For p>0, a vector v_2=2 i+(p+1) j is obtained by rotating the vector v_1=√3p i+ j by an angle θ about origin in counter clockwise direction. If tanθ…

We have,

$${\vec{v}}_{1}=\sqrt{3}p\hat{i}+\hat{j}$$v1​=3​pi^+j^​

$${\vec{v}}_{2}=2\hat{i}+(p+1)\hat{j}$$v2​=2i^+(p+1)j^​

$$\mathrm{tan}\theta =\frac{(\alpha \sqrt{3}-2)}{(4\sqrt{3}+3)}$$tanθ=(43​+3)(α3​−2)​

$$|{\vec{v}}_{1}|=|{\vec{v}}_{2}|$$∣v1​∣=∣v2​∣

$$\Rightarrow 3{p}^{2}+1=4+(p+1{)}^{2}$$⇒3p2+1=4+(p+1)2

$$\Rightarrow 2{p}^{2}-2p-4=0$$⇒2p2−2p−4=0

$$\Rightarrow {p}^{2}-p-2=0$$⇒p2−p−2=0

$\Rightarrow p=2,-1$⇒p=2,−1 (rejected)

Now,

$$\mathrm{cos}\theta =\frac{{\vec{v}}_{1}\cdot {\vec{v}}_{2}}{|{\vec{v}}_{1}|\cdot |{\vec{v}}_{2}|}=\frac{2\sqrt{3}p+(p+1)}{\sqrt{(p+1{)}^{2}+4}\sqrt{3{p}^{2}+1}}$$cosθ=∣v1​∣⋅∣v2​∣v1​⋅v2​​=(p+1)2+4​3p2+1​23​p+(p+1)​

$$\Rightarrow \mathrm{cos}\theta =\frac{4\sqrt{3}+3}{\sqrt{13}\sqrt{13}}=\frac{4\sqrt{3}+3}{13}$$⇒cosθ=13​13​43​+3​=1343​+3​

$$\Rightarrow \mathrm{tan}\theta =\frac{\sqrt{112-24\sqrt{3}}}{4\sqrt{3}+3}=\frac{\sqrt{{(6\sqrt{3}-2)}^{2}}}{4\sqrt{3}+3}$$⇒tanθ=43​+3112−243​​​=43​+3(63​−2)2​​

$$\Rightarrow \mathrm{tan}\theta =\frac{6\sqrt{3}-2}{4\sqrt{3}+3}=\frac{\alpha \sqrt{3}-2}{4\sqrt{3}+3}$$⇒tanθ=43​+363​−2​=43​+3α3​−2​

$$\Rightarrow \alpha =6$$⇒α=6