move triangle points in fixed directions until area is zero I have a triangle given by the points $x_0$, $x_1$, $x_2$, and three (non-normalized) vecotr $v_0$, $v_1$, $v_2$ into the direction of which the points are to be moved:

$x_0' = x_0 + t \cdot v_0$
$x_1' = x_1 + t \cdot v_1$
$x_2' = x_2 + t \cdot v_2$
Question: How far can I move the points ($t=?$) until the area of the resulting triangle is 0?
One way of doing is is to look at the area of the triangle given be the projection of the edges onto one another,
$$
A = \frac{1}{2} \sqrt{
        e_0^T e_1 \cdot e_1^T e_2
        + e_1^T e_2 \cdot e_2^T e_0
        + e_2^T e_0 \cdot e_0^T e_1
    }
$$
and setting the term in the root to 0. This yields a quartic equation for $t$. The nature of the problem has me suspect, however, that there is a simpler approach.
Any hints?
 A: I think you must have made a mistake in your formula somewhere, the solution is less complicated. I'm assuming a planar problem, that means $x_0,x_1,x_2$ and $v_0,v_1,v_2$ are all coplanar.
For a $\triangle ABC$ the area is one half of the absolute value of the cross product of any 2 sides, considered as vectors:
$$A_{\triangle ABC} = \frac12\lvert\vec{AB}\times\vec{AC}\rvert = \frac12\lvert\vec{BA}\times\vec{BC}\rvert = \frac12\lvert\vec{CA}\times\vec{CB}\rvert.$$
Since all of the considered vectors lie in the $x-y$-plane, the cross product is a vector in the $z$-direction only, which simplifies the calculation.
If we apply this formula to the time dependent vertices $x_0(t),x_1(t),x_2(t)$, we get
$$
\begin{split}
A_{\triangle x_0(t)x_1(t)x_2(t)} &= \frac12\left\|\overrightarrow{x_0(t)x_1(t)}\times\overrightarrow{x_0(t)x_2(t)}\right\|\\
 &=\frac12\left\|
\begin{pmatrix} 
x_{1x}-x_{0x} + t (v_{1x}-v_{0x}) \\
x_{1y}-x_{0y} + t (v_{1y}-v_{0y}) \\
0
\end{pmatrix}
\times
\begin{pmatrix} 
x_{2x}-x_{0x} + t (v_{2x}-v_{0x}) \\
x_{2y}-x_{0y} + t (v_{2y}-v_{0y}) \\
0
\end{pmatrix}
\right\|\\
&= \frac{1}{2}\left\|\begin{pmatrix}
0\\
0\\
(x_{1x}-x_{0x} + t (v_{1x}-v_{0x}))(x_{2y}-x_{0y} + t (v_{2y}-v_{0y})) - (x_{1y}-x_{0y} + t (v_{1y}-v_{0y}))(x_{2x}-x_{0x} + t (v_{2x}-v_{0x}))
\end{pmatrix}
\right\|.
\end{split}
$$
Setting $A_{\triangle x_0(t)x_1(t)x_2(t)} = 0$ yields a (generally) quadratic equation.
