Mixing
Can you find the z component in this system?
$begingroup$
I will use uppercase letters to denote known variables and lowercase letters to denote unkown ones.
Assume I have a vector $<X', Y', z'>$, and an invertible matrix $T$, and a vector $<x,y,z>$ constrained to a known plane $P=(N,V)$ ($N$ being the normal and $V$ being a point in the plane).
We know that $<X', Y', z'> = Ttimes<x,y,z>$.
Is it possible to either find $z'$ or $<x,y,z>$ in this system?
linear-algebra matrices linear-transformations systems-of-equations
asked Jan 21 at 15:51
MakoganMakogan
786217
$endgroup$
add a comment |
$begingroup$
I will use uppercase letters to denote known variables and lowercase letters to denote unkown ones.
Assume I have a vector $<X', Y', z'>$, and an invertible matrix $T$, and a vector $<x,y,z>$ constrained to a known plane $P=(N,V)$ ($N$ being the normal and $V$ being a point in the plane).
We know that $<X', Y', z'> = Ttimes<x,y,z>$.
Is it possible to either find $z'$ or $<x,y,z>$ in this system?
linear-algebra matrices linear-transformations systems-of-equations
asked Jan 21 at 15:51
MakoganMakogan
786217
$endgroup$
$begingroup$
Wouldn't you need to know either $T$, or $x$ and $y$ to answer this question?
$endgroup$
– R. Burton
Jan 21 at 16:18
$begingroup$
You do know $T$, it's an uppercase variable.
$endgroup$
– Makogan
Jan 21 at 16:58
$begingroup$
Then yes, it is possible.
$endgroup$
– R. Burton
Jan 21 at 16:59
add a comment |
$begingroup$
I will use uppercase letters to denote known variables and lowercase letters to denote unkown ones.
Assume I have a vector $<X', Y', z'>$, and an invertible matrix $T$, and a vector $<x,y,z>$ constrained to a known plane $P=(N,V)$ ($N$ being the normal and $V$ being a point in the plane).
We know that $<X', Y', z'> = Ttimes<x,y,z>$.
Is it possible to either find $z'$ or $<x,y,z>$ in this system?
linear-algebra matrices linear-transformations systems-of-equations
asked Jan 21 at 15:51
MakoganMakogan
786217
$endgroup$
I will use uppercase letters to denote known variables and lowercase letters to denote unkown ones.
Assume I have a vector $<X', Y', z'>$, and an invertible matrix $T$, and a vector $<x,y,z>$ constrained to a known plane $P=(N,V)$ ($N$ being the normal and $V$ being a point in the plane).
We know that $<X', Y', z'> = Ttimes<x,y,z>$.
Is it possible to either find $z'$ or $<x,y,z>$ in this system?
linear-algebra matrices linear-transformations systems-of-equations
linear-algebra matrices linear-transformations systems-of-equations
asked Jan 21 at 15:51
MakoganMakogan
786217
asked Jan 21 at 15:51
MakoganMakogan
786217
asked Jan 21 at 15:51
MakoganMakogan
786217
asked Jan 21 at 15:51
MakoganMakogan
786217
asked Jan 21 at 15:51
MakoganMakogan
786217
786217
$begingroup$
Wouldn't you need to know either $T$, or $x$ and $y$ to answer this question?
$endgroup$
– R. Burton
Jan 21 at 16:18
$begingroup$
You do know $T$, it's an uppercase variable.
$endgroup$
– Makogan
Jan 21 at 16:58
$begingroup$
Then yes, it is possible.
$endgroup$
– R. Burton
Jan 21 at 16:59
add a comment |
$begingroup$
Wouldn't you need to know either $T$, or $x$ and $y$ to answer this question?
$endgroup$
– R. Burton
Jan 21 at 16:18
$begingroup$
You do know $T$, it's an uppercase variable.
$endgroup$
– Makogan
Jan 21 at 16:58
$begingroup$
Then yes, it is possible.
$endgroup$
– R. Burton
Jan 21 at 16:59
$begingroup$
Wouldn't you need to know either $T$, or $x$ and $y$ to answer this question?
$endgroup$
– R. Burton
Jan 21 at 16:18
$begingroup$
Wouldn't you need to know either $T$, or $x$ and $y$ to answer this question?
$endgroup$
– R. Burton
Jan 21 at 16:18
$begingroup$
You do know $T$, it's an uppercase variable.
$endgroup$
– Makogan
Jan 21 at 16:58
$begingroup$
You do know $T$, it's an uppercase variable.
$endgroup$
– Makogan
Jan 21 at 16:58
$begingroup$
Then yes, it is possible.
$endgroup$
– R. Burton
Jan 21 at 16:59
$begingroup$
Then yes, it is possible.
$endgroup$
– R. Burton
Jan 21 at 16:59
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
From the fact that $r = (x,y,z)$ lies in the plane you get:
$$N^T(r-V)= 0$$
From the other equation you have:
$$q = Tr$$
Where $q = (X', Y', z')$. Note that $T$ is invertible so the equations are not linearly dependent. Then ultimately you get 4 equations for 4 unknowns.
answered Jan 21 at 17:22
lightxbulblightxbulb
1,040311
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From the fact that $r = (x,y,z)$ lies in the plane you get:
$$N^T(r-V)= 0$$
From the other equation you have:
$$q = Tr$$
Where $q = (X', Y', z')$. Note that $T$ is invertible so the equations are not linearly dependent. Then ultimately you get 4 equations for 4 unknowns.
answered Jan 21 at 17:22
lightxbulblightxbulb
1,040311
$endgroup$
add a comment |
$begingroup$
From the fact that $r = (x,y,z)$ lies in the plane you get:
$$N^T(r-V)= 0$$
From the other equation you have:
$$q = Tr$$
Where $q = (X', Y', z')$. Note that $T$ is invertible so the equations are not linearly dependent. Then ultimately you get 4 equations for 4 unknowns.
answered Jan 21 at 17:22
lightxbulblightxbulb
1,040311
$endgroup$
add a comment |
$begingroup$
From the fact that $r = (x,y,z)$ lies in the plane you get:
$$N^T(r-V)= 0$$
From the other equation you have:
$$q = Tr$$
Where $q = (X', Y', z')$. Note that $T$ is invertible so the equations are not linearly dependent. Then ultimately you get 4 equations for 4 unknowns.
answered Jan 21 at 17:22
lightxbulblightxbulb
1,040311
$endgroup$
From the fact that $r = (x,y,z)$ lies in the plane you get:
$$N^T(r-V)= 0$$
From the other equation you have:
$$q = Tr$$
Where $q = (X', Y', z')$. Note that $T$ is invertible so the equations are not linearly dependent. Then ultimately you get 4 equations for 4 unknowns.
answered Jan 21 at 17:22
lightxbulblightxbulb
1,040311
answered Jan 21 at 17:22
lightxbulblightxbulb
1,040311
answered Jan 21 at 17:22
lightxbulblightxbulb
1,040311
answered Jan 21 at 17:22
lightxbulblightxbulb
1,040311
1,040311
add a comment |
add a comment |
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$begingroup$
Wouldn't you need to know either $T$, or $x$ and $y$ to answer this question?
$endgroup$
– R. Burton
Jan 21 at 16:18
$begingroup$
You do know $T$, it's an uppercase variable.
$endgroup$
– Makogan
Jan 21 at 16:58
$begingroup$
Then yes, it is possible.
$endgroup$
– R. Burton
Jan 21 at 16:59