Determine homogeneous transformation matrix for reflection about the line $y = mx + b$, or specifically $y =...
up vote
3
down vote
favorite
Determine the homogeneous transformation matrix for reflection about the line
$y = mx + b$, or specifically $ y = 2x – 6$.
I use $mx - y +b =0$: $text{slope} = m$, $tan(theta)= m$
intersection with the axes:
$x =0$ is $y = -b$ and $y =0$ is $x = dfrac{b}{m}$
My question is, what can I do next?
linear-algebra
edited Aug 7 '15 at 6:27
coldnumber
3,2841820
asked Nov 4 '13 at 22:18
Adel Hassan
9317
add a comment |
up vote
3
down vote
favorite
Determine the homogeneous transformation matrix for reflection about the line
$y = mx + b$, or specifically $ y = 2x – 6$.
I use $mx - y +b =0$: $text{slope} = m$, $tan(theta)= m$
intersection with the axes:
$x =0$ is $y = -b$ and $y =0$ is $x = dfrac{b}{m}$
My question is, what can I do next?
linear-algebra
edited Aug 7 '15 at 6:27
coldnumber
3,2841820
asked Nov 4 '13 at 22:18
Adel Hassan
9317
1
What are your thoughts on the problem so far? What have you tried?
– Cameron Buie
Nov 4 '13 at 22:21
1
zebragraph.com/Geometers_Corner_files/Reflection.pdf
– Mickey Tin
Apr 2 '14 at 15:14
Your question is not clear. Do you want to reflect in the line $y=mx+b$ or $y=2x-6$?
– Rory Daulton
Dec 31 '14 at 4:29
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Determine the homogeneous transformation matrix for reflection about the line
$y = mx + b$, or specifically $ y = 2x – 6$.
I use $mx - y +b =0$: $text{slope} = m$, $tan(theta)= m$
intersection with the axes:
$x =0$ is $y = -b$ and $y =0$ is $x = dfrac{b}{m}$
My question is, what can I do next?
linear-algebra
edited Aug 7 '15 at 6:27
coldnumber
3,2841820
asked Nov 4 '13 at 22:18
Adel Hassan
9317
Determine the homogeneous transformation matrix for reflection about the line
$y = mx + b$, or specifically $ y = 2x – 6$.
I use $mx - y +b =0$: $text{slope} = m$, $tan(theta)= m$
intersection with the axes:
$x =0$ is $y = -b$ and $y =0$ is $x = dfrac{b}{m}$
My question is, what can I do next?
linear-algebra
linear-algebra
edited Aug 7 '15 at 6:27
coldnumber
3,2841820
asked Nov 4 '13 at 22:18
Adel Hassan
9317
edited Aug 7 '15 at 6:27
coldnumber
3,2841820
asked Nov 4 '13 at 22:18
Adel Hassan
9317
edited Aug 7 '15 at 6:27
coldnumber
3,2841820
edited Aug 7 '15 at 6:27
coldnumber
3,2841820
edited Aug 7 '15 at 6:27
coldnumber
3,2841820
3,2841820
asked Nov 4 '13 at 22:18
Adel Hassan
9317
asked Nov 4 '13 at 22:18
Adel Hassan
9317
asked Nov 4 '13 at 22:18
Adel Hassan
9317
9317
1
What are your thoughts on the problem so far? What have you tried?
– Cameron Buie
Nov 4 '13 at 22:21
1
zebragraph.com/Geometers_Corner_files/Reflection.pdf
– Mickey Tin
Apr 2 '14 at 15:14
Your question is not clear. Do you want to reflect in the line $y=mx+b$ or $y=2x-6$?
– Rory Daulton
Dec 31 '14 at 4:29
add a comment |
1
What are your thoughts on the problem so far? What have you tried?
– Cameron Buie
Nov 4 '13 at 22:21
1
zebragraph.com/Geometers_Corner_files/Reflection.pdf
– Mickey Tin
Apr 2 '14 at 15:14
Your question is not clear. Do you want to reflect in the line $y=mx+b$ or $y=2x-6$?
– Rory Daulton
Dec 31 '14 at 4:29
1
1
What are your thoughts on the problem so far? What have you tried?
– Cameron Buie
Nov 4 '13 at 22:21
What are your thoughts on the problem so far? What have you tried?
– Cameron Buie
Nov 4 '13 at 22:21
1
1
zebragraph.com/Geometers_Corner_files/Reflection.pdf
– Mickey Tin
Apr 2 '14 at 15:14
zebragraph.com/Geometers_Corner_files/Reflection.pdf
– Mickey Tin
Apr 2 '14 at 15:14
Your question is not clear. Do you want to reflect in the line $y=mx+b$ or $y=2x-6$?
– Rory Daulton
Dec 31 '14 at 4:29
Your question is not clear. Do you want to reflect in the line $y=mx+b$ or $y=2x-6$?
– Rory Daulton
Dec 31 '14 at 4:29
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
Write down the transform in vector form and you can derive it from scratch. The line can be written as $mx-y=-b$. The normal to the line is $vec{n}=(m,-1)$. Let's also introduce some initial point on the line -- let's say $vec{r}_0=(x,y)=(0,b)$. How do you reflect about a line through the center? Easy - subtract twice the projection to the normal. How about if it's not through the center? Move to the center, transform, and move back. The transform is:
$$vec{r} mapsto (vec{r}-vec{r}_0)-2hat{n}( hat{n}cdot(vec{r}-vec{r}_0))+vec{r}_0$$
I normalized the normal, $hat{n}=frac{vec{n}}{|n|}=frac{(m,-1)}{sqrt{m^2+1}}$.
This can be simplified to:
$$vec{r}-2hat{n}( hat{n}cdotvec{r})-2hat{n}( hat{n}cdotvec{r}_0)$$
The first two terms are the standard reflection matrix in 2D and represents the 2×2 block of your homogeneous 3×3 matrix. The last term is the offset that contributes to both components of the result, and is in the homogeneous coordinates obtained from the last column (which multiplies the constant third coordinate).
Observe that because $vec{r}_0$ is on the line, it satisfies the original equation $vec{n}cdotvec{r}_0=-b$, so our choice of the origin point is irrelevant.
Now, you can simply write the matrix out:
$$begin{bmatrix}x'\y'\1end{bmatrix}=begin{bmatrix}1-2n_x^2/|n|^2 & -2n_xn_y/|n|^2 & 2n_xb/|n|^2 \ -2n_xn_y/|n|^2 & 1-2n_y^2/|n|^2 & 2n_yb/|n|^2 \0 & 0& 1end{bmatrix}begin{bmatrix}x\y\1end{bmatrix}$$
In our specific case:
$$begin{bmatrix}1-2m^2/(m^2+1) & 2m/(m^2+1) & 2mb/(m^2+1) \ 2m/(m^2+1) & 1-2/(m^2+1) & -2b/(m^2+1) \0 & 0& 1end{bmatrix}$$
You can see it's way easier, if you use the canonical implicit form for the line. If you have
$$ax+by=c$$
where $a^2+b^2=1$, then the matrix simplifies to
$$begin{bmatrix}1-2a^2 & -2ab & -2ac \ -2ab & 1-2b^2 & -2bc \0 & 0& 1end{bmatrix}$$
or, if you want, $I-2(a,b,0)^T(a,b,c)$.
answered Feb 19 '16 at 8:19
orion
12.9k11836
add a comment |
up vote
-1
down vote
Note that any matrix $mathbf Acdot vec{0}=vec{0}$, so there is no matrix that can flip over $y=2x-6$ as it must map $vec 0 to (4 frac{4}{5},-2frac{2}{5})$. You might want to do something about that.
answered Nov 4 '13 at 22:50
Tim Ratigan
6,1151028
We're talking about homogeneous coordinates. All affine transforms can be written as homogeneous matrices. So we're working with vectors in the form (x,y,1).
– orion
Feb 19 '16 at 8:03
add a comment |
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2 Answers
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Write down the transform in vector form and you can derive it from scratch. The line can be written as $mx-y=-b$. The normal to the line is $vec{n}=(m,-1)$. Let's also introduce some initial point on the line -- let's say $vec{r}_0=(x,y)=(0,b)$. How do you reflect about a line through the center? Easy - subtract twice the projection to the normal. How about if it's not through the center? Move to the center, transform, and move back. The transform is:
$$vec{r} mapsto (vec{r}-vec{r}_0)-2hat{n}( hat{n}cdot(vec{r}-vec{r}_0))+vec{r}_0$$
I normalized the normal, $hat{n}=frac{vec{n}}{|n|}=frac{(m,-1)}{sqrt{m^2+1}}$.
This can be simplified to:
$$vec{r}-2hat{n}( hat{n}cdotvec{r})-2hat{n}( hat{n}cdotvec{r}_0)$$
The first two terms are the standard reflection matrix in 2D and represents the 2×2 block of your homogeneous 3×3 matrix. The last term is the offset that contributes to both components of the result, and is in the homogeneous coordinates obtained from the last column (which multiplies the constant third coordinate).
Observe that because $vec{r}_0$ is on the line, it satisfies the original equation $vec{n}cdotvec{r}_0=-b$, so our choice of the origin point is irrelevant.
Now, you can simply write the matrix out:
$$begin{bmatrix}x'\y'\1end{bmatrix}=begin{bmatrix}1-2n_x^2/|n|^2 & -2n_xn_y/|n|^2 & 2n_xb/|n|^2 \ -2n_xn_y/|n|^2 & 1-2n_y^2/|n|^2 & 2n_yb/|n|^2 \0 & 0& 1end{bmatrix}begin{bmatrix}x\y\1end{bmatrix}$$
In our specific case:
$$begin{bmatrix}1-2m^2/(m^2+1) & 2m/(m^2+1) & 2mb/(m^2+1) \ 2m/(m^2+1) & 1-2/(m^2+1) & -2b/(m^2+1) \0 & 0& 1end{bmatrix}$$
You can see it's way easier, if you use the canonical implicit form for the line. If you have
$$ax+by=c$$
where $a^2+b^2=1$, then the matrix simplifies to
$$begin{bmatrix}1-2a^2 & -2ab & -2ac \ -2ab & 1-2b^2 & -2bc \0 & 0& 1end{bmatrix}$$
or, if you want, $I-2(a,b,0)^T(a,b,c)$.
answered Feb 19 '16 at 8:19
orion
12.9k11836
add a comment |
up vote
0
down vote
Write down the transform in vector form and you can derive it from scratch. The line can be written as $mx-y=-b$. The normal to the line is $vec{n}=(m,-1)$. Let's also introduce some initial point on the line -- let's say $vec{r}_0=(x,y)=(0,b)$. How do you reflect about a line through the center? Easy - subtract twice the projection to the normal. How about if it's not through the center? Move to the center, transform, and move back. The transform is:
$$vec{r} mapsto (vec{r}-vec{r}_0)-2hat{n}( hat{n}cdot(vec{r}-vec{r}_0))+vec{r}_0$$
I normalized the normal, $hat{n}=frac{vec{n}}{|n|}=frac{(m,-1)}{sqrt{m^2+1}}$.
This can be simplified to:
$$vec{r}-2hat{n}( hat{n}cdotvec{r})-2hat{n}( hat{n}cdotvec{r}_0)$$
The first two terms are the standard reflection matrix in 2D and represents the 2×2 block of your homogeneous 3×3 matrix. The last term is the offset that contributes to both components of the result, and is in the homogeneous coordinates obtained from the last column (which multiplies the constant third coordinate).
Observe that because $vec{r}_0$ is on the line, it satisfies the original equation $vec{n}cdotvec{r}_0=-b$, so our choice of the origin point is irrelevant.
Now, you can simply write the matrix out:
$$begin{bmatrix}x'\y'\1end{bmatrix}=begin{bmatrix}1-2n_x^2/|n|^2 & -2n_xn_y/|n|^2 & 2n_xb/|n|^2 \ -2n_xn_y/|n|^2 & 1-2n_y^2/|n|^2 & 2n_yb/|n|^2 \0 & 0& 1end{bmatrix}begin{bmatrix}x\y\1end{bmatrix}$$
In our specific case:
$$begin{bmatrix}1-2m^2/(m^2+1) & 2m/(m^2+1) & 2mb/(m^2+1) \ 2m/(m^2+1) & 1-2/(m^2+1) & -2b/(m^2+1) \0 & 0& 1end{bmatrix}$$
You can see it's way easier, if you use the canonical implicit form for the line. If you have
$$ax+by=c$$
where $a^2+b^2=1$, then the matrix simplifies to
$$begin{bmatrix}1-2a^2 & -2ab & -2ac \ -2ab & 1-2b^2 & -2bc \0 & 0& 1end{bmatrix}$$
or, if you want, $I-2(a,b,0)^T(a,b,c)$.
answered Feb 19 '16 at 8:19
orion
12.9k11836
add a comment |
up vote
0
down vote
up vote
0
down vote
Write down the transform in vector form and you can derive it from scratch. The line can be written as $mx-y=-b$. The normal to the line is $vec{n}=(m,-1)$. Let's also introduce some initial point on the line -- let's say $vec{r}_0=(x,y)=(0,b)$. How do you reflect about a line through the center? Easy - subtract twice the projection to the normal. How about if it's not through the center? Move to the center, transform, and move back. The transform is:
$$vec{r} mapsto (vec{r}-vec{r}_0)-2hat{n}( hat{n}cdot(vec{r}-vec{r}_0))+vec{r}_0$$
I normalized the normal, $hat{n}=frac{vec{n}}{|n|}=frac{(m,-1)}{sqrt{m^2+1}}$.
This can be simplified to:
$$vec{r}-2hat{n}( hat{n}cdotvec{r})-2hat{n}( hat{n}cdotvec{r}_0)$$
The first two terms are the standard reflection matrix in 2D and represents the 2×2 block of your homogeneous 3×3 matrix. The last term is the offset that contributes to both components of the result, and is in the homogeneous coordinates obtained from the last column (which multiplies the constant third coordinate).
Observe that because $vec{r}_0$ is on the line, it satisfies the original equation $vec{n}cdotvec{r}_0=-b$, so our choice of the origin point is irrelevant.
Now, you can simply write the matrix out:
$$begin{bmatrix}x'\y'\1end{bmatrix}=begin{bmatrix}1-2n_x^2/|n|^2 & -2n_xn_y/|n|^2 & 2n_xb/|n|^2 \ -2n_xn_y/|n|^2 & 1-2n_y^2/|n|^2 & 2n_yb/|n|^2 \0 & 0& 1end{bmatrix}begin{bmatrix}x\y\1end{bmatrix}$$
In our specific case:
$$begin{bmatrix}1-2m^2/(m^2+1) & 2m/(m^2+1) & 2mb/(m^2+1) \ 2m/(m^2+1) & 1-2/(m^2+1) & -2b/(m^2+1) \0 & 0& 1end{bmatrix}$$
You can see it's way easier, if you use the canonical implicit form for the line. If you have
$$ax+by=c$$
where $a^2+b^2=1$, then the matrix simplifies to
$$begin{bmatrix}1-2a^2 & -2ab & -2ac \ -2ab & 1-2b^2 & -2bc \0 & 0& 1end{bmatrix}$$
or, if you want, $I-2(a,b,0)^T(a,b,c)$.
answered Feb 19 '16 at 8:19
orion
12.9k11836
Write down the transform in vector form and you can derive it from scratch. The line can be written as $mx-y=-b$. The normal to the line is $vec{n}=(m,-1)$. Let's also introduce some initial point on the line -- let's say $vec{r}_0=(x,y)=(0,b)$. How do you reflect about a line through the center? Easy - subtract twice the projection to the normal. How about if it's not through the center? Move to the center, transform, and move back. The transform is:
$$vec{r} mapsto (vec{r}-vec{r}_0)-2hat{n}( hat{n}cdot(vec{r}-vec{r}_0))+vec{r}_0$$
I normalized the normal, $hat{n}=frac{vec{n}}{|n|}=frac{(m,-1)}{sqrt{m^2+1}}$.
This can be simplified to:
$$vec{r}-2hat{n}( hat{n}cdotvec{r})-2hat{n}( hat{n}cdotvec{r}_0)$$
The first two terms are the standard reflection matrix in 2D and represents the 2×2 block of your homogeneous 3×3 matrix. The last term is the offset that contributes to both components of the result, and is in the homogeneous coordinates obtained from the last column (which multiplies the constant third coordinate).
Observe that because $vec{r}_0$ is on the line, it satisfies the original equation $vec{n}cdotvec{r}_0=-b$, so our choice of the origin point is irrelevant.
Now, you can simply write the matrix out:
$$begin{bmatrix}x'\y'\1end{bmatrix}=begin{bmatrix}1-2n_x^2/|n|^2 & -2n_xn_y/|n|^2 & 2n_xb/|n|^2 \ -2n_xn_y/|n|^2 & 1-2n_y^2/|n|^2 & 2n_yb/|n|^2 \0 & 0& 1end{bmatrix}begin{bmatrix}x\y\1end{bmatrix}$$
In our specific case:
$$begin{bmatrix}1-2m^2/(m^2+1) & 2m/(m^2+1) & 2mb/(m^2+1) \ 2m/(m^2+1) & 1-2/(m^2+1) & -2b/(m^2+1) \0 & 0& 1end{bmatrix}$$
You can see it's way easier, if you use the canonical implicit form for the line. If you have
$$ax+by=c$$
where $a^2+b^2=1$, then the matrix simplifies to
$$begin{bmatrix}1-2a^2 & -2ab & -2ac \ -2ab & 1-2b^2 & -2bc \0 & 0& 1end{bmatrix}$$
or, if you want, $I-2(a,b,0)^T(a,b,c)$.
answered Feb 19 '16 at 8:19
orion
12.9k11836
edited Feb 19 '16 at 8:32
answered Feb 19 '16 at 8:19
orion
12.9k11836
answered Feb 19 '16 at 8:19
orion
12.9k11836
answered Feb 19 '16 at 8:19
orion
12.9k11836
12.9k11836
add a comment |
add a comment |
up vote
-1
down vote
Note that any matrix $mathbf Acdot vec{0}=vec{0}$, so there is no matrix that can flip over $y=2x-6$ as it must map $vec 0 to (4 frac{4}{5},-2frac{2}{5})$. You might want to do something about that.
answered Nov 4 '13 at 22:50
Tim Ratigan
6,1151028
We're talking about homogeneous coordinates. All affine transforms can be written as homogeneous matrices. So we're working with vectors in the form (x,y,1).
– orion
Feb 19 '16 at 8:03
add a comment |
up vote
-1
down vote
Note that any matrix $mathbf Acdot vec{0}=vec{0}$, so there is no matrix that can flip over $y=2x-6$ as it must map $vec 0 to (4 frac{4}{5},-2frac{2}{5})$. You might want to do something about that.
answered Nov 4 '13 at 22:50
Tim Ratigan
6,1151028
We're talking about homogeneous coordinates. All affine transforms can be written as homogeneous matrices. So we're working with vectors in the form (x,y,1).
– orion
Feb 19 '16 at 8:03
add a comment |
up vote
-1
down vote
up vote
-1
down vote
Note that any matrix $mathbf Acdot vec{0}=vec{0}$, so there is no matrix that can flip over $y=2x-6$ as it must map $vec 0 to (4 frac{4}{5},-2frac{2}{5})$. You might want to do something about that.
answered Nov 4 '13 at 22:50
Tim Ratigan
6,1151028
Note that any matrix $mathbf Acdot vec{0}=vec{0}$, so there is no matrix that can flip over $y=2x-6$ as it must map $vec 0 to (4 frac{4}{5},-2frac{2}{5})$. You might want to do something about that.
answered Nov 4 '13 at 22:50
Tim Ratigan
6,1151028
answered Nov 4 '13 at 22:50
Tim Ratigan
6,1151028
answered Nov 4 '13 at 22:50
Tim Ratigan
6,1151028
answered Nov 4 '13 at 22:50
Tim Ratigan
6,1151028
6,1151028
We're talking about homogeneous coordinates. All affine transforms can be written as homogeneous matrices. So we're working with vectors in the form (x,y,1).
– orion
Feb 19 '16 at 8:03
add a comment |
We're talking about homogeneous coordinates. All affine transforms can be written as homogeneous matrices. So we're working with vectors in the form (x,y,1).
– orion
Feb 19 '16 at 8:03
We're talking about homogeneous coordinates. All affine transforms can be written as homogeneous matrices. So we're working with vectors in the form (x,y,1).
– orion
Feb 19 '16 at 8:03
We're talking about homogeneous coordinates. All affine transforms can be written as homogeneous matrices. So we're working with vectors in the form (x,y,1).
– orion
Feb 19 '16 at 8:03
add a comment |
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1
What are your thoughts on the problem so far? What have you tried?
– Cameron Buie
Nov 4 '13 at 22:21
1
zebragraph.com/Geometers_Corner_files/Reflection.pdf
– Mickey Tin
Apr 2 '14 at 15:14
Your question is not clear. Do you want to reflect in the line $y=mx+b$ or $y=2x-6$?
– Rory Daulton
Dec 31 '14 at 4:29