Intersection line and plane - vector to equation to matrix











up vote
0
down vote

favorite












I need to find the coordinates of the intersection of the following plane and line through (0,0,0):



enter image description here



Translation: "en" means "and"



I do this by writing out the equations, claiming them to be equal enter image description here



and thus creating a matrix to solve for $lambda, mu, sigma $:



enter image description here



So $lambda=3, mu=0, sigma=4$, but this isn't the answer as the equations are NOT equal for those values. I really do not see where I go wrong though...



I expected not to find any solution, as the line and the plane are parallel. Is the fact that the solution I find is not a real answer to the problem an indication of this? I do not think so. I think there is a mistake in rewriting the equations to the matrix, but I really do not see what goes wrong.



This all should be very easy :( But even after 1 hour I do not see my mistake. I would really appreciate some input on my line of thought.










share|cite|improve this question






















  • What system did you solve to get $textbf{x}$?
    – Archaick
    Apr 11 '15 at 18:38










  • The I rewrote the systems of equations to put all the terms with greek letters on the left (matrix A) and the numbers to the right (vector b). A times x should be b. So I sweeped matrix A | b. The first column of A is lambda, the second mu and the third sigma.
    – Robin
    Apr 11 '15 at 18:46












  • If $textbf{b}$ is a column vector with entries $lambda,mu,$ and $sigma$ in that order then $Atextbf{b}=textbf{0}$ will give you a solution for $textbf{b}$ which will give you $textbf{x}$.
    – Archaick
    Apr 11 '15 at 18:52










  • Do you see the images in the post? I used Matrix A and solution matrix b (not zero) to find x, thus in my system I have Ax = b. But the x which I find this way (row manipulation) is not an answer And there should be NO ANSWER POSSIBLE as the line and plane are parallel
    – Robin
    Apr 11 '15 at 19:14

















up vote
0
down vote

favorite












I need to find the coordinates of the intersection of the following plane and line through (0,0,0):



enter image description here



Translation: "en" means "and"



I do this by writing out the equations, claiming them to be equal enter image description here



and thus creating a matrix to solve for $lambda, mu, sigma $:



enter image description here



So $lambda=3, mu=0, sigma=4$, but this isn't the answer as the equations are NOT equal for those values. I really do not see where I go wrong though...



I expected not to find any solution, as the line and the plane are parallel. Is the fact that the solution I find is not a real answer to the problem an indication of this? I do not think so. I think there is a mistake in rewriting the equations to the matrix, but I really do not see what goes wrong.



This all should be very easy :( But even after 1 hour I do not see my mistake. I would really appreciate some input on my line of thought.










share|cite|improve this question






















  • What system did you solve to get $textbf{x}$?
    – Archaick
    Apr 11 '15 at 18:38










  • The I rewrote the systems of equations to put all the terms with greek letters on the left (matrix A) and the numbers to the right (vector b). A times x should be b. So I sweeped matrix A | b. The first column of A is lambda, the second mu and the third sigma.
    – Robin
    Apr 11 '15 at 18:46












  • If $textbf{b}$ is a column vector with entries $lambda,mu,$ and $sigma$ in that order then $Atextbf{b}=textbf{0}$ will give you a solution for $textbf{b}$ which will give you $textbf{x}$.
    – Archaick
    Apr 11 '15 at 18:52










  • Do you see the images in the post? I used Matrix A and solution matrix b (not zero) to find x, thus in my system I have Ax = b. But the x which I find this way (row manipulation) is not an answer And there should be NO ANSWER POSSIBLE as the line and plane are parallel
    – Robin
    Apr 11 '15 at 19:14















up vote
0
down vote

favorite









up vote
0
down vote

favorite











I need to find the coordinates of the intersection of the following plane and line through (0,0,0):



enter image description here



Translation: "en" means "and"



I do this by writing out the equations, claiming them to be equal enter image description here



and thus creating a matrix to solve for $lambda, mu, sigma $:



enter image description here



So $lambda=3, mu=0, sigma=4$, but this isn't the answer as the equations are NOT equal for those values. I really do not see where I go wrong though...



I expected not to find any solution, as the line and the plane are parallel. Is the fact that the solution I find is not a real answer to the problem an indication of this? I do not think so. I think there is a mistake in rewriting the equations to the matrix, but I really do not see what goes wrong.



This all should be very easy :( But even after 1 hour I do not see my mistake. I would really appreciate some input on my line of thought.










share|cite|improve this question













I need to find the coordinates of the intersection of the following plane and line through (0,0,0):



enter image description here



Translation: "en" means "and"



I do this by writing out the equations, claiming them to be equal enter image description here



and thus creating a matrix to solve for $lambda, mu, sigma $:



enter image description here



So $lambda=3, mu=0, sigma=4$, but this isn't the answer as the equations are NOT equal for those values. I really do not see where I go wrong though...



I expected not to find any solution, as the line and the plane are parallel. Is the fact that the solution I find is not a real answer to the problem an indication of this? I do not think so. I think there is a mistake in rewriting the equations to the matrix, but I really do not see what goes wrong.



This all should be very easy :( But even after 1 hour I do not see my mistake. I would really appreciate some input on my line of thought.







vectors






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Apr 11 '15 at 18:23









Robin

617




617












  • What system did you solve to get $textbf{x}$?
    – Archaick
    Apr 11 '15 at 18:38










  • The I rewrote the systems of equations to put all the terms with greek letters on the left (matrix A) and the numbers to the right (vector b). A times x should be b. So I sweeped matrix A | b. The first column of A is lambda, the second mu and the third sigma.
    – Robin
    Apr 11 '15 at 18:46












  • If $textbf{b}$ is a column vector with entries $lambda,mu,$ and $sigma$ in that order then $Atextbf{b}=textbf{0}$ will give you a solution for $textbf{b}$ which will give you $textbf{x}$.
    – Archaick
    Apr 11 '15 at 18:52










  • Do you see the images in the post? I used Matrix A and solution matrix b (not zero) to find x, thus in my system I have Ax = b. But the x which I find this way (row manipulation) is not an answer And there should be NO ANSWER POSSIBLE as the line and plane are parallel
    – Robin
    Apr 11 '15 at 19:14




















  • What system did you solve to get $textbf{x}$?
    – Archaick
    Apr 11 '15 at 18:38










  • The I rewrote the systems of equations to put all the terms with greek letters on the left (matrix A) and the numbers to the right (vector b). A times x should be b. So I sweeped matrix A | b. The first column of A is lambda, the second mu and the third sigma.
    – Robin
    Apr 11 '15 at 18:46












  • If $textbf{b}$ is a column vector with entries $lambda,mu,$ and $sigma$ in that order then $Atextbf{b}=textbf{0}$ will give you a solution for $textbf{b}$ which will give you $textbf{x}$.
    – Archaick
    Apr 11 '15 at 18:52










  • Do you see the images in the post? I used Matrix A and solution matrix b (not zero) to find x, thus in my system I have Ax = b. But the x which I find this way (row manipulation) is not an answer And there should be NO ANSWER POSSIBLE as the line and plane are parallel
    – Robin
    Apr 11 '15 at 19:14


















What system did you solve to get $textbf{x}$?
– Archaick
Apr 11 '15 at 18:38




What system did you solve to get $textbf{x}$?
– Archaick
Apr 11 '15 at 18:38












The I rewrote the systems of equations to put all the terms with greek letters on the left (matrix A) and the numbers to the right (vector b). A times x should be b. So I sweeped matrix A | b. The first column of A is lambda, the second mu and the third sigma.
– Robin
Apr 11 '15 at 18:46






The I rewrote the systems of equations to put all the terms with greek letters on the left (matrix A) and the numbers to the right (vector b). A times x should be b. So I sweeped matrix A | b. The first column of A is lambda, the second mu and the third sigma.
– Robin
Apr 11 '15 at 18:46














If $textbf{b}$ is a column vector with entries $lambda,mu,$ and $sigma$ in that order then $Atextbf{b}=textbf{0}$ will give you a solution for $textbf{b}$ which will give you $textbf{x}$.
– Archaick
Apr 11 '15 at 18:52




If $textbf{b}$ is a column vector with entries $lambda,mu,$ and $sigma$ in that order then $Atextbf{b}=textbf{0}$ will give you a solution for $textbf{b}$ which will give you $textbf{x}$.
– Archaick
Apr 11 '15 at 18:52












Do you see the images in the post? I used Matrix A and solution matrix b (not zero) to find x, thus in my system I have Ax = b. But the x which I find this way (row manipulation) is not an answer And there should be NO ANSWER POSSIBLE as the line and plane are parallel
– Robin
Apr 11 '15 at 19:14






Do you see the images in the post? I used Matrix A and solution matrix b (not zero) to find x, thus in my system I have Ax = b. But the x which I find this way (row manipulation) is not an answer And there should be NO ANSWER POSSIBLE as the line and plane are parallel
– Robin
Apr 11 '15 at 19:14












1 Answer
1






active

oldest

votes

















up vote
0
down vote













The system of equations you're trying to express as a matrix operation is:



$sigmaleft(
begin{array}{c}
1\
-2\
-4\
end{array}
right)=left(
begin{array}{c}
3\
0\
-2\
end{array}
right)+lambdaleft(
begin{array}{c}
1\
1\
1\
end{array}
right)+muleft(
begin{array}{c}
2\
-1\
-3\
end{array}
right)$



Hence, $
begin{pmatrix}
1 & 2 & -1 \
1 & -1 & 2 \
1 & -3 & 4
end{pmatrix} begin{pmatrix}
lambda \
mu \
sigma
end{pmatrix} =begin{pmatrix}
-3 \
0 \
2
end{pmatrix}$



Row reducing (what you were referring to as sweeping, I believe) will give you solutions for $lambda, mu,$ and $sigma$. In order to find $textbf{x}$, we use the fact that



$sigmabegin{pmatrix}
1 \
-2 \
-4
end{pmatrix}=textbf{x}$






share|cite|improve this answer





















  • Yes, thank you. That is what I did. But the solutions I find this way ($lambda=3, mu=0, sigma=4$) are not right, for they do not fit. I substracted row 1 (R1) from R2 and R3. Then I added 3*R3 to R2 and this way I got $sigma=-1, mu=0, lambda=-2$. Using the software Maple to solve it gave the answers in the first post though. Neither are actual answers to the original system. So something goes wrong during the calculation of these systems. It shouldn't, but it does... And I don't see where and it really bugs me.
    – Robin
    Apr 11 '15 at 19:57












  • Row reducing will yield that $0=1$, in the equation that would normally yield $sigma$; that is, that there is no solution. I noticed that there is a mistake in your matrix, maybe that accounts for it: check the third row, second column entry.
    – Archaick
    Apr 11 '15 at 20:16












  • Thank you so much! That was what I was looking for. You just made my day :D
    – Robin
    Apr 11 '15 at 20:21










  • No problem, glad to help.
    – Archaick
    Apr 11 '15 at 20:26











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',
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%2f1230310%2fintersection-line-and-plane-vector-to-equation-to-matrix%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








up vote
0
down vote













The system of equations you're trying to express as a matrix operation is:



$sigmaleft(
begin{array}{c}
1\
-2\
-4\
end{array}
right)=left(
begin{array}{c}
3\
0\
-2\
end{array}
right)+lambdaleft(
begin{array}{c}
1\
1\
1\
end{array}
right)+muleft(
begin{array}{c}
2\
-1\
-3\
end{array}
right)$



Hence, $
begin{pmatrix}
1 & 2 & -1 \
1 & -1 & 2 \
1 & -3 & 4
end{pmatrix} begin{pmatrix}
lambda \
mu \
sigma
end{pmatrix} =begin{pmatrix}
-3 \
0 \
2
end{pmatrix}$



Row reducing (what you were referring to as sweeping, I believe) will give you solutions for $lambda, mu,$ and $sigma$. In order to find $textbf{x}$, we use the fact that



$sigmabegin{pmatrix}
1 \
-2 \
-4
end{pmatrix}=textbf{x}$






share|cite|improve this answer





















  • Yes, thank you. That is what I did. But the solutions I find this way ($lambda=3, mu=0, sigma=4$) are not right, for they do not fit. I substracted row 1 (R1) from R2 and R3. Then I added 3*R3 to R2 and this way I got $sigma=-1, mu=0, lambda=-2$. Using the software Maple to solve it gave the answers in the first post though. Neither are actual answers to the original system. So something goes wrong during the calculation of these systems. It shouldn't, but it does... And I don't see where and it really bugs me.
    – Robin
    Apr 11 '15 at 19:57












  • Row reducing will yield that $0=1$, in the equation that would normally yield $sigma$; that is, that there is no solution. I noticed that there is a mistake in your matrix, maybe that accounts for it: check the third row, second column entry.
    – Archaick
    Apr 11 '15 at 20:16












  • Thank you so much! That was what I was looking for. You just made my day :D
    – Robin
    Apr 11 '15 at 20:21










  • No problem, glad to help.
    – Archaick
    Apr 11 '15 at 20:26















up vote
0
down vote













The system of equations you're trying to express as a matrix operation is:



$sigmaleft(
begin{array}{c}
1\
-2\
-4\
end{array}
right)=left(
begin{array}{c}
3\
0\
-2\
end{array}
right)+lambdaleft(
begin{array}{c}
1\
1\
1\
end{array}
right)+muleft(
begin{array}{c}
2\
-1\
-3\
end{array}
right)$



Hence, $
begin{pmatrix}
1 & 2 & -1 \
1 & -1 & 2 \
1 & -3 & 4
end{pmatrix} begin{pmatrix}
lambda \
mu \
sigma
end{pmatrix} =begin{pmatrix}
-3 \
0 \
2
end{pmatrix}$



Row reducing (what you were referring to as sweeping, I believe) will give you solutions for $lambda, mu,$ and $sigma$. In order to find $textbf{x}$, we use the fact that



$sigmabegin{pmatrix}
1 \
-2 \
-4
end{pmatrix}=textbf{x}$






share|cite|improve this answer





















  • Yes, thank you. That is what I did. But the solutions I find this way ($lambda=3, mu=0, sigma=4$) are not right, for they do not fit. I substracted row 1 (R1) from R2 and R3. Then I added 3*R3 to R2 and this way I got $sigma=-1, mu=0, lambda=-2$. Using the software Maple to solve it gave the answers in the first post though. Neither are actual answers to the original system. So something goes wrong during the calculation of these systems. It shouldn't, but it does... And I don't see where and it really bugs me.
    – Robin
    Apr 11 '15 at 19:57












  • Row reducing will yield that $0=1$, in the equation that would normally yield $sigma$; that is, that there is no solution. I noticed that there is a mistake in your matrix, maybe that accounts for it: check the third row, second column entry.
    – Archaick
    Apr 11 '15 at 20:16












  • Thank you so much! That was what I was looking for. You just made my day :D
    – Robin
    Apr 11 '15 at 20:21










  • No problem, glad to help.
    – Archaick
    Apr 11 '15 at 20:26













up vote
0
down vote










up vote
0
down vote









The system of equations you're trying to express as a matrix operation is:



$sigmaleft(
begin{array}{c}
1\
-2\
-4\
end{array}
right)=left(
begin{array}{c}
3\
0\
-2\
end{array}
right)+lambdaleft(
begin{array}{c}
1\
1\
1\
end{array}
right)+muleft(
begin{array}{c}
2\
-1\
-3\
end{array}
right)$



Hence, $
begin{pmatrix}
1 & 2 & -1 \
1 & -1 & 2 \
1 & -3 & 4
end{pmatrix} begin{pmatrix}
lambda \
mu \
sigma
end{pmatrix} =begin{pmatrix}
-3 \
0 \
2
end{pmatrix}$



Row reducing (what you were referring to as sweeping, I believe) will give you solutions for $lambda, mu,$ and $sigma$. In order to find $textbf{x}$, we use the fact that



$sigmabegin{pmatrix}
1 \
-2 \
-4
end{pmatrix}=textbf{x}$






share|cite|improve this answer












The system of equations you're trying to express as a matrix operation is:



$sigmaleft(
begin{array}{c}
1\
-2\
-4\
end{array}
right)=left(
begin{array}{c}
3\
0\
-2\
end{array}
right)+lambdaleft(
begin{array}{c}
1\
1\
1\
end{array}
right)+muleft(
begin{array}{c}
2\
-1\
-3\
end{array}
right)$



Hence, $
begin{pmatrix}
1 & 2 & -1 \
1 & -1 & 2 \
1 & -3 & 4
end{pmatrix} begin{pmatrix}
lambda \
mu \
sigma
end{pmatrix} =begin{pmatrix}
-3 \
0 \
2
end{pmatrix}$



Row reducing (what you were referring to as sweeping, I believe) will give you solutions for $lambda, mu,$ and $sigma$. In order to find $textbf{x}$, we use the fact that



$sigmabegin{pmatrix}
1 \
-2 \
-4
end{pmatrix}=textbf{x}$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Apr 11 '15 at 19:47









Archaick

1,781625




1,781625












  • Yes, thank you. That is what I did. But the solutions I find this way ($lambda=3, mu=0, sigma=4$) are not right, for they do not fit. I substracted row 1 (R1) from R2 and R3. Then I added 3*R3 to R2 and this way I got $sigma=-1, mu=0, lambda=-2$. Using the software Maple to solve it gave the answers in the first post though. Neither are actual answers to the original system. So something goes wrong during the calculation of these systems. It shouldn't, but it does... And I don't see where and it really bugs me.
    – Robin
    Apr 11 '15 at 19:57












  • Row reducing will yield that $0=1$, in the equation that would normally yield $sigma$; that is, that there is no solution. I noticed that there is a mistake in your matrix, maybe that accounts for it: check the third row, second column entry.
    – Archaick
    Apr 11 '15 at 20:16












  • Thank you so much! That was what I was looking for. You just made my day :D
    – Robin
    Apr 11 '15 at 20:21










  • No problem, glad to help.
    – Archaick
    Apr 11 '15 at 20:26


















  • Yes, thank you. That is what I did. But the solutions I find this way ($lambda=3, mu=0, sigma=4$) are not right, for they do not fit. I substracted row 1 (R1) from R2 and R3. Then I added 3*R3 to R2 and this way I got $sigma=-1, mu=0, lambda=-2$. Using the software Maple to solve it gave the answers in the first post though. Neither are actual answers to the original system. So something goes wrong during the calculation of these systems. It shouldn't, but it does... And I don't see where and it really bugs me.
    – Robin
    Apr 11 '15 at 19:57












  • Row reducing will yield that $0=1$, in the equation that would normally yield $sigma$; that is, that there is no solution. I noticed that there is a mistake in your matrix, maybe that accounts for it: check the third row, second column entry.
    – Archaick
    Apr 11 '15 at 20:16












  • Thank you so much! That was what I was looking for. You just made my day :D
    – Robin
    Apr 11 '15 at 20:21










  • No problem, glad to help.
    – Archaick
    Apr 11 '15 at 20:26
















Yes, thank you. That is what I did. But the solutions I find this way ($lambda=3, mu=0, sigma=4$) are not right, for they do not fit. I substracted row 1 (R1) from R2 and R3. Then I added 3*R3 to R2 and this way I got $sigma=-1, mu=0, lambda=-2$. Using the software Maple to solve it gave the answers in the first post though. Neither are actual answers to the original system. So something goes wrong during the calculation of these systems. It shouldn't, but it does... And I don't see where and it really bugs me.
– Robin
Apr 11 '15 at 19:57






Yes, thank you. That is what I did. But the solutions I find this way ($lambda=3, mu=0, sigma=4$) are not right, for they do not fit. I substracted row 1 (R1) from R2 and R3. Then I added 3*R3 to R2 and this way I got $sigma=-1, mu=0, lambda=-2$. Using the software Maple to solve it gave the answers in the first post though. Neither are actual answers to the original system. So something goes wrong during the calculation of these systems. It shouldn't, but it does... And I don't see where and it really bugs me.
– Robin
Apr 11 '15 at 19:57














Row reducing will yield that $0=1$, in the equation that would normally yield $sigma$; that is, that there is no solution. I noticed that there is a mistake in your matrix, maybe that accounts for it: check the third row, second column entry.
– Archaick
Apr 11 '15 at 20:16






Row reducing will yield that $0=1$, in the equation that would normally yield $sigma$; that is, that there is no solution. I noticed that there is a mistake in your matrix, maybe that accounts for it: check the third row, second column entry.
– Archaick
Apr 11 '15 at 20:16














Thank you so much! That was what I was looking for. You just made my day :D
– Robin
Apr 11 '15 at 20:21




Thank you so much! That was what I was looking for. You just made my day :D
– Robin
Apr 11 '15 at 20:21












No problem, glad to help.
– Archaick
Apr 11 '15 at 20:26




No problem, glad to help.
– Archaick
Apr 11 '15 at 20:26


















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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


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%2f1230310%2fintersection-line-and-plane-vector-to-equation-to-matrix%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

Berounka

Sphinx de Gizeh

Different font size/position of beamer's navigation symbols template's content depending on regular/plain...