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Q.
The distance between two parallel lines can be taken out by the formula
Three Dimensional Geometry
Solution:
If two lines $l_1$ and $l_2$ are parallel, then they are coplanar.
Let the lines be given by
$ r=a_1+\lambda b$ ...(i)
and $r=a_2+\mu b$....(ii)
where $a_1$ is the position vector of a point $S$ on $l_1$ and $a_2$ is the position vector of a point $T$ on $I_2$.
As $I_1, I_2$ are coplanar, if the foot of the perpendicular from $T$ on the line $l_1$ is $P$, then the distance between the lines, $l_1$ and $I_2=| TP |$.
Let $\theta$ be the angle between the vector ST and $b$.
Then, $ b \times S T=(|b||S T| \sin \theta) n$ where $\hat{n}$ is the unit vector perpendicular to the plane of the line $I_1$ and $I_2$
But $ ST = a _2- a _1$
Therefore, from Eq. (iii), we get
$ b \times\left( a _2- a _1\right)=| b | PT \hat{n} (\because PT - ST \sin \theta)$
$\text { i.e., } \left| b \times\left( a _2- a _1\right)\right|=| b | P T \cdot 1 ( as |\hat{n}|=1)$
Hence, the distance between the given parallel lines is
$d=|P T|=\left|\frac{b \times\left(a_2-a_1\right)}{|b|}\right|$
