Distance between parallel lines

Theorem 6.7 (Distance between parallel lines). The distance between parallel lines $\vec r=\vec a_1+\lambda\vec b$ and $\vec r=\vec a_2+\lambda\vec b$ (note that both lines share the same direction vector $\vec b$, which is exactly what makes them parallel) is $\left|(\vec a_2-\vec a_1)\times\hat b\right|$, where $\hat b=\vec b/|\vec b|$ is the unit vector along $\vec b$.

Proof (outline). Let $L_1$ pass through $A(\vec a_1)$ and $L_2$ pass through $B(\vec a_2)$, both lines running in the direction $\vec b$ (Fig. 6.7). Drop a perpendicular $BM$ from $B$ onto $L_1$, meeting $L_1$ at the foot $M$; this perpendicular length $BM$ is exactly the distance we want. In the right-angled triangle $AMB$, let $\theta=\angle BAM$ be the angle that $\overrightarrow{AB}$ makes with the common direction $\vec b$ at $A$. Then $\sin\theta=\dfrac{BM}{AB}$, so $BM=AB\sin\theta=AB\cdot|\hat b|\sin\theta$. But $AB\cdot|\hat b|\sin\theta$ is precisely the magnitude of the cross product $\overrightarrow{AB}\times\hat b=(\vec a_2-\vec a_1)\times\hat b$. Hence the distance between the two parallel lines is $d=BM=\left|(\vec a_2-\vec a_1)\times\hat b\right|$.

Ex.(17) Find the distance between parallel lines $\vec r=(2\hat i-\hat j+\hat k)+\lambda(2\hat i+\hat j-2\hat k)$ and $\vec r=(\hat i-\hat j+2\hat k)+\mu(2\hat i+\hat j-2\hat k)$

Solution. Here $\vec a_1=2\hat i-\hat j+\hat k$, $\vec a_2=\hat i-\hat j+2\hat k$, $\vec b=2\hat i+\hat j-2\hat k$.

Alternative method (distance of a point from a line). The distance between $\vec r=\vec a_1+\lambda\vec b$ and $\vec r=\vec a_2+\lambda\vec b$ equals the distance of the point $A(\vec a_1)$ from the line $\vec r=\vec a_2+\lambda\vec b$, given by

Here $\vec a_1-\vec a_2=\hat i-\hat k$, so $|\vec a_1-\vec a_2|=\sqrt2$; and $(\vec a_1-\vec a_2)\cdot\vec b=(\hat i-\hat k)\cdot(2\hat i+\hat j-2\hat k)=2+0+2=4$, with $|\vec b|=3$. So

matching the first method exactly.

Ex.(18) Find the distance between parallel lines $\dfrac{x}{2}=\dfrac{y}{-1}=\dfrac{z}{2}$ and $\dfrac{x-1}{2}=\dfrac{y-1}{-1}=\dfrac{z-1}{2}$ …