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Step 1: Equate the magnitudes of the two vectors
We are given
\overrightarrow{v}_1 = \sqrt{3} p \,\hat{i} + \hat{j},
and
\overrightarrow{v}_2 = 2\,\hat{i} + (p+1)\,\hat{j}
with the condition that
|\overrightarrow{v}_1| = |\overrightarrow{v}_2|.
Thus,
|\overrightarrow{v}_1|^2 = 3p^2 + 1, \quad |\overrightarrow{v}_2|^2 = 4 + (p+1)^2.
Setting these equal,
\[
3p^2 + 1 = 4 + (p+1)^2.
\]
Step 2: Solve for p
Expand and simplify:
\[
3p^2 + 1 = 4 + p^2 + 2p + 1
\quad\Longrightarrow\quad 3p^2 + 1 = p^2 + 2p + 5
\quad\Longrightarrow\quad 2p^2 - 2p - 4 = 0.
\]
Divide every term by 2:
\[
p^2 - p - 2 = 0.
\]
This quadratic factors to:
\[
(p - 2)(p + 1) = 0 \quad\Longrightarrow\quad p=2 \text{ or } p=-1.
\]
Because p>0 is given, we take p=2.
Step 3: Compute the dot product and cosine of the angle
With p=2, the vectors become
\[
\overrightarrow{v}_1 = \sqrt{3}\cdot 2\,\hat{i} + \hat{j}
= 2\sqrt{3}\,\hat{i} + \hat{j},
\]
\[
\overrightarrow{v}_2 = 2\,\hat{i} + (2+1)\,\hat{j}
= 2\,\hat{i} + 3\,\hat{j}.
\]
Their dot product is:
\[
\overrightarrow{v}_1 \cdot \overrightarrow{v}_2
= (2\sqrt{3})(2) + (1)(3)
= 4\sqrt{3} + 3.
\]
Next, their magnitudes:
\[
|\overrightarrow{v}_1|
= \sqrt{(2\sqrt{3})^2 + 1^2}
= \sqrt{12 + 1} = \sqrt{13},
\]
\[
|\overrightarrow{v}_2|
= \sqrt{2^2 + 3^2}
= \sqrt{4 + 9} = \sqrt{13}.
\]
Hence,
\[
\cos \theta
= \frac{\overrightarrow{v}_1 \cdot \overrightarrow{v}_2}{|\overrightarrow{v}_1|\;|\overrightarrow{v}_2|}
= \frac{\,4\sqrt{3} + 3\,}{\sqrt{13}\cdot \sqrt{13}}
= \frac{\,4\sqrt{3} + 3\,}{13}.
\]
Step 4: Identify the tangent of the angle
We are told that
\[
\tan \theta
= \frac{\alpha \sqrt{3} - 2}{4\sqrt{3} + 3}.
\]
From calculations (or by using known trigonometric identities related to \cos \theta = \frac{\,4\sqrt{3}+3\,}{13} ), it turns out that
\[
\tan \theta
= \frac{6\sqrt{3} - 2}{4\sqrt{3} + 3}.
\]
Comparing
\[
\frac{\alpha \sqrt{3} - 2}{4\sqrt{3} + 3}
\quad\text{with}\quad
\frac{6\sqrt{3} - 2}{4\sqrt{3} + 3},
\]
we see that
\[
\alpha = 6.
\]
Final Answer
The required value of \alpha is 6 .
