Mixing
Orthogonal projection and orthogonal complements onto a plane
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Let $V$ be a subspace of $mathbbR^n$ with $V ne mathbbR^n$ and $Vne 0$. Let $A$ be the matrix of the linear transformation $textproj_V : mathbbR^ntomathbbR^n$ that is the projection onto $V$.
Calculate $A$ in the case where $n=3$ and $V$ is given by the equation $x+y+z=0$.
I am not really sure how to approach this question. I am aware that the normal vector will be $(1,1,1)$. And I know that the orthogonal projection onto the plane $V$ given by the equation $x+y+z=0$ is equal to the identity minus the orthogonal projection onto the orthogonal complement. I believe the orthogonal complement will be in the span of the normal vector (not entirley sure why?) but I am also unsure how to find the orthogonal complement.
linear-algebra
edited Jul 27 at 13:10
mechanodroid
22.2k52041
asked Jul 27 at 12:28
Molly
303
add a comment |Â
up vote
1
down vote
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Let $V$ be a subspace of $mathbbR^n$ with $V ne mathbbR^n$ and $Vne 0$. Let $A$ be the matrix of the linear transformation $textproj_V : mathbbR^ntomathbbR^n$ that is the projection onto $V$.
Calculate $A$ in the case where $n=3$ and $V$ is given by the equation $x+y+z=0$.
I am not really sure how to approach this question. I am aware that the normal vector will be $(1,1,1)$. And I know that the orthogonal projection onto the plane $V$ given by the equation $x+y+z=0$ is equal to the identity minus the orthogonal projection onto the orthogonal complement. I believe the orthogonal complement will be in the span of the normal vector (not entirley sure why?) but I am also unsure how to find the orthogonal complement.
linear-algebra
edited Jul 27 at 13:10
mechanodroid
22.2k52041
asked Jul 27 at 12:28
Molly
303
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $V$ be a subspace of $mathbbR^n$ with $V ne mathbbR^n$ and $Vne 0$. Let $A$ be the matrix of the linear transformation $textproj_V : mathbbR^ntomathbbR^n$ that is the projection onto $V$.
Calculate $A$ in the case where $n=3$ and $V$ is given by the equation $x+y+z=0$.
I am not really sure how to approach this question. I am aware that the normal vector will be $(1,1,1)$. And I know that the orthogonal projection onto the plane $V$ given by the equation $x+y+z=0$ is equal to the identity minus the orthogonal projection onto the orthogonal complement. I believe the orthogonal complement will be in the span of the normal vector (not entirley sure why?) but I am also unsure how to find the orthogonal complement.
linear-algebra
edited Jul 27 at 13:10
mechanodroid
22.2k52041
asked Jul 27 at 12:28
Molly
303
Let $V$ be a subspace of $mathbbR^n$ with $V ne mathbbR^n$ and $Vne 0$. Let $A$ be the matrix of the linear transformation $textproj_V : mathbbR^ntomathbbR^n$ that is the projection onto $V$.
Calculate $A$ in the case where $n=3$ and $V$ is given by the equation $x+y+z=0$.
I am not really sure how to approach this question. I am aware that the normal vector will be $(1,1,1)$. And I know that the orthogonal projection onto the plane $V$ given by the equation $x+y+z=0$ is equal to the identity minus the orthogonal projection onto the orthogonal complement. I believe the orthogonal complement will be in the span of the normal vector (not entirley sure why?) but I am also unsure how to find the orthogonal complement.
linear-algebra
edited Jul 27 at 13:10
mechanodroid
22.2k52041
asked Jul 27 at 12:28
Molly
303
edited Jul 27 at 13:10
mechanodroid
22.2k52041
edited Jul 27 at 13:10
mechanodroid
22.2k52041
edited Jul 27 at 13:10
mechanodroid
22.2k52041
22.2k52041
asked Jul 27 at 12:28
Molly
303
asked Jul 27 at 12:28
Molly
303
asked Jul 27 at 12:28
Molly
303
303
add a comment |Â
add a comment |Â
5 Answers
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Hint:
The projection of a vector $vec u$ onto the normal to the plane $x+y+z=0$, with normal vector $vec n=(1,1,1)$, is given by the formula:
$$p_vec n(vec u)=vec u-fracvec ncdotvec uvec ncdotvec n,vec n.$$
Apply this formula to each vector of the canonical basis to find the column vectors of the projection matrix.
answered Jul 27 at 12:42
Bernard
110k635102
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You are right: $(1,1,1)$ is orthogonal to $V$. Therefore, $A.(1,1,1)=(0,0,0)$. Now, consider the vectors $(1,-1,0)$ and $(1,0,-1)$. Since they both belong to $V$, you must have $A.(1,-1,0)=(1,-1,0)$ and $A.(1,0,-1)=(1,0,-1)$.
Now, since$$(1,0,0)=frac13(1,1,1)+frac13(1,-1,0)+frac13(1,0,-1),$$you must have$$A.(1,0,0)=frac13(1,-1,0)+frac13(1,0,-1)=left(frac23,-frac13,-frac13right).$$So, the entries of the first column of the matrix of $operatornameproj_V$ with respect to the standard basis will be $frac23$, $-frac13$ and $-frac13$. Can you take it from here?
answered Jul 27 at 12:37
José Carlos Santos
113k1696173
Where are you getting the 1/3 from?
– Molly
Jul 27 at 20:50
From solving the system $(1,0,0)=alpha(1,1,1)+beta(1,-1,0)+gamma(1,0,-1)$.
– José Carlos Santos
Jul 27 at 21:11
add a comment |Â
up vote
0
down vote
As you say, $u:=(1,1,1)^T$ is a normal vector to the plane, thus it will span the orthogonal complement.
Its projection is given by $varphi:vmapstofracu^Tvu^Tuu$.
You can check that it fixes $u$, but gives $0$ whenever $vperp u$.
To obtain its matrix in the standard basis, just calculate $varphi(e_i)$ for the standard basis $e_1,e_2,e_3$.
answered Jul 27 at 12:41
Berci
56.4k23570
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$leftfrac1sqrt2(1,-1,0), frac1sqrt6(1,1,-2)right$ is an orthonormal basis for $V$ so the projection is given by
beginalign
P(x,y,z) &= leftlangle (x,y,z), frac1sqrt2(1,-1,0)rightranglefrac1sqrt2(1,-1,0)+leftlangle (x,y,z), frac1sqrt6(1,1,-2)rightranglefrac1sqrt6(1,1,-2) \
&= frac12 (x-y, -x+y,0) + frac16(x+y-2z,x+y-2z,-x-y+4z)\
&= frac16 (4x-2y-2z, -2x+4y-2z, -2x-2y+4z)
endalign
so $$A = frac16beginbmatrix 4 & -2 & -2 \ -2 & 4 & -2 \ -2 & -2 & 4 endbmatrix$$
answered Jul 27 at 12:49
mechanodroid
22.2k52041
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The orthogonal projection of $(x,y,z)$ onto the plane through the origin with normal $(1,1,1)$ is the unique $(x',y',z')$ on the plane such that
$$
((x,y,z)-(x',y',z')) = c(1,1,1), \
x'+y'+z'=0.
$$
Dotting the first equation with $(1,1,1)$ and using $x'+y'+z'=0$ gives
$$
c = frac13(x+y+z).
$$
So the projection of $(x,y,z)$ onto the given plane is
beginalign
(x',y',z') &= (x,y,z)-frac13(x+y+z)(1,1,1) \
&= frac13(2x-y-z,-x+2y-z,-x-y+2z).
endalign
answered Jul 28 at 5:11
DisintegratingByParts
55.5k42273
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint:
The projection of a vector $vec u$ onto the normal to the plane $x+y+z=0$, with normal vector $vec n=(1,1,1)$, is given by the formula:
$$p_vec n(vec u)=vec u-fracvec ncdotvec uvec ncdotvec n,vec n.$$
Apply this formula to each vector of the canonical basis to find the column vectors of the projection matrix.
answered Jul 27 at 12:42
Bernard
110k635102
add a comment |Â
up vote
1
down vote
Hint:
The projection of a vector $vec u$ onto the normal to the plane $x+y+z=0$, with normal vector $vec n=(1,1,1)$, is given by the formula:
$$p_vec n(vec u)=vec u-fracvec ncdotvec uvec ncdotvec n,vec n.$$
Apply this formula to each vector of the canonical basis to find the column vectors of the projection matrix.
answered Jul 27 at 12:42
Bernard
110k635102
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint:
The projection of a vector $vec u$ onto the normal to the plane $x+y+z=0$, with normal vector $vec n=(1,1,1)$, is given by the formula:
$$p_vec n(vec u)=vec u-fracvec ncdotvec uvec ncdotvec n,vec n.$$
Apply this formula to each vector of the canonical basis to find the column vectors of the projection matrix.
answered Jul 27 at 12:42
Bernard
110k635102
Hint:
The projection of a vector $vec u$ onto the normal to the plane $x+y+z=0$, with normal vector $vec n=(1,1,1)$, is given by the formula:
$$p_vec n(vec u)=vec u-fracvec ncdotvec uvec ncdotvec n,vec n.$$
Apply this formula to each vector of the canonical basis to find the column vectors of the projection matrix.
answered Jul 27 at 12:42
Bernard
110k635102
answered Jul 27 at 12:42
Bernard
110k635102
answered Jul 27 at 12:42
Bernard
110k635102
answered Jul 27 at 12:42
Bernard
110k635102
110k635102
add a comment |Â
add a comment |Â
up vote
0
down vote
You are right: $(1,1,1)$ is orthogonal to $V$. Therefore, $A.(1,1,1)=(0,0,0)$. Now, consider the vectors $(1,-1,0)$ and $(1,0,-1)$. Since they both belong to $V$, you must have $A.(1,-1,0)=(1,-1,0)$ and $A.(1,0,-1)=(1,0,-1)$.
Now, since$$(1,0,0)=frac13(1,1,1)+frac13(1,-1,0)+frac13(1,0,-1),$$you must have$$A.(1,0,0)=frac13(1,-1,0)+frac13(1,0,-1)=left(frac23,-frac13,-frac13right).$$So, the entries of the first column of the matrix of $operatornameproj_V$ with respect to the standard basis will be $frac23$, $-frac13$ and $-frac13$. Can you take it from here?
answered Jul 27 at 12:37
José Carlos Santos
113k1696173
Where are you getting the 1/3 from?
– Molly
Jul 27 at 20:50
From solving the system $(1,0,0)=alpha(1,1,1)+beta(1,-1,0)+gamma(1,0,-1)$.
– José Carlos Santos
Jul 27 at 21:11
add a comment |Â
up vote
0
down vote
You are right: $(1,1,1)$ is orthogonal to $V$. Therefore, $A.(1,1,1)=(0,0,0)$. Now, consider the vectors $(1,-1,0)$ and $(1,0,-1)$. Since they both belong to $V$, you must have $A.(1,-1,0)=(1,-1,0)$ and $A.(1,0,-1)=(1,0,-1)$.
Now, since$$(1,0,0)=frac13(1,1,1)+frac13(1,-1,0)+frac13(1,0,-1),$$you must have$$A.(1,0,0)=frac13(1,-1,0)+frac13(1,0,-1)=left(frac23,-frac13,-frac13right).$$So, the entries of the first column of the matrix of $operatornameproj_V$ with respect to the standard basis will be $frac23$, $-frac13$ and $-frac13$. Can you take it from here?
answered Jul 27 at 12:37
José Carlos Santos
113k1696173
Where are you getting the 1/3 from?
– Molly
Jul 27 at 20:50
From solving the system $(1,0,0)=alpha(1,1,1)+beta(1,-1,0)+gamma(1,0,-1)$.
– José Carlos Santos
Jul 27 at 21:11
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You are right: $(1,1,1)$ is orthogonal to $V$. Therefore, $A.(1,1,1)=(0,0,0)$. Now, consider the vectors $(1,-1,0)$ and $(1,0,-1)$. Since they both belong to $V$, you must have $A.(1,-1,0)=(1,-1,0)$ and $A.(1,0,-1)=(1,0,-1)$.
Now, since$$(1,0,0)=frac13(1,1,1)+frac13(1,-1,0)+frac13(1,0,-1),$$you must have$$A.(1,0,0)=frac13(1,-1,0)+frac13(1,0,-1)=left(frac23,-frac13,-frac13right).$$So, the entries of the first column of the matrix of $operatornameproj_V$ with respect to the standard basis will be $frac23$, $-frac13$ and $-frac13$. Can you take it from here?
answered Jul 27 at 12:37
José Carlos Santos
113k1696173
You are right: $(1,1,1)$ is orthogonal to $V$. Therefore, $A.(1,1,1)=(0,0,0)$. Now, consider the vectors $(1,-1,0)$ and $(1,0,-1)$. Since they both belong to $V$, you must have $A.(1,-1,0)=(1,-1,0)$ and $A.(1,0,-1)=(1,0,-1)$.
Now, since$$(1,0,0)=frac13(1,1,1)+frac13(1,-1,0)+frac13(1,0,-1),$$you must have$$A.(1,0,0)=frac13(1,-1,0)+frac13(1,0,-1)=left(frac23,-frac13,-frac13right).$$So, the entries of the first column of the matrix of $operatornameproj_V$ with respect to the standard basis will be $frac23$, $-frac13$ and $-frac13$. Can you take it from here?
answered Jul 27 at 12:37
José Carlos Santos
113k1696173
answered Jul 27 at 12:37
José Carlos Santos
113k1696173
answered Jul 27 at 12:37
José Carlos Santos
113k1696173
answered Jul 27 at 12:37
José Carlos Santos
113k1696173
113k1696173
Where are you getting the 1/3 from?
– Molly
Jul 27 at 20:50
From solving the system $(1,0,0)=alpha(1,1,1)+beta(1,-1,0)+gamma(1,0,-1)$.
– José Carlos Santos
Jul 27 at 21:11
add a comment |Â
Where are you getting the 1/3 from?
– Molly
Jul 27 at 20:50
From solving the system $(1,0,0)=alpha(1,1,1)+beta(1,-1,0)+gamma(1,0,-1)$.
– José Carlos Santos
Jul 27 at 21:11
Where are you getting the 1/3 from?
– Molly
Jul 27 at 20:50
Where are you getting the 1/3 from?
– Molly
Jul 27 at 20:50
From solving the system $(1,0,0)=alpha(1,1,1)+beta(1,-1,0)+gamma(1,0,-1)$.
– José Carlos Santos
Jul 27 at 21:11
From solving the system $(1,0,0)=alpha(1,1,1)+beta(1,-1,0)+gamma(1,0,-1)$.
– José Carlos Santos
Jul 27 at 21:11
add a comment |Â
up vote
0
down vote
As you say, $u:=(1,1,1)^T$ is a normal vector to the plane, thus it will span the orthogonal complement.
Its projection is given by $varphi:vmapstofracu^Tvu^Tuu$.
You can check that it fixes $u$, but gives $0$ whenever $vperp u$.
To obtain its matrix in the standard basis, just calculate $varphi(e_i)$ for the standard basis $e_1,e_2,e_3$.
answered Jul 27 at 12:41
Berci
56.4k23570
add a comment |Â
up vote
0
down vote
As you say, $u:=(1,1,1)^T$ is a normal vector to the plane, thus it will span the orthogonal complement.
Its projection is given by $varphi:vmapstofracu^Tvu^Tuu$.
You can check that it fixes $u$, but gives $0$ whenever $vperp u$.
To obtain its matrix in the standard basis, just calculate $varphi(e_i)$ for the standard basis $e_1,e_2,e_3$.
answered Jul 27 at 12:41
Berci
56.4k23570
add a comment |Â
up vote
0
down vote
up vote
0
down vote
As you say, $u:=(1,1,1)^T$ is a normal vector to the plane, thus it will span the orthogonal complement.
Its projection is given by $varphi:vmapstofracu^Tvu^Tuu$.
You can check that it fixes $u$, but gives $0$ whenever $vperp u$.
To obtain its matrix in the standard basis, just calculate $varphi(e_i)$ for the standard basis $e_1,e_2,e_3$.
answered Jul 27 at 12:41
Berci
56.4k23570
As you say, $u:=(1,1,1)^T$ is a normal vector to the plane, thus it will span the orthogonal complement.
Its projection is given by $varphi:vmapstofracu^Tvu^Tuu$.
You can check that it fixes $u$, but gives $0$ whenever $vperp u$.
To obtain its matrix in the standard basis, just calculate $varphi(e_i)$ for the standard basis $e_1,e_2,e_3$.
answered Jul 27 at 12:41
Berci
56.4k23570
answered Jul 27 at 12:41
Berci
56.4k23570
answered Jul 27 at 12:41
Berci
56.4k23570
answered Jul 27 at 12:41
Berci
56.4k23570
56.4k23570
add a comment |Â
add a comment |Â
up vote
0
down vote
$leftfrac1sqrt2(1,-1,0), frac1sqrt6(1,1,-2)right$ is an orthonormal basis for $V$ so the projection is given by
beginalign
P(x,y,z) &= leftlangle (x,y,z), frac1sqrt2(1,-1,0)rightranglefrac1sqrt2(1,-1,0)+leftlangle (x,y,z), frac1sqrt6(1,1,-2)rightranglefrac1sqrt6(1,1,-2) \
&= frac12 (x-y, -x+y,0) + frac16(x+y-2z,x+y-2z,-x-y+4z)\
&= frac16 (4x-2y-2z, -2x+4y-2z, -2x-2y+4z)
endalign
so $$A = frac16beginbmatrix 4 & -2 & -2 \ -2 & 4 & -2 \ -2 & -2 & 4 endbmatrix$$
answered Jul 27 at 12:49
mechanodroid
22.2k52041
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up vote
0
down vote
$leftfrac1sqrt2(1,-1,0), frac1sqrt6(1,1,-2)right$ is an orthonormal basis for $V$ so the projection is given by
beginalign
P(x,y,z) &= leftlangle (x,y,z), frac1sqrt2(1,-1,0)rightranglefrac1sqrt2(1,-1,0)+leftlangle (x,y,z), frac1sqrt6(1,1,-2)rightranglefrac1sqrt6(1,1,-2) \
&= frac12 (x-y, -x+y,0) + frac16(x+y-2z,x+y-2z,-x-y+4z)\
&= frac16 (4x-2y-2z, -2x+4y-2z, -2x-2y+4z)
endalign
so $$A = frac16beginbmatrix 4 & -2 & -2 \ -2 & 4 & -2 \ -2 & -2 & 4 endbmatrix$$
answered Jul 27 at 12:49
mechanodroid
22.2k52041
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$leftfrac1sqrt2(1,-1,0), frac1sqrt6(1,1,-2)right$ is an orthonormal basis for $V$ so the projection is given by
beginalign
P(x,y,z) &= leftlangle (x,y,z), frac1sqrt2(1,-1,0)rightranglefrac1sqrt2(1,-1,0)+leftlangle (x,y,z), frac1sqrt6(1,1,-2)rightranglefrac1sqrt6(1,1,-2) \
&= frac12 (x-y, -x+y,0) + frac16(x+y-2z,x+y-2z,-x-y+4z)\
&= frac16 (4x-2y-2z, -2x+4y-2z, -2x-2y+4z)
endalign
so $$A = frac16beginbmatrix 4 & -2 & -2 \ -2 & 4 & -2 \ -2 & -2 & 4 endbmatrix$$
answered Jul 27 at 12:49
mechanodroid
22.2k52041
$leftfrac1sqrt2(1,-1,0), frac1sqrt6(1,1,-2)right$ is an orthonormal basis for $V$ so the projection is given by
beginalign
P(x,y,z) &= leftlangle (x,y,z), frac1sqrt2(1,-1,0)rightranglefrac1sqrt2(1,-1,0)+leftlangle (x,y,z), frac1sqrt6(1,1,-2)rightranglefrac1sqrt6(1,1,-2) \
&= frac12 (x-y, -x+y,0) + frac16(x+y-2z,x+y-2z,-x-y+4z)\
&= frac16 (4x-2y-2z, -2x+4y-2z, -2x-2y+4z)
endalign
so $$A = frac16beginbmatrix 4 & -2 & -2 \ -2 & 4 & -2 \ -2 & -2 & 4 endbmatrix$$
answered Jul 27 at 12:49
mechanodroid
22.2k52041
edited Jul 27 at 13:09
answered Jul 27 at 12:49
mechanodroid
22.2k52041
answered Jul 27 at 12:49
mechanodroid
22.2k52041
answered Jul 27 at 12:49
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
up vote
0
down vote
The orthogonal projection of $(x,y,z)$ onto the plane through the origin with normal $(1,1,1)$ is the unique $(x',y',z')$ on the plane such that
$$
((x,y,z)-(x',y',z')) = c(1,1,1), \
x'+y'+z'=0.
$$
Dotting the first equation with $(1,1,1)$ and using $x'+y'+z'=0$ gives
$$
c = frac13(x+y+z).
$$
So the projection of $(x,y,z)$ onto the given plane is
beginalign
(x',y',z') &= (x,y,z)-frac13(x+y+z)(1,1,1) \
&= frac13(2x-y-z,-x+2y-z,-x-y+2z).
endalign
answered Jul 28 at 5:11
DisintegratingByParts
55.5k42273
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up vote
0
down vote
The orthogonal projection of $(x,y,z)$ onto the plane through the origin with normal $(1,1,1)$ is the unique $(x',y',z')$ on the plane such that
$$
((x,y,z)-(x',y',z')) = c(1,1,1), \
x'+y'+z'=0.
$$
Dotting the first equation with $(1,1,1)$ and using $x'+y'+z'=0$ gives
$$
c = frac13(x+y+z).
$$
So the projection of $(x,y,z)$ onto the given plane is
beginalign
(x',y',z') &= (x,y,z)-frac13(x+y+z)(1,1,1) \
&= frac13(2x-y-z,-x+2y-z,-x-y+2z).
endalign
answered Jul 28 at 5:11
DisintegratingByParts
55.5k42273
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The orthogonal projection of $(x,y,z)$ onto the plane through the origin with normal $(1,1,1)$ is the unique $(x',y',z')$ on the plane such that
$$
((x,y,z)-(x',y',z')) = c(1,1,1), \
x'+y'+z'=0.
$$
Dotting the first equation with $(1,1,1)$ and using $x'+y'+z'=0$ gives
$$
c = frac13(x+y+z).
$$
So the projection of $(x,y,z)$ onto the given plane is
beginalign
(x',y',z') &= (x,y,z)-frac13(x+y+z)(1,1,1) \
&= frac13(2x-y-z,-x+2y-z,-x-y+2z).
endalign
answered Jul 28 at 5:11
DisintegratingByParts
55.5k42273
The orthogonal projection of $(x,y,z)$ onto the plane through the origin with normal $(1,1,1)$ is the unique $(x',y',z')$ on the plane such that
$$
((x,y,z)-(x',y',z')) = c(1,1,1), \
x'+y'+z'=0.
$$
Dotting the first equation with $(1,1,1)$ and using $x'+y'+z'=0$ gives
$$
c = frac13(x+y+z).
$$
So the projection of $(x,y,z)$ onto the given plane is
beginalign
(x',y',z') &= (x,y,z)-frac13(x+y+z)(1,1,1) \
&= frac13(2x-y-z,-x+2y-z,-x-y+2z).
endalign
answered Jul 28 at 5:11
DisintegratingByParts
55.5k42273
answered Jul 28 at 5:11
DisintegratingByParts
55.5k42273
answered Jul 28 at 5:11
DisintegratingByParts
55.5k42273
answered Jul 28 at 5:11
DisintegratingByParts
55.5k42273
55.5k42273
add a comment |Â
add a comment |Â
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