\[F(x,y,z)=0, \; G(x,y,z)=0\]
have a point \[(x_1,y_1,z_1)\]
in common. What are the conditions on the partial derivatives for the surcease to have a common tangent?The tangent plane to
\[F(x,y,z)=0\]
at the point \[(x_1,y_1,z_1)\]
is \[\frac{ \partial F}{\partial x}|_{(x_1,y_1,z_1)} (x-x_1)+ \frac{ \partial F}{\partial y}|_{(x_1,y_1,z_1)} (y-y_1)+ \frac{ \partial F}{\partial z}|_{(x_1,y_1,z_1)} (z-z_1)=0\]
.The tangent plane to
\[G(x,y,z)=0\]
at the point \[(x_1,y_1,z_1)\]
is \[\frac{ \partial G}{\partial x}|_{(x_1,y_1,z_1)} (x-x_1)+ \frac{ \partial G}{\partial y}|_{(x_1,y_1,z_1)} (y-y_1)+ \frac{ \partial G}{\partial z}|_{(x_1,y_1,z_1)} (z-z_1)=0\]
.For these two planes to be the same plane the partial differentials mus be a multiple of each other, so that
\[\frac{\partial F}{\partial x}|_{(x_1,y_1,z_1)}=k \frac{\partial G}{\partial x}|_{(x_1,y_1,z_1)}, \; \frac{\partial F}{\partial y}|_{(x_1,y_1,z_1)}=k \frac{\partial G}{\partial y}|_{(x_1,y_1,z_1)}, \; \frac{\partial F}{\partial z}|_{(x_1,y_1,z_1)}=k \frac{\partial G}{\partial z}|_{(x_1,y_1,z_1)}\]
If the tangent planes are orthogonal, then the dot product of the gradient vectors to the surface must be zero, so
\[\begin{equation} \begin{aligned} \frac{\partial F}{\partial x}|_{(x_1,y_1,z_1)} \frac{\partial g}{\partial x}|_{(x_1,y_1,z_1)} &+ \frac{\partial F}{\partial y}|_{(x_1,y_1,z_1)} \frac{\partial G}{\partial y}|_{(x_1,y_1,z_1)} \\ &+ \frac{\partial F}{\partial z}|_{(x_1,y_1,z_1)} \frac{\partial G}{\partial z}|_{(x_1,y_1,z_1)}\\ &= (\nabla F) \cdot (\nabla G)=0 \end{aligned} \end{equation} \]
.