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Homeostasis logic/math problem

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Intersection of 3 planes in 3D space

Limiting value of a sequence when n tends to infinityDifference between planes intersecting along a common line and coincidingIntersection of 3 rotated non-orthogonal planesExample of straight line and a point in $Bbb R^3$ such that there are infinitely many planes passing through itCalculate point of intersection line of two planesFinding solution to system of equationsIntersection of 2D planes in 4D spaceWhat is a geometric interpretation of all these information?Why can't we find the intersection line of two planes just by using algebra?Linear independence in linear applicationsChanging $b$ of $AX =b$

2

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Q) Let $P_{1},P_{2},$ and $P_{3}$ denote, respectively, the planes defined by

$a_{1}x + b_{1}y + c_{1}z =alpha_{1}$

$a_{2}x + b_{2}y + c_{2}z = alpha _{2}$

$a_{3}x + b_{3}y + c_{3}z = alpha _{3}$

It is given that $P_{1},P_{2},$ and $P_{3}$ intersect exactly at one point when $alpha _{1}= alpha _{2}= alpha _{3}=1$ . If now $alpha _{1}=2, alpha _{2}=3 ;and ; alpha _{3}=4$ then the planes

(A) do not have any common point of intersection

(B) intersect at a unique point

(C) intersect along a straight line

(D) intersect along a plane

My Approach :- Since, System of equations $Ax=b$ has a solution (either unique or infinitely many) when vector b is in the column space of $A$. So, for given vector $b=(1,1,1)$ , $Ax=b$ gives unique solution it means vector $b$ is in the column space of $A$ and all $3$ vectors of $A$ i.e. $(a_{1},a_{2},a_{3}),(b_{1},b_{2},b_{3}),(c_{1},c_{2},c_{3})$ are linearly independent and So, matrix formed by these $3$ vectors is invertible.

Now, if vector $b$ changed into $(2,3,4)$ then how to sure that this vector will be in the same column space. If it is in the same column space then it will give unique solution because all the vectors of $A$ will still be linearly independent and if it is not in the column space of $A$ then there will be no solution , So these $3$ planes neither cut at a point nor a straight line(infinitely many solution). So, I am confused with option $(A)$ and $(B)$. Please help.

linear-algebra vector-spaces

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asked 3 hours ago

ankitankit

275

$endgroup$

add a comment | 

2

$begingroup$

Q) Let $P_{1},P_{2},$ and $P_{3}$ denote, respectively, the planes defined by

$a_{1}x + b_{1}y + c_{1}z =alpha_{1}$

$a_{2}x + b_{2}y + c_{2}z = alpha _{2}$

$a_{3}x + b_{3}y + c_{3}z = alpha _{3}$

It is given that $P_{1},P_{2},$ and $P_{3}$ intersect exactly at one point when $alpha _{1}= alpha _{2}= alpha _{3}=1$ . If now $alpha _{1}=2, alpha _{2}=3 ;and ; alpha _{3}=4$ then the planes

(A) do not have any common point of intersection

(B) intersect at a unique point

(C) intersect along a straight line

(D) intersect along a plane

My Approach :- Since, System of equations $Ax=b$ has a solution (either unique or infinitely many) when vector b is in the column space of $A$. So, for given vector $b=(1,1,1)$ , $Ax=b$ gives unique solution it means vector $b$ is in the column space of $A$ and all $3$ vectors of $A$ i.e. $(a_{1},a_{2},a_{3}),(b_{1},b_{2},b_{3}),(c_{1},c_{2},c_{3})$ are linearly independent and So, matrix formed by these $3$ vectors is invertible.

Now, if vector $b$ changed into $(2,3,4)$ then how to sure that this vector will be in the same column space. If it is in the same column space then it will give unique solution because all the vectors of $A$ will still be linearly independent and if it is not in the column space of $A$ then there will be no solution , So these $3$ planes neither cut at a point nor a straight line(infinitely many solution). So, I am confused with option $(A)$ and $(B)$. Please help.

linear-algebra vector-spaces

share|cite|improve this question

asked 3 hours ago

ankitankit

275

$endgroup$

add a comment | 

$begingroup$

Q) Let $P_{1},P_{2},$ and $P_{3}$ denote, respectively, the planes defined by

$a_{1}x + b_{1}y + c_{1}z =alpha_{1}$

$a_{2}x + b_{2}y + c_{2}z = alpha _{2}$

$a_{3}x + b_{3}y + c_{3}z = alpha _{3}$

It is given that $P_{1},P_{2},$ and $P_{3}$ intersect exactly at one point when $alpha _{1}= alpha _{2}= alpha _{3}=1$ . If now $alpha _{1}=2, alpha _{2}=3 ;and ; alpha _{3}=4$ then the planes

(A) do not have any common point of intersection

(B) intersect at a unique point

(C) intersect along a straight line

(D) intersect along a plane

My Approach :- Since, System of equations $Ax=b$ has a solution (either unique or infinitely many) when vector b is in the column space of $A$. So, for given vector $b=(1,1,1)$ , $Ax=b$ gives unique solution it means vector $b$ is in the column space of $A$ and all $3$ vectors of $A$ i.e. $(a_{1},a_{2},a_{3}),(b_{1},b_{2},b_{3}),(c_{1},c_{2},c_{3})$ are linearly independent and So, matrix formed by these $3$ vectors is invertible.

Now, if vector $b$ changed into $(2,3,4)$ then how to sure that this vector will be in the same column space. If it is in the same column space then it will give unique solution because all the vectors of $A$ will still be linearly independent and if it is not in the column space of $A$ then there will be no solution , So these $3$ planes neither cut at a point nor a straight line(infinitely many solution). So, I am confused with option $(A)$ and $(B)$. Please help.

linear-algebra vector-spaces

share|cite|improve this question

asked 3 hours ago

ankitankit

275

$endgroup$

share|cite|improve this question

asked 3 hours ago

ankitankit

275

share|cite|improve this question

asked 3 hours ago

ankitankit

275

asked 3 hours ago

ankitankit

275

asked 3 hours ago

ankitankit

275

3 Answers
3

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2

$begingroup$

The key thing you pointed out is that $Ax = b$ having a unique solution for some $b$ shows that the columns of $A$ are linearly independent. In particular, the columns of $A$ are three linearly independent vectors in $mathbb{R}^3$, so they actually form a full basis for $mathbb{R}^3$. By the definition of a basis, every vector in $mathbb{R}^3$, such as the vector $(2,3,4)$, can be uniquely written as a linear combination of $A$'s columns. This unique linear combination corresponds to the unique solution to $Ax = (2,3,4)$, so the answer is (B).

share|cite|improve this answer

answered 3 hours ago

LiamLiam

766

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add a comment | 

2

$begingroup$

The values of $alpha_i$ are irrelevant, only that the planes intersect at a unique point. The coefficient matrix is thus of full rank, and for any given $alpha_i$ said matrix can be inverted and left-multiplied with the $alpha_i$ vector to yield a unique solution for $x,y,z$. Thus B is the correct answer.

share|cite|improve this answer

answered 3 hours ago

Parcly TaxelParcly Taxel

42.1k1372101

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks @Parcly Taxel
  $endgroup$
  – ankit
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ankit Really? Why don't you upvote and accept my answer?
  $endgroup$
  – Parcly Taxel
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have already upvoted all the answers. I have less than 15 reputation that's why it is not showing. I have selected that answer because that includes the concept of basis and spanning a vector space which I forgot this concept while solving this problem. I really appreciate your and other's efforts while answering the answer of my query :)
  $endgroup$
  – ankit
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ankit Go ask another question.
  $endgroup$
  – Parcly Taxel
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/3124615/…
  $endgroup$
  – ankit
  2 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Since $Ax=b$ has unique solution for a particular $b$, we know that $A$ is invertible; i.e. $A^{-1}$ exists. Therefore, no matter what other $b$ you pick, there will still be the solution $x=A^{-1}b$ to the equation. Can you continue?

share|cite|improve this answer

answered 3 hours ago

YiFanYiFan

4,2291627

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  yes, Thank you !.
  $endgroup$
  – ankit
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can the downvoter please explain?
  $endgroup$
  – YiFan
  2 hours ago
  

  
  
  
  

add a comment | 

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2

$begingroup$

The key thing you pointed out is that $Ax = b$ having a unique solution for some $b$ shows that the columns of $A$ are linearly independent. In particular, the columns of $A$ are three linearly independent vectors in $mathbb{R}^3$, so they actually form a full basis for $mathbb{R}^3$. By the definition of a basis, every vector in $mathbb{R}^3$, such as the vector $(2,3,4)$, can be uniquely written as a linear combination of $A$'s columns. This unique linear combination corresponds to the unique solution to $Ax = (2,3,4)$, so the answer is (B).

share|cite|improve this answer

answered 3 hours ago

LiamLiam

766

$endgroup$

add a comment | 

2

$begingroup$

The key thing you pointed out is that $Ax = b$ having a unique solution for some $b$ shows that the columns of $A$ are linearly independent. In particular, the columns of $A$ are three linearly independent vectors in $mathbb{R}^3$, so they actually form a full basis for $mathbb{R}^3$. By the definition of a basis, every vector in $mathbb{R}^3$, such as the vector $(2,3,4)$, can be uniquely written as a linear combination of $A$'s columns. This unique linear combination corresponds to the unique solution to $Ax = (2,3,4)$, so the answer is (B).

share|cite|improve this answer

answered 3 hours ago

LiamLiam

766

$endgroup$

add a comment | 

$begingroup$

The key thing you pointed out is that $Ax = b$ having a unique solution for some $b$ shows that the columns of $A$ are linearly independent. In particular, the columns of $A$ are three linearly independent vectors in $mathbb{R}^3$, so they actually form a full basis for $mathbb{R}^3$. By the definition of a basis, every vector in $mathbb{R}^3$, such as the vector $(2,3,4)$, can be uniquely written as a linear combination of $A$'s columns. This unique linear combination corresponds to the unique solution to $Ax = (2,3,4)$, so the answer is (B).

share|cite|improve this answer

answered 3 hours ago

LiamLiam

766

$endgroup$

share|cite|improve this answer

answered 3 hours ago

LiamLiam

766

answered 3 hours ago

LiamLiam

766

answered 3 hours ago

LiamLiam

766

2

$begingroup$

The values of $alpha_i$ are irrelevant, only that the planes intersect at a unique point. The coefficient matrix is thus of full rank, and for any given $alpha_i$ said matrix can be inverted and left-multiplied with the $alpha_i$ vector to yield a unique solution for $x,y,z$. Thus B is the correct answer.

share|cite|improve this answer

answered 3 hours ago

Parcly TaxelParcly Taxel

42.1k1372101

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks @Parcly Taxel
  $endgroup$
  – ankit
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ankit Really? Why don't you upvote and accept my answer?
  $endgroup$
  – Parcly Taxel
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have already upvoted all the answers. I have less than 15 reputation that's why it is not showing. I have selected that answer because that includes the concept of basis and spanning a vector space which I forgot this concept while solving this problem. I really appreciate your and other's efforts while answering the answer of my query :)
  $endgroup$
  – ankit
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ankit Go ask another question.
  $endgroup$
  – Parcly Taxel
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/3124615/…
  $endgroup$
  – ankit
  2 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

The values of $alpha_i$ are irrelevant, only that the planes intersect at a unique point. The coefficient matrix is thus of full rank, and for any given $alpha_i$ said matrix can be inverted and left-multiplied with the $alpha_i$ vector to yield a unique solution for $x,y,z$. Thus B is the correct answer.

share|cite|improve this answer

answered 3 hours ago

Parcly TaxelParcly Taxel

42.1k1372101

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks @Parcly Taxel
  $endgroup$
  – ankit
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ankit Really? Why don't you upvote and accept my answer?
  $endgroup$
  – Parcly Taxel
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have already upvoted all the answers. I have less than 15 reputation that's why it is not showing. I have selected that answer because that includes the concept of basis and spanning a vector space which I forgot this concept while solving this problem. I really appreciate your and other's efforts while answering the answer of my query :)
  $endgroup$
  – ankit
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ankit Go ask another question.
  $endgroup$
  – Parcly Taxel
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/3124615/…
  $endgroup$
  – ankit
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

The values of $alpha_i$ are irrelevant, only that the planes intersect at a unique point. The coefficient matrix is thus of full rank, and for any given $alpha_i$ said matrix can be inverted and left-multiplied with the $alpha_i$ vector to yield a unique solution for $x,y,z$. Thus B is the correct answer.

share|cite|improve this answer

answered 3 hours ago

Parcly TaxelParcly Taxel

42.1k1372101

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Parcly TaxelParcly Taxel

42.1k1372101

answered 3 hours ago

Parcly TaxelParcly Taxel

42.1k1372101

answered 3 hours ago

Parcly TaxelParcly Taxel

42.1k1372101

1

$begingroup$

Since $Ax=b$ has unique solution for a particular $b$, we know that $A$ is invertible; i.e. $A^{-1}$ exists. Therefore, no matter what other $b$ you pick, there will still be the solution $x=A^{-1}b$ to the equation. Can you continue?

share|cite|improve this answer

answered 3 hours ago

YiFanYiFan

4,2291627

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  yes, Thank you !.
  $endgroup$
  – ankit
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can the downvoter please explain?
  $endgroup$
  – YiFan
  2 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

Since $Ax=b$ has unique solution for a particular $b$, we know that $A$ is invertible; i.e. $A^{-1}$ exists. Therefore, no matter what other $b$ you pick, there will still be the solution $x=A^{-1}b$ to the equation. Can you continue?

share|cite|improve this answer

answered 3 hours ago

YiFanYiFan

4,2291627

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  yes, Thank you !.
  $endgroup$
  – ankit
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can the downvoter please explain?
  $endgroup$
  – YiFan
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Since $Ax=b$ has unique solution for a particular $b$, we know that $A$ is invertible; i.e. $A^{-1}$ exists. Therefore, no matter what other $b$ you pick, there will still be the solution $x=A^{-1}b$ to the equation. Can you continue?

share|cite|improve this answer

answered 3 hours ago

YiFanYiFan

4,2291627

$endgroup$

share|cite|improve this answer

answered 3 hours ago

YiFanYiFan

4,2291627

answered 3 hours ago

YiFanYiFan

4,2291627

answered 3 hours ago

YiFanYiFan

4,2291627