Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (A) | Angle between \(\hat{\imath}-\hat{\jmath}\) and \(\hat{\jmath} + \hat{k}\) | (I) | 0 |
| (B) | Angle between \(2\hat{\jmath}-\hat{k}\) and \(\hat{\jmath} + 2\hat{k}\) | (II) | \(\displaystyle\frac{2\pi}{3}\) |
| (C) | Angle between \(\hat{\imath} + 2\hat{\jmath}\) and \(5\hat{\imath}+ 10\hat{\jmath}\) | (III) | \(\displaystyle\frac{\pi}{6}\) |
| (D) | Angle between \(\sqrt{3}\hat{\imath}+\hat{\jmath}\) and \(\hat{\imath}\) | (IV) | \(\displaystyle\frac{\pi}{2}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
The angle \(\theta\) between \(\vec{a}\) and \(\vec{b}\) comes from the dot product:
\(\displaystyle \cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}\)
(A) Angle between \(\hat{\imath}-\hat{\jmath}\) and \(\hat{\jmath} + \hat{k}\)
This matches (II) 2π/3:
\((1)(0) + (-1)(1) + (0)(1) = -1\)
\(\displaystyle \cos\theta = \frac{-1}{\sqrt{2}\,\sqrt{2}} = -\frac{1}{2},\quad \theta = \frac{2\pi}{3}\)
(B) Angle between \(2\hat{\jmath}-\hat{k}\) and \(\hat{\jmath} + 2\hat{k}\)
This matches (IV) π/2. The dot product is zero, so the vectors are perpendicular:
\((0)(0) + (2)(1) + (-1)(2) = 0\)
(C) Angle between \(\hat{\imath} + 2\hat{\jmath}\) and \(5\hat{\imath}+ 10\hat{\jmath}\)
This matches (I) 0. The second vector is a positive multiple of the first, so they point the same way:
\(5\hat{\imath} + 10\hat{\jmath} = 5(\hat{\imath} + 2\hat{\jmath})\)
(D) Angle between \(\sqrt{3}\hat{\imath}+\hat{\jmath}\) and \(\hat{\imath}\)
This matches (III) π/6:
\(\displaystyle \cos\theta = \frac{\sqrt{3}}{\sqrt{3 + 1}\times 1} = \frac{\sqrt{3}}{2},\quad \theta = \frac{\pi}{6}\)
So the matching is (A)-(II), (B)-(IV), (C)-(I), (D)-(III).