Find the shortest distance between tfracx+11=tfrac…

Find the shortest distance between $\tfrac{x+1}1=\tfrac y1=\tfrac{z-1}2$ and $\tfrac x1=\tfrac{y+1}3=\tfrac{z-2}4$.

Explanation of Method

The shortest distance between two skew lines.

Step 1: Solution

$\overrightarrow{M_1M_2}=\{1,-1,1\}$, $\vec p_1=\{1,1,2\}$, $\vec p_2=\{1,3,4\}$, and $\vec n=\vec p_1\times\vec p_2=\{-2,-2,2\}$: \[\begin{vmatrix}1&-1&1\\1&1&2\\1&3&4\end{vmatrix}=-2+2+2=2,\qquad |\vec n|=2\sqrt3 .\]

\[d=\frac{2}{2\sqrt3}=\frac{\sqrt3}{3}\approx0.577\]

The directions are not proportional and the determinant is $\ne0$, so the lines are skew.

$d=\tfrac{\sqrt3}{3}$ $\ell_1$ $\ell_2$

The skew lines of Problem 243 and their common perpendicular.

Result

\[\boxed{\,d=\frac{2}{2\sqrt3}=\frac{\sqrt3}{3}\approx0.577\,}\]