Hi there! could you elaborate some more? I'm not sure what you mean with fixed direction v. also if I understand correct, your vector $u$ is my vector $AP$. right? but $AP$ is a segment of $AB$, so the 2 coincide - there is not just one intersection point. right?

$v$ is the vector you project against, it's the $a$ in your notation. But the magnitude of $v$ doesn't change the projection; only the direction $v$ is pointing in matters, so sometimes we refer to a projection in a given direction, often specified by a unit vector.

hmm and "specifies $u$ up to a line` means? Which is this line? I still cannot see how the intersection (of what) will give me a result.. (sorry 3am here)

It means that if you know the projection of $u$ in the direction of $v$, then there is a line that you know $u$ must lie in; at least when you're in $2$ dimensions. In $3$ dimensions this is a plane. If you draw what a projection means geometrically in $2$ dimensions, then you should be able to find this line pretty easily, and in $3$ dimensions you just need to revolve this line to generate the plane. (By this I mean, go draw the picture or you won't see what I'm saying.)

I can only definitely see what you are saying if you draw the picture :) . Having said that I have drawn the picture and it looks like this for 2d tutorial.math.lamar.edu/Classes/CalcII/DotProduct.aspx (scroll down to projections), is this the line you refer to? (ie $b$ projected onto $a$ ?

or rather see this. math.lsa.umich.edu/~glarose/classes/calcIII/web/13_5 there is a plane too. The thing is in 3D space a multitude of vectors (all lying on the plane) will intersect with "the line" $u$ to which you refer. no?

Yes, that dotted line is the line I'm referring to. You can probably see that if you extend the dotted line and let $c$ be any vector along this line, then $c$ and $b$ have the same projections onto $a$. Now, if these points were to lie in $3$-dimensional space, then you can also revolve this dotted line to generate a plane, all of whose vectors will have the same projection onto $a$.

i mean one condition is that the dotted line is perp to $a$. so $ \vec dotline \cdot $a$ = 0

but this is not enough to determine the x,y,z of the dotted line...

and to find the intersection of the dotted line to $b$ I need to have find the dotted line first..

To determine the plane (since you are in 3-space) you need: a vector normal to the plane, and a point on the plane. The normal vector is easy: $a$ will do just fine. For the point on the plane, the projection will work.

As to finding $b$, like I said, this alone isn't enough information in general to find $b$, it will only tell you the plane that $b$ lies in.

But in the case of the problem with four points $ABCD$, you need to find where $AB$ intersects this plane. Since a plane and a line can intersect at only 0, 1, or infinitely many points, and you'll get 1 intersection so long as the points are not in some weird configuration, this is enough information to find the intersection.

After this point there isn't much I can do to help you other than restate everything I've said here.

But like I've said, you've got all the information you need to solve the problem, it's a matter of you actually filling in the details if you need an answer in terms of coordinates.