For p 0,a vector vec{v}_{2} = 2 hat{i} + (p | vedclass

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(A) Given $\vec{v}_{1} = \sqrt{3} p \hat{i} + \hat{j}$ and $\vec{v}_{2} = 2 \hat{i} + (p + 1) \hat{j}$.Since rotation preserves the magnitude of the v

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$6$

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(A) Given $\vec{v}_{1} = \sqrt{3} p \hat{i} + \hat{j}$ and $\vec{v}_{2} = 2 \hat{i} + (p + 1) \hat{j}$.Since rotation preserves the magnitude of the vector,$|\vec{v}_{1}| = |\vec{v}_{2}|$.$(\sqrt{3}p)^2 + 1^2 = 2^2 + (p + 1)^2$$3p^2 + 1 = 4 + p^2 + 2p + 1$$2p^2 - 2p - 4 = 0 \Rightarrow p^2 - p - 2 = 0$.Solving for $p$,$(p - 2)(p + 1) = 0$. Since $p > 0$,we have $p = 2$.Thus,$\vec{v}_{1} = 2\sqrt{3} \hat{i} + \hat{j}$ and $\vec{v}_{2} = 2 \hat{i} + 3 \hat{j}$.$|\vec{v}_{1}| = \sqrt{(2\sqrt{3})^2 + 1^2} = \sqrt{12 + 1} = \sqrt{13}$.$|\vec{v}_{2}| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$.Using $\vec{v}_{1} \cdot \vec{v}_{2} = |\vec{v}_{1}| |\vec{v}_{2}| \cos 	heta$,we get $(2\sqrt{3})(2) + (1)(3) = \sqrt{13} \cdot \sqrt{13} \cos 	heta$.$4\sqrt{3} + 3 = 13 \cos 	heta \Rightarrow \cos 	heta = \frac{4\sqrt{3} + 3}{13}$.Since $\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{(4\sqrt{3} + 3)^2}{169} = \frac{169 - (48 + 9 + 24\sqrt{3})}{169} = \frac{112 - 24\sqrt{3}}{169}$.$\sin 	heta = \frac{\sqrt{112 - 24\sqrt{3}}}{13} = \frac{\sqrt{4(28 - 6\sqrt{3})}}{13} = \frac{2\sqrt{(3\sqrt{3} - 1)^2}}{13} = \frac{2(3\sqrt{3} - 1)}{13} = \frac{6\sqrt{3} - 2}{13}$.$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{6\sqrt{3} - 2}{4\sqrt{3} + 3}$.Comparing with $\frac{\alpha \sqrt{3} - 2}{4\sqrt{3} + 3}$,we get $\alpha = 6$.

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