Equation of a ... 3D object??? (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?

EDIT
I thought this .gif is interesting!

 A: Yes you can, of course. 
A very simple exemple is the sphere, which is a 3D object:
$(x-x_0)^2+(y-y_0)^2+(z-z_0)^2= r^2$.
A: 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.
A: "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.  
