If vec{a} and vec{b} are two unit vectors inc | vedclass

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(A) Given that $|\vec{a}| = |\vec{b}| = 1$ and the angle $	heta = \frac{\pi}{3}$.Using the property of the magnitude of the sum of two vectors:$|\vec

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greater than $1$

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(A) Given that $|\vec{a}| = |\vec{b}| = 1$ and the angle $	heta = \frac{\pi}{3}$.Using the property of the magnitude of the sum of two vectors:$|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b})$$|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}| \cos 	heta$Substituting the given values:$|\vec{a} + \vec{b}|^2 = 1^2 + 1^2 + 2(1)(1) \cos \frac{\pi}{3}$$|\vec{a} + \vec{b}|^2 = 1 + 1 + 2 	imes \frac{1}{2}$$|\vec{a} + \vec{b}|^2 = 1 + 1 + 1 = 3$Taking the square root on both sides:$|\vec{a} + \vec{b}| = \sqrt{3}$Since $\sqrt{3} \approx 1.732$,which is greater than $1$,the value of $|\vec{a} + \vec{b}|$ is greater than $1$.

If $\vec{a}$ and $\vec{b}$ are two unit vectors inclined at an angle $\frac{\pi}{3}$,then the value of $|\vec{a}+\vec{b}|$ is

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