Why I can find a vector normal to the plane using it's arguments

Question

I know I can find a vector normal to the plane ax + by +cz = 0 which is (a,b,c) but I don't know why.

I would be pleased if you could explain the method used to find vectors normal to the plane.

Thank You!

Answer 1

Suppose you know $(x_0,y_0,z_0)$ is a point on the plane, and $\left<a,b,c\right>$ is a vector normal to the plane. Then if $(x,y,z)$ is any other point on the plane, then the two vectors $\left<a,b,c\right>$ and $\left<x-x_0,y-y_0,z-z_0\right>$ are perpendicular -- this is precisely what "normal vector" means. That, in turn, is equivalent to $$\left<a,b,c\right> \cdot \left<x-x_0,y-y_0,z-z_0\right> = 0$$ which is the same as $$a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$$

This can be rearranged in the form $$ax + by + cz = d$$ where $d$ is a constant that can be expressed in terms of the other parameters ($a,b,c,x_0,y_0,z_0$).

Now this whole procedure can be run in reverse: if you are given an equation of the form $$ax+by+cz=d$$ then if you find any point $(x_0,y_0,z_0)$ that satisfies the equation, you would have $$a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$$ which means that if $(x,y,z)$ is any other point satisfying the condition above, then the two vectors $\left<a,b,c\right>$ and $\left<x-x_0,y-y_0,z-z_0\right>$ are perpendicular to each other, which means that $\left<a,b,c\right>$ is a normal vector.

Answer 2

The user @symplectomorphic tell you why. Your equation is one which passes through the origin, so we may take $p = (0,0,0)$ to be a point on our plane. If $(a',b',c')$ is a normal vector to your plane then $(a',b',c') \cdot ((x,y,z) - p) = 0$. And now you see that after simplifying, it is the coefficients of the remaining variables $x,y,z$ which automatically give the coordinates of the normal vector.