Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Vector equation | Cartesian equation | ||
| (A) | \(\vec{r} = (\hat{\imath}+2\hat{\jmath}-3\hat{k})+\lambda(2\hat{\imath}-\hat{\jmath}+4\hat{k})\) | (I) | \(\displaystyle\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-3}\) |
| (B) | \(\vec{r} = (2\hat{\imath}-\hat{\jmath}+4\hat{k})+\lambda(\hat{\imath}+2\hat{\jmath}-3\hat{k})\) | (II) | \(\displaystyle\frac{x-1}{-2}=\frac{y-2}{1}=\frac{z+3}{4}\) |
| (C) | \(\vec{r} = (-\hat{\imath}-2\hat{\jmath}+3\hat{k})+\lambda(2\hat{\imath}-\hat{\jmath}+4\hat{k})\) | (III) | \(\displaystyle\frac{x-1}{2}=\frac{y-2}{-1}=\frac{z+3}{4}\) |
| (D) | \(\vec{r} = (\hat{\imath}+2\hat{\jmath}-3\hat{k})+\lambda(-2\hat{\imath}+\hat{\jmath}+4\hat{k})\) | (IV) | \(\displaystyle\frac{x+1}{2}=\frac{y+2}{-1}=\frac{z-3}{4}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
The line \(\vec{r} = (x_1\hat{\imath}+y_1\hat{\jmath}+z_1\hat{k})+\lambda(a\hat{\imath}+b\hat{\jmath}+c\hat{k})\) has the Cartesian form \(\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}\): the point gives the numerators and the direction gives the denominators.
(A) point (1, 2, −3), direction 2, −1, 4
This matches (III) \(\frac{x-1}{2}=\frac{y-2}{-1}=\frac{z+3}{4}\).
(B) point (2, −1, 4), direction 1, 2, −3
This matches (I) \(\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-3}\).
(C) point (−1, −2, 3), direction 2, −1, 4
This matches (IV) \(\frac{x+1}{2}=\frac{y+2}{-1}=\frac{z-3}{4}\).
(D) point (1, 2, −3), direction −2, 1, 4
This matches (II) \(\frac{x-1}{-2}=\frac{y-2}{1}=\frac{z+3}{4}\).