Plane with normal vector I have been struggling to understand all basic concepts of the plane from this websites. In distance from origin section of this websites, I read
if the unit normal vector $(a_1, b_1, c_1),$ then, the point $P_1$ on the plane becomes $(Da_1, Db_1, Dc_1),$ where D is the distance from the origin.See the below  image:

My questions are :
1.If unit normal vector is $(a_1, b_1, c_1),$ then, how the point $P_1$ on the plane becomes $(Da_1, Db_1, Dc_1) ?$
2.If unit normal is $(1/3,2/3,2/3)$ then $P_1$ becomes $(2/3,4/3,4/3)$
Where $D=2.$ We know that normal vector began on the plane at point $P_1$ and ends at $(1/3,2/3,2/3).$ My questions is how unit normal vector coordinates value less than $P_1$ coordinates value, because unit normal vector pointing outside of the plane it should be greater coordinates value than $P_1?$
 A: The coordinates of a point $P$ are the coordinates of the vector $\vec{OP}$, with $O$ the origin. The coordinates of a vector $\vec{AB}$ are the coordinates of its end point $B$ minus the coordinates of its starting point $A$:
$$\vec{AB} = \vec{OB} - \vec{OA}. $$

*

*We are told that $P_1$ is at distance $D$ of the origin $O$ and its direction is given by the vector $\vec{n}_1$. In other words, $\Vert\vec{OP_1}\Vert = D$ and $\vec{OP_1} = \alpha\vec{n}_1$ for some positive real number $\alpha$. Since $\vec{n}_1$ is a unit vector, we have
$$D = \Vert\vec{OP_1}\Vert = \alpha\Vert\vec{n}_1\Vert = \alpha,$$
which implies $\vec{OP_1} = D\vec{n}_1$. Thus the coordinates of $P_1$ are $(Da_1, Db_1, Dc_1)$ with $(a_1, b_1, c_1)$ the coordinates of $\vec{n_1}$.


*If $\vec{n}_1$ starts at $P_1$ and ends at a point $Q_1$, i.e. $\vec{n}_1 = \vec{P_1Q_1}$, then the coordinates of $Q_1$ are the sum of the coordinates of $P_1$ and $\vec{n}_1$, which gives $((D+1)a_1, (D+1)b_1, (D+1)c_1)$. If $\vec{n}_1$ is $(1/3, 2/3, 2/3)$ and $D=2$ then $P_1$ has coordinates $(2/3, 4/3, 4/3)$ and $\vec{n}_1$ ends at $Q_1$ with coordinates $(1, 2, 2)$.
