Why no slope for a plane?

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Similar to how lines have slopes defined in terms of $\Delta y$ and $\Delta x$, why can't planes have slopes defined in therms of $\Delta x$, $\Delta y$, and $\Delta z$?

Couldn't these could be represented in terms of a vector: $$\left[ \begin{array}{c} \Delta x \\ \Delta y \\ \Delta z \end{array} \right]$$

Why is instead of a slope of a plane is its normal vector used instead? Couldn't an equation of a plane be derived from the vector of $\Delta x$, and $\Delta y$ and $\Delta z$ instead of from the normal vector?

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A point and a normal vector is a convenient parameterization for a plane because it is coordinate-independent. Sometimes planes are parameterized by a point and a gradient ("slope") instead, for instance you can represent the plane

$$z = ax+by+c$$

with the point $(1,1,a+b+c)$ and gradient $\nabla_{x,y}z = (a,b)$. But notice that this requires choosing a preferred direction -- in this case, the $z$ direction. A normal vector does not require you to make such a choice.

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This is a great question... however all will be answered in Integral Calculus. :P A simple explanation is that you are now in 3D space. A plane is flat, so you use instead a "normal vector" which is perpendicular to the plane for these types of calculations.

Look up how gradients and planes are used to calculate slope, it'll explain everything.