Unit vectors $\vec{a}$ and $\vec{b}$ are inclined at an angle $2\theta$ and $|\vec{a}-\vec{b}|<1$. If $0\leq\theta<\pi$ then $\theta$ may belong to:
Step-by-Step Solution
Key Concept: Expand $|\vec{a}-\vec{b}|^2$ using the dot product formula to express the magnitude in terms of $\sin\theta$, then solve the resulting inequality.
Since $\vec{a}$ and $\vec{b}$ are unit vectors inclined at angle $2\theta$, we have $|\vec{a}| = |\vec{b}| = 1$ and $\vec{a} \cdot \vec{b} = \cos(2\theta)$. Computing $|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b} = 1 + 1 - 2\cos(2\theta) = 2(1-\cos(2\theta)) = 4\sin^2(\theta)$. Therefore $|\vec{a}-\vec{b}| = 2|\sin\theta|$. The condition $|\vec{a}-\vec{b}| < 1$ gives $2|\sin\theta| < 1$, so $|\sin\theta| < \frac{1}{2}$. For $0 \leq \theta < \pi$, this means $\sin\theta < \frac{1}{2}$, which holds when $\theta \in [0, \pi/6) \cup (5\pi/6, \pi)$.
Correct Answer: 1,2