For p0, v_2=2 i+(p+1) j is obtained by rotating v_1=√3p i+ j by angle counterclockwise.

Question

For $p>0$, $\vec v_2=2\hat i+(p+1)\hat j$ is obtained by rotating $\vec v_1=\sqrt3p\hat i+\hat j$ by angle $	heta$ counterclockwise. If $	an	heta = \dfrac{\alpha\sqrt3-2}{4\sqrt3+3}$, then $\alpha$ is \_\_\_\_\_\_\_\_\_.

Answer: 6

Accepted Answer

Equal magnitudes: $3p^2+1=4+(p+1)^2 \implies 2p^2-2p-4=0 \implies p^2-p-2=0 \implies p=2$ (rejecting $-1$).

$\vec v_1=(2\sqrt3,1)$, $\vec v_2=(2,3)$: $\cos\theta = \dfrac{\vec v_1\cdot\vec v_2}{|\vec v_1||\vec v_2|} = \dfrac{4\sqrt3+3}{13}$ (since $|\vec v_1|=|\vec v_2|=\sqrt{13}$).

Computing $\sin\theta$ from the cross-product magnitude and forming $\tan\theta$ gives
$$\tan\theta = \frac{6\sqrt3-2}{4\sqrt3+3} = \frac{\alpha\sqrt3-2}{4\sqrt3+3} \implies \alpha=6$$

Solutions · 1