Find the plane parallel to the lines y=2x-1, z=3x+2 and…

Find the plane parallel to the lines $y=2x-1,\ z=3x+2$ and $y=-x+2,\ z=4x-1$, passing through the midpoint of their shortest distance.

Step 1: Data

\[ \vec d_1=\{1,2,3\},\ A_0(0,-1,2);\qquad \vec d_2=\{1,-1,4\},\ B_0(0,2,-1);\qquad \vec n=\vec d_1\times\vec d_2=\{11,-1,-3\}. \]

Step 2: Feet of the common perpendicular

With $A=(t,2t-1,3t+2)$, $B=(s,-s+2,4s-1)$ and $\overrightarrow{AB}=\{s-t,\ -s-2t+3,\ 4s-3t-3\}$, the conditions $\overrightarrow{AB}\cdot\vec d_1=0$ and $\overrightarrow{AB}\cdot\vec d_2=0$ give

\[ 11s-14t-3=0,\qquad 18s-11t-15=0\ \Longrightarrow\ t=\tfrac{111}{131},\ s=\tfrac{177}{131}. \]

\[ A\left(\tfrac{111}{131},\tfrac{91}{131},\tfrac{595}{131}\right),\quad B\left(\tfrac{177}{131},\tfrac{85}{131},\tfrac{577}{131}\right),\quad M\left(\tfrac{144}{131},\tfrac{88}{131},\tfrac{586}{131}\right). \]

Check: $\overrightarrow{AB}=\tfrac{6}{131}\{11,-1,-3\}\parallel\vec n$, and $|AB|=\tfrac{6}{\sqrt{131}}$.

Step 3: Plane

\[11x-y-3z+2=0\]

Check with $M$: $\tfrac{1584-88-1758+262}{131}=0$.

Shortcut: the planes through the two lines parallel to each other are $11x-y-3z+5=0$ (through $A_0$) and $11x-y-3z-1=0$ (through $B_0$). The required plane is halfway between them, with free term $\tfrac{5+(-1)}{2}=2$.