Vector intersection I have 2 vectors and their start points.
i.e.
$\vec p_1, \vec v_1$ and $\vec p_2, \vec v_2$
Now I want check if vectors intersect. 
I found this alghoritm.
$\vec c = \vec p_2 - \vec p_1$
$\vec n_1 = $ perpendicular of $\vec v_1$
$\vec n_2 = $ perpendicular of $\vec p$
$d = (\vec n_2 \cdot v_2) / (\vec n_1 \cdot \vec v_2)$
If d is between 0 and 1 then there is intersection on point
$\vec p = \vec p1 + (\vec v_1 \cdot d)$
But this seems to not work.
Is there any other method?
First of all I need to know if 2 vectors intersect, and then get point of intersection.
 A: HINT : Let $P_1(x_1,y_1), v_1(a,b),P_2(x_2,y_2),v_2(c,d).$ Also, let $Q_1,Q_2$ be the end point respectilvely.
Two vectors intersect at the point $P$
$$\iff \text{There exists (k,l)$\in\mathbb R$ such that $\vec{P_1P}=k\vec{P_1Q_1},\vec{P_2P}=l\vec{P_2Q_2}$}$$
From here, you can get a condition about $x_1,y_1,a,b,x_2,y_2,c,d,k,l.$
A: Your method does not look right.
Here is how I would solve
We want
$$ p = p_1 + \lambda v_1 = p_2 + \mu v_2, ~~ 0\le \lambda, \mu \le 1$$
So
$$
c = p_1 - p_2 = \mu v_2 - \lambda v_1$$
Let $n_1$ be perpendicular to $v_1$ and $n_2$ perpendicular to $v_2$.
Then
$$
< n_1, c > = \mu <n_1, v_2>, ~~~ <n_2, c> = -\lambda <n_2,v_1>$$
So, calculate
$$
\mu= \frac{<n_1,c>}{<n_1, v_2>},~~~\lambda= \frac{<n_2,-c>}{<n_2, v_1>}$$
If both $\lambda$ and $\mu$ are between 0 and 1 you have the point of intersection.
Note: If the vectors are parallel, you have to go back to the original equation and see if $c$ is parallel to $v_1$ or not. If it is not, then there is no solution. If so, you can solve for $\alpha$ and $\mu$.
EDIT: -c in lambda
