Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Line | A point on the line | ||
| (A) | \(\displaystyle\frac{x-1}{2}=\frac{3-y}{1}=\frac{z+2}{4}\) | (I) | (4, 1, −2) |
| (B) | \(\vec{r} = (2-\lambda)\hat{\imath}+(1+3\lambda)\hat{\jmath}-4\hat{k}\) | (II) | (3, 2, 2) |
| (C) | \(2x-4 = 3y+3 = z\) | (III) | (8, 3, 12) |
| (D) | \(\vec{r} = (\hat{\imath}-2\hat{k})+\mu(3\hat{\imath}+\hat{\jmath})\) | (IV) | (1, 4, −4) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
A point lies on a line when one value of the parameter gives all three of its coordinates (for a Cartesian line: when all the ratios are equal).
(A) \(\frac{x-1}{2}=\frac{3-y}{1}=\frac{z+2}{4}\)
This matches (II) (3, 2, 2):
\( \displaystyle \frac{3-1}{2} = 1,\quad \frac{3-2}{1} = 1,\quad \frac{2+2}{4} = 1 \)
(B) \(\vec{r} = (2-\lambda)\hat{\imath}+(1+3\lambda)\hat{\jmath}-4\hat{k}\)
This matches (IV) (1, 4, −4). Every point has z = −4, and \(\lambda = 1\) gives
\(x = 2-1 = 1,\quad y = 1+3 = 4\)
(C) \(2x-4 = 3y+3 = z\)
This matches (III) (8, 3, 12):
\(2(8)-4 = 12,\quad 3(3)+3 = 12,\quad z = 12\)
(D) \(\vec{r} = (\hat{\imath}-2\hat{k})+\mu(3\hat{\imath}+\hat{\jmath})\)
This matches (I) (4, 1, −2). The general point is \((1+3\mu,\ \mu,\ -2)\), and \(\mu = 1\) gives (4, 1, −2).