Proving that the solution set of $ax+by+cz=0$ and $dx+ey+fz=0$ is a line in $mathbb{R}^3$












2












$begingroup$


I have a question here which says:



"The solution set of:



$$ax+by+cz=0$$
$$dx+ey+fz=0$$



is a line in $mathbb{R^3}$"



I am supposed to show that this is true or false. If it's true, I have to explain why and if it's false, I have to explain why it's false.



I don't think it's true. We COULD have a line if the planes intersected but if $a,b,c$ are scalar multiples of $d,e,f$, then we would just have the same plane since there are infinitely many solutions. Is that the right should process or did I mess up my reasoning?






linear-algebra geometry systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Awesome. Thanks :) .
    $endgroup$
    – Future Math person
    Nov 30 '18 at 3:45






  • 3




    $begingroup$
    There's also the possibility that the planes are parallel and therefore never intersect, in which case there are no solutions.
    $endgroup$
    – Robert Howard
    Nov 30 '18 at 3:45






  • 2




    $begingroup$
    @RobertHoward actually no solution is NOT a possibility here because of the homogeneous system, so either a line or a plane are the only possible solutions.
    $endgroup$
    – Anurag A
    Nov 30 '18 at 3:48






  • 1




    $begingroup$
    I caught it though so no worries!
    $endgroup$
    – Future Math person
    Nov 30 '18 at 4:03






  • 1




    $begingroup$
    @RobertHoward I’m astonished that your comment got three upvotes!
    $endgroup$
    – amd
    Nov 30 '18 at 5:06
















2












$begingroup$


I have a question here which says:



"The solution set of:



$$ax+by+cz=0$$
$$dx+ey+fz=0$$



is a line in $mathbb{R^3}$"



I am supposed to show that this is true or false. If it's true, I have to explain why and if it's false, I have to explain why it's false.



I don't think it's true. We COULD have a line if the planes intersected but if $a,b,c$ are scalar multiples of $d,e,f$, then we would just have the same plane since there are infinitely many solutions. Is that the right should process or did I mess up my reasoning?






linear-algebra geometry systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Awesome. Thanks :) .
    $endgroup$
    – Future Math person
    Nov 30 '18 at 3:45






  • 3




    $begingroup$
    There's also the possibility that the planes are parallel and therefore never intersect, in which case there are no solutions.
    $endgroup$
    – Robert Howard
    Nov 30 '18 at 3:45






  • 2




    $begingroup$
    @RobertHoward actually no solution is NOT a possibility here because of the homogeneous system, so either a line or a plane are the only possible solutions.
    $endgroup$
    – Anurag A
    Nov 30 '18 at 3:48






  • 1




    $begingroup$
    I caught it though so no worries!
    $endgroup$
    – Future Math person
    Nov 30 '18 at 4:03






  • 1




    $begingroup$
    @RobertHoward I’m astonished that your comment got three upvotes!
    $endgroup$
    – amd
    Nov 30 '18 at 5:06














2












2








2





$begingroup$


I have a question here which says:



"The solution set of:



$$ax+by+cz=0$$
$$dx+ey+fz=0$$



is a line in $mathbb{R^3}$"



I am supposed to show that this is true or false. If it's true, I have to explain why and if it's false, I have to explain why it's false.



I don't think it's true. We COULD have a line if the planes intersected but if $a,b,c$ are scalar multiples of $d,e,f$, then we would just have the same plane since there are infinitely many solutions. Is that the right should process or did I mess up my reasoning?






linear-algebra geometry systems-of-equations







share|cite|improve this question











$endgroup$




I have a question here which says:



"The solution set of:



$$ax+by+cz=0$$
$$dx+ey+fz=0$$



is a line in $mathbb{R^3}$"



I am supposed to show that this is true or false. If it's true, I have to explain why and if it's false, I have to explain why it's false.



I don't think it's true. We COULD have a line if the planes intersected but if $a,b,c$ are scalar multiples of $d,e,f$, then we would just have the same plane since there are infinitely many solutions. Is that the right should process or did I mess up my reasoning?






linear-algebra geometry systems-of-equations




linear-algebra geometry systems-of-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 5:08









Blue

48.4k870154




48.4k870154










asked Nov 30 '18 at 3:43









Future Math personFuture Math person

972817




972817












  • $begingroup$
    Awesome. Thanks :) .
    $endgroup$
    – Future Math person
    Nov 30 '18 at 3:45






  • 3




    $begingroup$
    There's also the possibility that the planes are parallel and therefore never intersect, in which case there are no solutions.
    $endgroup$
    – Robert Howard
    Nov 30 '18 at 3:45






  • 2




    $begingroup$
    @RobertHoward actually no solution is NOT a possibility here because of the homogeneous system, so either a line or a plane are the only possible solutions.
    $endgroup$
    – Anurag A
    Nov 30 '18 at 3:48






  • 1




    $begingroup$
    I caught it though so no worries!
    $endgroup$
    – Future Math person
    Nov 30 '18 at 4:03






  • 1




    $begingroup$
    @RobertHoward I’m astonished that your comment got three upvotes!
    $endgroup$
    – amd
    Nov 30 '18 at 5:06


















  • $begingroup$
    Awesome. Thanks :) .
    $endgroup$
    – Future Math person
    Nov 30 '18 at 3:45






  • 3




    $begingroup$
    There's also the possibility that the planes are parallel and therefore never intersect, in which case there are no solutions.
    $endgroup$
    – Robert Howard
    Nov 30 '18 at 3:45






  • 2




    $begingroup$
    @RobertHoward actually no solution is NOT a possibility here because of the homogeneous system, so either a line or a plane are the only possible solutions.
    $endgroup$
    – Anurag A
    Nov 30 '18 at 3:48






  • 1




    $begingroup$
    I caught it though so no worries!
    $endgroup$
    – Future Math person
    Nov 30 '18 at 4:03






  • 1




    $begingroup$
    @RobertHoward I’m astonished that your comment got three upvotes!
    $endgroup$
    – amd
    Nov 30 '18 at 5:06
















$begingroup$
Awesome. Thanks :) .
$endgroup$
– Future Math person
Nov 30 '18 at 3:45




$begingroup$
Awesome. Thanks :) .
$endgroup$
– Future Math person
Nov 30 '18 at 3:45




3




3




$begingroup$
There's also the possibility that the planes are parallel and therefore never intersect, in which case there are no solutions.
$endgroup$
– Robert Howard
Nov 30 '18 at 3:45




$begingroup$
There's also the possibility that the planes are parallel and therefore never intersect, in which case there are no solutions.
$endgroup$
– Robert Howard
Nov 30 '18 at 3:45




2




2




$begingroup$
@RobertHoward actually no solution is NOT a possibility here because of the homogeneous system, so either a line or a plane are the only possible solutions.
$endgroup$
– Anurag A
Nov 30 '18 at 3:48




$begingroup$
@RobertHoward actually no solution is NOT a possibility here because of the homogeneous system, so either a line or a plane are the only possible solutions.
$endgroup$
– Anurag A
Nov 30 '18 at 3:48




1




1




$begingroup$
I caught it though so no worries!
$endgroup$
– Future Math person
Nov 30 '18 at 4:03




$begingroup$
I caught it though so no worries!
$endgroup$
– Future Math person
Nov 30 '18 at 4:03




1




1




$begingroup$
@RobertHoward I’m astonished that your comment got three upvotes!
$endgroup$
– amd
Nov 30 '18 at 5:06




$begingroup$
@RobertHoward I’m astonished that your comment got three upvotes!
$endgroup$
– amd
Nov 30 '18 at 5:06










1 Answer
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$begingroup$

When absolutely nothing is provided about $a,b,c,d,e,f$ then for one you may even just take say all of them to be equal to zero, and then the system of solutions is the whole of $mathbb R^3$, which is not a line.



Of course, a plane is also obtainable : it is obtained exactly when the ratios $d:a,e:b,f:c$ are the same (else, the planes would not be parallel and hence would intersect at a line passing through zero). Then, the equations $ax+by+cz = 0$ and $dx+ey+fz = 0$ describe the same plane. This is clearly seen for example with $a,b,c,d,e,f$ all equal to $1$, or say $a,b,c=1$, $d,e,f = 2$. So, obviously the statement given can be false.



Note , however, that a solution must exist. This is because of the rank-nullity theorem, applied on the linear transformation $(x,y,z) to (ax+by+cz,dx+ey+fz)$. This is a map from $mathbb R^3$ to $mathbb R^2$. It has rank atmost $2$, therefore nullity at least one i.e. the null space has at least one line.






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    $begingroup$

    When absolutely nothing is provided about $a,b,c,d,e,f$ then for one you may even just take say all of them to be equal to zero, and then the system of solutions is the whole of $mathbb R^3$, which is not a line.



    Of course, a plane is also obtainable : it is obtained exactly when the ratios $d:a,e:b,f:c$ are the same (else, the planes would not be parallel and hence would intersect at a line passing through zero). Then, the equations $ax+by+cz = 0$ and $dx+ey+fz = 0$ describe the same plane. This is clearly seen for example with $a,b,c,d,e,f$ all equal to $1$, or say $a,b,c=1$, $d,e,f = 2$. So, obviously the statement given can be false.



    Note , however, that a solution must exist. This is because of the rank-nullity theorem, applied on the linear transformation $(x,y,z) to (ax+by+cz,dx+ey+fz)$. This is a map from $mathbb R^3$ to $mathbb R^2$. It has rank atmost $2$, therefore nullity at least one i.e. the null space has at least one line.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      When absolutely nothing is provided about $a,b,c,d,e,f$ then for one you may even just take say all of them to be equal to zero, and then the system of solutions is the whole of $mathbb R^3$, which is not a line.



      Of course, a plane is also obtainable : it is obtained exactly when the ratios $d:a,e:b,f:c$ are the same (else, the planes would not be parallel and hence would intersect at a line passing through zero). Then, the equations $ax+by+cz = 0$ and $dx+ey+fz = 0$ describe the same plane. This is clearly seen for example with $a,b,c,d,e,f$ all equal to $1$, or say $a,b,c=1$, $d,e,f = 2$. So, obviously the statement given can be false.



      Note , however, that a solution must exist. This is because of the rank-nullity theorem, applied on the linear transformation $(x,y,z) to (ax+by+cz,dx+ey+fz)$. This is a map from $mathbb R^3$ to $mathbb R^2$. It has rank atmost $2$, therefore nullity at least one i.e. the null space has at least one line.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        When absolutely nothing is provided about $a,b,c,d,e,f$ then for one you may even just take say all of them to be equal to zero, and then the system of solutions is the whole of $mathbb R^3$, which is not a line.



        Of course, a plane is also obtainable : it is obtained exactly when the ratios $d:a,e:b,f:c$ are the same (else, the planes would not be parallel and hence would intersect at a line passing through zero). Then, the equations $ax+by+cz = 0$ and $dx+ey+fz = 0$ describe the same plane. This is clearly seen for example with $a,b,c,d,e,f$ all equal to $1$, or say $a,b,c=1$, $d,e,f = 2$. So, obviously the statement given can be false.



        Note , however, that a solution must exist. This is because of the rank-nullity theorem, applied on the linear transformation $(x,y,z) to (ax+by+cz,dx+ey+fz)$. This is a map from $mathbb R^3$ to $mathbb R^2$. It has rank atmost $2$, therefore nullity at least one i.e. the null space has at least one line.






        share|cite|improve this answer









        $endgroup$



        When absolutely nothing is provided about $a,b,c,d,e,f$ then for one you may even just take say all of them to be equal to zero, and then the system of solutions is the whole of $mathbb R^3$, which is not a line.



        Of course, a plane is also obtainable : it is obtained exactly when the ratios $d:a,e:b,f:c$ are the same (else, the planes would not be parallel and hence would intersect at a line passing through zero). Then, the equations $ax+by+cz = 0$ and $dx+ey+fz = 0$ describe the same plane. This is clearly seen for example with $a,b,c,d,e,f$ all equal to $1$, or say $a,b,c=1$, $d,e,f = 2$. So, obviously the statement given can be false.



        Note , however, that a solution must exist. This is because of the rank-nullity theorem, applied on the linear transformation $(x,y,z) to (ax+by+cz,dx+ey+fz)$. This is a map from $mathbb R^3$ to $mathbb R^2$. It has rank atmost $2$, therefore nullity at least one i.e. the null space has at least one line.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 3:53









        астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

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