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(A) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$.The angle between them is $2	heta$.We know that $|\vec

Vedclass · Accepted Answer

$[0, \pi/6) \cup (5\pi/6, \pi]$

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(A) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$.The angle between them is $2	heta$.We know that $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b})$.$|\vec{a} - \vec{b}|^2 = 1^2 + 1^2 - 2|\vec{a}||\vec{b}| \cos(2	heta) = 2 - 2 \cos(2	heta)$.Using the trigonometric identity $1 - \cos(2	heta) = 2 \sin^2(	heta)$,we get:$|\vec{a} - \vec{b}|^2 = 2(1 - \cos(2	heta)) = 2(2 \sin^2(	heta)) = 4 \sin^2(	heta)$.Taking the square root on both sides,$|\vec{a} - \vec{b}| = 2|\sin(	heta)|$.Given the condition $|\vec{a} - \vec{b}| Since $0 \le 	heta \le \pi$,$\sin(	heta)$ is non-negative,so $\sin(	heta) This inequality holds for $	heta \in [0, \pi/6) \cup (5\pi/6, \pi]$.

If the angle between unit vectors $\vec{a}$ and $\vec{b}$ is $2\theta$,and $|\vec{a} - \vec{b}| < 1$ with $0 \le \theta \le \pi$,then in which interval does $\theta$ lie?

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