Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Equation of line | Information | ||
| (A) | \(\vec r = (3\hat{\imath} - 2\hat{\jmath} + \hat{k}) + \lambda(\hat{\jmath} - 2\hat{k})\) | (I) | Direction ratios are 2, 4, −1 |
| (B) | \(\displaystyle\frac{2-x}{1} = \frac{2y+1}{4}, z = 2\) | (II) | Perpendicular to \(2\hat{\imath} - \hat{\jmath} + \hat{k}\) |
| (C) | \(\displaystyle\frac{x}{1} = \frac{y-3}{2} = \frac{3-4z}{2}\) | (III) | Passing through the point (3, −2, 1) |
| (D) | \(\vec r = (3\hat{\imath} + 2\hat{\jmath} + \hat{k}) + \lambda(2\hat{\imath} + \hat{\jmath} - 3\hat{k})\) | (IV) | Direction ratios are −1, 2, 0 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
The denominators give the direction ratios only when x, y and z each have coefficient 1 in the numerators, so the Cartesian forms are rewritten first.
(A) \(\vec r = (3\hat{\imath} - 2\hat{\jmath} + \hat{k}) + \lambda(\hat{\jmath} - 2\hat{k})\)
This matches (III) Passing through the point (3, −2, 1).
At \(\lambda = 0\), the position vector is \(3\hat{\imath} - 2\hat{\jmath} + \hat{k}\), the point (3, −2, 1).
(B) \(\displaystyle\frac{2-x}{1} = \frac{2y+1}{4}, z = 2\)
This matches (IV) Direction ratios are −1, 2, 0.
\(\displaystyle \frac{2 - x}{1} = \frac{x - 2}{-1}\)
\(\displaystyle \frac{2y + 1}{4} = \frac{y + \frac{1}{2}}{2}\)
z stays equal to 2, so its direction ratio is 0. The direction ratios are −1, 2, 0.
(C) \(\displaystyle\frac{x}{1} = \frac{y-3}{2} = \frac{3-4z}{2}\)
This matches (I) Direction ratios are 2, 4, −1.
\(\displaystyle \frac{3 - 4z}{2} = \frac{-4\left(z - \frac{3}{4}\right)}{2}\)
\(\displaystyle = \frac{z - \frac{3}{4}}{-\frac{1}{2}}\)
The direction ratios are \(1, 2, -\frac{1}{2}\); multiplying by 2 gives 2, 4, −1.
(D) \(\vec r = (3\hat{\imath} + 2\hat{\jmath} + \hat{k}) + \lambda(2\hat{\imath} + \hat{\jmath} - 3\hat{k})\)
This matches (II) Perpendicular to \(2\hat{\imath} - \hat{\jmath} + \hat{k}\).
\((2\hat{\imath} + \hat{\jmath} - 3\hat{k})\cdot(2\hat{\imath} - \hat{\jmath} + \hat{k}) = 4 - 1 - 3 = 0\)
So the matching is (A)-(III), (B)-(IV), (C)-(I), (D)-(II).