Trouble understanding the tangent plane of a surface

Question

$$z = z_0 +a(x-x_o)+b(y-y_0)$$

where $a$ and $b$ are the partial derivates with respect to $x$ and $y$. I dont understad why $a$ and $b$ are there since they are tangent to the plane not normal?

Answer 1

$ a $ and $ b $ are numbers, not vectors. If you rewrite your equation as $$ a(x-x_0)+b(y-y_0)+(-1)(z-z_0) = 0, $$ you can see that the vector $ (a, b, -1) $ is perpendicular (normal) to your plane (which presumably meets the surface at the point $ (x_0, y_0, z_0) $). On the other hand, the lines $$ \{ (x_0, y_0, z_0) + s \cdot (1, 0, a) \mid s \in \mathbb R \} = \{ (x_0 + s, y_0, z_0 + as) \mid s \in \mathbb R \} $$ and $$ \{ (x_0, y_0, z_0) + t \cdot (0, 1, b) \mid t \in \mathbb R \} = \{ (x_0, y_0 + t, z_0 + bt) \mid t \in \mathbb R \} $$ are tangent to the surface (and therefore contained in the tangent plane). That is the geometric meaning of partial derivatives.