Answer to Question #104582 in Analytic Geometry for Deepak

Question

Does there exist a plane targent to x
2 −2y
2 +2z
2 = 8 and which passes through
2x+3y+2z = 8, x−y+2z = 5? Justify your answer.

Answer 1

$x^2-2y^2+2z^2=8$

The plane which passes through 

$2x+3y+2z=8\\ x-y+2z=5$

Point

$y=0\\ \left\{\begin{matrix} 2x+2z=8\\ x+2z=5 \end{matrix}\right.\\ x=3\\ z=1\\ A(3,0,1)$

directional vector line

$\vec{a}=\vec{n_1} \times\vec{n_2}\\ \vec{n_1}=(2,3,2)\\ \vec{n_2}=(1,-1,2)\\ \vec{a}=(\begin{vmatrix} 3 & 2 \\ -1 & 2 \end{vmatrix}; -\begin{vmatrix} 2 & 2 \\ 1 & 2 \end{vmatrix};\begin{vmatrix} 2& 3 \\ 1 & -1 \end{vmatrix})=\\ =(8;-2;-5)$

Take any point on

$x^2-2y^2+2z^2=8\\ M(x_0;y_0,z_0)\\ x_0^2-2y_0^2+2z_0^2=8\\ x_0^2=2y_0^2-2z_0^2+8$

The equation of the plane passing through the points $M, A$ and the vector $\vec{a}$

$\begin{vmatrix} x-3 & y-0 &z-1\\ x_0-3 & y_0-0&z_0-1\\ 8&-2&-5 \end{vmatrix}=0\\ (x-3)\begin{vmatrix} y_0 & z_0-1\\ -2 & -5 \end{vmatrix}-y\begin{vmatrix} x_0-3 & z_0-1 \\ 8 & -5 \end{vmatrix}+\\+(z-1)\begin{vmatrix} x_0-3 & y_0 \\ 8 & -2 \end{vmatrix}=0\\ (x-3)(-5y_0+2z_0-2)-y(-5x_0-8z_0+23)+\\ +(z-1)(-2x_0-8y_0+6)=0$

Answer: there plane a exist and have equation

$(x-3)(-5y_0+2z_0-2)-y(-5x_0-8z_0+23)+\\ +(z-1)(-2x_0-8y_0+6)=0$

where $x_0^2=2y_0^2-2z_0^2+8$