How is it that dot product does projection but so does division?

Question

Can anyone explain how it is that dot products can perform projection but so can division, and how the processes are related (or not)?

With dot product, you can project one vector onto another vector:

$\hat{v} = \hat{a} \cdot \hat{b}$

Where $b$ is normalized and you want to project $a$ onto that vector direction.

Division can be used to do projection too though. In the case of converting 3d coordinates to 2d coordinates (like in computer graphics for rendering 3d geometry on a 2d screen), that can be done by dividing x and y by z:

$x' = x / z$
$y' = y / z$

Where $x$, $y$ and $z$ are the 3d coordinates of a point, and $x'$ and $y'$ are the 2d coordinates of the point.

In the dot product case, one vector has to be normalized, is that where the equivalent division is happening maybe?

Thanks!

Answer 1

"Projection" can mean many things, and about the only feature they have in common is that it's about an (idempotent) transformation that moves points somewhere along straight lines. (And even that is not fully general -- although some "map projections" can be explained as moving points along straight lines from the surface of a sphere to an image plane, there are many that cannot).

1. The kind of projection done by $\vec y=\vec a\cdot\vec b$ is a projection that moves points from space onto a line, according to the rule that each point must move along a line that intersects the given line perpendicularly. If the given line goes through $(0,0,0)$ in the direction of $\vec b$ where $\|\vec b\|=1$, then the projection moves the point $\vec a$ to $(\vec a\cdot \vec b) \vec b$. (If $\vec b$ is not necessarily normalized, the correct formula is $\frac{\vec a \cdot \vec b}{\vec b\cdot\vec b} \vec b$).

2. The projection you get by computing $\frac xz$ and $\frac yz$ is a central perspective, where each point in space moves to an image plane, according to the rule that each point must move along a line that passes through the camera point. In your case the camera is at $(0,0,0)$ and the image plane is $z=1$, so the projection is actually $(x,y,z)\mapsto (\frac xz,\frac yz, 1)$.

As can be seen, these two projections do fairly different things -- there is not much in common between "the set of lines that intersect $\vec b$ perpendicularly" and "the set of lines that pass through the camera point", and the rule for where on the projecting line each point ends up is different too. So there is no particular reason to expect that the formulas for doing either would be similar.