The shortest distance (in units) between the lines \(\displaystyle\frac{1-x}{1}=\frac{2y-10}{2}=\frac{z+1}{1}\) and \(\displaystyle\frac{x-3}{-1}=\frac{y-5}{1}=\frac{z-0}{1}\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\sqrt{\displaystyle\frac{14}{3}}\)
In standard form, with \(\frac{1-x}{1} = \frac{x-1}{-1}\) and \(\frac{2y-10}{2} = \frac{y-5}{1}\), the first line is
\( \displaystyle \frac{x-1}{-1} = \frac{y-5}{1} = \frac{z+1}{1} \)
So it passes through \(\vec a_1 = \hat{\imath} + 5\hat{\jmath} - \hat{k}\) with direction \(\vec b = -\hat{\imath} + \hat{\jmath} + \hat{k}\). The second line passes through \(\vec a_2 = 3\hat{\imath} + 5\hat{\jmath}\) with the same direction \(-\hat{\imath} + \hat{\jmath} + \hat{k}\).
The directions are equal, so the lines are parallel, and the distance is
\( \displaystyle d = \frac{\left|\vec b\times(\vec a_2 - \vec a_1)\right|}{|\vec b|} \)
\( \vec a_2 - \vec a_1 = 2\hat{\imath} + 0\hat{\jmath} + \hat{k} \)
\( \displaystyle \vec b\times(\vec a_2 - \vec a_1) = \begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\ -1 & 1 & 1\\ 2 & 0 & 1\end{vmatrix} \)
\( = \hat{\imath}(1 - 0) - \hat{\jmath}(-1 - 2) + \hat{k}(0 - 2) \)
\( = \hat{\imath} + 3\hat{\jmath} - 2\hat{k} \)
\( \left|\hat{\imath} + 3\hat{\jmath} - 2\hat{k}\right| = \sqrt{1 + 9 + 4} = \sqrt{14},\quad |\vec b| = \sqrt{3} \)
\( \displaystyle d = \frac{\sqrt{14}}{\sqrt{3}} = \sqrt{\frac{14}{3}} \)
Tip: Check the direction ratios first. For parallel lines the skew-lines formula cannot be used, because \(\vec b_1\times\vec b_2 = \vec 0\) puts zero in its denominator.