Mixing
How to calculate the area of a triangle ABC when given three position vectors $a, b$, and $ c$ in 3D?
$begingroup$
Where $a = ( 1, 2, 3), b = (2, 1, 3) $, and $c = (3,1,2). $
triangle vectors area
edited Jan 9 at 2:38
Community♦
1
asked Apr 3 '14 at 14:21
user138913user138913
941210
$endgroup$
add a comment |
$begingroup$
Where $a = ( 1, 2, 3), b = (2, 1, 3) $, and $c = (3,1,2). $
triangle vectors area
edited Jan 9 at 2:38
Community♦
1
asked Apr 3 '14 at 14:21
user138913user138913
941210
$endgroup$
$begingroup$
We can find the lengths of the sides and apply Heron's Formula.
$endgroup$
– Indrayudh Roy
Apr 3 '14 at 14:26
add a comment |
$begingroup$
Where $a = ( 1, 2, 3), b = (2, 1, 3) $, and $c = (3,1,2). $
triangle vectors area
edited Jan 9 at 2:38
Community♦
1
asked Apr 3 '14 at 14:21
user138913user138913
941210
$endgroup$
Where $a = ( 1, 2, 3), b = (2, 1, 3) $, and $c = (3,1,2). $
triangle vectors area
triangle vectors area
edited Jan 9 at 2:38
Community♦
1
asked Apr 3 '14 at 14:21
user138913user138913
941210
edited Jan 9 at 2:38
Community♦
1
asked Apr 3 '14 at 14:21
user138913user138913
941210
edited Jan 9 at 2:38
Community♦
1
edited Jan 9 at 2:38
Community♦
1
edited Jan 9 at 2:38
Community♦
1
1
asked Apr 3 '14 at 14:21
user138913user138913
941210
asked Apr 3 '14 at 14:21
user138913user138913
941210
asked Apr 3 '14 at 14:21
user138913user138913
941210
941210
$begingroup$
We can find the lengths of the sides and apply Heron's Formula.
$endgroup$
– Indrayudh Roy
Apr 3 '14 at 14:26
add a comment |
$begingroup$
We can find the lengths of the sides and apply Heron's Formula.
$endgroup$
– Indrayudh Roy
Apr 3 '14 at 14:26
$begingroup$
We can find the lengths of the sides and apply Heron's Formula.
$endgroup$
– Indrayudh Roy
Apr 3 '14 at 14:26
$begingroup$
We can find the lengths of the sides and apply Heron's Formula.
$endgroup$
– Indrayudh Roy
Apr 3 '14 at 14:26
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Heron works of course but it would be simpler to take half the length of the cross product
$(b-a)times(c-a)$.
answered Apr 3 '14 at 14:34
jenajena
32412
$endgroup$
add a comment |
$begingroup$
Solution: Construct the vectors $hat{ab}$, $hat{ac}$ and take $frac{1}{2} |hat{ab} times hat{ac}|$. We take half of the resulting since the original gives the area of the parallelogram decsribed by the vectors.
answered Apr 3 '14 at 14:27
Mr.FryMr.Fry
3,89021223
$endgroup$
add a comment |
$begingroup$
use this formula:
$$S=sqrt{p(p-s_1)(p-s_2)(p-s_3)}$$
where $p=frac{s_1+s_2+s_3}{2}$, and $s_1,s_2,s_3$ are the lengths of the three sides. You can get $s_1,s_2,s_3$ from the three positions.
answered Apr 3 '14 at 14:29
MartialMartial
1,0041917
$endgroup$
add a comment |
$begingroup$
Compute the differences $a-b$ and $a-c$. then take the cross product of them $(a-c)times(a-c)$. The norm of this, divided by two is the area of the triangle.
answered Apr 3 '14 at 14:32
JackJack
23416
$endgroup$
add a comment |
$begingroup$
One way to do it would be to first find out the length of each of the 3 sides, by simply applying Pythagorus. e.g.
AB = $sqrt{(2-0)^2 + (1-1)^2 + (4-3)^2}=sqrt5$
You can do this for all of the sides, and it is possible to find the area of a triangle given the length of each of its sides.
answered Apr 3 '14 at 14:31
JangoJango
352214
$endgroup$
add a comment |
$begingroup$
Cross product works great as a black box, but it lacks geometric intuition.
For triangle ABC, we construct a line segment intersecting C, and normal to side AB. The triangle's area is $0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{CH} right |$
$$left | overrightarrow{CH} right |=left | overrightarrow{AC} right | cdot sin(θ)$$
where
$$θ = arccos(frac{overrightarrow{AB} cdot overrightarrow{AC}}{left | overrightarrow{AB} right | cdot left | overrightarrow{AC} right |})$$
Thus:
$$Area=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$
This is equal to:
$$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |$$
This is why
$$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$
answered Jan 4 at 18:19
zyczyc
1114
$endgroup$
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
});
}
});
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%2f738236%2fhow-to-calculate-the-area-of-a-triangle-abc-when-given-three-position-vectors-a%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Heron works of course but it would be simpler to take half the length of the cross product
$(b-a)times(c-a)$.
answered Apr 3 '14 at 14:34
jenajena
32412
$endgroup$
add a comment |
$begingroup$
Heron works of course but it would be simpler to take half the length of the cross product
$(b-a)times(c-a)$.
answered Apr 3 '14 at 14:34
jenajena
32412
$endgroup$
add a comment |
$begingroup$
Heron works of course but it would be simpler to take half the length of the cross product
$(b-a)times(c-a)$.
answered Apr 3 '14 at 14:34
jenajena
32412
$endgroup$
Heron works of course but it would be simpler to take half the length of the cross product
$(b-a)times(c-a)$.
answered Apr 3 '14 at 14:34
jenajena
32412
answered Apr 3 '14 at 14:34
jenajena
32412
answered Apr 3 '14 at 14:34
jenajena
32412
answered Apr 3 '14 at 14:34
jenajena
32412
32412
add a comment |
add a comment |
$begingroup$
Solution: Construct the vectors $hat{ab}$, $hat{ac}$ and take $frac{1}{2} |hat{ab} times hat{ac}|$. We take half of the resulting since the original gives the area of the parallelogram decsribed by the vectors.
answered Apr 3 '14 at 14:27
Mr.FryMr.Fry
3,89021223
$endgroup$
add a comment |
$begingroup$
Solution: Construct the vectors $hat{ab}$, $hat{ac}$ and take $frac{1}{2} |hat{ab} times hat{ac}|$. We take half of the resulting since the original gives the area of the parallelogram decsribed by the vectors.
answered Apr 3 '14 at 14:27
Mr.FryMr.Fry
3,89021223
$endgroup$
add a comment |
$begingroup$
Solution: Construct the vectors $hat{ab}$, $hat{ac}$ and take $frac{1}{2} |hat{ab} times hat{ac}|$. We take half of the resulting since the original gives the area of the parallelogram decsribed by the vectors.
answered Apr 3 '14 at 14:27
Mr.FryMr.Fry
3,89021223
$endgroup$
Solution: Construct the vectors $hat{ab}$, $hat{ac}$ and take $frac{1}{2} |hat{ab} times hat{ac}|$. We take half of the resulting since the original gives the area of the parallelogram decsribed by the vectors.
answered Apr 3 '14 at 14:27
Mr.FryMr.Fry
3,89021223
answered Apr 3 '14 at 14:27
Mr.FryMr.Fry
3,89021223
answered Apr 3 '14 at 14:27
Mr.FryMr.Fry
3,89021223
answered Apr 3 '14 at 14:27
Mr.FryMr.Fry
3,89021223
3,89021223
add a comment |
add a comment |
$begingroup$
use this formula:
$$S=sqrt{p(p-s_1)(p-s_2)(p-s_3)}$$
where $p=frac{s_1+s_2+s_3}{2}$, and $s_1,s_2,s_3$ are the lengths of the three sides. You can get $s_1,s_2,s_3$ from the three positions.
answered Apr 3 '14 at 14:29
MartialMartial
1,0041917
$endgroup$
add a comment |
$begingroup$
use this formula:
$$S=sqrt{p(p-s_1)(p-s_2)(p-s_3)}$$
where $p=frac{s_1+s_2+s_3}{2}$, and $s_1,s_2,s_3$ are the lengths of the three sides. You can get $s_1,s_2,s_3$ from the three positions.
answered Apr 3 '14 at 14:29
MartialMartial
1,0041917
$endgroup$
add a comment |
$begingroup$
use this formula:
$$S=sqrt{p(p-s_1)(p-s_2)(p-s_3)}$$
where $p=frac{s_1+s_2+s_3}{2}$, and $s_1,s_2,s_3$ are the lengths of the three sides. You can get $s_1,s_2,s_3$ from the three positions.
answered Apr 3 '14 at 14:29
MartialMartial
1,0041917
$endgroup$
use this formula:
$$S=sqrt{p(p-s_1)(p-s_2)(p-s_3)}$$
where $p=frac{s_1+s_2+s_3}{2}$, and $s_1,s_2,s_3$ are the lengths of the three sides. You can get $s_1,s_2,s_3$ from the three positions.
answered Apr 3 '14 at 14:29
MartialMartial
1,0041917
answered Apr 3 '14 at 14:29
MartialMartial
1,0041917
answered Apr 3 '14 at 14:29
MartialMartial
1,0041917
answered Apr 3 '14 at 14:29
MartialMartial
1,0041917
1,0041917
add a comment |
add a comment |
$begingroup$
Compute the differences $a-b$ and $a-c$. then take the cross product of them $(a-c)times(a-c)$. The norm of this, divided by two is the area of the triangle.
answered Apr 3 '14 at 14:32
JackJack
23416
$endgroup$
add a comment |
$begingroup$
Compute the differences $a-b$ and $a-c$. then take the cross product of them $(a-c)times(a-c)$. The norm of this, divided by two is the area of the triangle.
answered Apr 3 '14 at 14:32
JackJack
23416
$endgroup$
add a comment |
$begingroup$
Compute the differences $a-b$ and $a-c$. then take the cross product of them $(a-c)times(a-c)$. The norm of this, divided by two is the area of the triangle.
answered Apr 3 '14 at 14:32
JackJack
23416
$endgroup$
Compute the differences $a-b$ and $a-c$. then take the cross product of them $(a-c)times(a-c)$. The norm of this, divided by two is the area of the triangle.
answered Apr 3 '14 at 14:32
JackJack
23416
answered Apr 3 '14 at 14:32
JackJack
23416
answered Apr 3 '14 at 14:32
JackJack
23416
answered Apr 3 '14 at 14:32
JackJack
23416
23416
add a comment |
add a comment |
$begingroup$
One way to do it would be to first find out the length of each of the 3 sides, by simply applying Pythagorus. e.g.
AB = $sqrt{(2-0)^2 + (1-1)^2 + (4-3)^2}=sqrt5$
You can do this for all of the sides, and it is possible to find the area of a triangle given the length of each of its sides.
answered Apr 3 '14 at 14:31
JangoJango
352214
$endgroup$
add a comment |
$begingroup$
One way to do it would be to first find out the length of each of the 3 sides, by simply applying Pythagorus. e.g.
AB = $sqrt{(2-0)^2 + (1-1)^2 + (4-3)^2}=sqrt5$
You can do this for all of the sides, and it is possible to find the area of a triangle given the length of each of its sides.
answered Apr 3 '14 at 14:31
JangoJango
352214
$endgroup$
add a comment |
$begingroup$
One way to do it would be to first find out the length of each of the 3 sides, by simply applying Pythagorus. e.g.
AB = $sqrt{(2-0)^2 + (1-1)^2 + (4-3)^2}=sqrt5$
You can do this for all of the sides, and it is possible to find the area of a triangle given the length of each of its sides.
answered Apr 3 '14 at 14:31
JangoJango
352214
$endgroup$
One way to do it would be to first find out the length of each of the 3 sides, by simply applying Pythagorus. e.g.
AB = $sqrt{(2-0)^2 + (1-1)^2 + (4-3)^2}=sqrt5$
You can do this for all of the sides, and it is possible to find the area of a triangle given the length of each of its sides.
answered Apr 3 '14 at 14:31
JangoJango
352214
answered Apr 3 '14 at 14:31
JangoJango
352214
answered Apr 3 '14 at 14:31
JangoJango
352214
answered Apr 3 '14 at 14:31
JangoJango
352214
352214
add a comment |
add a comment |
$begingroup$
Cross product works great as a black box, but it lacks geometric intuition.
For triangle ABC, we construct a line segment intersecting C, and normal to side AB. The triangle's area is $0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{CH} right |$
$$left | overrightarrow{CH} right |=left | overrightarrow{AC} right | cdot sin(θ)$$
where
$$θ = arccos(frac{overrightarrow{AB} cdot overrightarrow{AC}}{left | overrightarrow{AB} right | cdot left | overrightarrow{AC} right |})$$
Thus:
$$Area=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$
This is equal to:
$$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |$$
This is why
$$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$
answered Jan 4 at 18:19
zyczyc
1114
$endgroup$
add a comment |
$begingroup$
Cross product works great as a black box, but it lacks geometric intuition.
For triangle ABC, we construct a line segment intersecting C, and normal to side AB. The triangle's area is $0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{CH} right |$
$$left | overrightarrow{CH} right |=left | overrightarrow{AC} right | cdot sin(θ)$$
where
$$θ = arccos(frac{overrightarrow{AB} cdot overrightarrow{AC}}{left | overrightarrow{AB} right | cdot left | overrightarrow{AC} right |})$$
Thus:
$$Area=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$
This is equal to:
$$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |$$
This is why
$$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$
answered Jan 4 at 18:19
zyczyc
1114
$endgroup$
add a comment |
$begingroup$
Cross product works great as a black box, but it lacks geometric intuition.
For triangle ABC, we construct a line segment intersecting C, and normal to side AB. The triangle's area is $0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{CH} right |$
$$left | overrightarrow{CH} right |=left | overrightarrow{AC} right | cdot sin(θ)$$
where
$$θ = arccos(frac{overrightarrow{AB} cdot overrightarrow{AC}}{left | overrightarrow{AB} right | cdot left | overrightarrow{AC} right |})$$
Thus:
$$Area=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$
This is equal to:
$$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |$$
This is why
$$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$
answered Jan 4 at 18:19
zyczyc
1114
$endgroup$
Cross product works great as a black box, but it lacks geometric intuition.
For triangle ABC, we construct a line segment intersecting C, and normal to side AB. The triangle's area is $0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{CH} right |$
$$left | overrightarrow{CH} right |=left | overrightarrow{AC} right | cdot sin(θ)$$
where
$$θ = arccos(frac{overrightarrow{AB} cdot overrightarrow{AC}}{left | overrightarrow{AB} right | cdot left | overrightarrow{AC} right |})$$
Thus:
$$Area=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$
This is equal to:
$$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |$$
This is why
$$0.5cdot left | overrightarrow{AB} times overrightarrow{AC} right |=0.5cdot left | overrightarrow{AB} right |cdot left | overrightarrow{AC} right |cdot sin(θ)$$
answered Jan 4 at 18:19
zyczyc
1114
edited Jan 4 at 18:25
answered Jan 4 at 18:19
zyczyc
1114
answered Jan 4 at 18:19
zyczyc
1114
answered Jan 4 at 18:19
zyczyc
1114
1114
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.
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%2f738236%2fhow-to-calculate-the-area-of-a-triangle-abc-when-given-three-position-vectors-a%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
$begingroup$
We can find the lengths of the sides and apply Heron's Formula.
$endgroup$
– Indrayudh Roy
Apr 3 '14 at 14:26