If \( \lambda \) and \( \mu \) are parameters, then
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | \( \vec{r}=2\hat{\imath}+\hat{\jmath}-3\hat{k}+\lambda(\hat{\imath}-\hat{\jmath}+2\hat{k}) \) parallel to the line | I. | \( \vec{r}=(\hat{\jmath}+\hat{k})+\mu(-2\hat{\imath}+\hat{\jmath}-2\hat{k}) \) |
| B. | \( \vec{r}=\hat{\imath}+\hat{\jmath}+\lambda(\hat{\imath}+\hat{\jmath}-2\hat{k}) \) perpendicular to the line | II. | \( \vec{r}=(2\hat{\imath}+\hat{k})+\mu(-2\hat{\imath}+\hat{\jmath}) \) |
| C. | \( \displaystyle\frac{x}{-2}=\displaystyle\frac{y-1}{1}=\displaystyle\frac{z-1}{-2} \) in vector form | III. | \( \vec{r}=(\hat{\imath}-\hat{\jmath}+2\hat{k})+\mu(2\hat{\imath}-2\hat{\jmath}+4\hat{k}) \) |
| D. | \( \displaystyle\frac{2-x}{2}=y,z=1 \) in vector form | IV. | \( \displaystyle\frac{x-1}{2}=\displaystyle\frac{y+1}{4}=\displaystyle\frac{z}{3} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A - III, B - IV, C - I, D - II
The direction vectors of the lines in List-II are I: (−2, 1, −2), II: (−2, 1, 0), III: (2, −2, 4), IV: (2, 4, 3).
(A) \( \vec{r}=2\hat{\imath}+\hat{\jmath}-3\hat{k}+\lambda(\hat{\imath}-\hat{\jmath}+2\hat{k}) \) parallel to the line
This matches (III).
The direction (2, −2, 4) = 2(1, −1, 2), so the lines are parallel. No other direction in List-II is a multiple of (1, −1, 2).
(B) \( \vec{r}=\hat{\imath}+\hat{\jmath}+\lambda(\hat{\imath}+\hat{\jmath}-2\hat{k}) \) perpendicular to the line
This matches (IV).
Perpendicular lines have directions with zero dot product:
\( \displaystyle (1,1,-2)\cdot(2,4,3)=2+4-6=0 \)
For the others: with I it is −2 + 1 + 4 = 3, with II −2 + 1 + 0 = −1, with III 2 − 2 − 8 = −8, none zero.
(C) \( \frac{x}{-2}=\frac{y-1}{1}=\frac{z-1}{-2} \) in vector form
This matches (I).
The line passes through (0, 1, 1) with direction (−2, 1, −2):
\( \displaystyle \vec{r}=(\hat{\jmath}+\hat{k})+\mu(-2\hat{\imath}+\hat{\jmath}-2\hat{k}) \)
(D) \( \frac{2-x}{2}=y,z=1 \) in vector form
This matches (II).
In symmetric form it is \( \frac{x-2}{-2}=\frac{y-0}{1}=\frac{z-1}{0} \). The line passes through (2, 0, 1) with direction (−2, 1, 0):
\( \displaystyle \vec{r}=(2\hat{\imath}+\hat{k})+\mu(-2\hat{\imath}+\hat{\jmath}) \)
So the matching is A - III, B - IV, C - I, D - II.