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
1






active

oldest

votes


















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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3019598%2fproving-that-the-solution-set-of-axbycz-0-and-dxeyfz-0-is-a-line-in-ma%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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












      $begingroup$

      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

        38.3k33376




        38.3k33376






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3019598%2fproving-that-the-solution-set-of-axbycz-0-and-dxeyfz-0-is-a-line-in-ma%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

            Image
            2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
            Read more

            ComboBox Display Member on multiple fields

            Image
            2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
            Read more

            Is it possible to collect Nectar points via Trainline?

            Image
            2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
            Read more