(Stupid question...)

Well we can represent a point as something like $P(a,b,c)$

We can represent a line as $\dfrac{x-a}{p}=\dfrac{x-b}{q}=\dfrac{x-c}{r}$

We can also represent a plane as $ax+by+cz=d$

So can we represent a $3D$ object also???

Or do we need to have have a 4D cartesian system or something like that?


I thought this .gif is interesting!



3 Answers 3


Yes you can, of course. A very simple exemple is the sphere, which is a 3D object:

$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2= r^2$.

  • $\begingroup$ What about cubes? $\endgroup$
    – NeilRoy
    Commented Jun 11, 2015 at 8:30
  • 1
    $\begingroup$ $|x-x_0|<a$, $|y-y_0|<a$, $|z-z_0|<a$ defines the inside of a cube of length $2a$. $\endgroup$
    – Martigan
    Commented Jun 11, 2015 at 8:36

We can represent 3D objects as a function of three variables. Here are some simple, finite geometric shapes, defined implicitly in Cartesian coordinates. The coefficients are used to scale the shape, and can be set to any value. You will end up with tall or flat prisms and pyramids by adjusting them away from their current values.

Sphere : $x^2+y^2+z^2 = a^2 $

Torus : $\big(\sqrt{x^2+y^2} - a\big)^2 + z^2 = b^2 $

Triangle Torus : $\left|\big|\big(\sqrt{x^2+y^2} - a\big)^2\big| + 2z\right| + \left|\big(\sqrt{x^2+y^2} - a\big)^2\right| = b$

Square Torus : $\left|\big(\sqrt{x^2+y^2} - a\big)^2-z\right| + \left|\big(\sqrt{x^2+y^2} - a\big)^2+z\right| = b$

Cone : $\left |\sqrt{x^2+y^2} +2z\right | + \sqrt{x^2+y^2} = a$

Cylinder : $\left |\sqrt{x^2+y^2} - z\right | + \left |\sqrt{x^2+y^2} + z\right | = a$

Tetrahedron : $\left|\big||x|+2y\big|+|x| + 2z\right| + \big||x|+2y\big|+|x| = a$

Triangle Prism : $\big |\big||x|+2y\big|+|x| - 2z\big | + \big |\big||x|+2y\big|+|x| + 2z\big | = a $

Square Pyramid : $\big ||x-y|+|x+y| + 3z\big | + |x-y|+|x+y| = a$

Cube : $\big ||x-y|+|x+y| - 2z\big | +\big ||x-y|+|x+y| + 2z\big | = a$

And, of course, these can easily be extended to 4D shapes as well (or any number of dimensions), that are just as elementary.

  • 1
    $\begingroup$ wow! amazing answer bro! $\endgroup$
    – NeilRoy
    Commented Jun 12, 2015 at 2:22
  • 2
    $\begingroup$ @NeilRoy Technically, these are just surfaces of 3D objects. (In topology, these are considered 2D, because they're basically curved planes.) If you want to include its interior, you probably have to deal with inequalities. (Similar to how, in 2D, $x^2+y^2=1$ gives you the boundary of a circle, but $x^2+y^2\le1$ gives you the circle and its interior. And $x^2+y^2<1$ gives you just the interior but not the boundary.) $\endgroup$ Commented Jun 12, 2015 at 2:51
  • $\begingroup$ @columbus8myhw Thanks for the info $\endgroup$
    – NeilRoy
    Commented Jun 12, 2015 at 3:18
  • $\begingroup$ In the pic you provided...In the 4D shapes what is the "Tiger" or "Glome"? Where can i find pics of those? $\endgroup$
    – NeilRoy
    Commented Jun 12, 2015 at 6:49
  • $\begingroup$ The tiger is a 4D torus-like object, made by a circle embedded into a Clifford torus: i.imgur.com/j7LWsnP.gif passing through a 3-plane at different angles. It has two major diameters, which can be adjusted to any value without self-intersection, unlike $T^3$. A glome is another name for $S^3$ , the hypersphere : i.imgur.com/BqXvcG4.gif passing through. $\endgroup$ Commented Jun 12, 2015 at 14:42

"3D objects" are just subsets of $\mathbb{R}^3$. So the question is : How to represent subsets of a (normed) vector space?

I think the manifolds are a important part of it.


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