Calculating the area of the parallelogramm given $4$ vertices.
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I want to calculate the area of a parallelogramm given the following four vertices:
$$vec{p}=begin{pmatrix}2 \ 0\3 end {pmatrix},vec{q}=begin{pmatrix}8 \ 1\1 end {pmatrix},vec{r}=begin{pmatrix}6 \ -2\-1 end {pmatrix},vec{s}=begin{pmatrix}12 \ -1\-3 end {pmatrix}$$
I know I need to find two sides of the parallelogramm and then take the magnitude of the cross product. Here is my problem:
Assume that I chose to compute $vec{q}-vec{p}$ and $vec{s}-vec{p}$. My picture could look like this:
However, since I don't know the spatial location of $p,q,r,s$ then the picture could also look like this:
So how do I determine which points to subtract in order to get the correct pair of vectors? Does it even matter which pair I take or will the area be the same because of the symmetry of the problem?
geometry
asked Jan 6 at 16:02
NullspaceNullspace
374110
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add a comment |
$begingroup$
I want to calculate the area of a parallelogramm given the following four vertices:
$$vec{p}=begin{pmatrix}2 \ 0\3 end {pmatrix},vec{q}=begin{pmatrix}8 \ 1\1 end {pmatrix},vec{r}=begin{pmatrix}6 \ -2\-1 end {pmatrix},vec{s}=begin{pmatrix}12 \ -1\-3 end {pmatrix}$$
I know I need to find two sides of the parallelogramm and then take the magnitude of the cross product. Here is my problem:
Assume that I chose to compute $vec{q}-vec{p}$ and $vec{s}-vec{p}$. My picture could look like this:
However, since I don't know the spatial location of $p,q,r,s$ then the picture could also look like this:
So how do I determine which points to subtract in order to get the correct pair of vectors? Does it even matter which pair I take or will the area be the same because of the symmetry of the problem?
geometry
asked Jan 6 at 16:02
NullspaceNullspace
374110
$endgroup$
$begingroup$
If the first picture is correct, then $vec{r}=vec{q}+vec{s}-vec{p}$
$endgroup$
– saulspatz
Jan 6 at 16:35
add a comment |
$begingroup$
I want to calculate the area of a parallelogramm given the following four vertices:
$$vec{p}=begin{pmatrix}2 \ 0\3 end {pmatrix},vec{q}=begin{pmatrix}8 \ 1\1 end {pmatrix},vec{r}=begin{pmatrix}6 \ -2\-1 end {pmatrix},vec{s}=begin{pmatrix}12 \ -1\-3 end {pmatrix}$$
I know I need to find two sides of the parallelogramm and then take the magnitude of the cross product. Here is my problem:
Assume that I chose to compute $vec{q}-vec{p}$ and $vec{s}-vec{p}$. My picture could look like this:
However, since I don't know the spatial location of $p,q,r,s$ then the picture could also look like this:
So how do I determine which points to subtract in order to get the correct pair of vectors? Does it even matter which pair I take or will the area be the same because of the symmetry of the problem?
geometry
asked Jan 6 at 16:02
NullspaceNullspace
374110
$endgroup$
I want to calculate the area of a parallelogramm given the following four vertices:
$$vec{p}=begin{pmatrix}2 \ 0\3 end {pmatrix},vec{q}=begin{pmatrix}8 \ 1\1 end {pmatrix},vec{r}=begin{pmatrix}6 \ -2\-1 end {pmatrix},vec{s}=begin{pmatrix}12 \ -1\-3 end {pmatrix}$$
I know I need to find two sides of the parallelogramm and then take the magnitude of the cross product. Here is my problem:
Assume that I chose to compute $vec{q}-vec{p}$ and $vec{s}-vec{p}$. My picture could look like this:
However, since I don't know the spatial location of $p,q,r,s$ then the picture could also look like this:
So how do I determine which points to subtract in order to get the correct pair of vectors? Does it even matter which pair I take or will the area be the same because of the symmetry of the problem?
geometry
geometry
asked Jan 6 at 16:02
NullspaceNullspace
374110
asked Jan 6 at 16:02
NullspaceNullspace
374110
asked Jan 6 at 16:02
NullspaceNullspace
374110
asked Jan 6 at 16:02
NullspaceNullspace
374110
asked Jan 6 at 16:02
NullspaceNullspace
374110
374110
$begingroup$
If the first picture is correct, then $vec{r}=vec{q}+vec{s}-vec{p}$
$endgroup$
– saulspatz
Jan 6 at 16:35
add a comment |
$begingroup$
If the first picture is correct, then $vec{r}=vec{q}+vec{s}-vec{p}$
$endgroup$
– saulspatz
Jan 6 at 16:35
$begingroup$
If the first picture is correct, then $vec{r}=vec{q}+vec{s}-vec{p}$
$endgroup$
– saulspatz
Jan 6 at 16:35
$begingroup$
If the first picture is correct, then $vec{r}=vec{q}+vec{s}-vec{p}$
$endgroup$
– saulspatz
Jan 6 at 16:35
add a comment |
1 Answer
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$begingroup$
It doesn't matter.
Either way gives you the same value.
For your second drawing, move the top half of the parallelogram to the right of the lower half.
It's the same area.
answered Jan 6 at 16:39
marty cohenmarty cohen
74.5k549129
$endgroup$
$begingroup$
Of course! I don't know why I didn't see that. Thank you very much for your help!
$endgroup$
– Nullspace
Jan 6 at 16:46
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It doesn't matter.
Either way gives you the same value.
For your second drawing, move the top half of the parallelogram to the right of the lower half.
It's the same area.
answered Jan 6 at 16:39
marty cohenmarty cohen
74.5k549129
$endgroup$
$begingroup$
Of course! I don't know why I didn't see that. Thank you very much for your help!
$endgroup$
– Nullspace
Jan 6 at 16:46
add a comment |
$begingroup$
It doesn't matter.
Either way gives you the same value.
For your second drawing, move the top half of the parallelogram to the right of the lower half.
It's the same area.
answered Jan 6 at 16:39
marty cohenmarty cohen
74.5k549129
$endgroup$
$begingroup$
Of course! I don't know why I didn't see that. Thank you very much for your help!
$endgroup$
– Nullspace
Jan 6 at 16:46
add a comment |
$begingroup$
It doesn't matter.
Either way gives you the same value.
For your second drawing, move the top half of the parallelogram to the right of the lower half.
It's the same area.
answered Jan 6 at 16:39
marty cohenmarty cohen
74.5k549129
$endgroup$
It doesn't matter.
Either way gives you the same value.
For your second drawing, move the top half of the parallelogram to the right of the lower half.
It's the same area.
answered Jan 6 at 16:39
marty cohenmarty cohen
74.5k549129
answered Jan 6 at 16:39
marty cohenmarty cohen
74.5k549129
answered Jan 6 at 16:39
marty cohenmarty cohen
74.5k549129
answered Jan 6 at 16:39
marty cohenmarty cohen
74.5k549129
74.5k549129
$begingroup$
Of course! I don't know why I didn't see that. Thank you very much for your help!
$endgroup$
– Nullspace
Jan 6 at 16:46
add a comment |
$begingroup$
Of course! I don't know why I didn't see that. Thank you very much for your help!
$endgroup$
– Nullspace
Jan 6 at 16:46
$begingroup$
Of course! I don't know why I didn't see that. Thank you very much for your help!
$endgroup$
– Nullspace
Jan 6 at 16:46
$begingroup$
Of course! I don't know why I didn't see that. Thank you very much for your help!
$endgroup$
– Nullspace
Jan 6 at 16:46
add a comment |
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$begingroup$
If the first picture is correct, then $vec{r}=vec{q}+vec{s}-vec{p}$
$endgroup$
– saulspatz
Jan 6 at 16:35