For p ,> ,0, a vector v_ 2 =2 i+(p+1) j is obtained by rotating t… - Vidyadip

MCQ

For $p\,>\,0$, a vector $\vec{v}_{2}=2 \hat{i}+(p+1) \hat{j}$ is obtained by rotating the vector $\vec{v}_{1}=\sqrt{3} p \hat{i}+\hat{j}$ by an angle $\theta$ about origin in counter clockwise direction. If $\tan \theta=\frac{(\alpha \sqrt{3}-2)}{4 \sqrt{3}+3}$, then the value of $\alpha$ is equal to $....$

✓
$6$
B
$5$
C
$4$
D
$3$

✓

Answer

Correct option: A.

$6$

a
$\left|\vec{V}_{1}\right|=\left|\overrightarrow{\mathrm{V}}_{2}\right|$

$3 \mathrm{P}^{2}+1=4+(\mathrm{P}+1)^{2}$

$2 \mathrm{P}^{2}-2 P-4=0$

$\Rightarrow \mathrm{P}^{2}-\mathrm{P}-2=0$

$\mathrm{P}=2,-1$ (rejected)

$\cos \theta=\frac{\overline{V}_{1} \cdot \vec{V}_{2}}{\left|\overrightarrow{\mathrm{V}}_{1}\right| \cdot \overrightarrow{\mathrm{V}_{2}} \mid}$

$\cos \theta=\frac{4 \sqrt{3}+3}{\sqrt{13} \sqrt{13}}=\frac{4 \sqrt{3}+3}{13}$

$\tan \theta=\frac{\sqrt{112-24 \sqrt{3}}}{4 \sqrt{3}+\sqrt{3}}=\frac{6 \sqrt{3}-2}{4 \sqrt{3}+3}=\frac{\alpha \sqrt{3}-2}{4 \sqrt{3}+3}$

$\Rightarrow \alpha=6$

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