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
991210
$endgroup$
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
991210
$endgroup$
$begingroup$
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
991210
$endgroup$
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
991210
edited Jan 9 at 2:38
Community♦
1
asked Apr 3 '14 at 14:21
user138913user138913
991210
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
991210
asked Apr 3 '14 at 14:21
user138913user138913
991210
asked Apr 3 '14 at 14:21
user138913user138913
991210
991210
$begingroup$
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
33412
$endgroup$
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
1,00411017
$endgroup$
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
367214
$endgroup$
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 |
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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
33412
$endgroup$
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
33412
$endgroup$
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
33412
$endgroup$
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
33412
answered Apr 3 '14 at 14:34
jenajena
33412
answered Apr 3 '14 at 14:34
jenajena
33412
answered Apr 3 '14 at 14:34
jenajena
33412
33412
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
1,00411017
$endgroup$
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
1,00411017
$endgroup$
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
1,00411017
$endgroup$
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
1,00411017
answered Apr 3 '14 at 14:29
MartialMartial
1,00411017
answered Apr 3 '14 at 14:29
MartialMartial
1,00411017
answered Apr 3 '14 at 14:29
MartialMartial
1,00411017
1,00411017
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
367214
$endgroup$
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
367214
$endgroup$
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
367214
$endgroup$
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
367214
answered Apr 3 '14 at 14:31
JangoJango
367214
answered Apr 3 '14 at 14:31
JangoJango
367214
answered Apr 3 '14 at 14:31
JangoJango
367214
367214
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 |
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$begingroup$
We can find the lengths of the sides and apply Heron's Formula.
$endgroup$
– Indrayudh Roy
Apr 3 '14 at 14:26