Consider the lines \(L_1: \displaystyle\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{6}\) and \(L_2: \vec{r} = (4\hat{\imath}-3\hat{\jmath}+2\hat{k})+\mu(-4\hat{\imath}-6\hat{\jmath}-12\hat{k})\).
(A) \(L_1\) is parallel to \(L_2\).
(B) The point (5, 4, 14) lies on \(L_1\).
(C) The distance between \(L_1\) and \(L_2\) is \(\sqrt{13}\).
(D) \(L_1\) and \(L_2\) intersect.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (C) only
The direction \(-4\hat{\imath}-6\hat{\jmath}-12\hat{k} = -2(2\hat{\imath}+3\hat{\jmath}+6\hat{k})\), so the lines are parallel. Take \(\vec{b} = 2\hat{\imath}+3\hat{\jmath}+6\hat{k}\), with \(|\vec{b}| = \sqrt{4+9+36} = 7\).
The lines pass through (1, −1, 2) and (4, −3, 2):
\(\vec{a}_2-\vec{a}_1 = 3\hat{\imath}-2\hat{\jmath}\)
\(\vec{b}\times(\vec{a}_2-\vec{a}_1) = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 3 & 6 \\ 3 & -2 & 0 \end{vmatrix}\)
\(= \hat{\imath}(0+12)-\hat{\jmath}(0-18)+\hat{k}(-4-9) = 12\hat{\imath}+18\hat{\jmath}-13\hat{k}\)
\(|12\hat{\imath}+18\hat{\jmath}-13\hat{k}| = \sqrt{144+324+169} = \sqrt{637} = 7\sqrt{13}\)
\( \displaystyle d = \frac{7\sqrt{13}}{7} = \sqrt{13} \)
For the point (5, 4, 14) in \(L_1\):
\( \displaystyle \frac{5-1}{2} = 2,\quad \frac{4+1}{3} = \frac{5}{3},\quad \frac{14-2}{6} = 2 \)
So the lines are parallel at distance \(\sqrt{13}\), and the fractions of \(L_1\) at (5, 4, 14) are 2, \(\frac{5}{3}\), 2.
Checking the statements
(A) \(L_1\) is parallel to \(L_2\) — Correct. −4, −6, −12 is −2 times 2, 3, 6.
(B) The point (5, 4, 14) lies on \(L_1\) — Incorrect. The three fractions 2, \(\frac{5}{3}\), 2 are not equal.
(C) The distance between \(L_1\) and \(L_2\) is \(\sqrt{13}\) — Correct. \(\frac{7\sqrt{13}}{7} = \sqrt{13}\).
(D) \(L_1\) and \(L_2\) intersect — Incorrect. They are parallel at distance \(\sqrt{13} \ne 0\), so they never meet.