0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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, $$(x,y,z)\in \Pi\Leftrightarrow ax+by+cz=d.$$

$\endgroup$
0
$\begingroup$

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 $\vec b$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.