Preface:
1. I have two straight line $r$ and $r'$, we know that
$$r: ax+by+c=0 \iff \mathbf{v}=(a,b)\perp r \tag 1$$ $$r: ax+by+c=0 \iff \mathbf{w}=(a',b')\perp r' \tag 2$$
If $r\parallel r'$ the coefficients, of $r'$, for example, are proportional to $r$, hence it must be
$$ \left|\begin{matrix} a & b \\ a' & b' \end{matrix}\right|=0 \iff ab'-a'b=0 \tag 3$$
If $r\perp r'$ must be
$$aa'+bb'=0 \tag 4$$
2. If $A\equiv(x_1,y_1)$ and $B\equiv(x_2,y_2)$ we know that $\overrightarrow{AB}=(x_2-x_1,y_2-y_1)$.
3. If $\mathbf{v}=(a,b) \parallel \mathbf{w}=(h,k)$ then $$ a=\lambda h, \quad b=\mu k \iff \frac ah=\frac bk \tag{4'}$$
Question:
Neglecting the trivial cases that two vectors $\mathbf{v}=(v_x,v_y)$ and $\mathbf{w}=(w_x,w_y)$ are both $\mathbf{0}$, we suppose that they are both $\neq \mathbf{0}$. It is easy to try that $$\mathbf{v}\bullet \mathbf{w}=v_xw_x+v_yw_y \tag 5$$ considering that $\theta$ is the convess angle
$$\mathbf{v}\bullet \mathbf{w}=vw\cos \theta, \quad \theta=\angle \mathbf{v}\mathbf{w} \tag 6$$
Now if $$\mathbf{v}\perp \mathbf{w} \iff \theta=\pi/2 \implies v_xw_x+v_yw_y=0 \iff (4)\tag 7$$
$$\mathbf{v}\parallel \mathbf{w} \iff \theta=0 \implies v_xw_x+v_yw_y=vw \tag 8$$
For the cross product $\times$
$$\mathbf{v}\times\mathbf{w}=(v_xw_y-v_yw_x)\hat{\bf z}\equiv \mathbf{u}=(u_x,u_y) \tag 9$$ where $\mathbf{u}$ is perpendicular to the plane of lies $\mathbf{v}$ and $\mathbf{w}$, and $|\mathbf{u}|=|\mathbf{v}\times\mathbf{w}|=vw\sin \theta$.
Now if $\mathbf{v}\times\mathbf{w}=0\iff \theta=0,\pi$ is it true that
$$\mathbf{v}\parallel \mathbf{w} \iff v_xw_y-v_yw_x=0\quad \color{red}{?} \tag{10}$$
For my humble opinion it is true. Infact the $(10) \iff (3)$.
Using the cross product what is the condition that $$\mathbf{v}\perp \mathbf{w}, \quad ? \tag{11}$$
I have this simple problem:
Assigned two vectors $\mathbf{v}=(4, k-2)$ and $\mathbf{w}=(k,6)$, you find for which $k\in \Bbb R$ values are parallel and perpendicular. The solution are $k=-4$ and $k=6$; and after $k=6/5$.
Using the $(8)$ (scalar product), I have
$$4k+6\left(k-2\right)=\sqrt{16+\left(k-2\right)^2}\cdot \sqrt{k^2+36}$$ that have only the solution $k=6$!!! and not also $k=-4$. Using the $(4)$ I have:
$$4k+6(k-2)=0 \iff k=6/5, \quad \text{ok!}$$
If from the $(9)$,
$$\mathbf{v}\times\mathbf{w}=(v_xw_y-v_yw_x)\hat{\bf z}$$
I find $$v_xw_y-v_yw_x=0 \implies \mathbf{v}\parallel\mathbf{w}$$
Hence, is it possible to find the condition of parallelism o perpendicularity using the cross vector, the $(9)$ and the $(11)$ that I not know? After is there a correlation between $(4')$ with the parallelism of the scalar product $(8)$ and cross product?