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?
 A: A normal vector $\textbf{N}$ to a plane $P$ is a vector such that for all $\textbf{v} \in P$, $\textbf{v} \perp \textbf{N} \Rightarrow \textbf{N} \cdot \textbf{v} = 0$. Given any point $p_0, \textbf{x} = (x,y,z) \in P$, we defined $\textbf{x} - p_0$ to be the vector which extends from $p_0$ to $\textbf{x}$. Hence, $\textbf{x} - p_0 \in P \Rightarrow \textbf{N} \cdot (\textbf{x} - p_0) = 0. $ Letting $\textbf{N} = (a,b,c)$, we see that any plane in $\mathbb{R}^3$ is given by the following equation: $ (a,b,c) \cdot (x-x_0,y-y_0,z-z_0) = 0  \iff ax + by + cz = D$, where we've set $D = ax_0 + by_0 + cz_0$ i.e $-D = d$. 
A: 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. 
A: $(x, y, z)$ is just some arbitrary vector in $\mathbb{R}^3$. Would you say that $(a, b, c)$ is orthogonal to all vectors in $\mathbb{R}^3$?
