Question

Find the shortest distance between the lines whose vector equations are

$\overrightarrow r=(\hat i+2\hat j+3\hat k)+\lambda(\hat i-3\hat j+2\hat k)$

and $\overrightarrow r=(4\hat i+5\hat j+6\hat k)+\mu(2\hat i+3\hat j+\hat k)$

Answer 1

The given lines are $\overrightarrow r=(\hat i+2\hat j+3\hat k)+\lambda(\hat i-3\hat j+2\hat k)$and $\overrightarrow r=(4\hat i+5\hat j+6\hat k)+\mu(2\hat i+3\hat j+\hat k)$

It is known that the shortest distance between the lines, $\overrightarrow r=\overrightarrow a_1+\lambda b_1$ and $\overrightarrow r=\overrightarrow a_2+\mu b_2$, is given by,
d=|(b1×b2).(a2-a2)/|b1×b2||...(1)

Comparing the given equations with $\overrightarrow r=\overrightarrow a_1+\lambda b_1$ and $\overrightarrow r=\overrightarrow a_2+\mu b_2$, we obtain

$a_1=\hat i+2\hat j+3\hat k$
$b_1=\hat i-3\hat j+2\hat k$
$a_2=4\hat i+5\hat j+6\hat k$
$b_2=2\hat i+3\hat j+\hat k$

$\overrightarrow a_2-\overrightarrow a_1$

=$(4\hat i+5\hat j+6\hat k)$-$(\hat i+2\hat j+3\hat k)$

=$3\hat i+3\hat j+3\hat k$

$\overrightarrow b_1*\overrightarrow b_2$

=$\begin{vmatrix}\hat i&\hat j&\hat k\\\1&-3&2\\\2&3&1\end{vmatrix}$

=(-3-6)$\hat i$-(1-4)$\hat j$+(3+6)$\hat k$

=-9$\hat i$+3$\hat j$+9$\hat k$

$\Rightarrow \mid \overrightarrow b_1*\overrightarrow b_2\mid$
=$\sqrt{(-9)^2+(3)^2+(9)^2}$
=$\sqrt{81+9+81}$
=$\sqrt{171}$
=$3\sqrt{19}$

$(\overrightarrow b_1*\overrightarrow b_2)$.$(\overrightarrow a_2-\overrightarrow a_1)$
=(-9$\hat i$+3$\hat j$+9$\hat k$).(3$\hat i$+3$\hat j$+3$\hat k$)
=-9×3+3×3+9×3
=9

Substituting all the values in equation (1), we obtain
d=|$\frac{9}{3\sqrt{19}}$|
=$\frac{3}{\sqrt {19}}$

Therefore, the shortest distance between the two given lines is $\frac{3}{\sqrt {19}}$ units.