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A common way to represent a plane is:

$$ a x + by + cz +d = 0 $$

Where (a, b, c) is the normal vector and corresponds to the orientation of the plane and $d$ is the distance from the origin following the normal direction. This would define the plane using 4 scalars (a, b, c, d)

If (a, b, c) are unit vector, so... wouldn't be possible to represent a plane with only 3 numbers?

$$ (p_{x}, p_{y}, p_{z}) = (a, b, c)*d $$

We could easily go back to the original form by doing:

$$ \frac{(p_{x}, p_{y}, p_{z})}{||(p_{x}, p_{y}, p_{z})||} = (a, b, c) $$

$$ ||(p_{x}, p_{y}, p_{z})|| = d $$

We would still be representing both orientation and distance of the plane but with only 3 scalars $(p_{x}, p_{y}, p_{z})$ which give us advantages when thinking about memory performance on computer algorithms since we are fully defining the geometric information of a plane but with less variables.

Why isn't this representation more frequently used since it is more compact comparing to the hessian normal form? Is there something I'm missing?

This made me think: Is there a more compact representation of a plane in $\mathfrak{R}^3$ that uses less then 3 scalars?

Leonardo Mariga
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  • The "common" way uses $12$ symbols. Your way uses $20$. Tell me how your way is more compact. – Gerry Myerson Dec 05 '20 at 04:44
  • I'm trying to find the most memory efficient way to store the geometric information of a plane. In the hessian normal form I need to store 4 variables (a, b, c, d), but assuming (a, b, c) unit vector, I would have to store only 3 variables (a.d, b.d, c.d) – Leonardo Mariga Dec 05 '20 at 04:52

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Sure, easiest way to see this is first starting in lower dimensions. If we have a point in $\mathbb{R}$, we can clearly parameterize this by a single value which satisfies the equation $x = c$. For a line in $\mathbb{R}^2$ we can again easily represent this with two real numbers, typically we write $y=mx+b$. Similarly in $\mathbb{R}^3$ we can parameterize a plane by $z= \alpha x + \beta y + \gamma$ where if we are given a plane $ax + by + cz = d$, we find that the coefficients $\alpha = -a/c$, $\beta = -b/c$ and $\gamma = d/c$ will define the same plane, simply by solving for $z$. $$$$ Note that we cannot parameterize a vertical line using the equation $y=mx+b$ for some $m,b\in\mathbb{R}$ but could instead use a polar parameterize where $(r,\theta)\mapsto \cos(\theta)x + \sin(\theta)y = r$. A similar problem occurs for the plane in three dimensions, it might be useful to work out a solution for yourself.

RyanK
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  • So, the only reason we use the hessian normal form is to be easier to understand? I can't think on any advantage other than that – Leonardo Mariga Dec 05 '20 at 05:14
  • Usually when studying mathematics, we like to explore many different ways of understanding a given topic, and often prefer intuitive formulations. The Hessian normal form allows for both intuitive understanding and eliminates the problem of not being able to define "to steep" of a plane. There might be other reasons that I am not aware of. – RyanK Dec 05 '20 at 05:31