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(C) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$. Since $a+b$ is a unit vector,we have $|a+b| = 1$. Squaring both sides,we get $

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$	heta=\frac{2 \pi}{3}$

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(C) Given that $a$ and $b$ are unit vectors,so $|a| = 1$ and $|b| = 1$. Since $a+b$ is a unit vector,we have $|a+b| = 1$. Squaring both sides,we get $|a+b|^2 = 1^2 = 1$. Using the property $|a+b|^2 = |a|^2 + |b|^2 + 2(a \cdot b)$,we have $1 + 1 + 2(a \cdot b) = 1$. This simplifies to $2 + 2(a \cdot b) = 1$,which gives $2(a \cdot b) = -1$,or $a \cdot b = -\frac{1}{2}$. Since $a \cdot b = |a||b| \cos 	heta$,we have $1 	imes 1 	imes \cos 	heta = -\frac{1}{2}$. Thus,$\cos 	heta = -\frac{1}{2}$. Since $\cos 	heta = -\frac{1}{2}$,the angle $	heta$ is $\frac{2\pi}{3}$.

Let $a$ and $b$ be two unit vectors and $\theta$ is the angle between them. Then,$a+b$ is a unit vector,if

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