The distance between the lines \(\vec{r}=\hat{\imath}-2\hat{\jmath}+3\hat{k}+\lambda(2\hat{\imath}+3\hat{\jmath}+6\hat{k})\) and \(\vec{r}=3\hat{\imath}-2\hat{\jmath}+1\hat{k}+\mu(4\hat{\imath}+6\hat{\jmath}+12\hat{k})\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{\sqrt{328}}{7}\)
The direction vectors are \(\vec{b}_1 = 2\hat{\imath}+3\hat{\jmath}+6\hat{k}\) and \(\vec{b}_2 = 4\hat{\imath}+6\hat{\jmath}+12\hat{k} = 2\vec{b}_1\), so the lines are parallel. For parallel lines \(\vec{r} = \vec{a}_1 + \lambda\vec{b}\) and \(\vec{r} = \vec{a}_2 + \mu\vec{b}\), the distance is
\( \displaystyle d = \frac{\left|\vec{b}\times\left(\vec{a}_2-\vec{a}_1\right)\right|}{|\vec{b}|} \)
Take \(\vec{b} = 2\hat{\imath}+3\hat{\jmath}+6\hat{k}\). With \(\vec{a}_1 = \hat{\imath}-2\hat{\jmath}+3\hat{k}\) and \(\vec{a}_2 = 3\hat{\imath}-2\hat{\jmath}+\hat{k}\):
\(\vec{a}_2-\vec{a}_1 = 2\hat{\imath}+0\hat{\jmath}-2\hat{k}\)
\( \displaystyle \vec{b}\times\left(\vec{a}_2-\vec{a}_1\right) = \begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\ 2 & 3 & 6\\ 2 & 0 & -2\end{vmatrix} \)
\( = \hat{\imath}(-6-0) - \hat{\jmath}(-4-12) + \hat{k}(0-6) \)
\( = -6\hat{\imath}+16\hat{\jmath}-6\hat{k} \)
The magnitudes:
\( \left|\vec{b}\times\left(\vec{a}_2-\vec{a}_1\right)\right| = \sqrt{36+256+36} = \sqrt{328} \)
\( |\vec{b}| = \sqrt{4+9+36} = 7 \)
So
\( \displaystyle d = \frac{\sqrt{328}}{7} \)
NCERT (2026-27 edition):
Distance Between Parallel Lines
