Intersection line and plane - vector to equation to matrix
I need to find the coordinates of the intersection of the following plane and line through (0,0,0):
Translation: "en" means "and"
I do this by writing out the equations, claiming them to be equal 
and thus creating a matrix to solve for $lambda, mu, sigma $:
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
asked Apr 11 '15 at 18:23
Robin
617
add a comment |
I need to find the coordinates of the intersection of the following plane and line through (0,0,0):
Translation: "en" means "and"
I do this by writing out the equations, claiming them to be equal
and thus creating a matrix to solve for $lambda, mu, sigma $:
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
asked Apr 11 '15 at 18:23
Robin
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
add a comment |
I need to find the coordinates of the intersection of the following plane and line through (0,0,0):
Translation: "en" means "and"
I do this by writing out the equations, claiming them to be equal
and thus creating a matrix to solve for $lambda, mu, sigma $:
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
asked Apr 11 '15 at 18:23
Robin
617
I need to find the coordinates of the intersection of the following plane and line through (0,0,0):
Translation: "en" means "and"
I do this by writing out the equations, claiming them to be equal
and thus creating a matrix to solve for $lambda, mu, sigma $:
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
vectors
asked Apr 11 '15 at 18:23
Robin
617
asked Apr 11 '15 at 18:23
Robin
617
asked Apr 11 '15 at 18:23
Robin
617
asked Apr 11 '15 at 18:23
Robin
617
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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
$sigmabegin{pmatrix}
1 \
-2 \
-4
end{pmatrix}=textbf{x}$
answered Apr 11 '15 at 19:47
Archaick
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
add a comment |
Your Answer
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1 Answer
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1 Answer
1
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oldest
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active
oldest
votes
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
$sigmabegin{pmatrix}
1 \
-2 \
-4
end{pmatrix}=textbf{x}$
answered Apr 11 '15 at 19:47
Archaick
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
add a comment |
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
$sigmabegin{pmatrix}
1 \
-2 \
-4
end{pmatrix}=textbf{x}$
answered Apr 11 '15 at 19:47
Archaick
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
add a comment |
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
$sigmabegin{pmatrix}
1 \
-2 \
-4
end{pmatrix}=textbf{x}$
answered Apr 11 '15 at 19:47
Archaick
1,781625
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
$sigmabegin{pmatrix}
1 \
-2 \
-4
end{pmatrix}=textbf{x}$
answered Apr 11 '15 at 19:47
Archaick
1,781625
answered Apr 11 '15 at 19:47
Archaick
1,781625
answered Apr 11 '15 at 19:47
Archaick
1,781625
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
add a comment |
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
add a comment |
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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