Question

Shortest distance between two skew lines can be evaluated by the formula, where lines are given by $r=a_1+\lambda b_1$ and $r=a_2+\mu b_2$.

• A. $d=\left|\frac{\left(b_1-b_2\right)\cdot \left(a_1\times a_2\right)}{|a_1\times a_2|}\right|$
• B. $d=\left|\frac{\left(b_1\cdot b_2\right)\times \left(a_2-a_1\right)}{|b_1||b_2|}\right|$
• C. $d=\left|\frac{\left(b_1\times b_2\right)\cdot \left(a_2-a_1\right)}{|b_1\times b_2|}\right|$
• D. $d=\left|\frac{\left(b_1\times b_2\right)\cdot \left(a_1\times a_2\right)}{|b_1\times b_2||a_1\times b_2|}\right|$

Answer 1

Answer: (C)

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,
$PQ=ST|\cos \theta |$
$\text{ But }\cos \theta =\left|\frac{PQ\cdot ST}{|PQ\Vert ST|}\right|$
$\cos \theta =\left|\frac{dn\hat{n}\cdot \left(a_2-a_1\right)}{dST}\right|\left(\because ST=a_2-a_1\right)$
$\cos \theta =\left|\frac{\left(b_1\times b_2\right)\cdot \left(a_2-a_1\right)}{ST|b_1\times b_2|}\right|\text{ [from Eq. (iii)] }$
Hence, the required shortest distance is
$d=PQ=ST|\cos \theta |\text{ or }d=\left|\frac{\left(b_1\times b_2\right)\cdot \left(a_2-a_1\right)}{|b_1\times b_2|}\right|$