There are two unit vectors \( \vec{a} \) and \( \vec{b} \) such that \( |\vec{a}+\vec{b}|=\sqrt{3} \) and \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \), then
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | \( |\vec{a}-\vec{b}| \) | I. | 0 |
| B. | \( \cos\theta \) | II. | 1 |
| C. | \( \theta \) | III. | \( \displaystyle\frac{1}{2} \) |
| D. | \( \sin3\theta \) | IV. | \( \displaystyle\frac{\pi}{3} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-II, B-III, C-IV, D-I
For unit vectors \( |\vec{a}+\vec{b}|^{2}=2+2\cos\theta \), and this equals 3.
\( \displaystyle 2+2\cos\theta=3 \)
\( \displaystyle \cos\theta=\frac{1}{2} \)
\( \displaystyle \theta=\frac{\pi}{3} \)
(A) \( |\vec{a}-\vec{b}|^{2}=2-2\cos\theta=1 \), so the value is 1, which matches (II).
(B) \( \cos\theta=\frac{1}{2} \), which matches (III).
(C) \( \theta=\frac{\pi}{3} \), which matches (IV).
(D) \( \sin3\theta=\sin\pi=0 \), which matches (I).
So the matching is (A)-(II), (B)-(III), (C)-(IV), (D)-(I).