Given two lines in space, either they are parallel, they intersect, or they are skew (lie in parallel planes). Determine whether the lines below, taken two at a time, are parallel, intersect, or are skew. If they intersect, find the point of intersection. Otherwise, find the distance between the two lines. L2 L3: x=1+t, y=-1+2t, z=2-2t, -∞

Let us consider the lines

$$\begin{gathered} \frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1} \\ \frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2} \end{gathered}$$

The lines are parallel if $a_1=a_2,b_1=b_2,c_1=c_2$

The distance between the lines is obtained as

$d=\frac{\left|\left|\begin{array}{ccc}i& j& k\\ x_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\end{array}\right|\right|}{\sqrt{a_1^2+b_1^2+c_1^2}}$

If the line is not parallel nor intersecting then the lines are skew lines

$d=\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{{\left(b_1{c}_2-b_2{c}_1\right)}^2+{\left(c_1{a}_2-c_2{a}_1\right)}^2+{\left(a_1{b}_2-a_2{b}_1\right)}^2}}$