Q.Find the shortest distance between the lines $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and $\dfrac{x-2}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5}$. Also find the equation of the line of shortest distance.

Using the standard skew-line formula with the two given lines' points and directions gives $SD=\frac{1}{\sqrt6}$; solving for the feet of the common perpendicular gives the connecting line.

Line 1: point $A_1=(1,2,3)$, direction $\vec d_1=(2,3,4)$. Line 2: point $A_2=(2,4,5)$, direction $\vec d_2=(3,4,5)$.

$\vec d_1\times\vec d_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\2&3&4\\3&4&5\end{vmatrix} = \hat i(15-16)-\hat j(10-12)+\hat k(8-9) = (-1,2,-1)$

$|\vec d_1\times\vec d_2|=\sqrt{1+4+1}=\sqrt6$. With $\overrightarrow{A_1A_2}=(1,2,2)$:

$SD = \dfrac{|\overrightarrow{A_1A_2}\cdot(\vec d_1\times\vec d_2)|}{|\vec d_1\times\vec d_2|} = \dfrac{|(1)(-1)+(2)(2)+(2)(-1)|}{\sqrt6} = \dfrac{|{-1+4-2}|}{\sqrt6} = \dfrac{1}{\sqrt6}$

Line of shortest distance: Let $P_1=(1+2t,2+3t,3+4t)$ on line 1 and $P_2=(2+3s,4+4s,5+5s)$ on line 2. Requiring $\overrightarrow{P_1P_2}\perp\vec d_1$ and $\overrightarrow{P_1P_2}\perp\vec d_2$ gives two equations:

$16+38s-29t=0,\qquad 21+50s-38t=0$

Solving simultaneously: $t=\dfrac13$, $s=-\dfrac16$.

So $P_1 = \left(1+\dfrac23,\,2+1,\,3+\dfrac43\right) = \left(\dfrac53,3,\dfrac{13}3\right)$ and $P_2=\left(\dfrac32,\dfrac{10}3,\dfrac{25}6\right)$.

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