I'm studying a chapter on vector analysis and I don't really understand why the following equations are switched for the two situations. (My course is not in English, so I apologize if I translated something wrong.)
Equations for the tangent plane and normal in a point on a surface
- The equation for a tangent plane in $p(x_0, y_0, z_0)$ on a surface $\varphi(x,y,z) = 0$:
$$\left(\frac{\partial \varphi}{\partial x}\right)_p (x-x_0) + \left(\frac{\partial \varphi}{\partial y}\right)_p (y-y_0) + \left(\frac{\partial \varphi}{\partial z}\right)_p (z-z_0) = 0$$
- The equation for the normal in $p(x_0, y_0, z_0)$ on a surface $\varphi(x,y,z) = 0$:
$$\frac{x-x_0}{\left(\frac{\partial \varphi}{\partial y}\right)_p} = \frac{y-y_0}{\left(\frac{\partial \varphi}{\partial y}\right)_p} = \frac{z-z_0}{\left(\frac{\partial \varphi}{\partial z}\right)_p} $$
Equations for the tangent and normal plane in a point on a curve
For $p(t=t_0) = \{f_1(t_0), f_2(t_0), f_3(t_0)\}$ with $t \in R$
- The equation for the tangent in $p$:
$$\frac{x-x_0}{f'_1(t_0)} = \frac{y-y_0}{f'_2(t_0)} = \frac{z-z_0}{f'_3(t_0)} $$
- The equation for the normal plane in $p$:
$$f'_1(t_0) (x-x_0) + f'_2(t_0)(y-y_0) + f'_3(t_0)(z-z_0) = 0$$
I tried to find a graphical explanation for this to understand how the equations were derived, but without much success.