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Existence of common perpendicular line between two affine subspaces of $mathbb{R}^n$

0

$begingroup$

I would like to prove the following result.

Let $alpha$ and $beta$ be two affine subspaces of $mathbb{R}^n$ considered as a Euclidean affine space. Then if $alpha$ and $beta$ have no common point then there exists a line that is intersecting and perpendicular to both of them.

What I have tried is as follows. As $alpha$ and $beta$ have no common point, $dim alpha+dimbetaleq n$. The case of equality can be treated quite easily since then $alpha$ is parallel to $beta$. Consider the case $dim alpha+dimbetaleq n-1$. Let $V$ and $W$ be directed vector spaces of $alpha$ and $beta$, then there is a vector space $U$ of dimension $1$ orthogonal to both $V$ and $W$. Of course the line that is perpendicular to $alpha$ and $beta$ must be directed by $U$, but I don't know how to choose $U$ so that the corresponding line intersects both $alpha$ and $beta$. What I think is to construct a hyperplane $gamma$ directed by $Voplus U$ and containing $alpha$. Then $gamma$ should intersect $beta$ at some point which is on the line we want. However it is not clear to me now.

Can someone help me? Thanks a lot!

linear-algebra geometry

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asked Mar 26 at 10:49

mappingmapping

1828

$endgroup$

add a comment | 

0

$begingroup$

I would like to prove the following result.

Let $alpha$ and $beta$ be two affine subspaces of $mathbb{R}^n$ considered as a Euclidean affine space. Then if $alpha$ and $beta$ have no common point then there exists a line that is intersecting and perpendicular to both of them.

What I have tried is as follows. As $alpha$ and $beta$ have no common point, $dim alpha+dimbetaleq n$. The case of equality can be treated quite easily since then $alpha$ is parallel to $beta$. Consider the case $dim alpha+dimbetaleq n-1$. Let $V$ and $W$ be directed vector spaces of $alpha$ and $beta$, then there is a vector space $U$ of dimension $1$ orthogonal to both $V$ and $W$. Of course the line that is perpendicular to $alpha$ and $beta$ must be directed by $U$, but I don't know how to choose $U$ so that the corresponding line intersects both $alpha$ and $beta$. What I think is to construct a hyperplane $gamma$ directed by $Voplus U$ and containing $alpha$. Then $gamma$ should intersect $beta$ at some point which is on the line we want. However it is not clear to me now.

Can someone help me? Thanks a lot!

linear-algebra geometry

share|cite|improve this question

asked Mar 26 at 10:49

mappingmapping

1828

$endgroup$

add a comment | 

$begingroup$

I would like to prove the following result.

Let $alpha$ and $beta$ be two affine subspaces of $mathbb{R}^n$ considered as a Euclidean affine space. Then if $alpha$ and $beta$ have no common point then there exists a line that is intersecting and perpendicular to both of them.

What I have tried is as follows. As $alpha$ and $beta$ have no common point, $dim alpha+dimbetaleq n$. The case of equality can be treated quite easily since then $alpha$ is parallel to $beta$. Consider the case $dim alpha+dimbetaleq n-1$. Let $V$ and $W$ be directed vector spaces of $alpha$ and $beta$, then there is a vector space $U$ of dimension $1$ orthogonal to both $V$ and $W$. Of course the line that is perpendicular to $alpha$ and $beta$ must be directed by $U$, but I don't know how to choose $U$ so that the corresponding line intersects both $alpha$ and $beta$. What I think is to construct a hyperplane $gamma$ directed by $Voplus U$ and containing $alpha$. Then $gamma$ should intersect $beta$ at some point which is on the line we want. However it is not clear to me now.

Can someone help me? Thanks a lot!

linear-algebra geometry

share|cite|improve this question

asked Mar 26 at 10:49

mappingmapping

1828

$endgroup$

share|cite|improve this question

asked Mar 26 at 10:49

mappingmapping

1828

share|cite|improve this question

asked Mar 26 at 10:49

mappingmapping

1828

asked Mar 26 at 10:49

mappingmapping

1828

asked Mar 26 at 10:49

mappingmapping

1828

1 Answer
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1

$begingroup$

Write $alpha=a+V$ and $beta=b+W$, where $a,b$ are points in $mathbb R^n$. Since $dim(V+W)<n$, we can write $mathbb R^n=(V+W)oplus (V+W)^perp$, where $U:=(V+W)^perp$ is a non-trivial subspace. Consider vector $vec{ab}= b-a$. Note that $vec{ab}notin V+W$, as otherwise $vec{ab}=vec v+vec w$ for some $vec vin V$ and $vec win W$, so $a+vec v=b-vec w$ is a common point of $alpha$ and $beta$. Write $vec{ab}= (vec v+vec w)+vec u$, where $vec vin V$, $vec win W$ and $vec uin U$, and note $vec uneq 0$.

Consider now $a'=a+vec vinalpha$, $b'=b-vec winbeta$, $L=mathrm{span}(vec u)$ and affine line $gamma=a'+L$. $gamma$ is directed by $L$, i.e. by $vec u$, so it is normal to $V+W$, i.e. it is normal to both $alpha$ and $beta$. Clearly, $gamma$ intersects $alpha$ in $a'$. Note $b'-a'= (b-a)-(vec v+vec w)=vec{ab}-(vec v+vec w)=vec u$, so $b'=a'+vec u$ is a common point of $beta$ and $gamma$.

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answered Mar 26 at 13:31

SMMSMM

3,268512

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice solution! Thank you very much!
  $endgroup$
  – mapping
  Mar 26 at 16:40
  

  
  
  
  

add a comment | 

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1

$begingroup$

Write $alpha=a+V$ and $beta=b+W$, where $a,b$ are points in $mathbb R^n$. Since $dim(V+W)<n$, we can write $mathbb R^n=(V+W)oplus (V+W)^perp$, where $U:=(V+W)^perp$ is a non-trivial subspace. Consider vector $vec{ab}= b-a$. Note that $vec{ab}notin V+W$, as otherwise $vec{ab}=vec v+vec w$ for some $vec vin V$ and $vec win W$, so $a+vec v=b-vec w$ is a common point of $alpha$ and $beta$. Write $vec{ab}= (vec v+vec w)+vec u$, where $vec vin V$, $vec win W$ and $vec uin U$, and note $vec uneq 0$.

Consider now $a'=a+vec vinalpha$, $b'=b-vec winbeta$, $L=mathrm{span}(vec u)$ and affine line $gamma=a'+L$. $gamma$ is directed by $L$, i.e. by $vec u$, so it is normal to $V+W$, i.e. it is normal to both $alpha$ and $beta$. Clearly, $gamma$ intersects $alpha$ in $a'$. Note $b'-a'= (b-a)-(vec v+vec w)=vec{ab}-(vec v+vec w)=vec u$, so $b'=a'+vec u$ is a common point of $beta$ and $gamma$.

share|cite|improve this answer

answered Mar 26 at 13:31

SMMSMM

3,268512

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice solution! Thank you very much!
  $endgroup$
  – mapping
  Mar 26 at 16:40
  

  
  
  
  

add a comment | 

1

$begingroup$

Write $alpha=a+V$ and $beta=b+W$, where $a,b$ are points in $mathbb R^n$. Since $dim(V+W)<n$, we can write $mathbb R^n=(V+W)oplus (V+W)^perp$, where $U:=(V+W)^perp$ is a non-trivial subspace. Consider vector $vec{ab}= b-a$. Note that $vec{ab}notin V+W$, as otherwise $vec{ab}=vec v+vec w$ for some $vec vin V$ and $vec win W$, so $a+vec v=b-vec w$ is a common point of $alpha$ and $beta$. Write $vec{ab}= (vec v+vec w)+vec u$, where $vec vin V$, $vec win W$ and $vec uin U$, and note $vec uneq 0$.

Consider now $a'=a+vec vinalpha$, $b'=b-vec winbeta$, $L=mathrm{span}(vec u)$ and affine line $gamma=a'+L$. $gamma$ is directed by $L$, i.e. by $vec u$, so it is normal to $V+W$, i.e. it is normal to both $alpha$ and $beta$. Clearly, $gamma$ intersects $alpha$ in $a'$. Note $b'-a'= (b-a)-(vec v+vec w)=vec{ab}-(vec v+vec w)=vec u$, so $b'=a'+vec u$ is a common point of $beta$ and $gamma$.

share|cite|improve this answer

answered Mar 26 at 13:31

SMMSMM

3,268512

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice solution! Thank you very much!
  $endgroup$
  – mapping
  Mar 26 at 16:40
  

  
  
  
  

add a comment | 

$begingroup$

Write $alpha=a+V$ and $beta=b+W$, where $a,b$ are points in $mathbb R^n$. Since $dim(V+W)<n$, we can write $mathbb R^n=(V+W)oplus (V+W)^perp$, where $U:=(V+W)^perp$ is a non-trivial subspace. Consider vector $vec{ab}= b-a$. Note that $vec{ab}notin V+W$, as otherwise $vec{ab}=vec v+vec w$ for some $vec vin V$ and $vec win W$, so $a+vec v=b-vec w$ is a common point of $alpha$ and $beta$. Write $vec{ab}= (vec v+vec w)+vec u$, where $vec vin V$, $vec win W$ and $vec uin U$, and note $vec uneq 0$.

Consider now $a'=a+vec vinalpha$, $b'=b-vec winbeta$, $L=mathrm{span}(vec u)$ and affine line $gamma=a'+L$. $gamma$ is directed by $L$, i.e. by $vec u$, so it is normal to $V+W$, i.e. it is normal to both $alpha$ and $beta$. Clearly, $gamma$ intersects $alpha$ in $a'$. Note $b'-a'= (b-a)-(vec v+vec w)=vec{ab}-(vec v+vec w)=vec u$, so $b'=a'+vec u$ is a common point of $beta$ and $gamma$.

share|cite|improve this answer

answered Mar 26 at 13:31

SMMSMM

3,268512

$endgroup$

share|cite|improve this answer

answered Mar 26 at 13:31

SMMSMM

3,268512

answered Mar 26 at 13:31

SMMSMM

3,268512

answered Mar 26 at 13:31

SMMSMM

3,268512