If $\vec{a}\times 2\vec{j}=0$, what can you say about $\vec{a}$?
$\vec{a}$ and $\vec{j}$ are parallel; paralelogram they span has area 0

There's gotta be more?? Please tell me.

If $\vec{b}\times\vec{j}=2\vec{i}$ and $\langle \vec{b}, \vec{j}\rangle=-1$, what's $\vec{b}$?
Are $\vec{i}$ and $\vec{j}$ meant as orthonormal vectors $(1,0,0)$ and $(0,1,0)$?

If so, is this right: $\vec{b}\times\vec{j}=(-b_3)\vec{i}+b_1\vec{k}=2\vec{i} \Rightarrow b_3=-2, b_1=0, b_2\in\mathbb{R}\Rightarrow \vec{b}=(0,p,-2);p\in\mathbb{R}$??? Geometrically sounds right, in right-handed system $\vec{-k}\times\vec{j}=\vec{i}$, but I want to be sure.

If $\vec{c}=(c_1,c_2,c_3)$ and $\vec{d}=(-1,2,0)$ what is $\vec{c}\times \vec{d}$?
$=(-2c_3,-c_3,2c_1+c_2)$

$\vec{a}=(1,1,1,1), \vec{b}=(1,-1,1,-1), \vec{c}=(1,i,-1,-i)$Are $\vec{c},\vec{c},\vec{c}$ in-dependent? Do they make a basis in $\mathbb{C}^4$
$\langle c,c\rangle=1-1+1-1=0$

$\langle a,b\rangle=1-1+1-1=0$
$\langle a,c\rangle=1+i-1-i=0$
$\langle b,c\rangle=1-i-1+i=0$
So, they are in-dependent.

But, they don't make a basis because "there's not enough of them" () for 4D vector space; they don't span the whole space? OK, what's more rigorous answer?

Let $X=Y=\mathbb{R}^2$ and $V$ be rotation of 90° around origin. Define kernel, image and rank of map $V$.
In kernel is only $\vec{0}$, right?
Image is the whole $\mathbb{R}^2$???
Rank of $V$ must be 2, otherwise it couldn't rotate (how could I better formalise my answer; if the 2 is correct of course)?

What should $X=Y=\mathbb{R}^2$ mean (what are $X$ and $Y$, obviously not points)?!

Let $Y$ be linear sub-space inside $\mathbb{K}^n$. Define orthogonal projection $\vec{y}$ of $\vec{z}\in\mathbb{K}^n$ to $Y.$
???
(Connects with previous quote.) Let also $\big{\vec{f_1},...,\vec{f_k}\big}$ be ortho-normal basis for $Y$. Express $\vec{y}$ projection (from previous) with this basis.
???