What does the magnitude and direction of a larger vector projected onto a smaller vector depend on? I'm trying to solve these questions:

Given $v = [3, -6, 2]$ and $w = [-1, 6, 5]$, find;

*

*$v \downarrow w$

*$w \downarrow v$

*What does the magnitude of $w \downarrow v$ depend on?

*What does the direction of $w \downarrow v$ depend on?


I already figured out the answer to 1 and 2, since they are pretty straightforward. But I don't really know the answer to 3 and 4.
 A: When one speaks of projecting $\vec a$ onto $\vec b$ (I will write $p_{\vec b}(\vec a)$), I think it is understood that you really mean "the projection of $\vec a$ in the direction of $\vec b$", that, is $\vec b$ is only used to get a direction, and its (nonzero) magnitude is irrelevant.
The direction of $p_{\vec b}(\vec a)$ depends only the direction of $\vec b$ (in fact, it  is the direction of $\vec b$).
The magnitude of $p_{\vec b}(\vec a)$ depends on the relative direction of $\vec a$ with respect to $\vec b$ (said differently, the angle between $\vec a$ and $\vec b$) as well as the magnitude of $\vec a$.
A: Let $p_\vec{y}$ be the projection on the line generated by $\vec{y}$, that is  $$p_\vec{y}:\begin{array}{l}\Bbb R^3 \to \Bbb R^3\\\vec{x}\mapsto \cfrac{\langle \vec{x} \mid \vec{y}\rangle}{\|\vec{x}\|\|\vec{y}\|}\vec{y} \end{array}$$
$\left\|p_\vec{y}(\vec{x})\right\|=\left|\cfrac{\langle \vec{x} \mid \vec{y}\rangle}{\|\vec{x}\|\|\vec{y}\|}\|\vec{y}\|\right|=\cfrac{\left|\cos\left( \vec{x} , \vec{y}\right)\|\vec{x}\|\|\vec{y}\|\right|}{\|\vec{x}\|\|\vec{y}\|}\|\vec{y}\|=\left|\cos\left(\vec{x},\vec{y}\right)\right|\|\vec y\|$ so it depends on the cosinus of the angle between $\vec x$ and $\vec y$.
Its direction depends on the sign of $\langle \vec{x} \mid \vec{y}\rangle$. If it is positive, it has the same direction as $y$ and otherwise, it has an opposite direction.
