Skew lines distance and parallel plane

Text

Given $r_1:\ (x,y,z)=(1+t,\,3,\,2-t)$ and $r_2:\ (x,y,z)=(2+2s,\,-1+s,\,4)$.

(a) Prove that $r_1$ and $r_2$ are skew.

(b) Compute the distance between the two lines.

(c) Find the plane through the origin parallel to both lines.

Solution

(a) The direction vectors are $\vec u=(1,0,-1)$ and $\vec v=(2,1,0)$, which are not parallel, so $\vec u\times\vec v=(1,-2,1)$. Taking $P_1=(1,3,2)$ on $r_1$ and $P_2=(2,-1,4)$ on $r_2$, we get $\overrightarrow{P_1P_2}=(1,-4,2)$; the scalar triple product equals $1+8+2=11\neq0$, so the two lines are not coplanar: they are skew.

(b) $d=\dfrac{|11|}{|\vec u\times\vec v|}=\dfrac{11}{\sqrt6}\approx4.491$.

(c) The plane through the origin with normal $(1,-2,1)$ (parallel to both lines, since $\vec u\times\vec v$ is normal to both $\vec u$ and $\vec v$) is $x-2y+z=0$. $\boxed{\text{skew};\ d=\dfrac{11}{\sqrt6}\approx4.491;\ \text{plane } x-2y+z=0}$