Relative position of two lines

Statement

Given $r_1:(1+t;\; 2-t;\; t)$ and $r_2:(1+2s;\; -1+s;\; 3+s)$, establish whether they are parallel, intersecting or skew. If they are skew, calculate their distance.

Solution

Points and directions: $A(1;2;0)$, $\vec v_1=(1;-1;1)$; $B(1;-1;3)$, $\vec v_2=(2;1;1)$.

Parallelism. $\tfrac{1}{2}\ne\tfrac{-1}{1}$: not parallel.

System $A+\vec v_1 t = B+\vec v_2 s$: $\begin{cases}1+t=1+2s\\2-t=-1+s\\t=3+s\end{cases}$ From the first $t=2s$; from the second $3-t=s$, that is $3-2s=s\Rightarrow s=1,\ t=2$. Check on the third: $t=3+s\Rightarrow 2=4$, false. The lines are skew.

Distance. $\vec{AB}=(0;-3;3)$ and $\vec v_1\times\vec v_2 = (-1\cdot 1-1\cdot 1,\; 1\cdot 2-1\cdot 1,\; 1\cdot 1-(-1)\cdot 2) = (-2;1;3),$ $|\vec v_1\times\vec v_2|=\sqrt{14}$, $\;\vec{AB}\cdot(\vec v_1\times\vec v_2)=0-3+9=6$. Hence $d = \frac{|6|}{\sqrt{14}} = \frac{3\sqrt{14}}{7}\approx \boxed{1{,}60}.$

Links

Topics: Analytic geometry in space
Concepts: Distance between skew lines · Scalar triple product · Line · Line in space · Skew lines
Skills: Analytic geometry · Reasoning by cases · Solving systems
Exercise types: Geometric problem