Oblique projection of a vector on a plane

[Evgeny.Makarov]

Suppose a plane contains the origin and has normal $n$. Is it true that the projection of a vector $u$ on the plane along vector $v$ is $(v\times u)\times n$, where $\times$ denotes the cross product? I can see that the direction is right, but I am not sure about the length. Links to textbooks and other sources are welcome.

 

[S.G. Janssens]

To make sure I understand you, by "oblique" do you mean that $u$ is not in the given plane, nor perpendicular to it? (And $v$ is in the plane?)

 

[Evgeny.Makarov]

By "oblique" I mean that the vector $v$, along which the projection is done, is not necessarily orthogonal to the plane. Vector $u$ is in general not in the plane; otherwise, its projection is $u$ itself. Vector $v$ is not in the plane; otherwise, projection along $v$ is not defined. By projection of $u$ along $v$ I mean a vector $u'$ such that $u'=u+\lambda v$ for some $\lambda\in\mathbb{R}$ and $u'\cdot n=0$.

 

[I like Serena]

Suppose a plane contains the origin and has normal $n$. Is it true that the projection of a vector $u$ on the plane along vector $v$ is $(v\times u)\times n$, where $\times$ denotes the cross product? I can see that the direction is right, but I am not sure about the length. Links to textbooks and other sources are welcome.

Counter example
Suppose $v$ is in the plane. Then the projection is not possible.
However, $(v\times u)\times n$ still yields a specific vector, which is a contradiction.

Formula for oblique projection
Let $p$ be the projection.
Then $p=u-\lambda v$ and $p\cdot n=0$.
Therefore $p\cdot n = u\cdot n - \lambda v\cdot n = 0$.
Solving for $\lambda$ and substituting back gives:
$$p=u-\frac{u\cdot n}{v\cdot n}v$$
This formula also works for a hyperplane in any number of dimensions.

 

[Evgeny.Makarov]

Counter example
Suppose $v$ is in the plane. Then the projection is not possible.
However, $(v\times u)\times n$ still yields a specific vector, which is a contradiction.

Let $\pi$ be the plane with normal $n$ and let $\text{proj}_\pi(u,v)$ be the projection of $u$ along $v$ on $\pi$. Obviously, $\text{proj}_\pi(u,v)$ is uniquely defined iff $v$ is not parallel to $\pi$. I am not so much interested in proving $\text{proj}_\pi(u,v)=(v\times u)\times n$ for all vectors $u$, $v$ and $n$. I'd like to prove this equality provided both sides are defined. In a similar way, $\dfrac{x}{y}\cdot y=x$ is also false in general, but it is still useful.

First, $(v\times u)\times n$ is perpendicular to $n$ and therefore lies in $\pi$. Also, $v\times u$ is perpendicular to the plane defined by $v$ and $u$, so $(v\times u)\times n$ lies in that plane, i.e., it equals $\mu u+\lambda v$ for some $\mu$ and $\lambda$. So it is a projection of some vector proportional to $u$ if not $u$ itself, i.e., it has the correct direction. (In fact, I plan to apply this to projective geometry, so only directions matter in the end.) But I still would like to know if $(v\times u)\times n$ has the right magnitude.

Formula for oblique projection
Let $p$ be the projection.
Then $p=u-\lambda v$ and $p\cdot n=0$.
Therefore $p\cdot n = u\cdot n - \lambda v\cdot n = 0$.
Solving for $\lambda$ and substituting back gives:
$$p=u-\frac{u\cdot n}{v\cdot n}v$$
This formula also works for a hyperplane in any number of dimensions.

Yes, thank you, and I have a question about this as well. Wikipedia shows that $p=u-(u\cdot n)v$ (this is case (S3) in the link above, where our $u$ is denoted by $p$ and its projection is $p'$) leads to the formula $p=(I-vn^T)u$ where vectors are viewed as columns and $A^T$ is matrix transpose, so $vn^T$ is a $3\times 3$ matrix. So the matrix of projection is $I-vn^T$. What I don't understand is where the denominator $v\cdot n$ from your formula went. Case (S1) assumes that $v\cdot n=1$; does this assumption apply to (S3) as well? But I have the book "Projective Geometry and Its Applications to Computer Graphics" by M.A. Penna and R.R. Pattersons, which says on page 113 that the matrix is indeed like this without assuming $v\cdot n=1$. Granted, matrices of projective transformations are defined up to a multiplicative constant, but here we have the identity matrix that is added and not just $vn^T$.

 

[I like Serena]

Let $\pi$ be the plane with normal $n$ and let $\text{proj}_\pi(u,v)$ be the projection of $u$ along $v$ on $\pi$. Obviously, $\text{proj}_\pi(u,v)$ is uniquely defined iff $v$ is not parallel to $\pi$. I am not so much interested in proving $\text{proj}_\pi(u,v)=(v\times u)\times n$ for all vectors $u$, $v$ and $n$. I'd like to prove this equality provided both sides are defined. In a similar way, $\dfrac{x}{y}\cdot y=x$ is also false in general, but it is still useful.

First, $(v\times u)\times n$ is perpendicular to $n$ and therefore lies in $\pi$. Also, $v\times u$ is perpendicular to the plane defined by $v$ and $u$, so $(v\times u)\times n$ lies in that plane, i.e., it equals $\mu u+\lambda v$ for some $\mu$ and $\lambda$. So it is a projection of some vector proportional to $u$ if not $u$ itself, i.e., it has the correct direction. (In fact, I plan to apply this to projective geometry, so only directions matter in the end.) But I still would like to know if $(v\times u)\times n$ has the right magnitude.

It has the wrong magnitude.
Let me clarify. Suppose $v$ is 'nearly' in $\pi$, then the projection vector has near-infinite length.
However, the magnitude of $(v\times u)\times n$ is bounded by the product of the individual magnitudes.

Yes, thank you, and I have a question about this as well. Wikipedia shows that $p=u-(u\cdot n)v$ (this is case (S3) in the link above, where our $u$ is denoted by $p$ and its projection is $p'$) leads to the formula $p=(I-vn^T)u$ where vectors are viewed as columns and $A^T$ is matrix transpose, so $vn^T$ is a $3\times 3$ matrix. So the matrix of projection is $I-vn^T$. What I don't understand is where the denominator $v\cdot n$ from your formula went. Case (S1) assumes that $v\cdot n=1$; does this assumption apply to (S3) as well? But I have the book "Projective Geometry and Its Applications to Computer Graphics" by M.A. Penna and R.R. Pattersons, which says on page 113 that the matrix is indeed like this without assuming $v\cdot n=1$. Granted, matrices of projective transformations are defined up to a multiplicative constant, but here we have the identity matrix that is added and not just $vn^T$.

I don't have that book, but I can guess what they did.
In computer graphics we want to do such a projection on a 'lot' of points, so it must be as computationally cheap as possible.
Generally, we pick $\|n\|=1$ in 'normal' orthogonal projections, which rids us from the division by $n\cdot n$ that we would otherwise have to do. Not to mention that it simplifies the projection formula.
In this case we want $v\cdot n=1$, meaning we will resize $v$ such that this is the case.
Generally that means the $\|v\|>1$.
The book would probably explain in some introductory section that $n$ and/or $v$ are 'suitably' picked or resized, and that this is assumed in the remainder of the book.

Anyway, looking at S3 on wikipedia, I believe that they forgot to mention that they indeed picked $v\cdot n=1$, making S3 a further specialization of S1.

As for the construction of the matrix, note that:
$$(vn^T)p = v(n^Tp)=v(p\cdot n) = (p\cdot n)v$$

 

[I like Serena]

FYI, I've updated the wiki page to include the S1 condition in S3, since I'm convinced it is plain wrong as it is now.
We'll see if anyone will try to revert or modify my changes.

 

[Evgeny.Makarov]

Thanks a lot. The only thing is that there should be no period before "And".

 

[I like Serena]

Thanks a lot. The only thing is that there should be no period before "And".

All right. I've changed the full stop to a comma.
Please modify it yourself if you feel the phrasing can be further improved. ;)