finding the vector components of a normal line

Question

given the equation of a plane: ax + by = c.

I have seen several places where it is being said that the normal vector to the plane has the components <a, b>. However, i am yet to find any satisfying proof for this.

Answer 1

hint: $ax + by + 0z = c\implies \vec{n} = <a,b,0> = <a,b>$ as viewed in the $2$ dimensional case.

Answer 2

Take any two points $M_0\,(x_0,y_0,z_0)$ and $M_1\,(x_1,y_1,z_1)$ in the place. From the equations satisfied by these points: $$ax_0+by_0=c=ax_1+by_1,$$ we deduce $$a(x_1-x_0)+b(y_1-y_0)=a(x_1-x_0)+b(y_1-y_0)+0(z_1-z_0)=0,$$ which means the vectors $\vec n\,(a,b,0)$ and $\overrightarrow{M_0M_1}$ are orthogonal.

Answer 3

Let $P(x_0,y_0,z_0)$ be a point of the plane.

the equation of the plane can be written as

$$a(x-x_0)+b(y-y_0)+c(z-z_0)=0$$

on the other hand

let $\vec{n}=(A,B,C)$ be the normal to the plane .

the plane can also be defined as the set of point $M(x,y,z)$ such that

$$\vec{PM}.\vec{n}=0$$ or

$$A(x-x_0)+B(y-y_0)+C(z-z_0)=0$$ thus

we can take

$$\vec{n}=(a,b,c).$$