Why must the plane go through the origin?

Question

In The Riddler's answer Motion in a plane , why must the plane go through the origin? Does the dot product equal 0 not just mean that there are two vectors $\vec r$ and $\vec c$ that are perpendicular?

Answer 1

$\vec c$ is a constant vector in the context of the answer, and The Riddler is pointing out that the set of all vectors $\vec r$ for which $\vec r\cdot\vec c=0$ forms a plane passing through the origin (and the plane is perpendicular to $\vec c$ ; that is, $\vec c$ is a normal vector to the plane). So if the particle’s position vector $\vec r$ always satisfies $\vec r\cdot\vec c=0$, the particle is confined to some plane through the origin.

Answer 2

Your question suggests to me that you don't have a good intuition for the definition. I think the following result may help you.

The following three formulations(there are more) of a plane $\Pi\subset \mathbb R^3$ are equivalent. I think of [1] as the more "obvious" definition, and [2],[3] as more geometric definitions.

[1] Basepoint plus Linear Span of two basis vectors

We say $\Pi$ is a plane passing through $p_0\in\mathbb R^3$ with tangent vectors $t_1,t_2\in\mathbb R^3$ if there exists 3 distinct points $\vec p_0,\vec p_1,\vec p_2\in\Pi$ that are not co-linear, such that if we define $\vec t_1 = \vec p_1-\vec p_0, \vec t_2 = \vec p_2 - \vec p_0$, then $$ \Pi =\{ \vec p_0 + \vec t_1\lambda + \vec t_2\mu : \lambda,\mu\in\mathbb R\}. $$

[2] Set of solutions of a linear equation

If there exists $a,b,c,d\in\mathbb R$ such that $$ \Pi = \{ (x,y,z) \in \mathbb R^3 : ax + by + cz = d \},$$ then we say that $\Pi$ is a plane defined as the solutions of $ax + by + cz = d$.

[3] Dot product with normal is a constant

We say $\Pi$ is a plane with normal $\vec n\in\mathbb R^3$ passing through $r_0\in\mathbb R^3$ if there exists a vector $\vec n\in\mathbb R^3\setminus \{0\}$ such that $$ \Pi = \{ \vec r \in \mathbb R^3 : \vec r \cdot \vec n = r_0 \cdot \vec n \}.$$

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Proof. [2] and [3] are obviously equivalent, as the LHS of [2] is the dot product of $(x,y,z)$ and $(a,b,c)$. Note that the $r_0$ in [3] can be an arbitrary point in $\Pi$. One obtains a normal from [1] by defining $$ n:= t_1\times t_2,$$ which leads to definitions [2],[3], and one obtains tangent vectors from [2] or [3] by finding 3 points in $\Pi$ that are not co-linear, leading to definition [1], QED.

Intuition for definition [3] from definition [1]

Consider the plane which has the $z$-coordinate fixed to have value 7. This is equivalent to saying that for any point $\vec r=(x,y,z)$ in the plane, $(x,y,z)\cdot (0,0,1) = 7$. In any of the other orthogonal directions i.e. $x,y$ you can move however you wish, and you will still remain in the plane.

For a more general plane, you start with one vector $\vec n\in\mathbb R^3$. Extend this to an orthogonal basis of $\mathbb R^3$ $\{\vec n, t_1, t_2\}$. Then if the plane is e.g. $$ \Pi : \vec r\cdot \vec n = 7,$$ this means every point in the plane is 7 "units" of $\vec n$ away from the origin, but you can move any amount you like in directions $\vec t_1, \vec t_2$, and still remain in the plane.

Hopefully it is now clear:

When does the plane pass through 0?

1. In [1], the plane passes through 0 if $\vec p_0 = \vec 0$ is an allowable choice in the definition of $\Pi$ (or if $\vec p_0$ is a linear combination of $t_1,t_2$).
2. In [2], the plane passes through 0 if $d=0$, since then $(x,y,z) = (0,0,0)$ is a solution.
3. In [3], the plane passes through 0 if $r_0=0$ is an allowable choice in the definition of $\Pi$.