What is the equation of a 3D cone with generalised tilt?

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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







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  • 2




    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






  • 1




    Charlene, the Wikipedia cone page does show an implicit form ($F(vecu)=0$),$$F(vecu) = vecucdotvecd - lvert vecu rvertlvert vecd rvert costheta$$for a right circular cone, where the axis is parallel to $vecd$, and aperture is $2theta$. So, if the vertex is at $vecc$, then$$F(vecu) = (vecu-vecc)cdotvecd - lvert vecu-veccrvert lvertvecdrvert costheta$$
    – Nominal Animal
    Jul 25 '16 at 2:43











  • 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 ($3times3$ 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 2:53














up vote
2
down vote

favorite












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







share|cite|improve this question

















  • 2




    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






  • 1




    Charlene, the Wikipedia cone page does show an implicit form ($F(vecu)=0$),$$F(vecu) = vecucdotvecd - lvert vecu rvertlvert vecd rvert costheta$$for a right circular cone, where the axis is parallel to $vecd$, and aperture is $2theta$. So, if the vertex is at $vecc$, then$$F(vecu) = (vecu-vecc)cdotvecd - lvert vecu-veccrvert lvertvecdrvert costheta$$
    – Nominal Animal
    Jul 25 '16 at 2:43











  • 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 ($3times3$ 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 2:53












up vote
2
down vote

favorite









up vote
2
down vote

favorite











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







share|cite|improve this question













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









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 25 '16 at 0:05









ervx

9,42531336




9,42531336









asked Jul 24 '16 at 15:38









Charlene

112




112







  • 2




    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






  • 1




    Charlene, the Wikipedia cone page does show an implicit form ($F(vecu)=0$),$$F(vecu) = vecucdotvecd - lvert vecu rvertlvert vecd rvert costheta$$for a right circular cone, where the axis is parallel to $vecd$, and aperture is $2theta$. So, if the vertex is at $vecc$, then$$F(vecu) = (vecu-vecc)cdotvecd - lvert vecu-veccrvert lvertvecdrvert costheta$$
    – Nominal Animal
    Jul 25 '16 at 2:43











  • 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 ($3times3$ 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 2:53












  • 2




    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






  • 1




    Charlene, the Wikipedia cone page does show an implicit form ($F(vecu)=0$),$$F(vecu) = vecucdotvecd - lvert vecu rvertlvert vecd rvert costheta$$for a right circular cone, where the axis is parallel to $vecd$, and aperture is $2theta$. So, if the vertex is at $vecc$, then$$F(vecu) = (vecu-vecc)cdotvecd - lvert vecu-veccrvert lvertvecdrvert costheta$$
    – Nominal Animal
    Jul 25 '16 at 2:43











  • 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 ($3times3$ 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 2:53







2




2




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





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




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




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




1




1




Charlene, the Wikipedia cone page does show an implicit form ($F(vecu)=0$),$$F(vecu) = vecucdotvecd - lvert vecu rvertlvert vecd rvert costheta$$for a right circular cone, where the axis is parallel to $vecd$, and aperture is $2theta$. So, if the vertex is at $vecc$, then$$F(vecu) = (vecu-vecc)cdotvecd - lvert vecu-veccrvert lvertvecdrvert costheta$$
– Nominal Animal
Jul 25 '16 at 2:43





Charlene, the Wikipedia cone page does show an implicit form ($F(vecu)=0$),$$F(vecu) = vecucdotvecd - lvert vecu rvertlvert vecd rvert costheta$$for a right circular cone, where the axis is parallel to $vecd$, and aperture is $2theta$. So, if the vertex is at $vecc$, then$$F(vecu) = (vecu-vecc)cdotvecd - lvert vecu-veccrvert lvertvecdrvert costheta$$
– Nominal Animal
Jul 25 '16 at 2:43













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 ($3times3$ 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 2:53




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 ($3times3$ 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 2:53










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Equation of circular cone with singular point at origin
beginalign*
beginpmatrix x & y & z endpmatrix
beginpmatrix a & h & g \ h & b & f \ g & f & c endpmatrix
beginpmatrix x \ y \ z endpmatrix &= 0 \
ax^2+by^2+cz^2+2(fyz+gzx+hxy) &=0
endalign*




where $beginpmatrix a & h & g \ h & b & f \ g & f & c endpmatrix$ has eigenvalues in form of $pm lambda^2$, $pm lambda^2$ and $mp mu^2$.






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    Equation of circular cone with singular point at origin
    beginalign*
    beginpmatrix x & y & z endpmatrix
    beginpmatrix a & h & g \ h & b & f \ g & f & c endpmatrix
    beginpmatrix x \ y \ z endpmatrix &= 0 \
    ax^2+by^2+cz^2+2(fyz+gzx+hxy) &=0
    endalign*




    where $beginpmatrix a & h & g \ h & b & f \ g & f & c endpmatrix$ has eigenvalues in form of $pm lambda^2$, $pm lambda^2$ and $mp mu^2$.






    share|cite|improve this answer



























      up vote
      0
      down vote














      Equation of circular cone with singular point at origin
      beginalign*
      beginpmatrix x & y & z endpmatrix
      beginpmatrix a & h & g \ h & b & f \ g & f & c endpmatrix
      beginpmatrix x \ y \ z endpmatrix &= 0 \
      ax^2+by^2+cz^2+2(fyz+gzx+hxy) &=0
      endalign*




      where $beginpmatrix a & h & g \ h & b & f \ g & f & c endpmatrix$ has eigenvalues in form of $pm lambda^2$, $pm lambda^2$ and $mp mu^2$.






      share|cite|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote










        Equation of circular cone with singular point at origin
        beginalign*
        beginpmatrix x & y & z endpmatrix
        beginpmatrix a & h & g \ h & b & f \ g & f & c endpmatrix
        beginpmatrix x \ y \ z endpmatrix &= 0 \
        ax^2+by^2+cz^2+2(fyz+gzx+hxy) &=0
        endalign*




        where $beginpmatrix a & h & g \ h & b & f \ g & f & c endpmatrix$ has eigenvalues in form of $pm lambda^2$, $pm lambda^2$ and $mp mu^2$.






        share|cite|improve this answer
















        Equation of circular cone with singular point at origin
        beginalign*
        beginpmatrix x & y & z endpmatrix
        beginpmatrix a & h & g \ h & b & f \ g & f & c endpmatrix
        beginpmatrix x \ y \ z endpmatrix &= 0 \
        ax^2+by^2+cz^2+2(fyz+gzx+hxy) &=0
        endalign*




        where $beginpmatrix a & h & g \ h & b & f \ g & f & c endpmatrix$ has eigenvalues in form of $pm lambda^2$, $pm lambda^2$ and $mp mu^2$.







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Jul 24 '16 at 16:42


























        answered Jul 24 '16 at 16:36









        Ng Chung Tak

        13k31130




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