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Q.
Shortest distance between two skew lines can be evaluated by the formula, where lines are given by $r = a _1+\lambda b _1$ and $r = a _2+\mu b _2$.
Three Dimensional Geometry
Solution:
Let $I_1$ and $I_2$ be two skew-lines with equations
$ r=a_1+\lambda b_1 $ ...(i)
and $ r=a_2+\mu b_2$....(ii)
Take any point $S$ on $l_1$ with position vector $a _1$ and $T$ on $I_2$ with position vector $a_2$. Then, the magnitude of the shortest distance vector will be equal to that of the projection of ST along the direction of the line of shortest distance.
If $PQ$ is the shortest distance vector between $I_1$ and $I_2$, then it being perpendicular to both $b_1$ and $b_2$, the unit vector $\hat{n}$ alnng $PQ$ wnulid therefore he
Then, $ PQ =d \hat{n}$
Where, $d$ is the magnitude of the shortest distance vector. Let $\theta$ be the angle between ST and PQ. Then,
$ PQ = ST |\cos \theta| $
$ \text { But } \cos \theta=\left|\frac{P Q \cdot S T}{|P Q \| S T|}\right|$
$ \cos \theta=\left|\frac{d n \hat{n} \cdot\left( a _2- a _1\right)}{d ST }\right| \left(\because ST = a _2- a _1\right) $
$ \cos \theta=\left|\frac{\left(b_1 \times b_2\right) \cdot\left(a_2-a_1\right)}{S T\left|b_1 \times b_2\right|}\right| \text { [from Eq. (iii)] }$
Hence, the required shortest distance is
$d=P Q=S T|\cos \theta| \text { or } d=\left|\frac{\left(b_1 \times b_2\right) \cdot\left(a_2-a_1\right)}{\left|b_1 \times b_2\right|}\right|$
