Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Pair of lines | Shortest distance | ||
| (A) | \(\vec{r} = \lambda(\hat{\imath}+\hat{k})\) and \(\vec{r} = (\hat{\imath}+3\hat{\jmath}+2\hat{k})+\mu(\hat{\imath}-\hat{k})\) | (I) | 0 |
| (B) | \(\vec{r} = 2\hat{\imath}+\lambda(\hat{\jmath}-\hat{k})\) and \(\vec{r} = 4\hat{\imath}+\mu(2\hat{\jmath}-2\hat{k})\) | (II) | 1 |
| (C) | \(\vec{r} = \hat{\imath}+\lambda\hat{\jmath}\) and \(\vec{r} = \hat{k}+\mu(\hat{\imath}+\hat{\jmath})\) | (III) | 2 |
| (D) | \(\vec{r} = (\hat{\imath}+\hat{\jmath}+\hat{k})+\lambda(\hat{\imath}-\hat{\jmath})\) and \(\vec{r} = (2\hat{\imath}+\hat{k})+\mu(\hat{\jmath}+\hat{k})\) | (IV) | 3 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(III), (C)-(II), (D)-(I)
For non-parallel lines the shortest distance is \(\displaystyle\frac{|(\vec{b}_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1)|}{|\vec{b}_1\times\vec{b}_2|}\); for parallel lines it is \(\displaystyle\frac{|\vec{b}\times(\vec{a}_2-\vec{a}_1)|}{|\vec{b}|}\).
(A) \(\vec{r} = \lambda(\hat{\imath}+\hat{k})\) and \(\vec{r} = (\hat{\imath}+3\hat{\jmath}+2\hat{k})+\mu(\hat{\imath}-\hat{k})\)
This matches (IV) 3:
\((\hat{\imath}+\hat{k})\times(\hat{\imath}-\hat{k}) = 2\hat{\jmath}\)
\( \displaystyle d = \frac{|2\hat{\jmath}\cdot(\hat{\imath}+3\hat{\jmath}+2\hat{k})|}{2} = \frac{6}{2} = 3 \)
(B) \(\vec{r} = 2\hat{\imath}+\lambda(\hat{\jmath}-\hat{k})\) and \(\vec{r} = 4\hat{\imath}+\mu(2\hat{\jmath}-2\hat{k})\)
This matches (III) 2. The lines are parallel, with \(\vec{b} = \hat{\jmath}-\hat{k}\) and \(\vec{a}_2-\vec{a}_1 = 2\hat{\imath}\):
\((\hat{\jmath}-\hat{k})\times 2\hat{\imath} = -2\hat{k}-2\hat{\jmath}\)
\( \displaystyle d = \frac{\sqrt{4+4}}{\sqrt{2}} = \frac{2\sqrt{2}}{\sqrt{2}} = 2 \)
(C) \(\vec{r} = \hat{\imath}+\lambda\hat{\jmath}\) and \(\vec{r} = \hat{k}+\mu(\hat{\imath}+\hat{\jmath})\)
This matches (II) 1:
\(\hat{\jmath}\times(\hat{\imath}+\hat{\jmath}) = -\hat{k}\)
\( \displaystyle d = \frac{|(-\hat{k})\cdot(-\hat{\imath}+\hat{k})|}{1} = 1 \)
(D) \(\vec{r} = (\hat{\imath}+\hat{\jmath}+\hat{k})+\lambda(\hat{\imath}-\hat{\jmath})\) and \(\vec{r} = (2\hat{\imath}+\hat{k})+\mu(\hat{\jmath}+\hat{k})\)
This matches (I) 0. The lines meet: \((1+\lambda,\ 1-\lambda,\ 1) = (2,\ \mu,\ 1+\mu)\) gives \(\lambda = 1\), \(\mu = 0\), and the point (2, 0, 1) is on both.