Match the LIST-I with LIST-II
| LIST-I Equation of Line | LIST-II Direction Ratio of line | ||
| A. | \( \displaystyle\frac{2-x}{1}=\displaystyle\frac{2y+1}{1}=\displaystyle\frac{z}{1} \) | I. | 2, 1, 1 |
| B. | \( x=2y+3,z=y+1 \) | II. | 1, −2, 2 |
| C. | \( \displaystyle\frac{x-1}{1}=\displaystyle\frac{y+1}{2},z=1 \) | III. | −2, 1, 2 |
| D. | \( \vec{r}=\hat{\imath}+\hat{\jmath}+\lambda\left(\hat{\imath}-2\hat{\jmath}+2\hat{k}\right) \) | IV. | 1, 2, 0 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-III, B-I, C-IV, D-II
When a line is written in the standard form \( \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} \), (a, b, c) are its direction ratios.
(A) \( \displaystyle\frac{2-x}{1}=\displaystyle\frac{2y+1}{1}=\displaystyle\frac{z}{1} \)
This matches III. −2, 1, 2.
\( \displaystyle \frac{x-2}{-1}=\frac{y+\frac{1}{2}}{\frac{1}{2}}=\frac{z}{1} \)
Direction ratios (−1, 1/2, 1), i.e. (−2, 1, 2) after multiplying by 2.
(B) \( x=2y+3,z=y+1 \)
This matches I. 2, 1, 1. Put y = t:
\( x=2t+3,\quad y=t,\quad z=t+1 \)
The coefficients of t give (2, 1, 1).
(C) \( \displaystyle\frac{x-1}{1}=\displaystyle\frac{y+1}{2},z=1 \)
This matches IV. 1, 2, 0. z stays equal to 1, so its direction ratio is 0; the others are 1 and 2.
(D) \( \vec{r}=\hat{\imath}+\hat{\jmath}+\lambda\left(\hat{\imath}-2\hat{\jmath}+2\hat{k}\right) \)
This matches II. 1, −2, 2. The vector multiplying \( \lambda \) gives the direction.