orthogonal projection of a vector onto a plane
$begingroup$
I am trying to find the orthogonal projection of a vector $vec u= (1,-1,2)$ onto a plane which has three points $vec a=(1,0,0)$ ,$vec b=(1,1,1)$, and $vec c=(0,0,1)$. I started by projecting $vec u$ onto $vec a$ and projecting the same vector onto $vec c$ and finally adding both projections ,but I am not getting what I am expecting. Can someone please help?
linear-algebra projection
edited Dec 28 '18 at 5:32
Andrei
12.4k21128
asked Dec 28 '18 at 1:18
husein mbhusein mb
82
$endgroup$
add a comment |
$begingroup$
I am trying to find the orthogonal projection of a vector $vec u= (1,-1,2)$ onto a plane which has three points $vec a=(1,0,0)$ ,$vec b=(1,1,1)$, and $vec c=(0,0,1)$. I started by projecting $vec u$ onto $vec a$ and projecting the same vector onto $vec c$ and finally adding both projections ,but I am not getting what I am expecting. Can someone please help?
linear-algebra projection
edited Dec 28 '18 at 5:32
Andrei
12.4k21128
asked Dec 28 '18 at 1:18
husein mbhusein mb
82
$endgroup$
$begingroup$
This plane does not pass through the origin.
$endgroup$
– amd
Dec 28 '18 at 1:26
add a comment |
$begingroup$
I am trying to find the orthogonal projection of a vector $vec u= (1,-1,2)$ onto a plane which has three points $vec a=(1,0,0)$ ,$vec b=(1,1,1)$, and $vec c=(0,0,1)$. I started by projecting $vec u$ onto $vec a$ and projecting the same vector onto $vec c$ and finally adding both projections ,but I am not getting what I am expecting. Can someone please help?
linear-algebra projection
edited Dec 28 '18 at 5:32
Andrei
12.4k21128
asked Dec 28 '18 at 1:18
husein mbhusein mb
82
$endgroup$
I am trying to find the orthogonal projection of a vector $vec u= (1,-1,2)$ onto a plane which has three points $vec a=(1,0,0)$ ,$vec b=(1,1,1)$, and $vec c=(0,0,1)$. I started by projecting $vec u$ onto $vec a$ and projecting the same vector onto $vec c$ and finally adding both projections ,but I am not getting what I am expecting. Can someone please help?
linear-algebra projection
linear-algebra projection
edited Dec 28 '18 at 5:32
Andrei
12.4k21128
asked Dec 28 '18 at 1:18
husein mbhusein mb
82
edited Dec 28 '18 at 5:32
Andrei
12.4k21128
asked Dec 28 '18 at 1:18
husein mbhusein mb
82
edited Dec 28 '18 at 5:32
Andrei
12.4k21128
edited Dec 28 '18 at 5:32
Andrei
12.4k21128
edited Dec 28 '18 at 5:32
Andrei
12.4k21128
12.4k21128
asked Dec 28 '18 at 1:18
husein mbhusein mb
82
asked Dec 28 '18 at 1:18
husein mbhusein mb
82
asked Dec 28 '18 at 1:18
husein mbhusein mb
82
82
$begingroup$
This plane does not pass through the origin.
$endgroup$
– amd
Dec 28 '18 at 1:26
add a comment |
$begingroup$
This plane does not pass through the origin.
$endgroup$
– amd
Dec 28 '18 at 1:26
$begingroup$
This plane does not pass through the origin.
$endgroup$
– amd
Dec 28 '18 at 1:26
$begingroup$
This plane does not pass through the origin.
$endgroup$
– amd
Dec 28 '18 at 1:26
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is easy to guess the equation of the plane as $x-y+z=1$ from which you see that the normal is $(1,-1,1)$. To find the projection choose $alpha$ such that $(1,-1,2)+alpha (1,-1,1)$ lies in the plane $x-y+z=1$. This gives $alpha =-1$ and the projection is nothing but $vec c=(0,0,1)$.
answered Dec 28 '18 at 5:37
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
add a comment |
$begingroup$
You can work out the normal to the plane by computing $$vec n=(vec a-vec b)times (vec b- vec c)$$where $times$ is the vector cross product. Then project your vector $vec u$ onto this normal to get $vec u_parallel$. Then the required projection onto the plane is $$vec u_perp=vec u-vec u_parallel+vec a$$ where the $vec a$ is added on to ensure the vector lies on the plane, rather than lying parallel to the plane, but starting at the origin.
answered Dec 28 '18 at 1:47
John DoeJohn Doe
11.2k11239
$endgroup$
$begingroup$
,Thank you for your answer it was very helpfull .
$endgroup$
– husein mb
Dec 28 '18 at 1:53
$begingroup$
The formula for the normal is wrong. For example try to find the normal to the plane defined by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. The normal should be along $(1,1,1)$. The correct way is $vec n=(vec a-vec b)times(vec a-vec c)$
$endgroup$
– Andrei
Dec 28 '18 at 4:36
$begingroup$
@Andrei yes, you're right, you need to pick two coplanar vectors.
$endgroup$
– John Doe
Dec 28 '18 at 4:38
1
$begingroup$
Better, but still wrong. The $vec u_perp$ obtained per your answer doesn’t lie on the plane. You appear to be making the same fundamental error as the OP in neglecting the fact that this plane doesn’t pass through the origin.
$endgroup$
– amd
Dec 28 '18 at 4:39
$begingroup$
@amd yep, thats right, thanks
$endgroup$
– John Doe
Dec 28 '18 at 4:49
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is easy to guess the equation of the plane as $x-y+z=1$ from which you see that the normal is $(1,-1,1)$. To find the projection choose $alpha$ such that $(1,-1,2)+alpha (1,-1,1)$ lies in the plane $x-y+z=1$. This gives $alpha =-1$ and the projection is nothing but $vec c=(0,0,1)$.
answered Dec 28 '18 at 5:37
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
add a comment |
$begingroup$
It is easy to guess the equation of the plane as $x-y+z=1$ from which you see that the normal is $(1,-1,1)$. To find the projection choose $alpha$ such that $(1,-1,2)+alpha (1,-1,1)$ lies in the plane $x-y+z=1$. This gives $alpha =-1$ and the projection is nothing but $vec c=(0,0,1)$.
answered Dec 28 '18 at 5:37
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
add a comment |
$begingroup$
It is easy to guess the equation of the plane as $x-y+z=1$ from which you see that the normal is $(1,-1,1)$. To find the projection choose $alpha$ such that $(1,-1,2)+alpha (1,-1,1)$ lies in the plane $x-y+z=1$. This gives $alpha =-1$ and the projection is nothing but $vec c=(0,0,1)$.
answered Dec 28 '18 at 5:37
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
$endgroup$
It is easy to guess the equation of the plane as $x-y+z=1$ from which you see that the normal is $(1,-1,1)$. To find the projection choose $alpha$ such that $(1,-1,2)+alpha (1,-1,1)$ lies in the plane $x-y+z=1$. This gives $alpha =-1$ and the projection is nothing but $vec c=(0,0,1)$.
answered Dec 28 '18 at 5:37
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
edited Dec 28 '18 at 5:57
answered Dec 28 '18 at 5:37
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Dec 28 '18 at 5:37
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
answered Dec 28 '18 at 5:37
Kavi Rama MurthyKavi Rama Murthy
62.7k42262
62.7k42262
add a comment |
add a comment |
$begingroup$
You can work out the normal to the plane by computing $$vec n=(vec a-vec b)times (vec b- vec c)$$where $times$ is the vector cross product. Then project your vector $vec u$ onto this normal to get $vec u_parallel$. Then the required projection onto the plane is $$vec u_perp=vec u-vec u_parallel+vec a$$ where the $vec a$ is added on to ensure the vector lies on the plane, rather than lying parallel to the plane, but starting at the origin.
answered Dec 28 '18 at 1:47
John DoeJohn Doe
11.2k11239
$endgroup$
$begingroup$
,Thank you for your answer it was very helpfull .
$endgroup$
– husein mb
Dec 28 '18 at 1:53
$begingroup$
The formula for the normal is wrong. For example try to find the normal to the plane defined by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. The normal should be along $(1,1,1)$. The correct way is $vec n=(vec a-vec b)times(vec a-vec c)$
$endgroup$
– Andrei
Dec 28 '18 at 4:36
$begingroup$
@Andrei yes, you're right, you need to pick two coplanar vectors.
$endgroup$
– John Doe
Dec 28 '18 at 4:38
1
$begingroup$
Better, but still wrong. The $vec u_perp$ obtained per your answer doesn’t lie on the plane. You appear to be making the same fundamental error as the OP in neglecting the fact that this plane doesn’t pass through the origin.
$endgroup$
– amd
Dec 28 '18 at 4:39
$begingroup$
@amd yep, thats right, thanks
$endgroup$
– John Doe
Dec 28 '18 at 4:49
add a comment |
$begingroup$
You can work out the normal to the plane by computing $$vec n=(vec a-vec b)times (vec b- vec c)$$where $times$ is the vector cross product. Then project your vector $vec u$ onto this normal to get $vec u_parallel$. Then the required projection onto the plane is $$vec u_perp=vec u-vec u_parallel+vec a$$ where the $vec a$ is added on to ensure the vector lies on the plane, rather than lying parallel to the plane, but starting at the origin.
answered Dec 28 '18 at 1:47
John DoeJohn Doe
11.2k11239
$endgroup$
$begingroup$
,Thank you for your answer it was very helpfull .
$endgroup$
– husein mb
Dec 28 '18 at 1:53
$begingroup$
The formula for the normal is wrong. For example try to find the normal to the plane defined by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. The normal should be along $(1,1,1)$. The correct way is $vec n=(vec a-vec b)times(vec a-vec c)$
$endgroup$
– Andrei
Dec 28 '18 at 4:36
$begingroup$
@Andrei yes, you're right, you need to pick two coplanar vectors.
$endgroup$
– John Doe
Dec 28 '18 at 4:38
1
$begingroup$
Better, but still wrong. The $vec u_perp$ obtained per your answer doesn’t lie on the plane. You appear to be making the same fundamental error as the OP in neglecting the fact that this plane doesn’t pass through the origin.
$endgroup$
– amd
Dec 28 '18 at 4:39
$begingroup$
@amd yep, thats right, thanks
$endgroup$
– John Doe
Dec 28 '18 at 4:49
add a comment |
$begingroup$
You can work out the normal to the plane by computing $$vec n=(vec a-vec b)times (vec b- vec c)$$where $times$ is the vector cross product. Then project your vector $vec u$ onto this normal to get $vec u_parallel$. Then the required projection onto the plane is $$vec u_perp=vec u-vec u_parallel+vec a$$ where the $vec a$ is added on to ensure the vector lies on the plane, rather than lying parallel to the plane, but starting at the origin.
answered Dec 28 '18 at 1:47
John DoeJohn Doe
11.2k11239
$endgroup$
You can work out the normal to the plane by computing $$vec n=(vec a-vec b)times (vec b- vec c)$$where $times$ is the vector cross product. Then project your vector $vec u$ onto this normal to get $vec u_parallel$. Then the required projection onto the plane is $$vec u_perp=vec u-vec u_parallel+vec a$$ where the $vec a$ is added on to ensure the vector lies on the plane, rather than lying parallel to the plane, but starting at the origin.
answered Dec 28 '18 at 1:47
John DoeJohn Doe
11.2k11239
edited Dec 28 '18 at 4:48
answered Dec 28 '18 at 1:47
John DoeJohn Doe
11.2k11239
answered Dec 28 '18 at 1:47
John DoeJohn Doe
11.2k11239
answered Dec 28 '18 at 1:47
John DoeJohn Doe
11.2k11239
11.2k11239
$begingroup$
,Thank you for your answer it was very helpfull .
$endgroup$
– husein mb
Dec 28 '18 at 1:53
$begingroup$
The formula for the normal is wrong. For example try to find the normal to the plane defined by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. The normal should be along $(1,1,1)$. The correct way is $vec n=(vec a-vec b)times(vec a-vec c)$
$endgroup$
– Andrei
Dec 28 '18 at 4:36
$begingroup$
@Andrei yes, you're right, you need to pick two coplanar vectors.
$endgroup$
– John Doe
Dec 28 '18 at 4:38
1
$begingroup$
Better, but still wrong. The $vec u_perp$ obtained per your answer doesn’t lie on the plane. You appear to be making the same fundamental error as the OP in neglecting the fact that this plane doesn’t pass through the origin.
$endgroup$
– amd
Dec 28 '18 at 4:39
$begingroup$
@amd yep, thats right, thanks
$endgroup$
– John Doe
Dec 28 '18 at 4:49
add a comment |
$begingroup$
,Thank you for your answer it was very helpfull .
$endgroup$
– husein mb
Dec 28 '18 at 1:53
$begingroup$
The formula for the normal is wrong. For example try to find the normal to the plane defined by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. The normal should be along $(1,1,1)$. The correct way is $vec n=(vec a-vec b)times(vec a-vec c)$
$endgroup$
– Andrei
Dec 28 '18 at 4:36
$begingroup$
@Andrei yes, you're right, you need to pick two coplanar vectors.
$endgroup$
– John Doe
Dec 28 '18 at 4:38
1
$begingroup$
Better, but still wrong. The $vec u_perp$ obtained per your answer doesn’t lie on the plane. You appear to be making the same fundamental error as the OP in neglecting the fact that this plane doesn’t pass through the origin.
$endgroup$
– amd
Dec 28 '18 at 4:39
$begingroup$
@amd yep, thats right, thanks
$endgroup$
– John Doe
Dec 28 '18 at 4:49
$begingroup$
,Thank you for your answer it was very helpfull .
$endgroup$
– husein mb
Dec 28 '18 at 1:53
$begingroup$
,Thank you for your answer it was very helpfull .
$endgroup$
– husein mb
Dec 28 '18 at 1:53
$begingroup$
The formula for the normal is wrong. For example try to find the normal to the plane defined by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. The normal should be along $(1,1,1)$. The correct way is $vec n=(vec a-vec b)times(vec a-vec c)$
$endgroup$
– Andrei
Dec 28 '18 at 4:36
$begingroup$
The formula for the normal is wrong. For example try to find the normal to the plane defined by $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. The normal should be along $(1,1,1)$. The correct way is $vec n=(vec a-vec b)times(vec a-vec c)$
$endgroup$
– Andrei
Dec 28 '18 at 4:36
$begingroup$
@Andrei yes, you're right, you need to pick two coplanar vectors.
$endgroup$
– John Doe
Dec 28 '18 at 4:38
$begingroup$
@Andrei yes, you're right, you need to pick two coplanar vectors.
$endgroup$
– John Doe
Dec 28 '18 at 4:38
1
1
$begingroup$
Better, but still wrong. The $vec u_perp$ obtained per your answer doesn’t lie on the plane. You appear to be making the same fundamental error as the OP in neglecting the fact that this plane doesn’t pass through the origin.
$endgroup$
– amd
Dec 28 '18 at 4:39
$begingroup$
Better, but still wrong. The $vec u_perp$ obtained per your answer doesn’t lie on the plane. You appear to be making the same fundamental error as the OP in neglecting the fact that this plane doesn’t pass through the origin.
$endgroup$
– amd
Dec 28 '18 at 4:39
$begingroup$
@amd yep, thats right, thanks
$endgroup$
– John Doe
Dec 28 '18 at 4:49
$begingroup$
@amd yep, thats right, thanks
$endgroup$
– John Doe
Dec 28 '18 at 4:49
add a comment |
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$begingroup$
This plane does not pass through the origin.
$endgroup$
– amd
Dec 28 '18 at 1:26