Proving that these four vectors form a parallelogramm
I am given four vectors in three-dimensional space and I want to check if they form a parallelogramm. The vectors are:
$$vec{p}=begin{pmatrix}2 \ 0\3 end {pmatrix},vec{q}=begin{pmatrix}8 \ 1\1 end {pmatrix},vec{r}=begin{pmatrix}6 \ -2\-1 end {pmatrix},vec{s}=begin{pmatrix}12 \ -1\-3 end {pmatrix}$$
I read this answer where Jack D'Aurizio states:
The convex quadrilateral $ABCD$ is a parallelogram if and only if the midpoint of $AC$ is the midpoint of $BD$, too. Hence you just need to compute $binom{4}{2}=6$ midpoints and check that two of them are the same point.
I tried to use this and computed all the line segment joining the vertices of the parallelogramm by subtracting the vectors.
$$vec{p}-vec{q}=(-6,-1,2)^T \ vec{p}-vec{r}=(-4,2,4)^T \ vec{p}-vec{s}=(-10,1,6)^T \ vec{q}-vec{r}=(2,3,2)^T \ vec{q}-vec{s}=(-4,2,4)^T \ vec{r}-vec{s}=(-6,-1,2)^T$$
I am not sure how to continue now. Also, I wasn't sure if the order of vector subtraction matters here. Any ideas?
geometry
asked Jan 6 at 11:22
NullspaceNullspace
326110
add a comment |
I am given four vectors in three-dimensional space and I want to check if they form a parallelogramm. The vectors are:
$$vec{p}=begin{pmatrix}2 \ 0\3 end {pmatrix},vec{q}=begin{pmatrix}8 \ 1\1 end {pmatrix},vec{r}=begin{pmatrix}6 \ -2\-1 end {pmatrix},vec{s}=begin{pmatrix}12 \ -1\-3 end {pmatrix}$$
I read this answer where Jack D'Aurizio states:
The convex quadrilateral $ABCD$ is a parallelogram if and only if the midpoint of $AC$ is the midpoint of $BD$, too. Hence you just need to compute $binom{4}{2}=6$ midpoints and check that two of them are the same point.
I tried to use this and computed all the line segment joining the vertices of the parallelogramm by subtracting the vectors.
$$vec{p}-vec{q}=(-6,-1,2)^T \ vec{p}-vec{r}=(-4,2,4)^T \ vec{p}-vec{s}=(-10,1,6)^T \ vec{q}-vec{r}=(2,3,2)^T \ vec{q}-vec{s}=(-4,2,4)^T \ vec{r}-vec{s}=(-6,-1,2)^T$$
I am not sure how to continue now. Also, I wasn't sure if the order of vector subtraction matters here. Any ideas?
geometry
asked Jan 6 at 11:22
NullspaceNullspace
326110
add a comment |
I am given four vectors in three-dimensional space and I want to check if they form a parallelogramm. The vectors are:
$$vec{p}=begin{pmatrix}2 \ 0\3 end {pmatrix},vec{q}=begin{pmatrix}8 \ 1\1 end {pmatrix},vec{r}=begin{pmatrix}6 \ -2\-1 end {pmatrix},vec{s}=begin{pmatrix}12 \ -1\-3 end {pmatrix}$$
I read this answer where Jack D'Aurizio states:
The convex quadrilateral $ABCD$ is a parallelogram if and only if the midpoint of $AC$ is the midpoint of $BD$, too. Hence you just need to compute $binom{4}{2}=6$ midpoints and check that two of them are the same point.
I tried to use this and computed all the line segment joining the vertices of the parallelogramm by subtracting the vectors.
$$vec{p}-vec{q}=(-6,-1,2)^T \ vec{p}-vec{r}=(-4,2,4)^T \ vec{p}-vec{s}=(-10,1,6)^T \ vec{q}-vec{r}=(2,3,2)^T \ vec{q}-vec{s}=(-4,2,4)^T \ vec{r}-vec{s}=(-6,-1,2)^T$$
I am not sure how to continue now. Also, I wasn't sure if the order of vector subtraction matters here. Any ideas?
geometry
asked Jan 6 at 11:22
NullspaceNullspace
326110
I am given four vectors in three-dimensional space and I want to check if they form a parallelogramm. The vectors are:
$$vec{p}=begin{pmatrix}2 \ 0\3 end {pmatrix},vec{q}=begin{pmatrix}8 \ 1\1 end {pmatrix},vec{r}=begin{pmatrix}6 \ -2\-1 end {pmatrix},vec{s}=begin{pmatrix}12 \ -1\-3 end {pmatrix}$$
I read this answer where Jack D'Aurizio states:
The convex quadrilateral $ABCD$ is a parallelogram if and only if the midpoint of $AC$ is the midpoint of $BD$, too. Hence you just need to compute $binom{4}{2}=6$ midpoints and check that two of them are the same point.
I tried to use this and computed all the line segment joining the vertices of the parallelogramm by subtracting the vectors.
$$vec{p}-vec{q}=(-6,-1,2)^T \ vec{p}-vec{r}=(-4,2,4)^T \ vec{p}-vec{s}=(-10,1,6)^T \ vec{q}-vec{r}=(2,3,2)^T \ vec{q}-vec{s}=(-4,2,4)^T \ vec{r}-vec{s}=(-6,-1,2)^T$$
I am not sure how to continue now. Also, I wasn't sure if the order of vector subtraction matters here. Any ideas?
geometry
geometry
asked Jan 6 at 11:22
NullspaceNullspace
326110
asked Jan 6 at 11:22
NullspaceNullspace
326110
asked Jan 6 at 11:22
NullspaceNullspace
326110
asked Jan 6 at 11:22
NullspaceNullspace
326110
asked Jan 6 at 11:22
NullspaceNullspace
326110
326110
add a comment |
add a comment |
1 Answer
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Notice that $$frac{vec{p}+vec{s}}2 = frac{vec{q}+vec{r}}2$$
Hence $PQSR$ is a parallelogram.
That is we have $vec{p}-vec{q} = vec{r}-vec{s}$ and $vec{p}-vec{r}=vec{q}-vec{s}$.
answered Jan 6 at 11:27
Siong Thye GohSiong Thye Goh
99.9k1465117
Thank you for your answer. This might be a stupid question but how do I know that: $$text{PQRS is a parallelogramm} iff frac{vec{p}+vec{s}}2 = frac{vec{q}+vec{r}}2$$
– Nullspace
Jan 6 at 11:45
I will check that out and come back if I have any more questions. Thanks again.
– Nullspace
Jan 6 at 11:50
add a comment |
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1 Answer
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1 Answer
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Notice that $$frac{vec{p}+vec{s}}2 = frac{vec{q}+vec{r}}2$$
Hence $PQSR$ is a parallelogram.
That is we have $vec{p}-vec{q} = vec{r}-vec{s}$ and $vec{p}-vec{r}=vec{q}-vec{s}$.
answered Jan 6 at 11:27
Siong Thye GohSiong Thye Goh
99.9k1465117
Thank you for your answer. This might be a stupid question but how do I know that: $$text{PQRS is a parallelogramm} iff frac{vec{p}+vec{s}}2 = frac{vec{q}+vec{r}}2$$
– Nullspace
Jan 6 at 11:45
I will check that out and come back if I have any more questions. Thanks again.
– Nullspace
Jan 6 at 11:50
add a comment |
Notice that $$frac{vec{p}+vec{s}}2 = frac{vec{q}+vec{r}}2$$
Hence $PQSR$ is a parallelogram.
That is we have $vec{p}-vec{q} = vec{r}-vec{s}$ and $vec{p}-vec{r}=vec{q}-vec{s}$.
answered Jan 6 at 11:27
Siong Thye GohSiong Thye Goh
99.9k1465117
Thank you for your answer. This might be a stupid question but how do I know that: $$text{PQRS is a parallelogramm} iff frac{vec{p}+vec{s}}2 = frac{vec{q}+vec{r}}2$$
– Nullspace
Jan 6 at 11:45
I will check that out and come back if I have any more questions. Thanks again.
– Nullspace
Jan 6 at 11:50
add a comment |
Notice that $$frac{vec{p}+vec{s}}2 = frac{vec{q}+vec{r}}2$$
Hence $PQSR$ is a parallelogram.
That is we have $vec{p}-vec{q} = vec{r}-vec{s}$ and $vec{p}-vec{r}=vec{q}-vec{s}$.
answered Jan 6 at 11:27
Siong Thye GohSiong Thye Goh
99.9k1465117
Notice that $$frac{vec{p}+vec{s}}2 = frac{vec{q}+vec{r}}2$$
Hence $PQSR$ is a parallelogram.
That is we have $vec{p}-vec{q} = vec{r}-vec{s}$ and $vec{p}-vec{r}=vec{q}-vec{s}$.
answered Jan 6 at 11:27
Siong Thye GohSiong Thye Goh
99.9k1465117
answered Jan 6 at 11:27
Siong Thye GohSiong Thye Goh
99.9k1465117
answered Jan 6 at 11:27
Siong Thye GohSiong Thye Goh
99.9k1465117
answered Jan 6 at 11:27
Siong Thye GohSiong Thye Goh
99.9k1465117
99.9k1465117
Thank you for your answer. This might be a stupid question but how do I know that: $$text{PQRS is a parallelogramm} iff frac{vec{p}+vec{s}}2 = frac{vec{q}+vec{r}}2$$
– Nullspace
Jan 6 at 11:45
I will check that out and come back if I have any more questions. Thanks again.
– Nullspace
Jan 6 at 11:50
add a comment |
Thank you for your answer. This might be a stupid question but how do I know that: $$text{PQRS is a parallelogramm} iff frac{vec{p}+vec{s}}2 = frac{vec{q}+vec{r}}2$$
– Nullspace
Jan 6 at 11:45
I will check that out and come back if I have any more questions. Thanks again.
– Nullspace
Jan 6 at 11:50
Thank you for your answer. This might be a stupid question but how do I know that: $$text{PQRS is a parallelogramm} iff frac{vec{p}+vec{s}}2 = frac{vec{q}+vec{r}}2$$
– Nullspace
Jan 6 at 11:45
Thank you for your answer. This might be a stupid question but how do I know that: $$text{PQRS is a parallelogramm} iff frac{vec{p}+vec{s}}2 = frac{vec{q}+vec{r}}2$$
– Nullspace
Jan 6 at 11:45
I will check that out and come back if I have any more questions. Thanks again.
– Nullspace
Jan 6 at 11:50
I will check that out and come back if I have any more questions. Thanks again.
– Nullspace
Jan 6 at 11:50
add a comment |
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