nes jų skaliarinė sandauga lygi nuliui:
a | t = 1 ⋅ m | t = 1 = 2 ⋅ ( − 2 3 ) + 3 ⋅ 2 9 + 4 ⋅ 1 6 = − 4 3 + 2 3 + 2 3 = 0 ; {\displaystyle \mathbf {a} |_{t=1}\cdot \mathbf {m} |_{t=1}=2\cdot (-{\frac {2}{3}})+3\cdot {\frac {2}{9}}+4\cdot {\frac {1}{6}}=-{\frac {4}{3}}+{\frac {2}{3}}+{\frac {2}{3}}=0;}
a | t = 5 ⋅ m | t = 5 = 10 ⋅ ( − 2 3 ) + 75 ⋅ 2 45 + 500 ⋅ 1 150 = − 20 3 + 10 3 + 10 3 = 0. {\displaystyle \mathbf {a} |_{t=5}\cdot \mathbf {m} |_{t=5}=10\cdot (-{\frac {2}{3}})+75\cdot {\frac {2}{45}}+500\cdot {\frac {1}{150}}=-{\frac {20}{3}}+{\frac {10}{3}}+{\frac {10}{3}}=0.} Update 1. Alternatyvus liestinei statmenas vektorius gautas pagal formulę n = { 2 − 3 ϕ ′ ( t ) ; 1 3 ψ ′ ( t ) ; 1 3 ω ′ ( t ) } {\displaystyle \mathbf {n} =\{{\frac {2}{-3\phi '(t)}};{\frac {1}{3\psi '(t)}};{\frac {1}{3\omega '(t)}}\}} yra paprastas triukas. Kadangi liestinės vektorius yra a = { ϕ ′ ( t ) ; ψ ′ ( t ) ; ω ′ ( t ) } , {\displaystyle \mathbf {a} =\{\phi '(t);\psi '(t);\omega '(t)\},} tai aišku, kad vektorių n ir a skaliarinė sandauga bus lygi nuliui. Todėl normalės vektorius gali būti ir n = { 1 ϕ ′ ( t ) ; − 2 ψ ′ ( t ) ; 1 ω ′ ( t ) } {\displaystyle \mathbf {n} =\{{\frac {1}{\phi '(t)}};{\frac {-2}{\psi '(t)}};{\frac {1}{\omega '(t)}}\}} ir n = { 1 ϕ ′ ( t ) ; 1 ψ ′ ( t ) ; − 2 ω ′ ( t ) } . {\displaystyle \mathbf {n} =\{{\frac {1}{\phi '(t)}};{\frac {1}{\psi '(t)}};{\frac {-2}{\omega '(t)}}\}.} Taip pat erdvinės kreivės (užrašytos parametriškai) liestinės vektoriui a statmenas vektorius bus ir pavyzdžiui vektorius m = { − 2 ϕ ′ ( t ) ; 1.5 ψ ′ ( t ) ; 0.5 ω ′ ( t ) } , {\displaystyle \mathbf {m} =\{{\frac {-2}{\phi '(t)}};{\frac {1.5}{\psi '(t)}};{\frac {0.5}{\omega '(t)}}\},} nes jų skaliarinė sandauga lygi nuliui:
a ⋅ m = ϕ ′ ( t ) ⋅ − 2 ϕ ′ ( t ) + ψ ′ ( t ) ⋅ 1.5 ψ ′ ( t ) + ω ′ ( t ) ⋅ 0.5 ω ′ ( t ) = − 2 + 1.5 + 0.5 = 0. {\displaystyle \mathbf {a} \cdot \mathbf {m} =\phi '(t)\cdot {\frac {-2}{\phi '(t)}}+\psi '(t)\cdot {\frac {1.5}{\psi '(t)}}+\omega '(t)\cdot {\frac {0.5}{\omega '(t)}}=-2+1.5+0.5=0.}
Tokiu budu erdvinės kreivės liestinės vektoriui a = { ϕ ′ ( t ) ; ψ ′ ( t ) ; ω ′ ( t ) } {\displaystyle \mathbf {a} =\{\phi '(t);\psi '(t);\omega '(t)\}} galima sudaryti begalo daug statmenų normalės vektorių n (kurių ortai skiriasi - normalės vektoriai neguli ant tos pačios tiesės). Kampo tarp vektorių radimas su sinusu
keisti
‖ a × b ‖ = ‖ a ‖ ‖ b ‖ sin ( θ ) , {\displaystyle \left\|\mathbf {a} \times \mathbf {b} \right\|=\left\|\mathbf {a} \right\|\left\|\mathbf {b} \right\|\sin(\theta ),}
sin θ = ‖ a × b ‖ ‖ a ‖ ‖ b ‖ , {\displaystyle \sin \theta ={\frac {\|\mathbf {a} \times \mathbf {b} \|}{\|\mathbf {a} \|\|\mathbf {b} \|}},} kur θ {\displaystyle \theta } yra kampas tarp vektorių a ir b .
Pavyzdžiui, duoti vektoriai a =(1; -2; 0), b =(3; 0; 0). ‖ 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 = | 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.
Pasitikriname:
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}}}.} 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 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}}.} Duoti vektoriai a =(1; 2; 3), b =(3; 5; 4). a × b = | i j k 1 2 3 3 5 4 | = | 2 3 5 4 | i − | 1 3 3 4 | j + | 1 2 3 5 | k = {\displaystyle \mathbf {a} \times \mathbf {b} ={\begin{vmatrix}i&j&k\\1&2&3\\3&5&4\end{vmatrix}}={\begin{vmatrix}2&3\\5&4\end{vmatrix}}i-{\begin{vmatrix}1&3\\3&4\end{vmatrix}}j+{\begin{vmatrix}1&2\\3&5\end{vmatrix}}k=} = ( − 1 ) 1 + 1 ⋅ ( 2 ⋅ 4 − 3 ⋅ 5 ) i + ( − 1 ) 1 + 2 ⋅ ( 1 ⋅ 4 − 3 ⋅ 3 ) j + ( − 1 ) 1 + 3 ⋅ ( 1 ⋅ 5 − 2 ⋅ 3 ) k = − 7 i + 5 j − 1 k = ( − 7 ; 5 ; − 1 ) . {\displaystyle =(-1)^{1+1}\cdot (2\cdot 4-3\cdot 5)i+(-1)^{1+2}\cdot (1\cdot 4-3\cdot 3)j+(-1)^{1+3}\cdot (1\cdot 5-2\cdot 3)k=-7i+5j-1k=(-7;5;-1).}
sin θ = ‖ a × b ‖ ‖ a ‖ ‖ b ‖ = ( − 7 ) 2 + 5 2 + ( − 1 ) 2 1 2 + 2 2 + 3 2 ⋅ 3 2 + 5 2 + 4 2 = 49 + 25 + 1 1 + 4 + 9 ⋅ 9 + 25 + 16 = {\displaystyle \sin \theta ={\frac {\|\mathbf {a} \times \mathbf {b} \|}{\|\mathbf {a} \|\|\mathbf {b} \|}}={\frac {\sqrt {(-7)^{2}+5^{2}+(-1)^{2}}}{{\sqrt {1^{2}+2^{2}+3^{2}}}\cdot {\sqrt {3^{2}+5^{2}+4^{2}}}}}={\frac {\sqrt {49+25+1}}{{\sqrt {1+4+9}}\cdot {\sqrt {9+25+16}}}}=} = 75 14 ⋅ 50 = 75 700 = 8.660254038 26.45751311 = 0.327326835. {\displaystyle ={\frac {\sqrt {75}}{{\sqrt {14}}\cdot {\sqrt {50}}}}={\frac {\sqrt {75}}{\sqrt {700}}}={\frac {8.660254038}{26.45751311}}=0.327326835.}
θ = arcsin 0.327326835 = 0.333473172 {\displaystyle \theta =\arcsin 0.327326835=0.333473172} radiano arba 19.10660535 laipsnio.
Patikriname kitu budu:
cos θ = a ⋅ b | | a | | ⋅ | | b | | = 1 ⋅ 3 + 2 ⋅ 5 + 3 ⋅ 4 1 2 + 2 2 + 3 2 ⋅ 3 2 + 5 2 + 4 2 = 3 + 10 + 12 14 ⋅ 50 = 25 700 = 0.944911182. {\displaystyle \cos \theta ={\frac {\mathbf {a} \cdot \mathbf {b} }{||\mathbf {a} ||\cdot ||\mathbf {b} ||}}={\frac {1\cdot 3+2\cdot 5+3\cdot 4}{{\sqrt {1^{2}+2^{2}+3^{2}}}\cdot {\sqrt {3^{2}+5^{2}+4^{2}}}}}={\frac {3+10+12}{{\sqrt {14}}\cdot {\sqrt {50}}}}={\frac {25}{\sqrt {700}}}=0.944911182.}
θ = arccos 0.944911182 = 0.333473172 {\displaystyle \theta =\arccos 0.944911182=0.333473172} radiano arba 19.10660535 laipsnio.
Pagal Pitagoro teoremą patikriname atsakymą. Tiek vektorius a , tiek vektorius b išeina iš taško (x; y; z)=(0; 0; 0). Vadinasi vektorius a ir vektorius b liečiasi tame pačiame taške, kurį pavadiname A . Taškas B turi koordinates (1; 2; 3), o taškas C turi koordinates (3; 5; 4). Tokiu budu ||a ||=AB=a, o ||b ||=AC=b. Turime trikampį ABC. Iš taško B(1; 2; 3) nuleista aukštinė h į trikampio kraštinę AC, susikirtimo tašką aukštinės h su kraštine AC, pavadinkime D . Kraštine AD=x, o kraštinė DC=||b ||-x=b-x. BC=c. c = ( 3 − 1 ) 2 + ( 5 − 2 ) 2 + ( 4 − 3 ) 2 = 2 2 + 3 2 + 1 2 = 4 + 9 + 1 = 14 = 3.741657387. {\displaystyle c={\sqrt {(3-1)^{2}+(5-2)^{2}+(4-3)^{2}}}={\sqrt {2^{2}+3^{2}+1^{2}}}={\sqrt {4+9+1}}={\sqrt {14}}=3.741657387.}
a = ‖ a ‖ = 1 2 + 2 2 + 3 2 = 1 + 4 + 9 = 14 = 3.741657387. {\displaystyle a=\|\mathbf {a} \|={\sqrt {1^{2}+2^{2}+3^{2}}}={\sqrt {1+4+9}}={\sqrt {14}}=3.741657387.}
b = ‖ b ‖ = 3 2 + 5 2 + 4 2 = 9 + 25 + 16 = 30 = 5.477225575. {\displaystyle b=\|\mathbf {b} \|={\sqrt {3^{2}+5^{2}+4^{2}}}={\sqrt {9+25+16}}={\sqrt {30}}=5.477225575.}
h = a 2 − x 2 = ( 14 ) 2 − x 2 = 14 − x 2 . {\displaystyle h={\sqrt {a^{2}-x^{2}}}={\sqrt {({\sqrt {14}})^{2}-x^{2}}}={\sqrt {14-x^{2}}}.}
h = c 2 − ( b − x ) 2 = ( 14 ) 2 − ( 30 − x ) 2 = 14 − ( 30 − 2 x 30 + x 2 ) . {\displaystyle h={\sqrt {c^{2}-(b-x)^{2}}}={\sqrt {({\sqrt {14}})^{2}-({\sqrt {30}}-x)^{2}}}={\sqrt {14-(30-2x{\sqrt {30}}+x^{2})}}.} a 2 − x 2 = c 2 − ( b − x ) 2 ; {\displaystyle {\sqrt {a^{2}-x^{2}}}={\sqrt {c^{2}-(b-x)^{2}}};}
a 2 − x 2 = c 2 − ( b − x ) 2 {\displaystyle a^{2}-x^{2}=c^{2}-(b-x)^{2}} ; a 2 − x 2 = c 2 − ( b 2 − 2 b x + x 2 ) ; {\displaystyle a^{2}-x^{2}=c^{2}-(b^{2}-2bx+x^{2});} a 2 − x 2 = c 2 − b 2 + 2 b x − x 2 ; {\displaystyle a^{2}-x^{2}=c^{2}-b^{2}+2bx-x^{2};} a 2 = c 2 − b 2 + 2 b x ; {\displaystyle a^{2}=c^{2}-b^{2}+2bx;} a 2 − c 2 + b 2 = 2 b x ; {\displaystyle a^{2}-c^{2}+b^{2}=2bx;} x = a 2 − c 2 + b 2 2 b = ( 14 ) 2 − ( 14 ) 2 + ( 30 ) 2 2 30 = 30 2 30 = 30 2 = 2.738612788. {\displaystyle x={\frac {a^{2}-c^{2}+b^{2}}{2b}}={\frac {({\sqrt {14}})^{2}-({\sqrt {14}})^{2}+({\sqrt {30}})^{2}}{2{\sqrt {30}}}}={\frac {30}{2{\sqrt {30}}}}={\frac {\sqrt {30}}{2}}=2.738612788.}
cos θ = x a = 2.738612788 14 = 2.738612788 3.741657387 = 0.731925054. {\displaystyle \cos \theta ={\frac {x}{a}}={\frac {2.738612788}{\sqrt {14}}}={\frac {2.738612788}{3.741657387}}=0.731925054.}
θ = arccos 0.731925054 = 0.749653438 {\displaystyle \theta =\arccos 0.731925054=0.749653438} arba 42.95197812 laipsnio.
h = a 2 − x 2 = ( 14 ) 2 − ( 30 2 ) 2 = 14 − 30 4 = 14 − 7.5 = 6.5 = 2.549509757. {\displaystyle h={\sqrt {a^{2}-x^{2}}}={\sqrt {({\sqrt {14}})^{2}-({\frac {\sqrt {30}}{2}})^{2}}}={\sqrt {14-{\frac {30}{4}}}}={\sqrt {14-7.5}}={\sqrt {6.5}}=2.549509757.} h = c 2 − ( b − x ) 2 = ( 14 ) 2 − ( 30 − 30 2 ) 2 = 14 − 30 4 = 14 − 7.5 = 6.5 = 2.549509757. {\displaystyle h={\sqrt {c^{2}-(b-x)^{2}}}={\sqrt {({\sqrt {14}})^{2}-({\sqrt {30}}-{\frac {\sqrt {30}}{2}})^{2}}}={\sqrt {14-{\frac {30}{4}}}}={\sqrt {14-7.5}}={\sqrt {6.5}}=2.549509757.}
sin θ = h a = 6.5 14 = 2.549509757 3.741657387 = 0.681385143. {\displaystyle \sin \theta ={\frac {h}{a}}={\frac {\sqrt {6.5}}{\sqrt {14}}}={\frac {2.549509757}{3.741657387}}=0.681385143.}
θ = arcsin 6.5 14 = arcsin 0.681385143 = 0.749653438 {\displaystyle \theta =\arcsin {\frac {\sqrt {6.5}}{\sqrt {14}}}=\arcsin 0.681385143=0.749653438} arba 42.95197812 laipsnio.
Trikampio ABC plotas yra
S Δ = 1 2 | | a × b | | = 1 2 ( − 7 ) 2 + 5 2 + ( − 1 ) 2 = 1 2 75 = 4 , 330127019. {\displaystyle S_{\Delta }={\frac {1}{2}}||a\times b||={\frac {1}{2}}{\sqrt {(-7)^{2}+5^{2}+(-1)^{2}}}={\frac {1}{2}}{\sqrt {75}}=4,330127019.}
Trikampio ABC plotą randame taikydami Herono formulę: p = P 2 = a + b + c 2 = 14 + 30 + 14 2 = 6 , 480270174. {\displaystyle p={\frac {P}{2}}={\frac {a+b+c}{2}}={\frac {{\sqrt {14}}+{\sqrt {30}}+{\sqrt {14}}}{2}}=6,480270174.}
S Δ = p ( p − a ) ( p − b ) ( p − c ) = {\displaystyle S_{\Delta }={\sqrt {p(p-a)(p-b)(p-c)}}=} = 6 , 480270174 ⋅ ( 6 , 480270174 − 14 ) ( 6 , 480270174 − 30 ) ( 6 , 480270174 − 14 ) = {\displaystyle ={\sqrt {6,480270174\cdot (6,480270174-{\sqrt {14}})(6,480270174-{\sqrt {30}})(6,480270174-{\sqrt {14}})}}=}
= 35 , 60196623 = 5 , 966738324. {\displaystyle ={\sqrt {35,60196623}}=5,966738324.}
Dar budas pasitikrinti trikampio ABC plotą:
S Δ = b ⋅ h 2 = 30 ⋅ 6.5 2 = 195 2 = 6 , 982120022. {\displaystyle S_{\Delta }={b\cdot h \over 2}={{\sqrt {30}}\cdot {\sqrt {6.5}} \over 2}={{\sqrt {195}} \over 2}=6,982120022.} Galbūt plotai ir kampai nesutampa skaičiuojant skirtingais būdais, nes trikampis ABC yra lygiašonis ir jam kosinusų teorema ar/ir kitos formulės netinka. Bet pasibraižius grafikus ir patikrinus kampus tarp vektorių įvairiais budais, buvo padaryta išvada, kad jokių budu atsakymas negali buti 42.95197812 laipsniai, o kažkurtai apie 18-20 laipsnių. Todėl kyla išvada, kad kosinusų teorema netinkama skaičiuoti kampams tų trikampių, kurie yra lygiašoniai.
Duoti vektoriai a =(4; 3; 0), b =(10; 0; 0). Rasime kampą tarp jų. a × b = | i j k 4 3 0 10 0 0 | = | 3 0 0 0 | i − | 4 0 10 0 | j + | 4 3 10 0 | k = 0 i + 0 j − 30 k = ( 0 ; 0 ; − 30 ) . {\displaystyle \mathbf {a} \times \mathbf {b} ={\begin{vmatrix}i&j&k\\4&3&0\\10&0&0\end{vmatrix}}={\begin{vmatrix}3&0\\0&0\end{vmatrix}}i-{\begin{vmatrix}4&0\\10&0\end{vmatrix}}j+{\begin{vmatrix}4&3\\10&0\end{vmatrix}}k=0i+0j-30k=(0;0;-30).}
sin θ = ‖ a × b ‖ ‖ a ‖ ‖ b ‖ = 0 2 + 0 2 + ( − 30 ) 2 4 2 + 3 2 + 0 2 ⋅ 10 2 + 0 2 + 0 2 = 900 25 ⋅ 100 = 30 5 ⋅ 10 = 3 5 = 0 , 6. {\displaystyle \sin \theta ={\frac {\|\mathbf {a} \times \mathbf {b} \|}{\|\mathbf {a} \|\|\mathbf {b} \|}}={\frac {\sqrt {0^{2}+0^{2}+(-30)^{2}}}{{\sqrt {4^{2}+3^{2}+0^{2}}}\cdot {\sqrt {10^{2}+0^{2}+0^{2}}}}}={\frac {\sqrt {900}}{{\sqrt {25}}\cdot {\sqrt {100}}}}={\frac {30}{5\cdot 10}}={\frac {3}{5}}=0,6.}
θ = arcsin 0 , 6 = 0.643501108 {\displaystyle \theta =\arcsin 0,6=0.643501108} radiano arba 36,86989765 laipsnio.
Patikriname kitu budu:
cos θ = a ⋅ b | | a | | ⋅ | | b | | = 4 ⋅ 10 + 3 ⋅ 0 + 0 ⋅ 0 4 2 + 3 2 + 0 2 ⋅ 10 2 + 0 2 + 0 2 = 40 25 ⋅ 100 = 40 5 ⋅ 10 = 4 5 = 0.8. {\displaystyle \cos \theta ={\frac {\mathbf {a} \cdot \mathbf {b} }{||\mathbf {a} ||\cdot ||\mathbf {b} ||}}={\frac {4\cdot 10+3\cdot 0+0\cdot 0}{{\sqrt {4^{2}+3^{2}+0^{2}}}\cdot {\sqrt {10^{2}+0^{2}+0^{2}}}}}={\frac {40}{{\sqrt {25}}\cdot {\sqrt {100}}}}={\frac {40}{5\cdot 10}}={\frac {4}{5}}=0.8.}
θ = arccos 0.8 = 0.643501108 {\displaystyle \theta =\arccos 0.8=0.643501108} radiano arba 36,86989765 laipsnio.Pavyzdžiui, duoti vektoriai a =(3; 5; 11), b =(7; 8; 2). a × b = | i j k 3 5 11 7 8 2 | = | 5 11 8 2 | i − | 3 11 7 2 | j + | 3 5 7 8 | k = {\displaystyle \mathbf {a} \times \mathbf {b} ={\begin{vmatrix}i&j&k\\3&5&11\\7&8&2\end{vmatrix}}={\begin{vmatrix}5&11\\8&2\end{vmatrix}}i-{\begin{vmatrix}3&11\\7&2\end{vmatrix}}j+{\begin{vmatrix}3&5\\7&8\end{vmatrix}}k=} = ( − 1 ) 1 + 1 ⋅ ( 5 ⋅ 2 − 8 ⋅ 11 ) i + ( − 1 ) 1 + 2 ⋅ ( 3 ⋅ 2 − 11 ⋅ 7 ) j + ( − 1 ) 1 + 3 ⋅ ( 3 ⋅ 8 − 5 ⋅ 7 ) k = − 7 i + 5 j − 1 k = ( − 78 ; 71 ; − 11 ) . {\displaystyle =(-1)^{1+1}\cdot (5\cdot 2-8\cdot 11)i+(-1)^{1+2}\cdot (3\cdot 2-11\cdot 7)j+(-1)^{1+3}\cdot (3\cdot 8-5\cdot 7)k=-7i+5j-1k=(-78;71;-11).}
sin θ = ‖ a × b ‖ ‖ a ‖ ‖ b ‖ = ( − 78 ) 2 + 71 2 + ( − 11 ) 2 3 2 + 5 2 + 11 2 ⋅ 7 2 + 8 2 + 2 2 = 6048 + 5041 + 121 9 + 25 + 121 ⋅ 49 + 64 + 4 = {\displaystyle \sin \theta ={\frac {\|\mathbf {a} \times \mathbf {b} \|}{\|\mathbf {a} \|\|\mathbf {b} \|}}={\frac {\sqrt {(-78)^{2}+71^{2}+(-11)^{2}}}{{\sqrt {3^{2}+5^{2}+11^{2}}}\cdot {\sqrt {7^{2}+8^{2}+2^{2}}}}}={\frac {\sqrt {6048+5041+121}}{{\sqrt {9+25+121}}\cdot {\sqrt {49+64+4}}}}=} = 11246 155 ⋅ 117 = 11246 18135 = 106 , 0471593 134 , 6662541 = 0.787481318. {\displaystyle ={\frac {\sqrt {11246}}{{\sqrt {155}}\cdot {\sqrt {117}}}}={\frac {\sqrt {11246}}{\sqrt {18135}}}={\frac {106,0471593}{134,6662541}}=0.787481318.}
θ = arcsin 0.786219889 = 0.906711738 {\displaystyle \theta =\arcsin 0.786219889=0.906711738} radiano arba 51,95075583 laipsnio.
Pagal Pitagoro teoremą patikriname atsakymą. Tiek vektorius a , tiek vektorius b išeina iš taško (x; y; z)=(0; 0; 0). Vadinasi vektorius a ir vektorius b liečiasi tame pačiame taške, kurį pavadiname A . Taškas B turi koordinates (3; 5; 11), o taškas C turi koordinates (7; 8; 2). Tokiu budu ||a ||=AB=a, o ||b ||=AC=b. Turime trikampį ABC. Iš taško C(7; 8; 2) nuleista aukštinė h į trikampio kraštinę AB=a, susikirtimo tašką aukštinės h su kraštine AB, pavadinkime D . Kraštinė AD=x, o kraštinė DB=||a ||-x=a-x. BC=c. c = ( 7 − 3 ) 2 + ( 8 − 5 ) 2 + ( 2 − 11 ) 2 = 4 2 + 3 2 + ( − 9 ) 2 = 16 + 9 + 81 = 106 = 10 , 29563014. {\displaystyle c={\sqrt {(7-3)^{2}+(8-5)^{2}+(2-11)^{2}}}={\sqrt {4^{2}+3^{2}+(-9)^{2}}}={\sqrt {16+9+81}}={\sqrt {106}}=10,29563014.}
a = ‖ a ‖ = 3 2 + 5 2 + 11 2 = 9 + 25 + 121 = 155 = 12 , 4498996. {\displaystyle a=\|\mathbf {a} \|={\sqrt {3^{2}+5^{2}+11^{2}}}={\sqrt {9+25+121}}={\sqrt {155}}=12,4498996.}
b = ‖ b ‖ = 7 2 + 8 2 + 2 2 = 49 + 64 + 4 = 117 = 10 , 81665383. {\displaystyle b=\|\mathbf {b} \|={\sqrt {7^{2}+8^{2}+2^{2}}}={\sqrt {49+64+4}}={\sqrt {117}}=10,81665383.}
h = b 2 − x 2 = ( 117 ) 2 − x 2 = 117 − x 2 . {\displaystyle h={\sqrt {b^{2}-x^{2}}}={\sqrt {({\sqrt {117}})^{2}-x^{2}}}={\sqrt {117-x^{2}}}.}
h = c 2 − ( a − x ) 2 = ( 106 ) 2 − ( 155 − x ) 2 = 14 − ( 155 − 2 x 155 + x 2 ) . {\displaystyle h={\sqrt {c^{2}-(a-x)^{2}}}={\sqrt {({\sqrt {106}})^{2}-({\sqrt {155}}-x)^{2}}}={\sqrt {14-(155-2x{\sqrt {155}}+x^{2})}}.} b 2 − x 2 = c 2 − a − x ) 2 ; {\displaystyle {\sqrt {b^{2}-x^{2}}}={\sqrt {c^{2}-a-x)^{2}}};}
b 2 − x 2 = c 2 − ( a − x ) 2 {\displaystyle b^{2}-x^{2}=c^{2}-(a-x)^{2}} ; b 2 − x 2 = c 2 − ( a 2 − 2 a x + x 2 ) ; {\displaystyle b^{2}-x^{2}=c^{2}-(a^{2}-2ax+x^{2});} b 2 − x 2 = c 2 − a 2 + 2 a x − x 2 ; {\displaystyle b^{2}-x^{2}=c^{2}-a^{2}+2ax-x^{2};} b 2 = c 2 − a 2 + 2 a x ; {\displaystyle b^{2}=c^{2}-a^{2}+2ax;} b 2 − c 2 + a 2 = 2 a x ; {\displaystyle b^{2}-c^{2}+a^{2}=2ax;} x = b 2 − c 2 + a 2 2 a = ( 117 ) 2 − ( 106 ) 2 + ( 155 ) 2 2 155 = 117 − 106 + 155 2 155 = 166 2 155 = 83 155 = 6.66672043. {\displaystyle x={\frac {b^{2}-c^{2}+a^{2}}{2a}}={\frac {({\sqrt {117}})^{2}-({\sqrt {106}})^{2}+({\sqrt {155}})^{2}}{2{\sqrt {155}}}}={\frac {117-106+155}{2{\sqrt {155}}}}={\frac {166}{2{\sqrt {155}}}}={\frac {83}{\sqrt {155}}}=6.66672043.}
cos θ = x b = 83 155 117 = 83 18135 = 0.616338521. {\displaystyle \cos \theta ={\frac {x}{b}}={\frac {\frac {83}{\sqrt {155}}}{\sqrt {117}}}={\frac {83}{\sqrt {18135}}}=0.616338521.}
θ = arccos 0.616338521 = 0.906711738 {\displaystyle \theta =\arccos 0.616338521=0.906711738} arba 51.95075583 laipsnio.
h = b 2 − x 2 = ( 117 ) 2 − ( 83 155 ) 2 = {\displaystyle h={\sqrt {b^{2}-x^{2}}}={\sqrt {({\sqrt {117}})^{2}-({\frac {83}{\sqrt {155}}})^{2}}}=} = 117 − 6889 155 = 117 − 44.44516129 = 72.55483871 = 8.517912814. {\displaystyle ={\sqrt {117-{\frac {6889}{155}}}}={\sqrt {117-44.44516129}}={\sqrt {72.55483871}}=8.517912814.}
h = c 2 − ( a − x ) 2 = ( 106 ) 2 − ( 155 − 83 155 ) 2 = 106 − ( 12.4498996 − 6.66672043 ) 2 = {\displaystyle h={\sqrt {c^{2}-(a-x)^{2}}}={\sqrt {({\sqrt {106}})^{2}-({\sqrt {155}}-{\frac {83}{\sqrt {155}}})^{2}}}={\sqrt {106-(12.4498996-6.66672043)^{2}}}=}
106 − 5.783179168 2 = 106 − 33.44516129 = 72.55483871 = 8.517912814. {\displaystyle {\sqrt {106-5.783179168^{2}}}={\sqrt {106-33.44516129}}={\sqrt {72.55483871}}=8.517912814.}
sin θ = h b = 8.517912814 117 = 8.517912814 10.81665383 = 0.787481318. {\displaystyle \sin \theta ={\frac {h}{b}}={\frac {8.517912814}{\sqrt {117}}}={\frac {8.517912814}{10.81665383}}=0.787481318.}
θ = arcsin 0.787481318 = 0.906711738 {\displaystyle \theta =\arcsin 0.787481318=0.906711738} arba 51.95075583 laipsnio.
Trikampio ABC plotas yra
S Δ = 1 2 | | a × b | | = 1 2 ( − 78 ) 2 + 71 2 + ( − 11 ) 2 = 1 2 11246 = 53.02357966. {\displaystyle S_{\Delta }={\frac {1}{2}}||a\times b||={\frac {1}{2}}{\sqrt {(-78)^{2}+71^{2}+(-11)^{2}}}={\frac {1}{2}}{\sqrt {11246}}=53.02357966.}
Trikampio ABC plotą randame taikydami Herono formulę: p = P 2 = a + b + c 2 = 155 + 117 + 106 2 = 16.78109178. {\displaystyle p={\frac {P}{2}}={\frac {a+b+c}{2}}={\frac {{\sqrt {155}}+{\sqrt {117}}+{\sqrt {106}}}{2}}=16.78109178.}
S Δ = p ( p − a ) ( p − b ) ( p − c ) = {\displaystyle S_{\Delta }={\sqrt {p(p-a)(p-b)(p-c)}}=} = 16.78109178 ⋅ ( 16.78109178 − 155 ) ( 16.78109178 − 117 ) ( 16.78109178 − 106 ) = {\displaystyle ={\sqrt {16.78109178\cdot (16.78109178-{\sqrt {155}})(16.78109178-{\sqrt {117}})(16.78109178-{\sqrt {106}})}}=}
= 16.78109178 ⋅ 4.331192185 ⋅ 5.964437956 ⋅ 6.485461642 = {\displaystyle ={\sqrt {16.78109178\cdot 4.331192185\cdot 5.964437956\cdot 6.485461642}}=}
= 2811.5 = 53.02357966. {\displaystyle ={\sqrt {2811.5}}=53.02357966.}
Dar budas pasitikrinti trikampio ABC plotą:
S Δ = a ⋅ h 2 = 155 ⋅ 8.517912814 2 = 53.02357966. {\displaystyle S_{\Delta }={a\cdot h \over 2}={{\sqrt {155}}\cdot 8.517912814 \over 2}=53.02357966.} Vektorinė vektorių sandauga
keisti
Grafinis vektorinės vektorių sandaugos pavaizdavimas Dviejų vektorių vektorinės sandaugos rezultatas yra vektorius status tiems dviems vektoriams. Jei duoti vektoriai a = ( a 1 ; a 2 ; a 3 ) {\displaystyle \mathbf {a} =(a_{1};a_{2};a_{3})} ir b = ( b 1 ; b 2 ; b 3 ) , {\displaystyle \mathbf {b} =(b_{1};b_{2};b_{3}),} tai vektorinė sandauga vektorių a ir b duos trečią vektorių n , {\displaystyle \mathbf {n} ,} kuris bus status vektoriui a ir vektoriui b .
n = a × b = | i j k a 1 a 2 a 3 b 1 b 2 b 3 | = | a 2 a 3 b 2 b 3 | i − | a 1 a 3 b 1 b 3 | j + | a 1 a 2 b 1 b 2 | k = {\displaystyle \mathbf {n} =\mathbf {a} \times \mathbf {b} ={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\a_{1}&a_{2}&a_{3}\\b_{1}&b_{2}&b_{3}\\\end{vmatrix}}={\begin{vmatrix}a_{2}&a_{3}\\b_{2}&b_{3}\end{vmatrix}}\mathbf {i} -{\begin{vmatrix}a_{1}&a_{3}\\b_{1}&b_{3}\end{vmatrix}}\mathbf {j} +{\begin{vmatrix}a_{1}&a_{2}\\b_{1}&b_{2}\end{vmatrix}}\mathbf {k} =}
= i a 2 b 3 + j a 3 b 1 + k a 1 b 2 − i a 3 b 2 − j a 1 b 3 − k a 2 b 1 = {\displaystyle =\mathbf {i} a_{2}b_{3}+\mathbf {j} a_{3}b_{1}+\mathbf {k} a_{1}b_{2}-\mathbf {i} a_{3}b_{2}-\mathbf {j} a_{1}b_{3}-\mathbf {k} a_{2}b_{1}=}
= i ( a 2 b 3 − a 3 b 2 ) + j ( a 3 b 1 − a 1 b 3 ) + k ( a 1 b 2 − a 2 b 1 ) = {\displaystyle =\mathbf {i} (a_{2}b_{3}-a_{3}b_{2})+\mathbf {j} (a_{3}b_{1}-a_{1}b_{3})+\mathbf {k} (a_{1}b_{2}-a_{2}b_{1})=}
= ( a 2 b 3 − a 3 b 2 ; a 3 b 1 − a 1 b 3 ; a 1 b 2 − a 2 b 1 ) . {\displaystyle =(a_{2}b_{3}-a_{3}b_{2};a_{3}b_{1}-a_{1}b_{3};a_{1}b_{2}-a_{2}b_{1}).} Vektorinė sandauga a × b gali būti interpretuojamas kaip plotas lygiagretainio , sudaryto iš kraštinių (arba tiesių) ||a || ir ||b ||. 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 Δ = 1 2 | | a × b | | = 5 2 . {\displaystyle S_{\Delta }={\frac {1}{2}}||a\times b||=5{\sqrt {2}}.}
Taikydami Herono formulę patikrinsime ar trikampio plotas S Δ {\displaystyle S_{\Delta }} surastas teisingai.
| | 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.} 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.}
Dedamųjų daugyba :
i × j = − ( j × i ) = k ; {\displaystyle i\times j=-(j\times i)=k;}
j × k = − ( k × j ) = i ; {\displaystyle j\times k=-(k\times j)=i;}
k × i = − ( i × k ) = j . {\displaystyle k\times i=-(i\times k)=j.}
i × i = j × j = k × k = 0. {\displaystyle i\times i=j\times j=k\times k=0.}
Rasime a × b , {\displaystyle a\times b,} jei a=2i-3j+5k=(2; -3; 5), b=4i+2j-6k=(4; 2; -6). a × b = ( 2 i − 3 j + 5 k ) × ( 4 i + 2 j − 6 k ) = 8 i i + 4 i j − 12 i k − 12 j k − 6 j j + 18 j k + 20 k i + 10 k j − 30 k k = {\displaystyle a\times b=(2i-3j+5k)\times (4i+2j-6k)=8ii+4ij-12ik-12jk-6jj+18jk+20ki+10kj-30kk=}
= 4 k + 12 j + 12 k + 18 i + 20 j − 10 i = 8 i + 32 j + 16 k = ( 8 ; 32 ; 16 ) . {\displaystyle =4k+12j+12k+18i+20j-10i=8i+32j+16k=(8;32;16).}
a × b = | i j k 2 − 3 5 4 2 − 6 | = | − 3 5 2 − 6 | i − | 2 5 4 − 6 | j + | 2 − 3 4 2 | k = 8 i + 32 j + 16 k = ( 8 ; 32 ; 16 ) . {\displaystyle a\times b={\begin{vmatrix}i&j&k\\2&-3&5\\4&2&-6\end{vmatrix}}={\begin{vmatrix}-3&5\\2&-6\end{vmatrix}}i-{\begin{vmatrix}2&5\\4&-6\end{vmatrix}}j+{\begin{vmatrix}2&-3\\4&2\end{vmatrix}}k=8i+32j+16k=(8;32;16).}
Apskaičiuosime trikampio su viršūnėmis taškuose A(-1; 0; 2), B(1; -2; 5), C(3; 0; -4) plotą. a=AB=(1-(-1); -2-0; 5-2)=(2; -2; 3); b=AC=(3-(-1); 0-0; -4-2)=(4; 0; -6); a × b = | i j k 2 − 2 3 4 0 − 6 | = | − 2 3 0 − 6 | i − | 2 3 4 − 6 | j + | 2 − 2 4 0 | k = 12 i + 24 j + 8 k = ( 12 ; 24 ; 8 ) . {\displaystyle a\times b={\begin{vmatrix}i&j&k\\2&-2&3\\4&0&-6\end{vmatrix}}={\begin{vmatrix}-2&3\\0&-6\end{vmatrix}}i-{\begin{vmatrix}2&3\\4&-6\end{vmatrix}}j+{\begin{vmatrix}2&-2\\4&0\end{vmatrix}}k=12i+24j+8k=(12;24;8).} | | a × b | | = 12 2 + 24 2 + 8 2 = 784 = 28. S Δ = 1 2 | | a × b | | = 1 2 ⋅ 28 = 14. {\displaystyle ||a\times b||={\sqrt {12^{2}+24^{2}+8^{2}}}={\sqrt {784}}=28.\;S_{\Delta }={\frac {1}{2}}||a\times b||={\frac {1}{2}}\cdot 28=14.}
Trikampio ABC viršunės yra taškai A(1; -1; 2), B(5; -6; 2) ir C(1; 3; -1). Apskaičiuosime šio trikampio plotą ir aukštinės h , nuleistos iš viršunės B į kraštinę AC, ilgį. Žinome, kad S Δ A B C = 1 2 | | A B × A C | | . {\displaystyle S_{\Delta ABC}={\frac {1}{2}}||AB\times AC||.} Randame vektorių AB ir AC koordinates bei vektorinę sandaugą: AB=(4; 5; 0), AC=(0; 4; -3),
A B × A C = | i j k 4 − 5 0 0 4 − 3 | = | − 5 0 4 − 3 | i − | 4 0 0 − 3 | j + | 4 − 5 0 4 | k = 15 i − ( − 12 ) j + 16 k = ( 15 ; 12 ; 16 ) . {\displaystyle AB\times AC={\begin{vmatrix}i&j&k\\4&-5&0\\0&4&-3\end{vmatrix}}={\begin{vmatrix}-5&0\\4&-3\end{vmatrix}}i-{\begin{vmatrix}4&0\\0&-3\end{vmatrix}}j+{\begin{vmatrix}4&-5\\0&4\end{vmatrix}}k=15i-(-12)j+16k=(15;12;16).} Apskaičiuojame lygiagretainio plotą: | | A B × A C | | = 15 2 + 12 2 + 16 2 = 625 = 25. {\displaystyle ||AB\times AC||={\sqrt {15^{2}+12^{2}+16^{2}}}={\sqrt {625}}=25.} Tada trikampio ABC plotas bus lygus S Δ A B C = 1 2 ⋅ 25 = 12.5. {\displaystyle S_{\Delta ABC}={\frac {1}{2}}\cdot 25=12.5.} Norėdami rasti trikampio aukšinę h , pritaikykime kitą trikampio ploto formulę: S Δ A B C = 1 2 | | A C | | ⋅ h . {\displaystyle S_{\Delta ABC}={\frac {1}{2}}||AC||\cdot h.} Sulyginę formules, gauname: 1 2 | | A B × A C | | = 1 2 | | A C | | ⋅ h . {\displaystyle {\frac {1}{2}}||AB\times AC||={\frac {1}{2}}||AC||\cdot h.} Iš čia trikampio ABC aukšinė h = | | A B × A C | | | | A C | | = 25 5 = 5 , {\displaystyle h={\frac {||AB\times AC||}{||AC||}}={\frac {25}{5}}=5,} kadangi | | A C | | = 0 2 + 4 2 + ( − 3 ) 2 = 0 + 16 + 9 = 5. {\displaystyle ||AC||={\sqrt {0^{2}+4^{2}+(-3)^{2}}}={\sqrt {0+16+9}}=5.}
Apskaičiuosime trikampio plotą, kai žinomi jo viršunių taškai B(5; 2; 6), C(-1; 3; 4) ir D(7; 3; -1). Trikampio kraštinių vektoriai yra šie: BC=(5-(-1); 2-3; 6-4)=(6; -1; 2); BD=(5-7; 2-3; 6-(-1))=(-2; -1; 7); CD=(-1-7; 3-3; 4-(-1))=(-8; 0; 5). Trikampio kraštinių ilgiai yra šie:
| | B C | | = 6 2 + ( − 1 ) 2 + 2 2 = 36 + 1 + 4 = 41 = 6.403124237 ; {\displaystyle ||BC||={\sqrt {6^{2}+(-1)^{2}+2^{2}}}={\sqrt {36+1+4}}={\sqrt {41}}=6.403124237;}
| | B D | | = ( − 2 ) 2 + ( − 1 ) 2 + 7 2 = 4 + 1 + 49 = 54 = 7.348469228 ; {\displaystyle ||BD||={\sqrt {(-2)^{2}+(-1)^{2}+7^{2}}}={\sqrt {4+1+49}}={\sqrt {54}}=7.348469228;}
| | C D | | = ( − 8 ) 2 + 0 2 + 5 2 = 64 + 0 + 25 = 89 = 9.433981132 ; {\displaystyle ||CD||={\sqrt {(-8)^{2}+0^{2}+5^{2}}}={\sqrt {64+0+25}}={\sqrt {89}}=9.433981132;}
Kadangi
B C × B D = | i j k 6 − 1 2 − 2 − 1 7 | = i ⋅ ( − 1 ) ⋅ 7 + j ⋅ 2 ⋅ ( − 2 ) + k ⋅ 6 ⋅ ( − 1 ) − i ⋅ 2 ⋅ ( − 1 ) − j ⋅ 6 ⋅ 7 − k ⋅ ( − 1 ) ⋅ ( − 2 ) = {\displaystyle BC\times BD={\begin{vmatrix}i&j&k\\6&-1&2\\-2&-1&7\end{vmatrix}}=i\cdot (-1)\cdot 7+j\cdot 2\cdot (-2)+k\cdot 6\cdot (-1)-i\cdot 2\cdot (-1)-j\cdot 6\cdot 7-k\cdot (-1)\cdot (-2)=} = − 7 i + 2 i − 4 j − 42 j − 6 k − 2 k = − 5 i − 46 j − 8 k = ( − 5 ; − 46 ; − 8 ) , {\displaystyle =-7i+2i-4j-42j-6k-2k=-5i-46j-8k=(-5;-46;-8),} tai trikampio plotas lygus:
S Δ = 1 2 ⋅ | | B C × B D | | = 1 2 ⋅ ( − 5 ) 2 + ( − 46 ) 2 + ( − 8 ) 2 = 25 + 2116 + 64 2 = 2205 2 = 23.47871376. {\displaystyle S_{\Delta }={1 \over 2}\cdot ||BC\times BD||={1 \over 2}\cdot {\sqrt {(-5)^{2}+(-46)^{2}+(-8)^{2}}}={{\sqrt {25+2116+64}} \over 2}={{\sqrt {2205}} \over 2}=23.47871376.}
Pagal Herono formulę pasitikriname ar trikampio plotas gautas teisingai. Randame trikampio pusperimetrį
p = | | B C | | + | | B D | | + | | C D | | 2 = 41 + 54 + 89 2 = 11.5927873. {\displaystyle p={||BC||+||BD||+||CD|| \over 2}={{\sqrt {41}}+{\sqrt {54}}+{\sqrt {89}} \over 2}=11.5927873.}
S Δ = p ( p − | | B C | | ) ( p − | | B D | | ) ( p − | | C D | | ) = {\displaystyle S_{\Delta }={\sqrt {p(p-||BC||)(p-||BD||)(p-||CD||)}}=}
= 11.5927873 ( 11.5927873 − 41 ) ( 11.5927873 − 54 ) ( 11.5927873 − 89 ) = {\displaystyle ={\sqrt {11.5927873(11.5927873-{\sqrt {41}})(11.5927873-{\sqrt {54}})(11.5927873-{\sqrt {89}})}}=}
= 11.5927873 ⋅ 5.189663062 ⋅ 4.244318071 ⋅ 2.158806167 = 551.2500001 = 23.47871377. {\displaystyle ={\sqrt {11.5927873\cdot 5.189663062\cdot 4.244318071\cdot 2.158806167}}={\sqrt {551.2500001}}=23.47871377.}
Trikampio plotą galima surasti ir su tokia formule: S Δ = 1 2 ⋅ ( x 1 ⋅ y 2 − x 2 ⋅ y 1 ) 2 + ( x 1 ⋅ z 2 − x 2 ⋅ z 1 ) 2 + ( y 1 ⋅ z 2 − y 2 ⋅ z 1 ) 2 = 1 2 ⋅ ( 6 ⋅ ( − 1 ) − ( − 2 ) ⋅ ( − 1 ) ) 2 + ( 6 ⋅ 7 − ( − 2 ) ⋅ 2 ) 2 + ( ( − 1 ) ⋅ 7 − ( − 1 ) ⋅ 2 ) 2 = {\displaystyle S_{\Delta }={\frac {1}{2}}\cdot {\sqrt {(x_{1}\cdot y_{2}-x_{2}\cdot y_{1})^{2}+(x_{1}\cdot z_{2}-x_{2}\cdot z_{1})^{2}+(y_{1}\cdot z_{2}-y_{2}\cdot z_{1})^{2}}}={\frac {1}{2}}\cdot {\sqrt {(6\cdot (-1)-(-2)\cdot (-1))^{2}+(6\cdot 7-(-2)\cdot 2)^{2}+((-1)\cdot 7-(-1)\cdot 2)^{2}}}=}
= 1 2 ⋅ ( − 6 − 2 ) 2 + ( 42 + 4 ) 2 + ( − 7 + 2 ) 2 = 1 2 ⋅ ( − 8 ) 2 + 46 2 + ( − 5 ) 2 = 1 2 ⋅ 64 + 2116 + 25 = 1 2 2205 = 23.47871376 , {\displaystyle ={\frac {1}{2}}\cdot {\sqrt {(-6-2)^{2}+(42+4)^{2}+(-7+2)^{2}}}={\frac {1}{2}}\cdot {\sqrt {(-8)^{2}+46^{2}+(-5)^{2}}}={\frac {1}{2}}\cdot {\sqrt {64+2116+25}}={\frac {1}{2}}{\sqrt {2205}}=23.47871376,}
kur B C = ( x 1 ; y 1 ; z 1 ) = ( 6 ; − 1 ; 2 ) {\displaystyle BC=(x_{1};y_{1};z_{1})=(6;-1;2)} , B D = ( x 2 ; y 2 ; z 2 ) = ( − 2 ; − 1 ; 7 ) {\displaystyle BD=(x_{2};y_{2};z_{2})=(-2;-1;7)} . Apskaičiuosime lygiagretainio plotą, kai turime vektorius OA =a =(5; 3; 0) ir OB =b =(4; 7; 0). Koordinačių pradžios taškas yra O (0; 0; 0). a × b = | i j k 5 3 0 4 7 0 | = | 3 0 7 0 | i − | 5 0 4 0 | j + | 5 3 4 7 | k = i ( 3 ⋅ 0 − 0 ⋅ 7 ) − j ( 5 ⋅ 0 − 0 ⋅ 4 ) + k ( 5 ⋅ 7 − 3 ⋅ 4 ) = 0 i + 0 j + ( 35 − 12 ) k = 0 i + 0 j + 23 k = ( 0 ; 0 ; 23 ) . {\displaystyle a\times b={\begin{vmatrix}i&j&k\\5&3&0\\4&7&0\end{vmatrix}}={\begin{vmatrix}3&0\\7&0\end{vmatrix}}i-{\begin{vmatrix}5&0\\4&0\end{vmatrix}}j+{\begin{vmatrix}5&3\\4&7\end{vmatrix}}k=i(3\cdot 0-0\cdot 7)-j(5\cdot 0-0\cdot 4)+k(5\cdot 7-3\cdot 4)=0i+0j+(35-12)k=0i+0j+23k=(0;0;23).}
S = ‖ a × b ‖ = 0 2 + 0 2 + 23 2 = 529 = 23. {\displaystyle S=\|a\times b\|={\sqrt {0^{2}+0^{2}+23^{2}}}={\sqrt {529}}=23.}
Dvimatėse koordinatėse galima taikyti ir trumpesnę formulę lygiagretainio arba trikampio plotui:
S = | x 1 ⋅ y 2 − x 2 ⋅ y 1 | = | 5 ⋅ 7 − 4 ⋅ 3 | = | 35 − 12 | = | 23 | = 23 ; {\displaystyle S=|x_{1}\cdot y_{2}-x_{2}\cdot y_{1}|=|5\cdot 7-4\cdot 3|=|35-12|=|23|=23;}
S Δ = | x 1 ⋅ y 2 − x 2 ⋅ y 1 | 2 = 23 2 = 11.5. {\displaystyle S_{\Delta }={\frac {|x_{1}\cdot y_{2}-x_{2}\cdot y_{1}|}{2}}={\frac {23}{2}}=11.5.}
Trimatėms koordinatėms alternatyvi formulė yra tokia, kad surasti lygiagretainio plotą:
S = ( x 1 ⋅ y 2 − x 2 ⋅ y 1 ) 2 + ( x 1 ⋅ z 2 − x 2 ⋅ z 1 ) 2 + ( y 1 ⋅ z 2 − y 2 ⋅ z 1 ) 2 . {\displaystyle S={\sqrt {(x_{1}\cdot y_{2}-x_{2}\cdot y_{1})^{2}+(x_{1}\cdot z_{2}-x_{2}\cdot z_{1})^{2}+(y_{1}\cdot z_{2}-y_{2}\cdot z_{1})^{2}}}.}
Turime dvi tiesių atkarpas: OA ir OB . Taškas O (0; 0; 0) yra koordinačių pradžios taškas. Taškas A (3; 5; 1) yra vektorius a =(3; 5; 1). Taškas B (5; 3; 1) yra vektorius b =(5; 3; 1). Vektorius a yra tiesės atkarpa nuo taško O (0; 0; 0) iki taško A (3; 5; 1). Vektorius b yra tiesės atkarpa nuo taško O (0; 0; 0) iki taško B (5; 3; 1). Sudauginę vektorine vektorių sandauga vektorius a ir b , gausime jiems statų vektorių c . Taigi, taško C , kuris su tašku O sudaro atkarpą statmeną atkarpoms OA ir OB , koordinatės yra: c = a × b = | i j k 3 5 1 5 3 1 | = | 5 1 3 1 | i − | 3 1 5 1 | j + | 3 5 5 3 | k = {\displaystyle \mathbf {c} =\mathbf {a} \times \mathbf {b} ={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\3&5&1\\5&3&1\\\end{vmatrix}}={\begin{vmatrix}5&1\\3&1\end{vmatrix}}\mathbf {i} -{\begin{vmatrix}3&1\\5&1\end{vmatrix}}\mathbf {j} +{\begin{vmatrix}3&5\\5&3\end{vmatrix}}\mathbf {k} =}
= i ( 5 ⋅ 1 − 1 ⋅ 3 ) − j ( 3 ⋅ 1 − 1 ⋅ 5 ) + k ( 3 ⋅ 3 − 5 ⋅ 5 ) = i ( 5 − 3 ) − j ( 3 − 5 ) + k ( 9 − 25 ) = {\displaystyle =\mathbf {i} (5\cdot 1-1\cdot 3)-\mathbf {j} (3\cdot 1-1\cdot 5)+\mathbf {k} (3\cdot 3-5\cdot 5)=\mathbf {i} (5-3)-\mathbf {j} (3-5)+\mathbf {k} (9-25)=}
= 2 i + 2 j − 16 k = ( 2 ; 2 ; − 16 ) . {\displaystyle =2\mathbf {i} +2\mathbf {j} -16\mathbf {k} =(2;2;-16).}
Gavome tašką C (2; 2; -16). Įsitikiname, kad kampas tarp vektoriaus a =(3; 5; 1) ir vektoriaus c =(2; 2; -16) yra lygus 90 laipsnių:
cos α = a ⋅ b ‖ a ‖ ⋅ ‖ c ‖ = 3 ⋅ 2 + 5 ⋅ 2 + 1 ⋅ ( − 16 ) 3 2 + 5 2 + 1 2 ⋅ 2 2 + 2 2 + ( − 16 ) 2 = 6 + 10 − 16 9 + 25 + 1 ⋅ 4 + 4 + 256 = 0 ; {\displaystyle \cos \alpha ={\frac {\mathbf {a} \cdot \mathbf {b} }{\|\mathbf {a} \|\cdot \|\mathbf {c} \|}}={\frac {3\cdot 2+5\cdot 2+1\cdot (-16)}{{\sqrt {3^{2}+5^{2}+1^{2}}}\cdot {\sqrt {2^{2}+2^{2}+(-16)^{2}}}}}={\frac {6+10-16}{{\sqrt {9+25+1}}\cdot {\sqrt {4+4+256}}}}=0;}
α = arccos ( 0 ) = π 2 = 1 , 570796327 {\displaystyle \alpha =\arccos(0)={\frac {\pi }{2}}=1,570796327} radiano arba α = 90 ∘ . {\displaystyle \alpha =90^{\circ }.}
Turime dvi tiesių atkarpas: OA ir OB . Taškas O (0; 0; 0) yra koordinačių pradžios taškas. Taškas A (3; 5; 1) yra vektorius a =(3; 5; 1). Taškas B (4; 3; 2) yra vektorius b =(4; 3; 2). Vektorius a yra tiesės atkarpa nuo taško O (0; 0; 0) iki taško A (3; 5; 1). Vektorius b yra tiesės atkarpa nuo taško O (0; 0; 0) iki taško B (4; 3; 2). Sudauginę vektorine vektorių sandauga vektorius a ir b , gausime jiems statų vektorių c . Taigi, taško C , kuris su tašku O sudaro atkarpą statmeną atkarpoms OA ir OB , koordinatės yra: c = a × b = | i j k 3 5 1 4 3 2 | = | 5 1 3 2 | i − | 3 1 4 2 | j + | 3 5 4 3 | k = {\displaystyle \mathbf {c} =\mathbf {a} \times \mathbf {b} ={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\3&5&1\\4&3&2\\\end{vmatrix}}={\begin{vmatrix}5&1\\3&2\end{vmatrix}}\mathbf {i} -{\begin{vmatrix}3&1\\4&2\end{vmatrix}}\mathbf {j} +{\begin{vmatrix}3&5\\4&3\end{vmatrix}}\mathbf {k} =}
= i ( 5 ⋅ 2 − 1 ⋅ 3 ) − j ( 3 ⋅ 2 − 1 ⋅ 4 ) + k ( 3 ⋅ 3 − 5 ⋅ 4 ) = i ( 10 − 3 ) − j ( 6 − 4 ) + k ( 9 − 20 ) = {\displaystyle =\mathbf {i} (5\cdot 2-1\cdot 3)-\mathbf {j} (3\cdot 2-1\cdot 4)+\mathbf {k} (3\cdot 3-5\cdot 4)=\mathbf {i} (10-3)-\mathbf {j} (6-4)+\mathbf {k} (9-20)=}
= 7 i − 2 j − 11 k = ( 7 ; − 2 ; − 11 ) . {\displaystyle =7\mathbf {i} -2\mathbf {j} -11\mathbf {k} =(7;-2;-11).}
Gavome tašką C (7; -2; -11).
Įsitikiname, kad kampas α {\displaystyle \alpha } tarp vektoriaus a =(3; 5; 1) ir vektoriaus c =(7; -2; -11) yra lygus 90 laipsnių:
cos α = a ⋅ b ‖ a ‖ ⋅ ‖ c ‖ = 3 ⋅ 7 + 5 ⋅ ( − 2 ) + 1 ⋅ ( − 11 ) 3 2 + 5 2 + 1 2 ⋅ 7 2 + ( − 2 ) 2 + ( − 11 ) 2 = 21 − 10 − 11 9 + 25 + 1 ⋅ 49 + 4 + 121 = 0 ; {\displaystyle \cos \alpha ={\frac {\mathbf {a} \cdot \mathbf {b} }{\|\mathbf {a} \|\cdot \|\mathbf {c} \|}}={\frac {3\cdot 7+5\cdot (-2)+1\cdot (-11)}{{\sqrt {3^{2}+5^{2}+1^{2}}}\cdot {\sqrt {7^{2}+(-2)^{2}+(-11)^{2}}}}}={\frac {21-10-11}{{\sqrt {9+25+1}}\cdot {\sqrt {49+4+121}}}}=0;}
α = arccos ( 0 ) = π 2 = 1 , 570796327 {\displaystyle \alpha =\arccos(0)={\frac {\pi }{2}}=1,570796327} radiano arba α = 90 ∘ . {\displaystyle \alpha =90^{\circ }.} Įsitikiname, kad kampas β {\displaystyle \beta } tarp vektoriaus b =(4; 3; 2) ir vektoriaus c =(7; -2; -11) yra lygus 90 laipsnių:
cos β = a ⋅ b ‖ a ‖ ⋅ ‖ c ‖ = 4 ⋅ 7 + 3 ⋅ ( − 2 ) + 2 ⋅ ( − 11 ) 4 2 + 3 2 + 2 2 ⋅ 7 2 + ( − 2 ) 2 + ( − 11 ) 2 = 28 − 6 − 22 16 + 9 + 4 ⋅ 49 + 4 + 121 = 0 ; {\displaystyle \cos \beta ={\frac {\mathbf {a} \cdot \mathbf {b} }{\|\mathbf {a} \|\cdot \|\mathbf {c} \|}}={\frac {4\cdot 7+3\cdot (-2)+2\cdot (-11)}{{\sqrt {4^{2}+3^{2}+2^{2}}}\cdot {\sqrt {7^{2}+(-2)^{2}+(-11)^{2}}}}}={\frac {28-6-22}{{\sqrt {16+9+4}}\cdot {\sqrt {49+4+121}}}}=0;}
β = arccos ( 0 ) = π 2 = 1 , 570796327 {\displaystyle \beta =\arccos(0)={\frac {\pi }{2}}=1,570796327} radiano arba β = 90 ∘ . {\displaystyle \beta =90^{\circ }.} Mišri vektorių sandauga
keisti
Mišri vektorių sandauga (a b c ) yra apibrėžiama:
( a b c ) = a ⋅ ( b × c ) . {\displaystyle (\mathbf {a} \ \mathbf {b} \ \mathbf {c} )=\mathbf {a} \cdot (\mathbf {b} \times \mathbf {c} ).}
Mišriają sandaugą taip pat galima užrašyti taip:
V = | ( a × b ) ⋅ c | {\displaystyle V=|(\mathbf {a} \times \mathbf {b} )\cdot \mathbf {c} |} ,čia V yra lygiagretainio gretasienio tūris.
Piramidės tūris yra:
V p i r . = 1 6 | ( a × b ) ⋅ c | . {\displaystyle V_{pir.}={\frac {1}{6}}|(\mathbf {a} \times \mathbf {b} )\cdot \mathbf {c} |.}
Duoti vektoriai a =(4; 9; 0), b =(7; 5; 0), c =(2; 3; 10). Rasime piramidės, kurią sudaro šie vektoriai, tūrį. ( a × b ) ⋅ c = | a x a y a z b z b y b z c x c y c z | = a x | b y b z c y c z | − a y | b x b z c x c z | + a z | b x b y c x c y | = {\displaystyle (\mathbf {a} \times \mathbf {b} )\cdot \mathbf {c} ={\begin{vmatrix}a_{x}&a_{y}&a_{z}\\b_{z}&b_{y}&b_{z}\\c_{x}&c_{y}&c_{z}\end{vmatrix}}=a_{x}{\begin{vmatrix}b_{y}&b_{z}\\c_{y}&c_{z}\end{vmatrix}}-a_{y}{\begin{vmatrix}b_{x}&b_{z}\\c_{x}&c_{z}\end{vmatrix}}+a_{z}{\begin{vmatrix}b_{x}&b_{y}\\c_{x}&c_{y}\end{vmatrix}}=}
= | 4 9 0 7 5 0 2 3 10 | = 4 | 5 0 3 10 | − 9 | 7 0 2 10 | + 0 ⋅ | 7 5 2 3 | = 4 ⋅ 50 − 9 ⋅ 70 + 0 = 200 − 630 = − 430. {\displaystyle ={\begin{vmatrix}4&9&0\\7&5&0\\2&3&10\end{vmatrix}}=4{\begin{vmatrix}5&0\\3&10\end{vmatrix}}-9{\begin{vmatrix}7&0\\2&10\end{vmatrix}}+0\cdot {\begin{vmatrix}7&5\\2&3\end{vmatrix}}=4\cdot 50-9\cdot 70+0=200-630=-430.}
Piramidės su viršūnėmis O(0; 0; 0), A(4; 9; 0), B(7; 5; 0), C(2; 3; 10) tūris yra:
V p i r . = 1 6 ⋅ | ( a × b ) ⋅ c | = 1 6 ⋅ | − 430 | = 430 6 = 215 3 = 71.66666667. {\displaystyle V_{pir.}={\frac {1}{6}}\cdot |(\mathbf {a} \times \mathbf {b} )\cdot \mathbf {c} |={\frac {1}{6}}\cdot |-430|={\frac {430}{6}}={\frac {215}{3}}=71.66666667.} Duoti vektoriai a =(4; 9; 0), b =(7; 5; 0), c =(0; 0; 10). Rasime piramidės, kurią sudaro šie vektoriai, tūrį. ( a × b ) ⋅ c = | a x a y a z b z b y b z c x c y c z | = a x | b y b z c y c z | − a y | b x b z c x c z | + a z | b x b y c x c y | = {\displaystyle (\mathbf {a} \times \mathbf {b} )\cdot \mathbf {c} ={\begin{vmatrix}a_{x}&a_{y}&a_{z}\\b_{z}&b_{y}&b_{z}\\c_{x}&c_{y}&c_{z}\end{vmatrix}}=a_{x}{\begin{vmatrix}b_{y}&b_{z}\\c_{y}&c_{z}\end{vmatrix}}-a_{y}{\begin{vmatrix}b_{x}&b_{z}\\c_{x}&c_{z}\end{vmatrix}}+a_{z}{\begin{vmatrix}b_{x}&b_{y}\\c_{x}&c_{y}\end{vmatrix}}=}
= | 4 9 0 7 5 0 0 0 10 | = 4 | 5 0 0 10 | − 9 | 7 0 0 10 | + 0 ⋅ | 7 5 0 0 | = 4 ⋅ 50 − 9 ⋅ 70 + 0 = 200 − 630 = − 430. {\displaystyle ={\begin{vmatrix}4&9&0\\7&5&0\\0&0&10\end{vmatrix}}=4{\begin{vmatrix}5&0\\0&10\end{vmatrix}}-9{\begin{vmatrix}7&0\\0&10\end{vmatrix}}+0\cdot {\begin{vmatrix}7&5\\0&0\end{vmatrix}}=4\cdot 50-9\cdot 70+0=200-630=-430.}
Piramidės su viršūnėmis O(0; 0; 0), A(4; 9; 0), B(7; 5; 0), C(0; 0; 10) tūris yra:
V p i r . = 1 6 ⋅ | ( a × b ) ⋅ c | = 1 6 ⋅ | − 430 | = 430 6 = 215 3 = 71.66666667. {\displaystyle V_{pir.}={\frac {1}{6}}\cdot |(\mathbf {a} \times \mathbf {b} )\cdot \mathbf {c} |={\frac {1}{6}}\cdot |-430|={\frac {430}{6}}={\frac {215}{3}}=71.66666667.}
Duoti vektoriai a=(1; 2; 0), b=(1; -2; 0), c=(0; 0; 3), kurių pradžios koordinatės yra (0; 0; 0). Rasime lygiagretainio 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 0 1 − 2 0 0 0 3 | = 3 ⋅ ( − 1 ) 3 + 3 | 1 2 1 − 2 | = 3 ( 1 ⋅ ( − 2 ) − 2 ⋅ 1 ) = − 12. {\displaystyle ={\begin{vmatrix}1&2&0\\1&-2&0\\0&0&3\end{vmatrix}}=3\cdot (-1)^{3+3}{\begin{vmatrix}1&2\\1&-2\end{vmatrix}}=3(1\cdot (-2)-2\cdot 1)=-12.} Gretasienio tūris yra |-12|=12. Taip pat galima skaičiuot taip:
V = | 1 2 0 1 − 2 0 0 0 3 | = 1 ⋅ ( − 2 ) ⋅ 3 + 2 ⋅ 0 ⋅ 0 + 0 ⋅ 1 ⋅ 0 − 1 ⋅ 0 ⋅ 0 − 2 ⋅ 1 ⋅ 3 − 0 ⋅ ( − 2 ) ⋅ 0 = − 6 + 0 + 0 − 0 − 6 − 0 = − 12. {\displaystyle V={\begin{vmatrix}1&2&0\\1&-2&0\\0&0&3\end{vmatrix}}=1\cdot (-2)\cdot 3+2\cdot 0\cdot 0+0\cdot 1\cdot 0-1\cdot 0\cdot 0-2\cdot 1\cdot 3-0\cdot (-2)\cdot 0=-6+0+0-0-6-0=-12.}
Patikriname ar atsakymas bus toks pat naudojant vektorine sandauga (vektorių a ir b) sudauginta su statmeno vektoriaus c ilgiu:
a × b = | i j k 1 2 0 1 − 2 0 | = {\displaystyle a\times b={\begin{vmatrix}i&j&k\\1&2&0\\1&-2&0\end{vmatrix}}=}
= i ⋅ 2 ⋅ 0 + j ⋅ 0 ⋅ 1 + k ⋅ 1 ⋅ ( − 2 ) − i ⋅ 0 ⋅ ( − 2 ) − j ⋅ 1 ⋅ 0 − k ⋅ 2 ⋅ 1 = {\displaystyle =i\cdot 2\cdot 0+j\cdot 0\cdot 1+k\cdot 1\cdot (-2)-i\cdot 0\cdot (-2)-j\cdot 1\cdot 0-k\cdot 2\cdot 1=} = 0 i − 0 i + 0 j − 0 j − 2 k − 2 k = 0 i + 0 j − 4 k = ( 0 ; 0 ; − 4 ) . {\displaystyle =0i-0i+0j-0j-2k-2k=0i+0j-4k=(0;0;-4).}
| | a × b | | = 0 2 + 0 2 + ( − 4 ) 2 = 0 + 0 + 16 = 16 = 4. {\displaystyle ||a\times b||={\sqrt {0^{2}+0^{2}+(-4)^{2}}}={\sqrt {0+0+16}}={\sqrt {16}}=4.}
| | c | | = 0 2 + 0 2 + 3 2 = 9 = 3. {\displaystyle ||c||={\sqrt {0^{2}+0^{2}+3^{2}}}={\sqrt {9}}=3.}
V = | | a × b | | ⋅ | | c | | = 4 ⋅ 3 = 12. {\displaystyle V=||a\times b||\cdot ||c||=4\cdot 3=12.}
Patikriname taikydami Herono formulę.
| | a | | = 1 2 + 2 2 + 0 2 = 5 = 2 , 236067978. {\displaystyle ||a||={\sqrt {1^{2}+2^{2}+0^{2}}}={\sqrt {5}}=2,236067978.}
| | b | | = 1 2 + ( − 2 ) 2 + 0 = 5 = 2 , 236067978. {\displaystyle ||b||={\sqrt {1^{2}+(-2)^{2}+0}}={\sqrt {5}}=2,236067978.} Atstumas tarp taškų a=(1; 2; 0) ir b=(1; -2; 0) yra lygus:
f = ( 1 − 1 ) 2 + ( 2 − ( − 2 ) ) 2 + ( 0 − 0 ) 2 = 16 = 4. {\displaystyle f={\sqrt {(1-1)^{2}+(2-(-2))^{2}+(0-0)^{2}}}={\sqrt {16}}=4.}
p = a + b + f 2 = 5 + 5 + 4 2 = 5 + 2 = 4.236067978. {\displaystyle p={a+b+f \over 2}={{\sqrt {5}}+{\sqrt {5}}+4 \over 2}={\sqrt {5}}+2=4.236067978.}
S Δ = p ( p − a ) ( p − b ) ( p − f ) = 4.236067978 ( 5 + 2 − 5 ) ( 5 + 2 − 5 ) ( 5 + 2 − 4 ) = {\displaystyle S_{\Delta }={\sqrt {p(p-a)(p-b)(p-f)}}={\sqrt {4.236067978({\sqrt {5}}+2-{\sqrt {5}})({\sqrt {5}}+2-{\sqrt {5}})({\sqrt {5}}+2-4)}}=}
= ( 5 + 2 ) ⋅ 2 ⋅ 2 ⋅ ( 5 − 2 ) = ( 5 − 4 ) ⋅ 2 ⋅ 2 = 4 = 2. {\displaystyle ={\sqrt {({\sqrt {5}}+2)\cdot 2\cdot 2\cdot ({\sqrt {5}}-2)}}={\sqrt {(5-4)\cdot 2\cdot 2}}={\sqrt {4}}=2.} S = 2 S Δ = 2 ⋅ 2 = 4. {\displaystyle S=2S_{\Delta }=2\cdot 2=4.}
V = S ⋅ | | c | | = 4 ⋅ 3 = 12. {\displaystyle V=S\cdot ||c||=4\cdot 3=12.}
Rasime piramidės su 4 viršunėmis, kurios pagrindas yra trikampis, tūrį: V = 1 6 | ( a × b ) ⋅ c | = 12 6 = 2. {\displaystyle V={\frac {1}{6}}|(a\times b)\cdot c|={\frac {12}{6}}=2.} Piramidės tūris yra 1 6 | ( a × b ) ⋅ c | {\displaystyle {\frac {1}{6}}|(a\times b)\cdot c|} todėl, kad piramidės pagrindo plotas yra puse (S=ab/2) lygiagretainio ploto, o kadangi gretasienio tūris yra V=abh=Sh ir piramidės (kurios pagrindas trikampis) tūris yra V=(ab/2)*h/3=abh/6=Sh/3, tai dėl to piramidės tūris yra V=abh/6 arba 1/6 gretasienio tūrio. Piramidės, kurios pagrindas yra keturkampis, tūris yra V = 1 3 | ( a × b ) ⋅ c | . {\displaystyle V={\frac {1}{3}}|(a\times b)\cdot c|.}
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 tūrisV 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.}
Pavyzdis . Duoti vektoriai A =a ={5; 3; 0}, B =b ={4; 11; 0}, C =c ={3; 7; 6}. Visi jie išeina iš koordinačių pradžios taško O (0; 0; 0), todėl visi trys vektoriai liečiasi tame pačiame taške (0; 0; 0). Rasime piramidės tūrį ir patikrinsimę jį (žinodami piramidės aukštį h = c z = 6 {\displaystyle h=c_{z}=6} ). Piramidės tūris, kurios pagriną sudaro OAB trikampis, yra:V p i r . = 1 6 ⋅ | ( a × b ) ⋅ c | = 1 6 ⋅ | 258 | = 43. {\displaystyle V_{pir.}={\frac {1}{6}}\cdot |(\mathbf {a} \times \mathbf {b} )\cdot \mathbf {c} |={\frac {1}{6}}\cdot |258|=43.}
kur ( a × b ) ⋅ c = | 5 3 0 4 11 0 3 7 6 | = 5 ⋅ ( − 1 ) 1 + 1 | 11 0 7 6 | + 3 ⋅ ( − 1 ) 1 + 2 | 4 0 3 6 | + 0 ⋅ ( − 1 ) 1 + 3 | 4 11 3 7 | = {\displaystyle (\mathbf {a} \times \mathbf {b} )\cdot \mathbf {c} ={\begin{vmatrix}5&3&0\\4&11&0\\3&7&6\end{vmatrix}}=5\cdot (-1)^{1+1}{\begin{vmatrix}11&0\\7&6\end{vmatrix}}+3\cdot (-1)^{1+2}{\begin{vmatrix}4&0\\3&6\end{vmatrix}}+0\cdot (-1)^{1+3}{\begin{vmatrix}4&11\\3&7\end{vmatrix}}=}
= 5 ⋅ ( − 1 ) 2 ⋅ ( 11 ⋅ 6 − 0 ⋅ 7 ) + 3 ⋅ ( − 1 ) 3 ⋅ ( 4 ⋅ 6 − 0 ⋅ 3 ) + 0 = 5 ⋅ 66 − 3 ⋅ 24 = 330 − 72 = 258. {\displaystyle =5\cdot (-1)^{2}\cdot (11\cdot 6-0\cdot 7)+3\cdot (-1)^{3}\cdot (4\cdot 6-0\cdot 3)+0=5\cdot 66-3\cdot 24=330-72=258.}
Toliau, surandame trikampio OAB plotą: a × b = | i j k 5 3 0 4 11 0 | = i ⋅ ( − 1 ) 1 + 1 | 3 0 11 0 | + j ⋅ ( − 1 ) 1 + 2 | 5 0 4 0 | + k ⋅ ( − 1 ) 1 + 3 | 5 3 4 11 | = {\displaystyle \mathbf {a} \times \mathbf {b} ={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\5&3&0\\4&11&0\end{vmatrix}}=\mathbf {i} \cdot (-1)^{1+1}{\begin{vmatrix}3&0\\11&0\end{vmatrix}}+\mathbf {j} \cdot (-1)^{1+2}{\begin{vmatrix}5&0\\4&0\end{vmatrix}}+\mathbf {k} \cdot (-1)^{1+3}{\begin{vmatrix}5&3\\4&11\end{vmatrix}}=}
= i ⋅ ( 3 ⋅ 0 − 0 ⋅ 11 ) − j ⋅ ( 5 ⋅ 0 − 0 ⋅ 4 ) + k ⋅ ( 5 ⋅ 11 − 3 ⋅ 4 ) = 0 i − 0 j ⋅ + k ⋅ ( 55 − 12 ) = 0 i + 0 j + 43 k = ( 0 ; 0 ; 43 ) ; {\displaystyle =\mathbf {i} \cdot (3\cdot 0-0\cdot 11)-\mathbf {j} \cdot (5\cdot 0-0\cdot 4)+\mathbf {k} \cdot (5\cdot 11-3\cdot 4)=0\mathbf {i} -0\mathbf {j} \cdot +\mathbf {k} \cdot (55-12)=0\mathbf {i} +0\mathbf {j} +43\mathbf {k} =(0;0;43);}
S Δ O A B = 1 2 ⋅ ‖ a × b ‖ = 1 2 ⋅ 0 2 + 0 2 + 43 2 = 43 2 = 21.5. {\displaystyle S_{\Delta OAB}={\frac {1}{2}}\cdot \|\mathbf {a} \times \mathbf {b} \|={\frac {1}{2}}\cdot {\sqrt {0^{2}+0^{2}+43^{2}}}={\frac {43}{2}}=21.5.}
Piramidės OABC tūris yra:
V p i r . = 1 3 ⋅ S Δ O A B ⋅ h = 21.5 ⋅ 6 3 = 43. {\displaystyle V_{pir.}={\frac {1}{3}}\cdot S_{\Delta OAB}\cdot h={\frac {21.5\cdot 6}{3}}=43.}
Pavyzdis . Duoti vektoriai a ={8; 6; 2}, b ={5; 9; 3}, c ={1; 2; 7}. Apskaičiuosime piramidės, kurią sudaro šie vektoriai, tūrį V .Sprendimas . Piramidės tūris yra lygus 1/6 mišrios vektorių a ={8; 6; 2}, b ={5; 9; 3}, c ={1; 2; 7} sandaugos. Taigi:
( a × b ) ⋅ c = | a x a y a z b z b y b z c x c y c z | = | 8 6 2 5 9 3 1 2 7 | = 8 ⋅ ( − 1 ) 1 + 3 | 9 3 2 7 | + 6 ⋅ ( − 1 ) 1 + 2 | 5 3 1 7 | + 2 ⋅ ( − 1 ) 1 + 3 | 5 9 1 2 | = {\displaystyle (\mathbf {a} \times \mathbf {b} )\cdot \mathbf {c} ={\begin{vmatrix}a_{x}&a_{y}&a_{z}\\b_{z}&b_{y}&b_{z}\\c_{x}&c_{y}&c_{z}\end{vmatrix}}={\begin{vmatrix}8&6&2\\5&9&3\\1&2&7\end{vmatrix}}=8\cdot (-1)^{1+3}{\begin{vmatrix}9&3\\2&7\end{vmatrix}}+6\cdot (-1)^{1+2}{\begin{vmatrix}5&3\\1&7\end{vmatrix}}+2\cdot (-1)^{1+3}{\begin{vmatrix}5&9\\1&2\end{vmatrix}}=}
= 8 ⋅ ( 9 ⋅ 7 − 3 ⋅ 2 ) − 6 ⋅ ( 5 ⋅ 7 − 3 ⋅ 1 ) + 2 ⋅ ( 5 ⋅ 2 − 9 ⋅ 1 ) = 8 ⋅ ( 63 − 6 ) − 6 ⋅ ( 35 − 3 ) + 2 ⋅ ( 10 − 9 ) = 8 ⋅ 57 − 6 ⋅ 32 + 2 ⋅ 1 = 285 − 192 + 2 = 95. {\displaystyle =8\cdot (9\cdot 7-3\cdot 2)-6\cdot (5\cdot 7-3\cdot 1)+2\cdot (5\cdot 2-9\cdot 1)=8\cdot (63-6)-6\cdot (35-3)+2\cdot (10-9)=8\cdot 57-6\cdot 32+2\cdot 1=285-192+2=95.}
V = 1 6 ⋅ | ( a × b ) ⋅ c | = 1 6 ⋅ | 95 | = 95 6 = 15.833333333. {\displaystyle V={\frac {1}{6}}\cdot |(\mathbf {a} \times \mathbf {b} )\cdot \mathbf {c} |={\frac {1}{6}}\cdot |95|={\frac {95}{6}}=15.833333333.} Kolinearūs ir komplanarūs vektoriai
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Vektoriai yra kolinearūs, jeigu a × b = 0. {\displaystyle a\times b=0.} Dvimačiai vektoriai yra kolinearūs, kai yra lygiagretūs.
Vektoriai yra komplanarūs, jeigu ( a × b ) ⋅ c = 0. {\displaystyle (\mathbf {a} \times \mathbf {b} )\cdot \mathbf {c} =0.} Trimatėje erdvėje vektoriai yra komplanarūs, kai priklauso tai pačiai ploštumai.
Pavyzdys . Ar gali keturi taškai A(1; 2; 3), B(2; 4; 1), C(1; -3; 6) ir D(4; -2; 3) priklausyti vienai plokštumai?Sprendimas. Taškai A , B , C ir D priklausys plokštumai g , kai vektoriai AB , AC ir AD bus komplanarūs. Randame šų vektorių koordinates:
a =AB=B-A=(2-1; 4-2; 1-3)={1; 2; -2},
b =AC=C-A=(1-1; -3-2; 6-3)={0; -5; 3},
c =AD=D-A=(4-1; -2-2; 3-3)={3; -4; 0}.
Apskaičiuojame mišriąją jų sandaugą:
( 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 (\mathbf {a} \times \mathbf {b} )\cdot \mathbf {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 0 − 5 3 3 − 4 0 | = 3 ⋅ ( − 1 ) 3 + 1 | 2 − 2 − 5 3 | + ( − 4 ) ⋅ ( − 1 ) 3 + 2 | 1 − 2 0 3 | + 0 ⋅ ( − 1 ) 3 + 3 | 1 2 0 − 5 | = {\displaystyle ={\begin{vmatrix}1&2&-2\\0&-5&3\\3&-4&0\end{vmatrix}}=3\cdot (-1)^{3+1}{\begin{vmatrix}2&-2\\-5&3\end{vmatrix}}+(-4)\cdot (-1)^{3+2}{\begin{vmatrix}1&-2\\0&3\end{vmatrix}}+0\cdot (-1)^{3+3}{\begin{vmatrix}1&2\\0&-5\end{vmatrix}}=}
= 3 ⋅ ( 2 ⋅ 3 − ( − 2 ) ⋅ ( − 5 ) ) + 4 ⋅ ( 1 ⋅ 3 − ( − 2 ) ⋅ 0 ) + 0 = 3 ⋅ ( 6 − 10 ) + 4 ⋅ 3 + 0 = 3 ⋅ ( − 4 ) + 12 = − 12 + 12 = 0. {\displaystyle =3\cdot (2\cdot 3-(-2)\cdot (-5))+4\cdot (1\cdot 3-(-2)\cdot 0)+0=3\cdot (6-10)+4\cdot 3+0=3\cdot (-4)+12=-12+12=0.}
Kadangi mišrioji tijų vektorių sandauga lygi nuliui, tai tie vektoriai yra komplanarūs, o taškai A , B , C ir D priklauso vienai plokštumai g .
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. Lagrandžo tapatumas
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Jėga ir vektoriai
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Duotos jėgos F projekcijos F x = 4 {\displaystyle F_{x}=4} , F y = 4 {\displaystyle F_{y}=4} , F z = − 4 2 . {\displaystyle F_{z}=-4{\sqrt {2}}.} Rasime jėgos dydį ||F || ir jos veikimo kryptį. Jėgos dydis yra: | | F | | = F x 2 + F y 2 + F z 2 = 4 2 + 4 2 + ( − 4 2 ) 2 = 16 + 16 + 32 = 64 = 8 {\displaystyle ||F||={\sqrt {F_{x}^{2}+F_{y}^{2}+F_{z}^{2}}}={\sqrt {4^{2}+4^{2}+(-4{\sqrt {2}})^{2}}}={\sqrt {16+16+32}}={\sqrt {64}}=8} . Rasime krypties kosinusus: cos α = F x | | F | | = 4 8 = 1 2 {\displaystyle \cos \alpha ={\frac {F_{x}}{||F||}}={\frac {4}{8}}={\frac {1}{2}}} , cos β = F y | | F | | = 4 8 = 1 2 {\displaystyle \cos \beta ={\frac {F_{y}}{||F||}}={\frac {4}{8}}={\frac {1}{2}}} , cos γ = F z | | F | | = − 4 2 8 = − 2 2 {\displaystyle \cos \gamma ={\frac {F_{z}}{||F||}}={\frac {-4{\sqrt {2}}}{8}}=-{\frac {\sqrt {2}}{2}}} . Iš čia randame kampus α = arccos 1 2 = 60 0 , β = arccos 1 2 = 60 0 {\displaystyle \alpha =\arccos {\frac {1}{2}}=60^{0},\;\beta =\arccos {\frac {1}{2}}=60^{0}\;} , γ = arccos − 2 2 = 135 0 . {\displaystyle \gamma =\arccos {\frac {-{\sqrt {2}}}{2}}=135^{0}.} Vadinasi, jėga ||F || veikia vektoriaus, sudarančio su koordinačių ašimis kampus α = 60 0 , β = 60 0 , γ = 135 0 , {\displaystyle \alpha =60^{0},\;\beta =60^{0},\;\gamma =135^{0},} kryptimi.
Vektorius a su ašimis Oy ir Oz sudaro kampus β = γ = 60 0 . {\displaystyle \beta =\gamma =60^{0}.} Rasime kampą α , {\displaystyle \alpha ,} kurį vektorius a sudaro su Ox ašimi. Kadangi cos 2 α + cos 2 β + cos 2 γ = 1 , {\displaystyle \cos ^{2}\alpha +\cos ^{2}\beta +\cos ^{2}\gamma =1,} tai cos 2 α = 1 − cos 2 β − cos 2 γ = 1 − 1 4 − 1 4 = 1 2 . {\displaystyle \cos ^{2}\alpha =1-\cos ^{2}\beta -\cos ^{2}\gamma =1-{\frac {1}{4}}-{\frac {1}{4}}={\frac {1}{2}}.} Iš čia cos α = 1 2 = ± 2 2 . {\displaystyle \cos \alpha ={\sqrt {\frac {1}{2}}}=\pm {\frac {\sqrt {2}}{2}}.} Tada α = arccos 2 2 = 45 0 {\displaystyle \alpha =\arccos {\frac {\sqrt {2}}{2}}=45^{0}} arba α = arccos 2 − 2 = 135 0 . {\displaystyle \alpha =\arccos {\frac {\sqrt {2}}{-2}}=135^{0}.}
Jėga F veikia vektoriaus, sudarančio su koordinačių ašimis kampus α = β = 120 0 , γ = 45 0 , {\displaystyle \alpha =\beta =120^{0},\;\gamma =45^{0},} kryptimi. Rasime jėgos F projekcijas, jei ||F ||=6. F x = F y = | | F | | cos α = cos β = 6 cos 120 0 = − 3 ; F z = | | F | | cos γ = 6 cos 45 0 = 3 2 . {\displaystyle F_{x}=F_{y}=||F||\cos \alpha =\cos \beta =6\cos 120^{0}=-3;\;F_{z}=||F||\cos \gamma =6\cos 45^{0}=3{\sqrt {2}}.} Jėgos F dedamosios F x = − 3 i ; F y = − 3 j ; F z = − 3 2 k . {\displaystyle F_{x}=-3i;\;F_{y}=-3j;\;F_{z}=-3{\sqrt {2}}k.}
Kampas tarp vektoriaus a → = ( x 0 ; y 0 ; z 0 ) {\displaystyle {\vec {a}}=(x_{0};y_{0};z_{0})} ir vektoriaus b → = ( x 1 ; y 1 ; z 1 ) {\displaystyle {\vec {b}}=(x_{1};y_{1};z_{1})} yra 90 laipsnių, jeigu jų skaliarinė sandaug lygi nuliui:
a → ⋅ b → = x 0 x 1 + y 0 y 1 + z 0 z 1 = 0. {\displaystyle {\vec {a}}\cdot {\vec {b}}=x_{0}x_{1}+y_{0}y_{1}+z_{0}z_{1}=0.}
Vektoriai nebūtinai turi išeiti iš to paties taško. Pasinaudoje šia savybe galime sudaryti plokštumos lygtį. Tarkime žinomas vienas plokštumos taškas M 0 ( x 0 ; y 0 ; z 0 ) . {\displaystyle M_{0}(x_{0};y_{0};z_{0}).} Plokštumos taškas M 0 ( x 0 ; y 0 ; z 0 ) {\displaystyle M_{0}(x_{0};y_{0};z_{0})} jungiasi su betkuriuo kitu plokštumos tašku kurio koordinatės M(x; y; z) . Tuomet galima sudaryti begalybę vektorių gulinčių ant tos pačios plokštumos ir iš to, kad M yra kintantis taškas užrašome tam tikro vektoriaus koordinates M 0 M → = ( x − x 0 ; y − y 0 ; z − z 0 ) {\displaystyle {\vec {M_{0}M}}=(x-x_{0};y-y_{0};z-z_{0})} gulinčio ant plokštumos. Kad visi gauti vektoriai su bet kokiomis taško M koordinatėmis x , y , z gulėtų ant tos pačios plokštumos, reikia, kad būtų tenkinama sąlyga:
M 0 M → ⋅ O N → = ( x − x 0 ) ⋅ ( x 1 − 0 ) + ( y − y 0 ) ⋅ ( y 1 − 0 ) + ( z − z 0 ) ⋅ ( z 1 − 0 ) = 0 , {\displaystyle {\vec {M_{0}M}}\cdot {\vec {ON}}=(x-x_{0})\cdot (x_{1}-0)+(y-y_{0})\cdot (y_{1}-0)+(z-z_{0})\cdot (z_{1}-0)=0,}
čia O N → = ( x 1 − 0 ; y 1 − 0 ; z 1 − 0 ) = ( x 1 ; y 1 ; z 1 ) {\displaystyle {\vec {ON}}=(x_{1}-0;y_{1}-0;z_{1}-0)=(x_{1};y_{1};z_{1})} yra vektorius statmenas vektoriui M 0 M → = ( x − x 0 ; y − y 0 ; z − z 0 ) ; {\displaystyle {\vec {M_{0}M}}=(x-x_{0};y-y_{0};z-z_{0});} taškas O (0; 0; 0) yra koordinačių pradžios taškas; taškas N ( x 1 ; y 1 ; z 1 ) {\displaystyle N(x_{1};y_{1};z_{1})} kartu su koordinačių pradžios tašku O (0; 0; 0) sudaro plokštumos normalės vektorių n → = ( x 1 − 0 ; y 1 − 0 ; z 1 − 0 ) = ( x 1 ; y 1 ; z 1 ) {\displaystyle {\vec {n}}=(x_{1}-0;y_{1}-0;z_{1}-0)=(x_{1};y_{1};z_{1})} stameną plokštumai ( x − x 0 ) ⋅ x 1 + ( y − y 0 ) ⋅ y 1 + ( z − z 0 ) ⋅ z 1 = 0. {\displaystyle (x-x_{0})\cdot x_{1}+(y-y_{0})\cdot y_{1}+(z-z_{0})\cdot z_{1}=0.} Tokiu budu sudauginę vieną žinomą vektorių O N → {\displaystyle {\vec {ON}}} ir vieną kintamą vektorių M 0 M → {\displaystyle {\vec {M_{0}M}}} ir prilyginę jų skaliarinę sandaugą nuliui (kad visi galimi vektoriai iš vektoriaus M 0 M → {\displaystyle {\vec {M_{0}M}}} būtų statūs vektoriui O N → {\displaystyle {\vec {ON}}} ), gavome plokštumos lygtį, kurios normalės vektorius yra n → = ( x 1 ; y 1 ; z 1 ) . {\displaystyle {\vec {n}}=(x_{1};y_{1};z_{1}).}