Re: središče krogle
Objavljeno: 8.12.2017 16:36
No prav.
Da ne ponavljam osnovnih enačb. Od prve enačbe sem odštel 2. in 3. in dobil:
x\(_0\)(x\(_2\) -x\(_1\)) + y\(_0\)(y\(_2\) -y\(_1\)) + z\(_0\)(z\(_2\) - z\(_1\)) + \(\frac{1}{2}\)(\(x_2^2\) - \(x_1^2\) + \(y_2^2\) - \(y_1^2\) + \(z_2^2\) - \(z_1^2\)) = 0
x\(_0\)(x\(_3\) -x\(_1\)) + y\(_0\)(y\(_3\) -y\(_1\)) + z\(_0\)(z\(_3\) - z\(_1\)) + \(\frac{1}{2}\)(\(x_3^2\) - \(x_1^2\) + \(y_3^2\) - \(y_1^2\) + \(z_3^2\) - \(z_1^2\)) = 0
Enačbi predstavljata dve ravnini:
A\(_1\)x\(_0\) + B\(_1\)y\(_0\) + C\(_1\)z\(_0\) + D\(_1\) = 0
A\(_2\)x\(_0\) + B\(_2\)y\(_0\) + C\(_2\)z\(_0\) + D\(_2\) = 0
Vektorja normal na ti ravnini sta: N\(_1\)(A\(_1\), B\(_1\), C\(_1\)) in N\(_2\)(A\(_2\), B\(_2\), C\(_2\))
Vektorski produkt obeh normal N\(_1\) x N\(_2 \) mi da vektor premice p: (a, b, c). =(-50, -175, 375)
Sedaj na tej premici določim neko točko pri z = 0 in izračunam x in y:
[A\(_1\) B\(_1\) ; A\(_2\) B\(_2\)]\(^{-1}\) x [ D\(_1\) ; D\(_2\) ] = [x ; y ]
x = 19,833, y = 24,66, z = 0
Vrednosti vstavim v parametrske oblike enačbe premice v katerih izrazim x\(_0\), y\(_0\) in z\(_0\)
x\(_0\) = x - at; y\(_0\) = y - at; z\(_0\) = - at
S temi izrazi potem v enačbi krogle nadomestim x\(_0\), y\(_0\) in z\(_0\) :
(x\(_1\) - (x-at))\(^2\) + (y\(_1\)-(y-bt))\(^2\) + (z\(_1\)+at)\(^2\) - \(R^2\) = 0
Iz tega izračunam t:
t\(_{1/2}\) = \(\frac{-(a(x_1-x)+b(y_1-y)+cz_1)\pm \sqrt{(a(x_1-x)+b(y_1-y)+cz_1)^2-(a^2+b^2+c^2)(x_1^2+y_1^2+z_1^2-R^2)}}{(a^2+b^2+c^2)}\)
Dobim: t\(_1\) = -6983,3 in \( t_2\) = -6983,4
Sem pa računal v excelu.
Da ne ponavljam osnovnih enačb. Od prve enačbe sem odštel 2. in 3. in dobil:
x\(_0\)(x\(_2\) -x\(_1\)) + y\(_0\)(y\(_2\) -y\(_1\)) + z\(_0\)(z\(_2\) - z\(_1\)) + \(\frac{1}{2}\)(\(x_2^2\) - \(x_1^2\) + \(y_2^2\) - \(y_1^2\) + \(z_2^2\) - \(z_1^2\)) = 0
x\(_0\)(x\(_3\) -x\(_1\)) + y\(_0\)(y\(_3\) -y\(_1\)) + z\(_0\)(z\(_3\) - z\(_1\)) + \(\frac{1}{2}\)(\(x_3^2\) - \(x_1^2\) + \(y_3^2\) - \(y_1^2\) + \(z_3^2\) - \(z_1^2\)) = 0
Enačbi predstavljata dve ravnini:
A\(_1\)x\(_0\) + B\(_1\)y\(_0\) + C\(_1\)z\(_0\) + D\(_1\) = 0
A\(_2\)x\(_0\) + B\(_2\)y\(_0\) + C\(_2\)z\(_0\) + D\(_2\) = 0
Vektorja normal na ti ravnini sta: N\(_1\)(A\(_1\), B\(_1\), C\(_1\)) in N\(_2\)(A\(_2\), B\(_2\), C\(_2\))
Vektorski produkt obeh normal N\(_1\) x N\(_2 \) mi da vektor premice p: (a, b, c). =(-50, -175, 375)
Sedaj na tej premici določim neko točko pri z = 0 in izračunam x in y:
[A\(_1\) B\(_1\) ; A\(_2\) B\(_2\)]\(^{-1}\) x [ D\(_1\) ; D\(_2\) ] = [x ; y ]
x = 19,833, y = 24,66, z = 0
Vrednosti vstavim v parametrske oblike enačbe premice v katerih izrazim x\(_0\), y\(_0\) in z\(_0\)
x\(_0\) = x - at; y\(_0\) = y - at; z\(_0\) = - at
S temi izrazi potem v enačbi krogle nadomestim x\(_0\), y\(_0\) in z\(_0\) :
(x\(_1\) - (x-at))\(^2\) + (y\(_1\)-(y-bt))\(^2\) + (z\(_1\)+at)\(^2\) - \(R^2\) = 0
Iz tega izračunam t:
t\(_{1/2}\) = \(\frac{-(a(x_1-x)+b(y_1-y)+cz_1)\pm \sqrt{(a(x_1-x)+b(y_1-y)+cz_1)^2-(a^2+b^2+c^2)(x_1^2+y_1^2+z_1^2-R^2)}}{(a^2+b^2+c^2)}\)
Dobim: t\(_1\) = -6983,3 in \( t_2\) = -6983,4
Sem pa računal v excelu.