vector normal to a plane There is this "shortcut" we learned that helps us find a vector perpendicular to a plane. Say, $ax+by+cz+d=0$ is the plane equation, then the vector $(a,b,c)$ is normal to this plane.
But why is this? Why does $d$ contribute nothing to the normal vector?
For example, let $(x,y,z)$ be a point on this plane, then it must satisfy the plane equation $ax+by+cz+d=0$, but when you take the dot product with the vector $(a,b,c)$, we get $(x,y,z)\cdot(a,b,c)=ax+by+cz=-d$, which isn't necessarily zero?

Let $M(x,y,z)$ and $A(x_1,y_1,z_1)$ placed in the plain.
Thus, $(a,b,c)\perp AM$ it's $\vec{(a,b,c)}\vec{(x-x_1,y-y_1,z-z_1)}=0$, which is
$$a(x-x_1)+b(y-y_1)+c(z-z_1)=0.$$
Now, let $-ax_1-by_1-cz_1=d$.
Thus, we got an equation of the plain:
$$ax+by+cz+d=0,$$
where $(a,b,c)$ is a normal to the plan. 
