Consider the lines \(L_1: \vec{r} = \lambda(\hat{\imath}+\hat{\jmath})\), \(L_2: \vec{r} = \hat{k}+\mu(\hat{\imath}-\hat{\jmath})\) and \(L_3: \vec{r} = (\hat{\imath}+\hat{\jmath}+\hat{k})+t\hat{k}\).
(A) \(L_1\) and \(L_2\) are skew lines.
(B) \(L_1\) and \(L_3\) intersect.
(C) \(L_2\) and \(L_3\) intersect.
(D) The shortest distance between \(L_1\) and \(L_2\) is 2.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
No two of the directions \(\hat{\imath}+\hat{\jmath}\), \(\hat{\imath}-\hat{\jmath}\), \(\hat{k}\) are parallel, so each pair either intersects or is skew. General points are
\(L_1: (\lambda, \lambda, 0)\qquad L_2: (\mu, -\mu, 1)\qquad L_3: (1, 1, 1+t)\)
\(L_1\) and \(L_2\): every point of \(L_1\) has z = 0 and every point of \(L_2\) has z = 1, so they never meet. Their shortest distance is
\((\hat{\imath}+\hat{\jmath})\times(\hat{\imath}-\hat{\jmath}) = -2\hat{k}\)
\( \displaystyle d = \frac{|(-2\hat{k})\cdot\hat{k}|}{|-2\hat{k}|} = \frac{2}{2} = 1 \)
\(L_1\) and \(L_3\): \((\lambda, \lambda, 0) = (1, 1, 1+t)\) gives \(\lambda = 1\), \(t = -1\), the point (1, 1, 0).
\(L_2\) and \(L_3\): \((\mu, -\mu, 1) = (1, 1, 1+t)\) needs \(\mu = 1\) and \(-\mu = 1\) together, which is impossible.
So \(L_1, L_2\) are skew at distance 1, \(L_1, L_3\) meet at (1, 1, 0), and \(L_2, L_3\) do not meet.
Checking the statements
(A) \(L_1\) and \(L_2\) are skew lines — Correct. Not parallel, and z = 0 on one, z = 1 on the other.
(B) \(L_1\) and \(L_3\) intersect — Correct. They meet at (1, 1, 0).
(C) \(L_2\) and \(L_3\) intersect — Incorrect. \(\mu = 1\) and \(\mu = -1\) cannot both hold, so they are skew.
(D) The shortest distance between \(L_1\) and \(L_2\) is 2 — Incorrect. It is \(\frac{2}{2} = 1\).