The distance between the lines \(\displaystyle\frac{x-2}{1}=\frac{y+1}{-2}=\frac{z-1}{2}\) and \(\displaystyle\frac{x+1}{-2}=\frac{y-3}{4}=\frac{z-2}{-4}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\sqrt{17}\)
The direction ratios −2, 4, −4 are −2 times 1, −2, 2, so the lines are parallel. Take \(\vec{b} = \hat{\imath}-2\hat{\jmath}+2\hat{k}\).
The lines pass through (2, −1, 1) and (−1, 3, 2):
\(\vec{a}_2-\vec{a}_1 = -3\hat{\imath}+4\hat{\jmath}+\hat{k}\)
\(\vec{b}\times(\vec{a}_2-\vec{a}_1) = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 1 & -2 & 2 \\ -3 & 4 & 1 \end{vmatrix}\)
\(= \hat{\imath}(-2-8)-\hat{\jmath}(1+6)+\hat{k}(4-6) = -10\hat{\imath}-7\hat{\jmath}-2\hat{k}\)
\(|\vec{b}\times(\vec{a}_2-\vec{a}_1)| = \sqrt{100+49+4} = \sqrt{153} = 3\sqrt{17}\)
\(|\vec{b}| = \sqrt{1+4+4} = 3\)
The distance between parallel lines is
\( \displaystyle d = \frac{3\sqrt{17}}{3} = \sqrt{17} \)
Why the other options are incorrect
Option (1) \(2\sqrt{17}\) — Incorrect. This takes the cross product with the second direction \(-2\hat{\imath}+4\hat{\jmath}-4\hat{k}\) (giving \(6\sqrt{17}\)) but divides by the length 3 of the first; the same vector must be used in both places.
Option (3) \(\displaystyle\frac{\sqrt{17}}{2}\) — Incorrect. This takes the cross product with \(\hat{\imath}-2\hat{\jmath}+2\hat{k}\) but divides by the length 6 of \(-2\hat{\imath}+4\hat{\jmath}-4\hat{k}\).
Option (4) \(3\sqrt{17}\) — Incorrect. This is \(|\vec{b}\times(\vec{a}_2-\vec{a}_1)|\) before dividing by \(|\vec{b}| = 3\).