Calculate area of triangle in space given points
up vote
1
down vote
favorite
Problem 5. a) Find the area of the space triangle with vertices $P_0, P_1, P_2$:
$$ P_0 = (2,1,0), P_1=(1,0,1), P_2=(2,-1,1) $$
My current solution is to use $frac{1}{2}Big|(P_1-P_0)×(P_2-P_0)Big|$, which is great but leaves me wanting for a more elegant solution similar to taking the determinant for triangles in the plane.
Is there a better solution, either in terms of elegance or ease of mental calculation?
vectors area
edited Apr 13 '17 at 12:20
Community♦
1
asked Aug 17 '16 at 16:46
Zaz
5451725
add a comment |
up vote
1
down vote
favorite
Problem 5. a) Find the area of the space triangle with vertices $P_0, P_1, P_2$:
$$ P_0 = (2,1,0), P_1=(1,0,1), P_2=(2,-1,1) $$
My current solution is to use $frac{1}{2}Big|(P_1-P_0)×(P_2-P_0)Big|$, which is great but leaves me wanting for a more elegant solution similar to taking the determinant for triangles in the plane.
Is there a better solution, either in terms of elegance or ease of mental calculation?
vectors area
edited Apr 13 '17 at 12:20
Community♦
1
asked Aug 17 '16 at 16:46
Zaz
5451725
The linked question answers in 2D, you could do the same for 3D.
– StubbornAtom
Aug 17 '16 at 16:49
1
Isn't $left( {{P_1} - {P_0}} right) times left( {{P_2} - {P_0}} right)$ a vector?
– ITA
Aug 17 '16 at 16:51
@IvanAbraham: Whoops. Indeed it is. Edited.
– Zaz
Aug 17 '16 at 17:30
@StubbornAtom: That's what I thought, but I haven't yet been able to wrap my head around how you would do that.
– Zaz
Aug 17 '16 at 17:32
$$ begin{bmatrix} 1 & 1 & 1 \ x_0 & x_1 & x_2 \ y_0 & y_1 & y_2 \ z_0 & z_1 & z_2 end{bmatrix} $$ isn't square, so has no determinant.
– Zaz
Aug 17 '16 at 17:34
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Problem 5. a) Find the area of the space triangle with vertices $P_0, P_1, P_2$:
$$ P_0 = (2,1,0), P_1=(1,0,1), P_2=(2,-1,1) $$
My current solution is to use $frac{1}{2}Big|(P_1-P_0)×(P_2-P_0)Big|$, which is great but leaves me wanting for a more elegant solution similar to taking the determinant for triangles in the plane.
Is there a better solution, either in terms of elegance or ease of mental calculation?
vectors area
edited Apr 13 '17 at 12:20
Community♦
1
asked Aug 17 '16 at 16:46
Zaz
5451725
Problem 5. a) Find the area of the space triangle with vertices $P_0, P_1, P_2$:
$$ P_0 = (2,1,0), P_1=(1,0,1), P_2=(2,-1,1) $$
My current solution is to use $frac{1}{2}Big|(P_1-P_0)×(P_2-P_0)Big|$, which is great but leaves me wanting for a more elegant solution similar to taking the determinant for triangles in the plane.
Is there a better solution, either in terms of elegance or ease of mental calculation?
vectors area
vectors area
edited Apr 13 '17 at 12:20
Community♦
1
asked Aug 17 '16 at 16:46
Zaz
5451725
edited Apr 13 '17 at 12:20
Community♦
1
asked Aug 17 '16 at 16:46
Zaz
5451725
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
asked Aug 17 '16 at 16:46
Zaz
5451725
asked Aug 17 '16 at 16:46
Zaz
5451725
asked Aug 17 '16 at 16:46
Zaz
5451725
5451725
The linked question answers in 2D, you could do the same for 3D.
– StubbornAtom
Aug 17 '16 at 16:49
1
Isn't $left( {{P_1} - {P_0}} right) times left( {{P_2} - {P_0}} right)$ a vector?
– ITA
Aug 17 '16 at 16:51
@IvanAbraham: Whoops. Indeed it is. Edited.
– Zaz
Aug 17 '16 at 17:30
@StubbornAtom: That's what I thought, but I haven't yet been able to wrap my head around how you would do that.
– Zaz
Aug 17 '16 at 17:32
$$ begin{bmatrix} 1 & 1 & 1 \ x_0 & x_1 & x_2 \ y_0 & y_1 & y_2 \ z_0 & z_1 & z_2 end{bmatrix} $$ isn't square, so has no determinant.
– Zaz
Aug 17 '16 at 17:34
add a comment |
The linked question answers in 2D, you could do the same for 3D.
– StubbornAtom
Aug 17 '16 at 16:49
1
Isn't $left( {{P_1} - {P_0}} right) times left( {{P_2} - {P_0}} right)$ a vector?
– ITA
Aug 17 '16 at 16:51
@IvanAbraham: Whoops. Indeed it is. Edited.
– Zaz
Aug 17 '16 at 17:30
@StubbornAtom: That's what I thought, but I haven't yet been able to wrap my head around how you would do that.
– Zaz
Aug 17 '16 at 17:32
$$ begin{bmatrix} 1 & 1 & 1 \ x_0 & x_1 & x_2 \ y_0 & y_1 & y_2 \ z_0 & z_1 & z_2 end{bmatrix} $$ isn't square, so has no determinant.
– Zaz
Aug 17 '16 at 17:34
The linked question answers in 2D, you could do the same for 3D.
– StubbornAtom
Aug 17 '16 at 16:49
The linked question answers in 2D, you could do the same for 3D.
– StubbornAtom
Aug 17 '16 at 16:49
1
1
Isn't $left( {{P_1} - {P_0}} right) times left( {{P_2} - {P_0}} right)$ a vector?
– ITA
Aug 17 '16 at 16:51
Isn't $left( {{P_1} - {P_0}} right) times left( {{P_2} - {P_0}} right)$ a vector?
– ITA
Aug 17 '16 at 16:51
@IvanAbraham: Whoops. Indeed it is. Edited.
– Zaz
Aug 17 '16 at 17:30
@IvanAbraham: Whoops. Indeed it is. Edited.
– Zaz
Aug 17 '16 at 17:30
@StubbornAtom: That's what I thought, but I haven't yet been able to wrap my head around how you would do that.
– Zaz
Aug 17 '16 at 17:32
@StubbornAtom: That's what I thought, but I haven't yet been able to wrap my head around how you would do that.
– Zaz
Aug 17 '16 at 17:32
$$ begin{bmatrix} 1 & 1 & 1 \ x_0 & x_1 & x_2 \ y_0 & y_1 & y_2 \ z_0 & z_1 & z_2 end{bmatrix} $$ isn't square, so has no determinant.
– Zaz
Aug 17 '16 at 17:34
$$ begin{bmatrix} 1 & 1 & 1 \ x_0 & x_1 & x_2 \ y_0 & y_1 & y_2 \ z_0 & z_1 & z_2 end{bmatrix} $$ isn't square, so has no determinant.
– Zaz
Aug 17 '16 at 17:34
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
I like your solution, but another way to do it would be to compute the lengths of the sides using the distance formula, and then use Heron's formula.
http://www.mathopenref.com/heronsformula.html
answered Aug 17 '16 at 16:49
Alfred Yerger
10.2k2147
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',
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
});
}
});
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%2f1895248%2fcalculate-area-of-triangle-in-space-given-points%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
up vote
0
down vote
I like your solution, but another way to do it would be to compute the lengths of the sides using the distance formula, and then use Heron's formula.
http://www.mathopenref.com/heronsformula.html
answered Aug 17 '16 at 16:49
Alfred Yerger
10.2k2147
add a comment |
up vote
0
down vote
I like your solution, but another way to do it would be to compute the lengths of the sides using the distance formula, and then use Heron's formula.
http://www.mathopenref.com/heronsformula.html
answered Aug 17 '16 at 16:49
Alfred Yerger
10.2k2147
add a comment |
up vote
0
down vote
up vote
0
down vote
I like your solution, but another way to do it would be to compute the lengths of the sides using the distance formula, and then use Heron's formula.
http://www.mathopenref.com/heronsformula.html
answered Aug 17 '16 at 16:49
Alfred Yerger
10.2k2147
I like your solution, but another way to do it would be to compute the lengths of the sides using the distance formula, and then use Heron's formula.
http://www.mathopenref.com/heronsformula.html
answered Aug 17 '16 at 16:49
Alfred Yerger
10.2k2147
answered Aug 17 '16 at 16:49
Alfred Yerger
10.2k2147
answered Aug 17 '16 at 16:49
Alfred Yerger
10.2k2147
answered Aug 17 '16 at 16:49
Alfred Yerger
10.2k2147
10.2k2147
add a comment |
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
To learn more, see our tips on writing great answers.
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%2f1895248%2fcalculate-area-of-triangle-in-space-given-points%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
The linked question answers in 2D, you could do the same for 3D.
– StubbornAtom
Aug 17 '16 at 16:49
1
Isn't $left( {{P_1} - {P_0}} right) times left( {{P_2} - {P_0}} right)$ a vector?
– ITA
Aug 17 '16 at 16:51
@IvanAbraham: Whoops. Indeed it is. Edited.
– Zaz
Aug 17 '16 at 17:30
@StubbornAtom: That's what I thought, but I haven't yet been able to wrap my head around how you would do that.
– Zaz
Aug 17 '16 at 17:32
$$ begin{bmatrix} 1 & 1 & 1 \ x_0 & x_1 & x_2 \ y_0 & y_1 & y_2 \ z_0 & z_1 & z_2 end{bmatrix} $$ isn't square, so has no determinant.
– Zaz
Aug 17 '16 at 17:34