Sr Examen

Otras calculadoras

Forma canónica/ x^2+3*y^2+3*z^2-2*x*y-2*y*z-2*x*z=0

x^2+3*y^2+3*z^2-2*x*y-2*y*z-2*x*z=0 forma canónica

El profesor se sorprenderá mucho al ver tu solución correcta😉

v

Gráfico:

x: [, ]
y: [, ]
z: [, ]

Calidad:

 (Cantidad de puntos en el eje)

Tipo de trazado:

Solución

Ha introducido [src] 
 2      2      2                            
x  + 3*y  + 3*z  - 2*x*y - 2*x*z - 2*y*z = 0
x2−2xy−2xz+3y2−2yz+3z2=0x^{2} - 2 x y - 2 x z + 3 y^{2} - 2 y z + 3 z^{2} = 0 
x^2 - 2*x*y - 2*x*z + 3*y^2 - 2*y*z + 3*z^2 = 0
Método de invariantes
Se da la ecuación de superficie de 2 grado:
x2−2xy−2xz+3y2−2yz+3z2=0x^{2} - 2 x y - 2 x z + 3 y^{2} - 2 y z + 3 z^{2} = 0
Esta ecuación tiene la forma:
a11x2+2a12xy+2a13xz+2a14x+a22y2+2a23yz+2a24y+a33z2+2a34z+a44=0a_{11} x^{2} + 2 a_{12} x y + 2 a_{13} x z + 2 a_{14} x + a_{22} y^{2} + 2 a_{23} y z + 2 a_{24} y + a_{33} z^{2} + 2 a_{34} z + a_{44} = 0
donde
a11=1a_{11} = 1
a12=−1a_{12} = -1
a13=−1a_{13} = -1
a14=0a_{14} = 0
a22=3a_{22} = 3
a23=−1a_{23} = -1
a24=0a_{24} = 0
a33=3a_{33} = 3
a34=0a_{34} = 0
a44=0a_{44} = 0
Las invariantes de esta ecuación al transformar las coordenadas son los determinantes:
I1=a11+a22+a33I_{1} = a_{11} + a_{22} + a_{33}
     |a11  a12|   |a22  a23|   |a11  a13|
I2 = |        | + |        | + |        |
     |a12  a22|   |a23  a33|   |a13  a33|

I3=∣a11a12a13a12a22a23a13a23a33∣I_{3} = \left|\begin{matrix}a_{11} & a_{12} & a_{13}\\a_{12} & a_{22} & a_{23}\\a_{13} & a_{23} & a_{33}\end{matrix}\right|
I4=∣a11a12a13a14a12a22a23a24a13a23a33a34a14a24a34a44∣I_{4} = \left|\begin{matrix}a_{11} & a_{12} & a_{13} & a_{14}\\a_{12} & a_{22} & a_{23} & a_{24}\\a_{13} & a_{23} & a_{33} & a_{34}\\a_{14} & a_{24} & a_{34} & a_{44}\end{matrix}\right|
I(λ)=∣a11−λa12a13a12a22−λa23a13a23a33−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}a_{11} - \lambda & a_{12} & a_{13}\\a_{12} & a_{22} - \lambda & a_{23}\\a_{13} & a_{23} & a_{33} - \lambda\end{matrix}\right|
     |a11  a14|   |a22  a24|   |a33  a34|
K2 = |        | + |        | + |        |
     |a14  a44|   |a24  a44|   |a34  a44|

     |a11  a12  a14|   |a22  a23  a24|   |a11  a13  a14|
     |             |   |             |   |             |
K3 = |a12  a22  a24| + |a23  a33  a34| + |a13  a33  a34|
     |             |   |             |   |             |
     |a14  a24  a44|   |a24  a34  a44|   |a14  a34  a44|

sustituimos coeficientes
I1=7I_{1} = 7
     |1   -1|   |3   -1|   |1   -1|
I2 = |      | + |      | + |      |
     |-1  3 |   |-1  3 |   |-1  3 |

I3=∣1−1−1−13−1−1−13∣I_{3} = \left|\begin{matrix}1 & -1 & -1\\-1 & 3 & -1\\-1 & -1 & 3\end{matrix}\right|
I4=∣1−1−10−13−10−1−1300000∣I_{4} = \left|\begin{matrix}1 & -1 & -1 & 0\\-1 & 3 & -1 & 0\\-1 & -1 & 3 & 0\\0 & 0 & 0 & 0\end{matrix}\right|
I(λ)=∣1−λ−1−1−13−λ−1−1−13−λ∣I{\left(\lambda \right)} = \left|\begin{matrix}1 - \lambda & -1 & -1\\-1 & 3 - \lambda & -1\\-1 & -1 & 3 - \lambda\end{matrix}\right|
     |1  0|   |3  0|   |3  0|
K2 = |    | + |    | + |    |
     |0  0|   |0  0|   |0  0|

     |1   -1  0|   |3   -1  0|   |1   -1  0|
     |         |   |         |   |         |
K3 = |-1  3   0| + |-1  3   0| + |-1  3   0|
     |         |   |         |   |         |
     |0   0   0|   |0   0   0|   |0   0   0|

I1=7I_{1} = 7
I2=12I_{2} = 12
I3=0I_{3} = 0
I4=0I_{4} = 0
I(λ)=−λ3+7λ2−12λI{\left(\lambda \right)} = - \lambda^{3} + 7 \lambda^{2} - 12 \lambda
K2=0K_{2} = 0
K3=0K_{3} = 0
Como
I3=0∧I4=0∧I2≠0I_{3} = 0 \wedge I_{4} = 0 \wedge I_{2} \neq 0
entonces por razón de tipos de rectas:
hay que
Formulamos la ecuación característica para nuestra superficie:
−I1λ2+I2λ−I3+λ3=0- I_{1} \lambda^{2} + I_{2} \lambda - I_{3} + \lambda^{3} = 0
o
λ3−7λ2+12λ=0\lambda^{3} - 7 \lambda^{2} + 12 \lambda = 0
λ1=4\lambda_{1} = 4
λ2=3\lambda_{2} = 3
λ3=0\lambda_{3} = 0
entonces la forma canónica de la ecuación será
(x~2λ1+y~2λ2)+K3I2=0\left(\tilde x^{2} \lambda_{1} + \tilde y^{2} \lambda_{2}\right) + \frac{K_{3}}{I_{2}} = 0
4x~2+3y~2=04 \tilde x^{2} + 3 \tilde y^{2} = 0
x~2(12)2+y~2(33)2=0\frac{\tilde x^{2}}{\left(\frac{1}{2}\right)^{2}} + \frac{\tilde y^{2}}{\left(\frac{\sqrt{3}}{3}\right)^{2}} = 0
es la ecuación para el tipo dos planos intersectantes imaginarios
- está reducida a la forma canónica
    © Sr Examen – Калькулятор