How to find shortest distance between two skew lines in 3D?

Question

If given 2 lines $\alpha$ and $\beta$, that are created by

1. 2 points: A and B
2. 2 plane intersection

I want to find shortest distance between them.

$$\left\{\begin{array}{c} P_1=x_1X+y_1Y+z_1Z+C=0 \\ P_2=x_2X+y_2Y+z_2Z+C=0\end{array}\right.$$
$$A=\left(x_3;y_3;z_3\right)$$
$$B=\left(x_4;y_4;z_4\right)$$
$$\alpha =n_1\times n_2=\left(\left|\begin{array}{cc} y_1 & z_1 \\ y_2 & z_2\end{array}\right|;-\left|\begin{array}{cc} x_1 & z_1 \\ x_2 & z_2\end{array}\right|;\left|\begin{array}{cc} x_1 & y_1 \\ x_2 & y_2\end{array}\right|\right)$$
$$\beta =$$

From here I tried:

The question of "shortest distance" is only interesting in the skew case. Let's say p0 and p1 are points on the lines L0 and L1, respectively. Also d0 and d1 are the direction vectors of L0 and L1, respectively. The shortest distance is (p0 - p1) * , in which * is dot product, and is the normalized cross product. The point on L0 that is nearest to L1 is p0 + d0(((p1 - p0) * k) / (d0 * k)), in which k is d1 x d0 x d1.

Read more: http://wiki.answers.com/Q/If_the_shortest_distance_between_two_points_is_a_straight_line_what_is_the_shortest_distance_between_two_straight_lines#ixzz17fAWKFst

I tried, but failed.

Answer 1

Per this wikipedia article, if your lines are $\vec{X}=\vec{X_1}+t\vec{D_1}$ and $\vec{X}=\vec{X_2}+t\vec{D_2}$, the distance between them is $$\left|\frac{\vec{D_1}\times\vec{D_2}}{|\vec{D_1}\times\vec{D_2}|}\cdot(\vec{X_1}-\vec{X_2})\right|.$$

Answer 2

Say you have two lines $\vec L_1 = \vec X_1 + t\vec D_1$ and $\vec L_2 = \vec X_2 + t\vec D_2$.

Start with $(\vec X_1 - \vec X_2)$, which is a skew (non-perpendicular) segment from one line to the other. The distance from $\vec L_1$ to $\vec L_2$ is the component of $(\vec X_1 - \vec X_2)$ that is perpendicular to the lines $\vec L_1$ and $\vec L_2$. We can find this direction by taking the cross product $(\vec L_1 \times \vec L_2)$. After normalizing by dividing by the norm $|\vec L_1 \times \vec L_2|$, take the dot product with $(\vec X_1 - \vec X_2)$ to find the length of the component.

The distance is therefore $$ \left|\frac{\vec{D_1}\times\vec{D_2}}{|\vec{D_1}\times\vec{D_2}|}\cdot(\vec{X_1}-\vec{X_2})\right| = \frac{\left|(\vec{D_1}\times\vec{D_2})\cdot(\vec{X_1}-\vec{X_2})\right|}{|\vec{D_1}\times\vec{D_2}|}. $$