What is the normal vector to the plane ax+by+cz=d?

Question

If d=0, then we obviously see that the equation is the Euclidean inner product with (a,b,c) and (x,y,z) that equals zero - and so (a,b,c) is the normal vector to the plane.

What if $d \ne 0$?

Then I don't have the same intuitive way of finding the normal - the "inner product" wouldn't be 0, and we couldn't conclude the orthogonality of the vectors.

Thanks,

Answer 1

If $(x_0,y_0z_0)$ and $(x_1,y_1,z_1)$ are any two points in the plane $ax+by+cz=d$, we have $ax_0+by_0+cz_0=ax_1+by_1+cz_1$, whence $$a(x_1-x_0)+b(y_1-y_0)+c(z_1-z_0)=0.$$

Answer 2

Since the gradient $\nabla f(x,y,z)$ is perpendicular to the level sets of $f(x,y,z)$, we can use the gradient $\nabla(ax+by+cz-d)=(a,b,c)$ which is perpendicular to the plane.