Question

Find the shortest distance between the lines
$\overrightarrow r=(\hat i+2\hat j+\hat k)+\lambda(\hat i-\hat j+\hat k)$ and
$\overrightarrow r=2\hat i-\hat j-\hat k+\mu\,(2\hat i+2\hat j+2\hat k)$

Answer 1

The equations of the given lines are
$\overrightarrow r=(\hat i+2\hat j+\hat k)+\lambda(\hat i-\hat j+\hat k)$
$\overrightarrow r=2\hat i-\hat j-\hat k+\mu\,(2\hat i+2\hat j+2\hat k)$

It is known that the shortest distance between the lines,

$\overrightarrow r= \overrightarrow a_1+\lambda\overrightarrow b_1$and $\overrightarrow r=\overrightarrow a_2+\mu\overrightarrow b_2$ is given by,
d=|(b1→×b2→).(a2→-a2→)/b1→×b2→|....(1)

Comparing the given equations, we obtain
$\overrightarrow a_1$=$\hat i+2\hat j+\hat k$
$\overrightarrow b_1$=$\hat i-\hat j+\hat k$

$\overrightarrow a_2$=$2\hat i-\hat j-\hat k$
$\overrightarrow b_2$=$2\hat i+\hat j+2\hat k$

$\overrightarrow a_2$-$\overrightarrow a_1$=($2\hat i-\hat j-\hat k$)-($\hat i+2\hat j+\hat k$)=$\hat i-3\hat j-2\hat k$

$\overrightarrow b_1*\overrightarrow b_2= \begin{vmatrix}\hat i&\hat j&\hat k\\\1&-1&1\\\2&1&2\end{vmatrix}$

$\overrightarrow b_1*\overrightarrow b_2$=$(-2-1)\hat i-(2-2)\hat j+(1+2)\hat k=3\hat i+3\hat k$

$\Rightarrow$ |$\overrightarrow b_1*\overrightarrow b_2$|

=$\sqrt{(-3)^2+(3)^2}$
=$\sqrt{9+9}$
=$\sqrt{18}$
=3$\sqrt2$

Substituting all the values in equation(1), we obtain
d=$\begin{vmatrix}\frac{(-3\hat i+3\hat k).(\hat i-3\hat j-2\hat k)}{3\sqrt2}\end{vmatrix}$

$\Rightarrow d=\begin{vmatrix}\frac{(-3.1+3(-2))}{3\sqrt2}\end{vmatrix}$

$\Rightarrow d=\begin{vmatrix}\frac{-9}{3\sqrt2}\end{vmatrix}$

$\Rightarrow d=\frac{3}{2}$

=$\frac{3\sqrt2}{2}$

Therefore, the shortest distance between the two lines is $\frac{3\sqrt2}{2}$ units.