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What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space

ervx
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Charlene
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    Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, and what have you tried? Thank you. – Andrew D. Hwang Jul 24 '16 at 15:53
  • Hi, I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin (0,0,0) - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space - hope this is clearer! Also, the form of the answer doesn't really matter to me at this point – Charlene Jul 24 '16 at 16:02
  • Just a comment for now: The general elliptical cone with vertex $(x_{0}, y_{0}, z_{0})$ has equation $Q(x - x_{0}, y - y_{0}, z - z_{0}) = 0$ for some homogeneous quadratic polynomial $Q$ whose ($3 \times 3$ symmetric) coefficient matrix is invertible but neither positive- nor negative-definite. Briefly, there's no single single formula analogous to, say, the Cartesian equation of a sphere with specified center and radius. – Andrew D. Hwang Jul 24 '16 at 16:36
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    Charlene, the Wikipedia cone page does show an implicit form ($F(\vec{u})=0$),$$F(\vec{u}) = \vec{u}\cdot\vec{d} - \lvert \vec{u} \rvert\lvert \vec{d} \rvert \cos\theta$$for a right circular cone, where the axis is parallel to $\vec{d}$, and aperture is $2\theta$. So, if the vertex is at $\vec{c}$, then$$F(\vec{u}) = (\vec{u}-\vec{c})\cdot\vec{d} - \lvert \vec{u}-\vec{c}\rvert \lvert\vec{d}\rvert \cos\theta$$ – Nominal Animal Jul 25 '16 at 02:43
  • You can work with cones when their vertex is at origin, and axis is parallel to one of the Cartesian coordinate axes, and not lose generality -- you can always use a rotation matrix ($3\times3$ orthogonal matrix) to rotate the axis to the correct direction, and translation to move the vertex to the correct position. Since the matrix is orthogonal, its inverse is its transpose, so you only really need to find the matrix that rotates one of the Cartesian axes to the desired cone axis direction. I find this easier than the coefficient matrix approach, but then again, I'm not a mathematician... – Nominal Animal Jul 25 '16 at 02:53

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Equation of circular cone with singular point at origin \begin{align*} \begin{pmatrix} x & y & z \end{pmatrix} \begin{pmatrix} a & h & g \\ h & b & f \\ g & f & c \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} &= 0 \\ ax^2+by^2+cz^2+2(fyz+gzx+hxy) &=0 \end{align*}

where $\begin{pmatrix} a & h & g \\ h & b & f \\ g & f & c \end{pmatrix}$ has eigenvalues in form of $\pm \lambda^2$, $\pm \lambda^2$ and $\mp \mu^2$.

Ng Chung Tak
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