What is the equation of a 3D cone with generalised tilt?
2
$begingroup$
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
3d
edited Jul 25 '16 at 0:05
ervx
10.3k31338
asked Jul 24 '16 at 15:38
CharleneCharlene
113
$endgroup$
add a comment |
2
$begingroup$
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
3d
edited Jul 25 '16 at 0:05
ervx
10.3k31338
asked Jul 24 '16 at 15:38
CharleneCharlene
113
$endgroup$
2
$begingroup$
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.
$endgroup$
– Andrew D. Hwang
Jul 24 '16 at 15:53
$begingroup$
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
$endgroup$
– Charlene
Jul 24 '16 at 16:02
$begingroup$
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.
$endgroup$
– Andrew D. Hwang
Jul 24 '16 at 16:36
1
$begingroup$
Charlene, the Wikipedia cone page does show an implicit form ($F(vec{u})=0$),$$F(vec{u}) = vec{u}cdotvec{d} - lvert vec{u} rvertlvert vec{d} rvert costheta$$for a right circular cone, where the axis is parallel to $vec{d}$, and aperture is $2theta$. So, if the vertex is at $vec{c}$, then$$F(vec{u}) = (vec{u}-vec{c})cdotvec{d} - lvert vec{u}-vec{c}rvert lvertvec{d}rvert costheta$$
$endgroup$
– Nominal Animal
Jul 25 '16 at 2:43
$begingroup$
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...
$endgroup$
– Nominal Animal
Jul 25 '16 at 2:53
add a comment |
2
2
2
$begingroup$
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
3d
edited Jul 25 '16 at 0:05
ervx
10.3k31338
asked Jul 24 '16 at 15:38
CharleneCharlene
113
$endgroup$
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
3d
3d
edited Jul 25 '16 at 0:05
ervx
10.3k31338
asked Jul 24 '16 at 15:38
CharleneCharlene
113
edited Jul 25 '16 at 0:05
ervx
10.3k31338
asked Jul 24 '16 at 15:38
CharleneCharlene
113
edited Jul 25 '16 at 0:05
ervx
10.3k31338
edited Jul 25 '16 at 0:05
ervx
10.3k31338
edited Jul 25 '16 at 0:05
ervx
10.3k31338
10.3k31338
asked Jul 24 '16 at 15:38
CharleneCharlene
113
asked Jul 24 '16 at 15:38
CharleneCharlene
113
asked Jul 24 '16 at 15:38
CharleneCharlene
113
113
2
$begingroup$
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.
$endgroup$
– Andrew D. Hwang
Jul 24 '16 at 15:53
$begingroup$
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
$endgroup$
– Charlene
Jul 24 '16 at 16:02
$begingroup$
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.
$endgroup$
– Andrew D. Hwang
Jul 24 '16 at 16:36
1
$begingroup$
Charlene, the Wikipedia cone page does show an implicit form ($F(vec{u})=0$),$$F(vec{u}) = vec{u}cdotvec{d} - lvert vec{u} rvertlvert vec{d} rvert costheta$$for a right circular cone, where the axis is parallel to $vec{d}$, and aperture is $2theta$. So, if the vertex is at $vec{c}$, then$$F(vec{u}) = (vec{u}-vec{c})cdotvec{d} - lvert vec{u}-vec{c}rvert lvertvec{d}rvert costheta$$
$endgroup$
– Nominal Animal
Jul 25 '16 at 2:43
$begingroup$
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...
$endgroup$
– Nominal Animal
Jul 25 '16 at 2:53
add a comment |
2
$begingroup$
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.
$endgroup$
– Andrew D. Hwang
Jul 24 '16 at 15:53
$begingroup$
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
$endgroup$
– Charlene
Jul 24 '16 at 16:02
$begingroup$
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.
$endgroup$
– Andrew D. Hwang
Jul 24 '16 at 16:36
1
$begingroup$
Charlene, the Wikipedia cone page does show an implicit form ($F(vec{u})=0$),$$F(vec{u}) = vec{u}cdotvec{d} - lvert vec{u} rvertlvert vec{d} rvert costheta$$for a right circular cone, where the axis is parallel to $vec{d}$, and aperture is $2theta$. So, if the vertex is at $vec{c}$, then$$F(vec{u}) = (vec{u}-vec{c})cdotvec{d} - lvert vec{u}-vec{c}rvert lvertvec{d}rvert costheta$$
$endgroup$
– Nominal Animal
Jul 25 '16 at 2:43
$begingroup$
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...
$endgroup$
– Nominal Animal
Jul 25 '16 at 2:53
2
2
$begingroup$
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.
$endgroup$
– Andrew D. Hwang
Jul 24 '16 at 15:53
$begingroup$
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.
$endgroup$
– Andrew D. Hwang
Jul 24 '16 at 15:53
$begingroup$
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
$endgroup$
– Charlene
Jul 24 '16 at 16:02
$begingroup$
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
$endgroup$
– Charlene
Jul 24 '16 at 16:02
$begingroup$
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.
$endgroup$
– Andrew D. Hwang
Jul 24 '16 at 16:36
$begingroup$
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.
$endgroup$
– Andrew D. Hwang
Jul 24 '16 at 16:36
1
1
$begingroup$
Charlene, the Wikipedia cone page does show an implicit form ($F(vec{u})=0$),$$F(vec{u}) = vec{u}cdotvec{d} - lvert vec{u} rvertlvert vec{d} rvert costheta$$for a right circular cone, where the axis is parallel to $vec{d}$, and aperture is $2theta$. So, if the vertex is at $vec{c}$, then$$F(vec{u}) = (vec{u}-vec{c})cdotvec{d} - lvert vec{u}-vec{c}rvert lvertvec{d}rvert costheta$$
$endgroup$
– Nominal Animal
Jul 25 '16 at 2:43
$begingroup$
Charlene, the Wikipedia cone page does show an implicit form ($F(vec{u})=0$),$$F(vec{u}) = vec{u}cdotvec{d} - lvert vec{u} rvertlvert vec{d} rvert costheta$$for a right circular cone, where the axis is parallel to $vec{d}$, and aperture is $2theta$. So, if the vertex is at $vec{c}$, then$$F(vec{u}) = (vec{u}-vec{c})cdotvec{d} - lvert vec{u}-vec{c}rvert lvertvec{d}rvert costheta$$
$endgroup$
– Nominal Animal
Jul 25 '16 at 2:43
$begingroup$
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...
$endgroup$
– Nominal Animal
Jul 25 '16 at 2:53
$begingroup$
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...
$endgroup$
– Nominal Animal
Jul 25 '16 at 2:53
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
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 $mp mu^2$.
answered Jul 24 '16 at 16:36
Ng Chung TakNg Chung Tak
14.6k31334
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1869611%2fwhat-is-the-equation-of-a-3d-cone-with-generalised-tilt%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
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 $mp mu^2$.
answered Jul 24 '16 at 16:36
Ng Chung TakNg Chung Tak
14.6k31334
$endgroup$
add a comment |
1
$begingroup$
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 $mp mu^2$.
answered Jul 24 '16 at 16:36
Ng Chung TakNg Chung Tak
14.6k31334
$endgroup$
add a comment |
1
1
1
$begingroup$
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 $mp mu^2$.
answered Jul 24 '16 at 16:36
Ng Chung TakNg Chung Tak
14.6k31334
$endgroup$
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 $mp mu^2$.
answered Jul 24 '16 at 16:36
Ng Chung TakNg Chung Tak
14.6k31334
edited Jul 24 '16 at 16:42
answered Jul 24 '16 at 16:36
Ng Chung TakNg Chung Tak
14.6k31334
answered Jul 24 '16 at 16:36
Ng Chung TakNg Chung Tak
14.6k31334
answered Jul 24 '16 at 16:36
Ng Chung TakNg Chung Tak
14.6k31334
14.6k31334
add a comment |
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1869611%2fwhat-is-the-equation-of-a-3d-cone-with-generalised-tilt%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
$begingroup$
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.
$endgroup$
– Andrew D. Hwang
Jul 24 '16 at 15:53
$begingroup$
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
$endgroup$
– Charlene
Jul 24 '16 at 16:02
$begingroup$
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.
$endgroup$
– Andrew D. Hwang
Jul 24 '16 at 16:36
1
$begingroup$
Charlene, the Wikipedia cone page does show an implicit form ($F(vec{u})=0$),$$F(vec{u}) = vec{u}cdotvec{d} - lvert vec{u} rvertlvert vec{d} rvert costheta$$for a right circular cone, where the axis is parallel to $vec{d}$, and aperture is $2theta$. So, if the vertex is at $vec{c}$, then$$F(vec{u}) = (vec{u}-vec{c})cdotvec{d} - lvert vec{u}-vec{c}rvert lvertvec{d}rvert costheta$$
$endgroup$
– Nominal Animal
Jul 25 '16 at 2:43
$begingroup$
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...
$endgroup$
– Nominal Animal
Jul 25 '16 at 2:53