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What point of the plane is closest to the origin? [closed]

Linear algebra question about a plane in $mathbbR^3$ not through the originFind the position vector of the point R that is closest to the origin on the plane a'x + b'y + c'z = eClosest point in $y = sqrtx$ to the origin is at $x=-1/2$?The perpendicular distance from the origin to point in the planepoint inclusion in a half-plane 3DHow can I find the location on the Z axis where two skew lines pass closest to each other on the XY plane?Find closest Point to Another PointPoint in $e^x$ that is closest to a lineIf $B$ is the point on a plane that is closest to $A$, then $AB$ is perpendicular to the plane.Finding the General equation of a plane in $mathbb R^3$ given the middle point between 2 vectors

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3

$begingroup$

What point of the plane $3x+2y+z-6=0$ in $3$-dimensional space is closest to the origin when the distance is measured by each of the following three norms: the $1$-norm, the $2$-norm, the $infty$-norm?

For $2$-norm, I know that we can calculate the minimum distance between the origin and the plane by formula and then set the line that is cross the origin and perpendicular to the plane, then we can get the point we want. But when it comes to 1-norm and $infty$-norm, it does not work.

linear-algebra optimization

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edited Mar 13 at 11:55

Rodrigo de Azevedo

13.1k41960

asked Mar 13 at 8:04

KristyKristy

454

$endgroup$

closed as off-topic by JP McCarthy, uniquesolution, Gibbs, Song, mau Mar 13 at 15:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – JP McCarthy, uniquesolution, Gibbs, Song, mau

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
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  $begingroup$
  Welcome to MSE! People don't really like to be given assignments here. The general spirit is this: if you want to get your homework done for you, you must demonstrate a minimal amount of effort. Hint: Copying the question correctly is less than minimal.
  $endgroup$
  – uniquesolution
  Mar 13 at 8:35
  
  

  
  
  
  

add a comment | 

4

3

$begingroup$

What point of the plane $3x+2y+z-6=0$ in $3$-dimensional space is closest to the origin when the distance is measured by each of the following three norms: the $1$-norm, the $2$-norm, the $infty$-norm?

For $2$-norm, I know that we can calculate the minimum distance between the origin and the plane by formula and then set the line that is cross the origin and perpendicular to the plane, then we can get the point we want. But when it comes to 1-norm and $infty$-norm, it does not work.

linear-algebra optimization

share|cite|improve this question

edited Mar 13 at 11:55

Rodrigo de Azevedo

13.1k41960

asked Mar 13 at 8:04

KristyKristy

454

$endgroup$

closed as off-topic by JP McCarthy, uniquesolution, Gibbs, Song, mau Mar 13 at 15:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – JP McCarthy, uniquesolution, Gibbs, Song, mau

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to MSE! People don't really like to be given assignments here. The general spirit is this: if you want to get your homework done for you, you must demonstrate a minimal amount of effort. Hint: Copying the question correctly is less than minimal.
  $endgroup$
  – uniquesolution
  Mar 13 at 8:35
  
  

  
  
  
  

add a comment | 

$begingroup$

What point of the plane $3x+2y+z-6=0$ in $3$-dimensional space is closest to the origin when the distance is measured by each of the following three norms: the $1$-norm, the $2$-norm, the $infty$-norm?

For $2$-norm, I know that we can calculate the minimum distance between the origin and the plane by formula and then set the line that is cross the origin and perpendicular to the plane, then we can get the point we want. But when it comes to 1-norm and $infty$-norm, it does not work.

linear-algebra optimization

share|cite|improve this question

edited Mar 13 at 11:55

Rodrigo de Azevedo

13.1k41960

asked Mar 13 at 8:04

KristyKristy

454

$endgroup$

share|cite|improve this question

edited Mar 13 at 11:55

Rodrigo de Azevedo

13.1k41960

asked Mar 13 at 8:04

KristyKristy

454

share|cite|improve this question

edited Mar 13 at 11:55

Rodrigo de Azevedo

13.1k41960

asked Mar 13 at 8:04

KristyKristy

454

edited Mar 13 at 11:55

Rodrigo de Azevedo

13.1k41960

edited Mar 13 at 11:55

Rodrigo de Azevedo

13.1k41960

asked Mar 13 at 8:04

KristyKristy

454

asked Mar 13 at 8:04

KristyKristy

454

1 Answer
1

active

oldest

votes

5

$begingroup$

For the $1$-norm the surfaces of equal distance from the origin (the equivalent of spheres) are octahedrons with axes aligned with the co-ordinate axes. As the "radius" increases from $0$, the first surface to intersect a given plane will intersect it at a vertex (and possibly simultaneously at more than one vertex for some planes). So the minimum distance will lie along one of the lines $x=y=0$, $x=z=0$ or $y=z=0$.

For the $infty$-norm the surfaces of equal distance from the origin are cubes with faces aligned with the co-ordinate axes. Again, as the "radius" increases from $0$, the first surface to intersect a given plane will intersect it at a vertex of the cube. So the minimum distance will lie along the one of the lines $x=y=z$, $x=-y=z$ or $x=y=-z$.

share|cite|improve this answer

edited Mar 13 at 12:19

answered Mar 13 at 12:13

gandalf61gandalf61

9,109825

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Algebraically, if $Ax+By+Cz=D,$ it cannot be that each of $|x|,|y|,|z$| is less than $R=frac +,$ otherwise $ |Ax+By+Cz|le |A|R+|B|R+|C|R<|D|.$
  $endgroup$
  – DanielWainfleet
  Mar 13 at 13:12
  
  

  
  
  
  

add a comment | 

5

$begingroup$

For the $1$-norm the surfaces of equal distance from the origin (the equivalent of spheres) are octahedrons with axes aligned with the co-ordinate axes. As the "radius" increases from $0$, the first surface to intersect a given plane will intersect it at a vertex (and possibly simultaneously at more than one vertex for some planes). So the minimum distance will lie along one of the lines $x=y=0$, $x=z=0$ or $y=z=0$.

For the $infty$-norm the surfaces of equal distance from the origin are cubes with faces aligned with the co-ordinate axes. Again, as the "radius" increases from $0$, the first surface to intersect a given plane will intersect it at a vertex of the cube. So the minimum distance will lie along the one of the lines $x=y=z$, $x=-y=z$ or $x=y=-z$.

share|cite|improve this answer

edited Mar 13 at 12:19

answered Mar 13 at 12:13

gandalf61gandalf61

9,109825

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Algebraically, if $Ax+By+Cz=D,$ it cannot be that each of $|x|,|y|,|z$| is less than $R=frac +,$ otherwise $ |Ax+By+Cz|le |A|R+|B|R+|C|R<|D|.$
  $endgroup$
  – DanielWainfleet
  Mar 13 at 13:12
  
  

  
  
  
  

add a comment | 

5

$begingroup$

For the $1$-norm the surfaces of equal distance from the origin (the equivalent of spheres) are octahedrons with axes aligned with the co-ordinate axes. As the "radius" increases from $0$, the first surface to intersect a given plane will intersect it at a vertex (and possibly simultaneously at more than one vertex for some planes). So the minimum distance will lie along one of the lines $x=y=0$, $x=z=0$ or $y=z=0$.

For the $infty$-norm the surfaces of equal distance from the origin are cubes with faces aligned with the co-ordinate axes. Again, as the "radius" increases from $0$, the first surface to intersect a given plane will intersect it at a vertex of the cube. So the minimum distance will lie along the one of the lines $x=y=z$, $x=-y=z$ or $x=y=-z$.

share|cite|improve this answer

edited Mar 13 at 12:19

answered Mar 13 at 12:13

gandalf61gandalf61

9,109825

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Algebraically, if $Ax+By+Cz=D,$ it cannot be that each of $|x|,|y|,|z$| is less than $R=frac +,$ otherwise $ |Ax+By+Cz|le |A|R+|B|R+|C|R<|D|.$
  $endgroup$
  – DanielWainfleet
  Mar 13 at 13:12
  
  

  
  
  
  

add a comment | 

$begingroup$

For the $1$-norm the surfaces of equal distance from the origin (the equivalent of spheres) are octahedrons with axes aligned with the co-ordinate axes. As the "radius" increases from $0$, the first surface to intersect a given plane will intersect it at a vertex (and possibly simultaneously at more than one vertex for some planes). So the minimum distance will lie along one of the lines $x=y=0$, $x=z=0$ or $y=z=0$.

For the $infty$-norm the surfaces of equal distance from the origin are cubes with faces aligned with the co-ordinate axes. Again, as the "radius" increases from $0$, the first surface to intersect a given plane will intersect it at a vertex of the cube. So the minimum distance will lie along the one of the lines $x=y=z$, $x=-y=z$ or $x=y=-z$.

share|cite|improve this answer

edited Mar 13 at 12:19

answered Mar 13 at 12:13

gandalf61gandalf61

9,109825

$endgroup$

share|cite|improve this answer

edited Mar 13 at 12:19

answered Mar 13 at 12:13

gandalf61gandalf61

9,109825

answered Mar 13 at 12:13

gandalf61gandalf61

9,109825

answered Mar 13 at 12:13

gandalf61gandalf61

9,109825