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(A) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$. We are given that $\sqrt{3}\vec{a}-\vec{b}$ is also a

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$30$

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(A) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$. We are given that $\sqrt{3}\vec{a}-\vec{b}$ is also a unit vector,so $|\sqrt{3}\vec{a}-\vec{b}| = 1$. Squaring both sides,we get $|\sqrt{3}\vec{a}-\vec{b}|^2 = 1^2$. Using the property $|\vec{u}-\vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2(\vec{u} \cdot \vec{v})$,we have: $3|\vec{a}|^2 + |\vec{b}|^2 - 2\sqrt{3}(\vec{a} \cdot \vec{b}) = 1$. Substituting $|\vec{a}| = 1$ and $|\vec{b}| = 1$: $3(1)^2 + (1)^2 - 2\sqrt{3}(|\vec{a}||\vec{b}|\cos 	heta) = 1$. $3 + 1 - 2\sqrt{3}(1)(1)\cos 	heta = 1$. $4 - 2\sqrt{3}\cos 	heta = 1$. $2\sqrt{3}\cos 	heta = 3$. $\cos 	heta = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$. Since $\cos 	heta = \frac{\sqrt{3}}{2}$,the angle $	heta = 30^{\circ}$.

If $\vec{a}$ and $\vec{b}$ are unit vectors,then the angle between $\vec{a}$ and $\vec{b}$ for $\sqrt{3}\vec{a}-\vec{b}$ to be a unit vector is (in $^{\circ}$)

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