Let L1 and L2 lines such that L1 : \( \vec{r}=\hat{\imath}+\hat{\jmath}+\lambda(2\hat{\imath}-\hat{\jmath}+\hat{k}) \) , L2: \( \vec{r}=2\hat{\imath}+\hat{\jmath}-\hat{k}+\mu(4\hat{\imath}-2\hat{\jmath}+2\hat{k}) \) and \( \theta \) be the acute angle between L1 and L2, then
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | \( \cos\theta \) | I. | \( \displaystyle\frac{\sqrt{66}}{6} \) |
| B. | \( \sin\theta \) | II. | 3 |
| C. | the shortest distance (in units) between L1 and L2 | III. | 1 |
| D. | (sum of squares of direction cosines of L1 ) +2 | IV. | 0 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-III, B-IV, C-I, D-II
The direction ratios are \( \vec{b}_1=(2,-1,1) \) and \( \vec{b}_2=(4,-2,2)=2\vec{b}_1 \), so the lines are parallel and the angle between them is 0.
A. \( \cos\theta \)
This matches III. 1.
\( \displaystyle \cos\theta=\frac{|\vec{b}_1\cdot\vec{b}_2|}{|\vec{b}_1||\vec{b}_2|} \)
\( \displaystyle =\frac{8+2+2}{\sqrt{6}\cdot\sqrt{24}} \)
\( \displaystyle =\frac{12}{12} \)
\( \displaystyle =1 \)
B. \( \sin\theta \)
This matches IV. 0.
\( \displaystyle \cos\theta=1\ \Rightarrow\ \theta=0 \)
\( \displaystyle \sin\theta=0 \)
C. the shortest distance (in units) between L1 and L2
This matches I. \( \frac{\sqrt{66}}{6} \).
For parallel lines \( \vec{r}=\vec{a}_1+\lambda\vec{b} \) and \( \vec{r}=\vec{a}_2+\mu\vec{b} \), the distance is
\( \displaystyle d=\frac{|\vec{b}\times(\vec{a}_2-\vec{a}_1)|}{|\vec{b}|} \)
With \( \vec{b}=2\hat{\imath}-\hat{\jmath}+\hat{k} \):
\( \displaystyle \vec{a}_2-\vec{a}_1=(2\hat{\imath}+\hat{\jmath}-\hat{k})-(\hat{\imath}+\hat{\jmath}) \)
\( \displaystyle =\hat{\imath}-\hat{k} \)
\( \displaystyle \vec{b}\times(\vec{a}_2-\vec{a}_1)=\begin{vmatrix}\hat{\imath} & \hat{\jmath} & \hat{k}\\2 & -1 & 1\\1 & 0 & -1\end{vmatrix} \)
\( \displaystyle =\hat{\imath}(1-0)-\hat{\jmath}(-2-1)+\hat{k}(0+1) \)
\( \displaystyle =\hat{\imath}+3\hat{\jmath}+\hat{k} \)
\( \displaystyle d=\frac{\sqrt{1+9+1}}{\sqrt{4+1+1}} \)
\( \displaystyle =\frac{\sqrt{11}}{\sqrt{6}} \)
\( \displaystyle =\frac{\sqrt{66}}{6} \)
D. (sum of squares of direction cosines of L1 ) +2
This matches II. 3.
\( \displaystyle l^2+m^2+n^2=1 \)
\( \displaystyle 1+2=3 \)
So the matching is A-III, B-IV, C-I, D-II.
Tip: Always compare the direction ratios first: if they are proportional, use the parallel-lines distance formula; the skew-lines formula would give 0/0 here.