Q.Find the distance between the parallel lines $\dfrac{x}{2}=\dfrac{y}{-1}=\dfrac{z}{2}$ and $\dfrac{x-1}{2}=\dfrac{y-1}{-1}=\dfrac{z-1}{2}$

For two parallel lines $\vec r=\vec a_1+\lambda\vec b$ and $\vec r=\vec a_2+\lambda\vec b$ -- sharing the same direction vector $\vec b$, which is exactly what makes them parallel -- the distance between them is

This comes from dropping a perpendicular $BM$ from a point $B(\vec a_2)$ on the second line onto the first line at foot $M$: in the right triangle formed with $A(\vec a_1)$, $BM=AB\sin\theta$ where $\theta$ is the angle between $\overrightarrow{AB}$ and $\vec b$, and $AB\,|\hat b|\sin\theta$ is exactly $|\overrightarrow{AB}\times\hat b|$. An equally valid alternative method treats the problem as the distance of the point $A(\vec a_1)$ from the second line, using $d=\sqrt{|\vec a_1-\vec a_2|^2-\left[\dfrac{(\vec a_1-\vec a_2)\cdot\vec b}{|\vec b|}\right]^2}$ (the same Pythagorean idea used for a point- …