# Parametric equation of a cone through 3D space [closed]

Does anyone know how to find the parametric equation of a cone with circumference centre (x,y,z) and vertex (a,b,c) with a radius R? I can only find an explanation that has the centre of circumference with z-coordinate 0.

Now suppose we want to build a cone between point $$O$$ (for apex) and base center point $$P$$. Further suppose that both $$O$$ and $$P$$ are in $$\mathbb{R}^{3}$$ with no constraint about axes. Given point $$O$$ and $$P$$ and $$R$$, where $$R$$ is the radius of the cone's base about $$P$$, what is the parametric equation of the cone?

The base is represented by a circle about $$P$$ and the circle is normal to the line from $$O$$ to $$P.$$ The general parametric equation for a circle in $$\mathbb{R}^{3}$$ is $$Circle:\quad\left[\begin{array}{c} x\\ y\\ z \end{array}\right]=\mathbf{P}+R\cdot cos\theta\cdot\mathbf{u}+R\cdot sin\theta\cdot\mathbf{v}\tag{Eq:1}$$

In this general circle parametrization, vectors $$\mathbf{u}$$ and $$\mathbf{v}$$ are two orthogonal vectors that lie in the plane of the circle. If, for example, the circle were in the $$xy$$ plane, then two such vectors would be $$\mathbf{u}=(1,0,0)$$ and $$\mathbf{v}=(0,1,0)$$. Then the circle equation would exactly match the definition that we are accustomed to seeing in the $$xy$$ plane.

The particular plane of interest is easily obtained from its normal, which is given by $$\mathbf{d=\mathbf{P-O}}$$. The plane of the cone base is defined as $$\mathbf{d\cdot}\left[\left(\begin{array}{c} x\\ y\\ z \end{array}\right)-\mathbf{P}\right]=0$$. In order to graph the plane, we would either have to expand the dot product to get into Cartesian coordinates, or find 3 points in the plane so that we could multiply two of them by parameters. But, we don't need to graph the plane, we just need a normal to it and $$\mathbf{d}$$ is one. An orthogonal to $$\mathbf{d}$$ will be a vector in the direction of the plane and it can be converted to a unit vector. An orthogonal to any vector, such as $$\mathbf{d}$$ is given by $$\mathbf{n}=\left(\begin{array}{c} d_{y}\\ -d_{x}\\ 0 \end{array}\right)$$ The two orthogonal unit vectors for the plane are given by $$\mathbf{u}=\frac{\mathbf{d\times n}}{\vert\mathbf{d\times n}\vert}\qquad\mathbf{v}=\frac{\mathbf{n}}{\vert\mathbf{n}\vert}$$

If we were to graph $$O$$ and $$P$$ and the circle, with $$R=1$$, $$O=(0,-1,2)$$, $$P=(2,2,-1)$$ it would appear as in the figure. The equation of the circle is: $$$$\left[\begin{array}{c} x\\ y\\ z \end{array}\right]=\left(\begin{array}{c} P_{x}\\ P_{y}\\ P_{z} \end{array}\right)+\left(\begin{array}{c} R\cdot cos\theta\cdot u_{x}\\ R\cdot cos\theta\cdot u_{y}\\ R\cdot cos\theta\cdot u_{z} \end{array}\right)+\left(\begin{array}{c} R\cdot sin\theta\cdot v_{x}\\ R\cdot sin\theta\cdot v_{y}\\ R\cdot sin\theta\cdot v_{z} \end{array}\right)\quad0\le\theta\le2\pi\tag{Eq:2}$$$$

The equation of the line segment is $$\left[\begin{array}{c} x\\ y\\ z \end{array}\right]=\left(\begin{array}{c} O_{x}\\ O_{y}\\ O_{z} \end{array}\right)+t\cdot\left(\begin{array}{c} d_{x}\\ d_{y}\\ d_{z} \end{array}\right)\tag{Eq:3}$$

In a $$\textbf{cone}$$, the radius would be ever expanding as some variable changed from $$O$$ to $$P.$$ $$H$$ = $$\vert P-O|=|d|$$. As $$h$$ goes from $$0$$ to $$H$$ the fraction, $$h/H$$ in the first term governs how far we have traversed the line, while the same fraction in the second and third term govern the radius of the cone along the line. Finally, the general parametric equation of a cone. $$\left[\begin{array}{c} x\\ y\\ z \end{array}\right]=\left(\begin{array}{c} O_{x}+{\displaystyle \frac{h}{H}d_{x}}\\ O_{y}+{\displaystyle \frac{h}{H}d_{y}}\\ O_{z}+{\displaystyle \frac{h}{H}d_{z}} \end{array}\right)+\left(\begin{array}{c} R\cdot{\displaystyle \frac{h}{H}\cdot cos\theta\cdot u_{x}}\\ R\cdot{\displaystyle \frac{h}{H}\cdot cos\theta\cdot u_{y}}\\ R\cdot{\displaystyle \frac{h}{H}\cdot cos\theta\cdot u_{z}} \end{array}\right)+\left(\begin{array}{c} R\cdot{\displaystyle \frac{h}{H}\cdot sin\theta\cdot v_{x}}\\ R\cdot{\displaystyle \frac{h}{H}\cdot sin\theta\cdot v_{y}}\\ R\cdot{\displaystyle \frac{h}{H}\cdot sin\theta\cdot v_{z}} \end{array}\right)\quad\begin{array}{c} 0\le h\le H\\ 0\le\theta\le2\pi \end{array}\label{eq:Cone_Equation_3d}$$