The shortest distance between the lines \(\displaystyle\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}\) and \(\displaystyle\frac{x-2}{4} = \frac{y-4}{6} = \frac{z-5}{8}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\sqrt{\displaystyle\frac{5}{29}}\)
The direction ratios of the second line are \((4, 6, 8) = 2(2, 3, 4)\), so the two lines are parallel. The point (2, 4, 5) of the second line does not lie on the first line (\(\frac{2-1}{2} \ne \frac{4-2}{3}\)), so they are distinct.
Let A(1, 2, 3) lie on the first line, B(2, 4, 5) on the second, and \(\vec b = 2\hat{\imath} + 3\hat{\jmath} + 4\hat{k}\).
\(\overrightarrow{AB} = \hat{\imath} + 2\hat{\jmath} + 2\hat{k}\)
The distance between parallel lines is
\(\displaystyle d = \frac{|\vec b \times \overrightarrow{AB}|}{|\vec b|}\)
\(\displaystyle \vec b \times \overrightarrow{AB} = \begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\ 2 & 3 & 4\\ 1 & 2 & 2\end{vmatrix}\)
\(= (6 - 8)\hat{\imath} - (4 - 4)\hat{\jmath} + (4 - 3)\hat{k}\)
\(= -2\hat{\imath} + \hat{k}\)
\(|\vec b \times \overrightarrow{AB}| = \sqrt{4 + 0 + 1} = \sqrt{5}\)
\(|\vec b| = \sqrt{4 + 9 + 16} = \sqrt{29}\)
\(\displaystyle d = \frac{\sqrt{5}}{\sqrt{29}} = \sqrt{\frac{5}{29}}\)