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Given a vector $r=\langle r_1,r_2,r_3\rangle$ it is known that $n=\langle n_1,n_2, n_3\rangle$, a vector perpendicular to it, is constant. It transpires that the plane which contains the vector $r$ passes through the origin.

The proof given for this is that $r\cdot n=0 \Rightarrow r_1n_1+r_2n_2+r_3n_3=0$ and $x=y=z=0$ satisfies the equation.

Why does this prove that the plane passes through the origin?

Is $r\cdot n$ the plane equation?

Doesn't any plane that go through the origin by that logic?

This answer was given in the comments to the answer of this question: Prove that a particle is traveling on a plane from its velocity and acceleration in space

This is the visualization: enter image description here

Here the plane is drawn to pass through origin but I could've drawn it to pass through any other point for example.

$\mathbf{EDIT:}$ at the end the explanation is quite simple: any vector is defined to start at the origin. Because the position vector travels on a plane and its normal vector is constant then, the motion has to start at the origin. If someone wants to post an answer I will gladly accept it.

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    $\begingroup$ $r\cdot n=0$ is the equation of a plane through the origin. $r\cdot n=1$ is not. $\endgroup$ Apr 28, 2017 at 20:46
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    $\begingroup$ Every line or vector perpendicular to the plane's normal vector is parallel to the plane. If the plane does not pass through the origin, the vector from the origin to a point in the plane meets the plane at some non-zero angle. The position vector is not parallel to the plane, so it cannot be perpendicular to the plane's normal vector. $\endgroup$
    – David K
    Apr 28, 2017 at 21:00
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    $\begingroup$ Conversely, if you know that there is some point in the plane whose position vector is perpendicular to the normal vector of the plane, the origin cannot be off the plane, so it must be on the plane. $\endgroup$
    – David K
    Apr 28, 2017 at 21:03
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    $\begingroup$ Your last comment is simply false. For example: the plane $z=2,$ normal vector $n=\langle 0,0,1\rangle$. One point in the plane is $(x,y,z)=(0,0,2),$ and its position vector is $r=\langle 0,0,2\rangle$ which is not perpendicular to $n=\langle 0,0,1\rangle$ (in fact it is parallel). $\endgroup$
    – David K
    Apr 28, 2017 at 21:12
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    $\begingroup$ When you say the plane containing the vector $r$, it seems that you're thinking of a “plane of vectors”, i.e., a two-dimensional subspace of the vector space $\mathbf{R}^3$, and subspaces always contain the zero vector. But if you think of a “plane of points”, then of course you can translate it in space and still keep $n$ as a vector normal to the plane and $r$ as a vector parallel to the plane. (But in that case you really shouldn't say that $r$ is contained in the plane.) $\endgroup$ Apr 29, 2017 at 8:21

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The comments give one way of understanding, but I figured I could maybe explain the original proof better.

Suppose our plane has normal vector $n$ (which is normal to the plane at every point on it) and an example point on the plane: $p$. These two vectors, one a direction and the other a point, determine any plane.

Then, our plane equation is $$ n\cdot(x-p)=0 $$

that is, every $x$ that satisfies this equation is a member of the plane. This is equivalent to $n\cdot x = n\cdot p$. Notice that $x=p$ is a member of the plane. But, $x=p+n$ is not a member, nor is $x=\vec{0}$, assuming $p\ne \vec{0}$.

Another way to write this is as a point set $\Pi$: $$ \Pi = \{ \;x\in\mathbb{R}^3\;|\;n\cdot x= n\cdot p\;\} $$ i.e. these are the set of points making up the plane. Again, notice that if $p=\vec{0}$, then the origin is a point in $\Pi$. But if $p$ does not vanish, then the origin is not in the plane.

The reason is that $p$ is a translation or shifting parameter. That is, a plane has an orientation parameter, $n$, which "rotates" it, and a position parameter $p$, which slides the plane around. When $p=\vec{0}$, we have slid the plane so that it intersects that origin. The plane equation in this case is $n\cdot x = 0$.


Here's a different approach. Every plane is determined by giving 3 unique points. Let's take $p,a,b$. Define $T_1=a-p$ and $T_2=b-p$. We can suppose $T_1$ and $T_2$ are orthogonal; if they are not, we can use Gram-Schmidt orthonormalization.

Now, suppose we walk around between $p$ to $a$ or $b$. This is the same as adding some multiple of $T_1$ or $T_2$ to $p$. So a parametric equation for the plane is $$ x(s,t) = p + sT_1 + tT_2 $$ so that if you input any $s$ and $t$, your output is a point on the plane. See also here. Notice that a normal vector is simply $n=(T_1\times T_2)/||T_1\times T_2||_2$. So we get equivalence to $n\cdot(x-p)=0$ as before.

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  • $\begingroup$ Thank you for a great explanation! $\endgroup$
    – Yos
    May 11, 2017 at 8:12

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