enter image description here

Let's say a plane is defined using normal vector, i.e., $\vec{n_0}$ and a point, i.e., $p_0$ in the plane as shown in the above figure. So we can determine the corresponding equation of the plane as $ax+by+cz+d =0$ where $d = -p_0^T\vec{n_0}$

The following two expressions I count not understand what do they represent

  1. $\frac{-\vec{n_0}\cdot d}{|\vec{n_0}|^2}$

  2. Let assume there is another plane is given by ($\vec{n_1},p_1$), then compute $ \vec{n_1} \times \vec{n_0} \times \vec{n_1}$

These are I encounter when I try to understand how the reference plane intersects with a set of other planes. I could not share the original work due to copyright issue. Any idea on this to understand what actually those represent really appreciated.

  • 1
    $\begingroup$ It might however be helpful to see this original work. $\endgroup$
    – Peter
    Commented Dec 25, 2020 at 12:25
  • $\begingroup$ @Peter I was not allowed to share it (: $\endgroup$
    – GPrathap
    Commented Dec 25, 2020 at 12:34
  • $\begingroup$ Your sentence "then $\vec{n_1} \times \vec{n_0} \times \vec{n_1}$" isn't finished ! Compute $(\vec{n_1} \times \vec{n_0}) \times \vec{n_1}$ and see if it exceptionnaly the same vector as $\vec{n_1} \times (\vec{n_0} \times \vec{n_1})$ because the cross product operation isn't associative in general $\endgroup$
    – Jean Marie
    Commented Dec 25, 2020 at 19:18
  • 1
    $\begingroup$ @JeanMarie - But by its anti-commutivity, $$(\vec{n_1} \times \vec{n_0}) \times \vec{n_1} = -\vec{n_1} \times (\vec{n_1} \times \vec{n_0}) = \vec{n_1} \times (-\vec{n_1} \times \vec{n_0}) = \vec{n_1} \times (\vec{n_0} \times \vec{n_1})$$ So in this limited case associativity does hold. $\endgroup$ Commented Dec 26, 2020 at 1:12

1 Answer 1

  1. $\dfrac{-\vec{n_0}}{\|\vec{n_0}\|}$ is the unit vector pointed in the opposite direction as $\vec{n_0}$. And $\dfrac d{\|\vec{n_0}\|}$ is the distance from the origin to the plane $\vec{n_0}\cdot \vec v = d$. Thus their product $$\dfrac{-d\vec{n_0}}{\|\vec{n_0}\|^2}$$ would be the vector pointing from the projection of the origin onto the plane back to the origin itself.
  2. For any two non-parallel vectors $\vec v, \vec w$, $\vec v \times \vec w$ is a vector perpendicular to both, with length $\|\vec v\|\|\vec w\|\sin \theta$, where $\theta$ is the angle between the two. So
    • $\|\vec{n_1} \times \vec{n_0} \times \vec{n_1}\| = \|\vec n_0\|\|\vec n_1\|^2\sin \theta$.
    • $\vec{n_1} \times \vec{n_0} \times \vec{n_1}$ is perpendicular to $\vec{n_0} \times \vec{n_1}$, and therefore must be in the same plane as the origin, $\vec{n_0}$ and $\vec{n_1}$. It is also perpendicular to $\vec{n_1}$. Some playing around with the RH rule should also convince you that in the plane of $\vec{n_0}$ and $\vec{n_1}$, it lies on the same side of the line through $\vec{n_1}$ as $\vec{n_0}$. In fact, it is the projection of $\vec{n_0}$ onto the line perpendicular to $\vec{n_1}$ in the common plane of the two, multiplied by the the length of $\vec{n_1}$.

Note that in the equation of a plane $\vec n \cdot \vec r = D$, if we divide by the $\|\vec n\|$, we get $$\dfrac{\vec n}{\|\vec n\|}\cdot \vec r = \dfrac{D}{\|\vec n\|}$$ which is another equation of the same time, except that $\vec n$ is replaced with the vector $\dfrac{\vec n}{\|\vec n\|}$, which has length $1$, and the constant is replaced with another constant $\dfrac{D}{\|\vec n\|}$. That constant turns out to be exactly the distance from the plane to the origin, so it is more meaningful that $D$ itself.

For this reason we generally prefer to always write plane equations with the normal being a unit vector. And if we make that restriction the two interpretations become more simple. If $\vec{n_0}$ is a unit vector, then $d\vec{n_0}$ is the projection of the origin onto the plane. And $-d\vec{n_0}$ is its opposite.

When $\|\vec{n_0}\| = \|\vec{n_1}\| = 1$, then $\vec{n_1} \times \vec{n_0} \times \vec{n_1}$ is exactly the projection of $\vec{n_0}$ onto the line perpendicular to $\vec{n_1}$ in their common plane.

  • $\begingroup$ [+1] Very didactic answer. $\endgroup$
    – Jean Marie
    Commented Dec 26, 2020 at 3:13
  • $\begingroup$ Thank you for my christmas gift :) $\endgroup$
    – GPrathap
    Commented Dec 26, 2020 at 7:27
  • $\begingroup$ @Paul Sinclair (Nothing to do with the present question): let me say that I like very much the logo you have chosen "Mental Arithmetics" of the wonderful painter Bogdanov-Belsky. $\endgroup$
    – Jean Marie
    Commented Dec 28, 2020 at 10:43
  • 1
    $\begingroup$ @JeanMarie - I've loved that painting ever since I first came across it in this forum. In addition to identifying strongly with the central figure, I feel like I've known everybody in that classroom, from the disconnected (in this case by privilege) boy who gravitates to authority, to the early responder whispering their answer, to the two chatterers, to the boy who appears to have given up in the back (or maybe answered even earlier), and of course, staring at the ceiling. $\endgroup$ Commented Dec 28, 2020 at 16:11
  • $\begingroup$ @Paul Sinclair Thanks for the reference to this forum. Yes, it is also to replaced in the context of a Russia at the eve of the XXth century with intellectuals and artists attempting to show that the "low level" of peasantry could be improved in particular by education.. $\endgroup$
    – Jean Marie
    Commented Dec 28, 2020 at 16:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .