For the lines \(L_1: \vec{r} = (\hat{\imath}-\hat{k})+\lambda(2\hat{\imath}-2\hat{\jmath}+\hat{k})\) and \(L_2: \vec{r} = (2\hat{\imath}+3\hat{\jmath}+\hat{k})+\mu(\hat{\imath}+2\hat{\jmath}+2\hat{k})\), write \(L_1\) as \(\vec{r} = \vec{a}_1+\lambda\vec{b}_1\) and \(L_2\) as \(\vec{r} = \vec{a}_2+\mu\vec{b}_2\). Match List-I with List-II.
| List-I | List-II | ||
|---|---|---|---|
| (A) | \(\vec{b}_1\cdot\vec{b}_2\) | (I) | −3 |
| (B) | \(|\vec{b}_1\times\vec{b}_2|\) | (II) | 0 |
| (C) | \((\vec{b}_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1)\) | (III) | 9 |
| (D) | Shortest distance between \(L_1\) and \(L_2\) | (IV) | \(\displaystyle\frac{1}{3}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
Here
\(\vec{b}_1 = 2\hat{\imath}-2\hat{\jmath}+\hat{k},\quad \vec{b}_2 = \hat{\imath}+2\hat{\jmath}+2\hat{k}\)
\(\vec{a}_2-\vec{a}_1 = \hat{\imath}+3\hat{\jmath}+2\hat{k}\)
(A) \(\vec{b}_1\cdot\vec{b}_2\)
This matches (II) 0:
\((2)(1)+(-2)(2)+(1)(2) = 2-4+2 = 0\)
(B) \(|\vec{b}_1\times\vec{b}_2|\)
This matches (III) 9:
\(\vec{b}_1\times\vec{b}_2 = \begin{vmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & -2 & 1 \\ 1 & 2 & 2 \end{vmatrix}\)
\(= \hat{\imath}(-4-2)-\hat{\jmath}(4-1)+\hat{k}(4+2) = -6\hat{\imath}-3\hat{\jmath}+6\hat{k}\)
\(|\vec{b}_1\times\vec{b}_2| = \sqrt{36+9+36} = 9\)
(C) \((\vec{b}_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1)\)
This matches (I) −3:
\((-6)(1)+(-3)(3)+(6)(2) = -6-9+12 = -3\)
(D) Shortest distance between \(L_1\) and \(L_2\)
This matches (IV) \(\frac{1}{3}\), by the shortest distance formula:
\( \displaystyle d = \frac{|-3|}{9} = \frac{1}{3} \)