The distance between the two lines \( L_1 \) and \( L_2 \) given by
\( L_1:\displaystyle\frac{x-1}{2}=\frac{y-2}{3}=\frac{-z-4}{-6} \) and \( L_2:\displaystyle\frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6} \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{\sqrt{293}}{7} \)
Since \( \frac{-z-4}{-6}=\frac{z+4}{6} \), the two lines in standard form are
\( \displaystyle L_1:\ \frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6} \)
\( \displaystyle L_2:\ \frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6} \)
Both have direction \( \vec{b}=2\hat{\imath}+3\hat{\jmath}+6\hat{k} \), so the lines are parallel, with \( |\vec{b}|=\sqrt{4+9+36}=7 \).
Points: \( A(1,2,-4) \) on \( L_1 \) and \( B(3,3,-5) \) on \( L_2 \), so \( \vec{AB}=2\hat{\imath}+\hat{\jmath}-\hat{k} \).
Distance between parallel lines:
\( \displaystyle d=\frac{|\vec{AB}\times\vec{b}|}{|\vec{b}|} \)
\( \displaystyle \vec{AB}\times\vec{b}=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\2 & 1 & -1\\2 & 3 & 6\end{vmatrix} \)
\( \displaystyle =(6+3)\hat{\imath}-(12+2)\hat{\jmath}+(6-2)\hat{k} \)
\( \displaystyle =9\hat{\imath}-14\hat{\jmath}+4\hat{k} \)
\( \displaystyle |\vec{AB}\times\vec{b}|=\sqrt{81+196+16}=\sqrt{293} \)
\( \displaystyle d=\frac{\sqrt{293}}{7} \)
Why the other options are incorrect
Option (1) \( \sqrt{293} \) — Incorrect. \( \sqrt{293} \) is \( |\vec{AB}\times\vec{b}| \), not yet divided by \( |\vec{b}|=7 \).
Option (2) 12 — Incorrect. The distance is \( \frac{\sqrt{293}}{7}\approx 2.45 \), not 12.
Option (4) 13 — Incorrect. The distance cannot exceed \( |\vec{AB}|=\sqrt{6}\approx 2.45 \), so it cannot be 13.
Tip: Before reading direction ratios, rewrite every fraction with +z; here \( \frac{-z-4}{-6} \) hides a direction ratio of +6, which makes the lines parallel.
NCERT (2026-27 edition):
Shortest Distance Between Two Lines
