# How is the “normal equation” defined in a plane in $\mathbb{R}^3$?

A plane in $\mathbb{R}^3$ and above is defined by one point $P$ contained in the plane and two direction vectors $\vec{a}$ and $\vec{b}$ that are not parallel.

A point $X$ in this plane is one such that $$X = P + s\cdot\vec{a}+t\cdot\vec{b}$$

For a certain $s,t \in \mathbb{R}$. This is called vectorial equation.

In $\mathbb{R}^3$ there is another way to define planes. With one point $P$ in the plane and only one direction vector $\vec{m}$, rather than two. This vector is perpendicular to the plane.

I'm guessing that the "vectorial equation" I described above does not apply in this scenario, as it requires two direction vectors and I only have one here.

I heard that in this case there is a "normal equation", however I am unsure how is it defined at all.

There is a definition in my book, but I don't quite grasp it:

All the points $(x,y,z)$ in a plane that contains $P = (p_1,p_2,p_3)$ and is perpendicular to the vector $\vec{n}=(a,b,c)$, are the only ones that satisfy:

$$ax+by+cz = d$$

Where $d = ap_1+bp_2+cp_3$.

Can you better explain this? I easily get the vectorial equation, but the normal equation is still fairly confusing to me.

Consider your left shoulder socket as the point $P$, and suppose that $\vec{n}$ is the vector that goes directly up as if you are raising your left arm to vertical. Now imagine a Saturn Ball Toy around your head... because $\vec{n}$ is normal, the saturn rings, extended out, form your plane.

From this position, consider a point $Q(x,y,z)$ in space. Point at it with your right arm forming the vector $\vec{PQ}$ (don't mind the reality and imagine that your right arm is pivoting around your left shoulder socket). Now from your viewpoint $Q(x,y,z)$ can either lie 'above' or 'below' the plane.

If the acute angle between $\vec{n}$ and $PQ$ is less than 90$^\circ$ then $Q$ lies 'above' the plane. the acute angle between $\vec{n}$ and $\vec{PQ}$ is greater than 90$^\circ$ then $Q$ lies 'below' the plane. It is only if $\vec{PQ}$ is perpendicular to $\vec{n}$ will $Q$ lie on your saturn-ring plane defined by $\vec{n}$ and $P$.

Now when are vectors perpendicular? So we want $\vec{n}\cdot \vec{PQ}\overset{!}{=}0$ for $Q$ to be on the plane:

\begin{align} \vec{n}\cdot\vec{PQ}&=0 \\\Rightarrow \vec{n}\cdot(\vec{Q}-\vec{P})&=0 \\\Rightarrow (a,b,c)\cdot(x-p_1,y-p_2,z-p_3)&=0 \\ \Rightarrow ax-ap_1+by-bp_2+cz-cp_3&=0 \\ \Rightarrow ax+by+cz&=ap_1+bp_2+cp_3=:d, \end{align}

and so we end up, with $\Pi$ your plane and $d$ as defined above, $$(x,y,z)\in \Pi\Leftrightarrow ax+by+cz=d.$$

The "vectorial" equation is of the parametric type. It means that by varying the two independent parameters, $s$ and $t$, you span the whole plane and get a double infinity of $X$ points.

The "normal" equation is of the implicit type. It means that for the whole space, only the points belonging to the plane do fulfill the given equation $\vec n.\vec X=d=\vec n.\vec P$, for some $P$ (belonging to the plane, as $X=P$ verifies the equation).

By the way, you can check the compatibility of the two representations by plugging $X$ from the vectorial equation into the normal equation:

$$\vec n.(\vec P+s.\vec a+t.\vec b)=\vec n.\vec P,$$ or $$s.\vec n.\vec a+t.\vec n.\vec b=0,$$ implying that both $\vec n.\vec a$ and $\vec n.\vec b$ are zero. $\vec n$ is perpendicular to $\vec a$ and $\vec b$.

The same situation arises for lines in 2D. On the one hand we can write a vector equation for a line $r=\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}+t\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b}$, which is an explicit description of all points on the line in terms of an initial point a and the direction b. We can also write down an implicit equation for the line y = mx+c where if we know x then we can calculate y. We can move between the two by noting that $r=\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{a}+t\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{b}$ means the x and y coordinates on the line are $x = a_1+tb_1$ and $y = a_2+tb_2$. Eliminating t between these two expressions will give the implicit form. Going back the other way we can take x = t and any point on the line is $r = (t, mt+c) = (0,c)+t(1,m)$.
The two forms of the equation of a plane are similar, an explicit vector form and an implicit scalar form. Starting with the vector form we can write x, y and z in terms of s and t. eliminating s and t between these equations gives the normal form. going back the other way put x = s, y = t then we can find a formula for z. The resulting triple (x, y, z) is our vector form.
Geometrically (as opposed to algebraically above), the normal formis simply an expression of the fact that a vector lying parallel to the plane is perpendicular to one normal to it i.e. their dot product is zero. If r = (x, y, z) is a general point on the plane and B = (a, b, c) a specific point on it then A-B = (x-a, y-b, z-c) is a vector parallel to the plane. If N is a vector normal to the plane then the dot product of N with A-B must be zero. This is your nornal form of the plane.