Problem: How can I show that the tangent plane to the hyperboloid $x^2+y^2-z^2=-1$ at point $(2,2,3)$ has equation $2(x-2)+2(y-2)-3(z-3)=0$
Attempt: Is the idea here is to slice the hyperboloid say at $x=2$ and $y=2$ to find to vectors say $v$ and $w$ and then taking there crossproduct gives the desired equation. But i'm getting nowhere. Is my approach correct? Any idea and hints would be much much appreciated!
Take the hyperboloid of equation (you should have defined the precise equation in your question) $$\Gamma\equiv x^2+y^2-z^2=-1.$$ Obviously $(2,2,3)\in\Gamma$. A plane in $\mathbb R^3$ is defined by a point and a orthogonal vector.
Since you want the plane tangent to the surface of the hyperboloid we'll take as orthogonal vector the gradient of the function $f(x,y)=\sqrt{x^2+y^2+1}$ evaluated in $(2,2,3)$, which is given by $$\nabla f(x,y)\vert_{(2,2)}=\Bigg(\dfrac{x}{\sqrt{x^2+y^2+1}},\dfrac{y}{\sqrt{x^2+y^2+1}}\Bigg)\Bigg\vert_{(2,2)}=\bigg(\frac{2}{3},\frac{2}{3}\bigg)$$ so the plane will have equation $$z-f(2,3)=\Big\langle\big(x-2,y-2\big),\big((\nabla f\vert_{(2,2)})_x,(\nabla f\vert_{(2,2)})_y\big) \Big\rangle$$ $$\pi\equiv \frac{2}{3}(x-2)+\frac{2}{3}(y-2)-(z-3)=0$$ $$\fbox{$2(x-2)+2(y-2)-3(z-3)=0$}$$