Why is the vector equation of a plane defined as n • (r-r0) = 0?

[That Neuron]

Just curious about a certain facet of the vector description of a plane. My query is as to why it is defined as n • (r-r₀) = 0. Great, that's because any two vectors with a dot product of 0 must be orthogonal to each other and if we have a point on a infinite plane with an associated vector we can define the plane perfectly. BUT here is my problem (Which will be resolved after a few minutes on here, I'm sure) If we can define the vector (r-r₀) as a single vector, say a, then why it seems to me that a multitude of different n vectors could be orthogonal to vector a, and thus it seems like a poor definition of a plane.

Just wondering if anyone can clear this up for me.


:)

Thanks.

 

[HallsofIvy]

If we can define the vector (r-r₀) as a single vector, say a, then why it seems to me that a multitude of different n vectors could be orthogonal to vector a, and thus it seems like a poor definition of a plane.

That's NOT a "definition" of a plane but it is a property of any plane. Every plane is two dimensional- given any point in the plane there exist an infinite number of lines through that point lying in the plane. Further, there exist a unique line (normal to the plane) through that poit and perpendicular to all those lines.

 

[That Neuron]

That's NOT a "definition" of a plane but it is a property of any plane. Every plane is two dimensional- given any point in the plane there exist an infinite number of lines through that point lying in the plane. Further, there exist a unique line (normal to the plane) through that poit and perpendicular to all those lines.

Right, BUT my book says that it is a definition (Or at least says that it determines a single plane) which is weird.

Also, how can we derive the standard cartesian equation a(x-x₀) + b(y-y₀) + c(z-z₀) from the vector equation?

 

[Fredrik]

If we can define the vector (r-r₀) as a single vector, say a, then why it seems to me that a multitude of different n vectors could be orthogonal to vector a, and thus it seems like a poor definition of a plane.

That "multitude" of vectors is the set
$$P=\left\{\mathbf r\in\mathbb R^3:\mathbf n\cdot\mathbf r=\mathbf 0\right\}.$$ This is a plane that contains $\mathbf 0$. Now let $\mathbf r_0\in\mathbb R^3$ be arbitrary and consider the set
$$Q=\left\{\mathbf r\in\mathbb R^3:\mathbf n\cdot(\mathbf{r-r_0})=\mathbf 0\right\}.$$ For all $\mathbf r\in Q$, $\mathbf r-\mathbf{r_0}$ is in the plane Q. This observation should make it easier to see that Q is a plane that contains $\mathbf r_0$. Q is what you get if you take P and add $\mathbf r_0$ to every vector in it.

Also, how can we derive the standard cartesian equation a(x-x₀) + b(y-y₀) + c(z-z₀) from the vector equation?

\begin{align}
&\mathbf r =(x,y,z)\\
&\mathbf r_0 =(x_0,y_0,z_0)\\
&\mathbf n =(a,b,c)
\end{align}

 

[That Neuron]

Oh, I see the mistake in my logic, I was looking at the cartesian equation being ambiguous as a result of the vector equation, I suppose that the vector equation is ambiguous until we add the actual vector, then we get a unique equation for the plane.

Don't know what I was thinking. There's only one plane orthogonal to a particular 'normal' vector, even if there are several vectors orthogonal to a single plane.