Question

помогите пожалуйста решить

Answer 1

Ответ:

$x(y+z)(xy-z)+8=0\ \ ,\ \ \ M_0(2;1;3)\\\\(xy+xz)(xy-z)+8=0\\\\x^2y^2-xyz+x^2yz-xz^2+8=0\\\\F(x,y,z)=x^2y^2-xyz+x^2yz-xz^2+8\\\\F'_{x}=2xy^2-yz+2xyz-z^2\ \ ,\ \ F'_{x}(M_0)=4-3+12-9=4\\\\F'_{y}=2x^2y-xz+x^2z\ \ ,\ \ \ F'_{y}(M_0)=8-6+12=14\\\\F'_{z}=-xy+x^2y-2xz\ \ ,\ \ \ F'_z}(M_0)=-2+4-12=-10\\\\\\a)\ \ F'_{x}(M_0)\cdot (x-x_0)+F'_{y}(M_0)\cdot (y-y_0)+F'_{z}(M_0)\cdot (z-z_0)=0\ \ \ kasatel\flat naya\ ploskost\\\\4(x-2)+14(y-1)-10(z-3)=0\ \Big|:2\\\\2(x-2)+7(y-1)-5(z-3)=0\\\\\boxed{\ 2x+7y-5z+4=0\ }$

$b)\ \ \dfrac{x-x_0}{F'_{x}(M_0)}=\dfrac{y-y_0}{F'_{y}(M_0)}=\dfrac{z-z_0}{F'_{z}(M_0)}\ \ \ normal\flat\\\\\\{}\ \ \ \ \ \dfrac{x-2}{4}=\dfrac{y-1}{14}=\dfrac{z-3}{-10}\ \ \Rightarrow \ \ \ \boxed{\ \dfrac{x-2}{2}=\dfrac{y-1}{7}=\dfrac{z-3}{-5}\ }$