Key Concepts and Formulas

• Vector Magnitude: The magnitude of a vector $\overrightarrow{v} = x\widehat{i} + y\widehat{j}$ is $|\overrightarrow{v}| = \sqrt{x^2 + y^2}$. Rotation about the origin preserves the magnitude of a vector.
• Dot Product: For vectors $\overrightarrow{A}$ and $\overrightarrow{B}$, $\overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \cos \theta$, where $\theta$ is the angle between them.
• Cross Product (2D interpretation): For vectors $\overrightarrow{v_1} = x_1\widehat{i} + y_1\widehat{j}$ and $\overrightarrow{v_2} = x_2\widehat{i} + y_2\widehat{j}$, the $z$-component of the cross product is $x_1y_2 - x_2y_1$. If this value is positive, the rotation from $\overrightarrow{v_1}$ to $\overrightarrow{v_2}$ is counter-clockwise, and $\sin \theta = \frac{x_1y_2 - x_2y_1}{|\overrightarrow{v_1}||\overrightarrow{v_2}|}$.
• Trigonometric Identity: $\sin^2 \theta + \cos^2 \theta = 1$.
• Simplification of Nested Radicals: $\sqrt{A \pm 2\sqrt{B}} = \sqrt{x} \pm \sqrt{y}$, where $x+y=A$ and $xy=B$.

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Step-by-Step Solution

Step 1: Determine the value of 'p' by equating vector magnitudes.

The problem states that vector $\overrightarrow{v_2}$ is obtained by rotating vector $\overrightarrow{v_1}$ about the origin. Rotation about the origin is an isometry, meaning it preserves the length (magnitude) of the vector. Therefore, $|\overrightarrow{v_1}| = |\overrightarrow{v_2}|$.

Given vectors: $\overrightarrow{v_1} = \sqrt{3}p\widehat{i} + \widehat{j}$ $\overrightarrow{v_2} = 2\widehat{i} + (p+1)\widehat{j}$

Calculate the square of the magnitudes: $|\overrightarrow{v_1}|^2 = (\sqrt{3}p)^2 + (1)^2 = 3p^2 + 1$ $|\overrightarrow{v_2}|^2 = (2)^2 + (p+1)^2 = 4 + (p^2 + 2p + 1) = p^2 + 2p + 5$

Equate the squared magnitudes: $3p^2 + 1 = p^2 + 2p + 5$

Rearrange into a quadratic equation: $2p^2 - 2p - 4 = 0$

Divide by 2: $p^2 - p - 2 = 0$

Factor the quadratic equation: $(p-2)(p+1) = 0$

This yields two possible values for $p$: $p=2$ and $p=-1$. The problem states that $p > 0$. Therefore, we select $p=2$.

Step 2: Substitute $p=2$ into the vectors and calculate $\cos \theta$.

Now that we have the value of $p$, we can write the explicit forms of the vectors.

$\overrightarrow{v_1} = \sqrt{3}(2)\widehat{i} + \widehat{j} = 2\sqrt{3}\widehat{i} + \widehat{j}$ $\overrightarrow{v_2} = 2\widehat{i} + (2+1)\widehat{j} = 2\widehat{i} + 3\widehat{j}$

Calculate the magnitudes of these vectors: $|\overrightarrow{v_1}| = \sqrt{(2\sqrt{3})^2 + 1^2} = \sqrt{12 + 1} = \sqrt{13}$ $|\overrightarrow{v_2}| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}$ As expected, the magnitudes are equal.

Calculate the dot product $\overrightarrow{v_1} \cdot \overrightarrow{v_2}$: $\overrightarrow{v_1} \cdot \overrightarrow{v_2} = (2\sqrt{3})(2) + (1)(3) = 4\sqrt{3} + 3$

Use the dot product formula to find $\cos \theta$: $\cos \theta = \frac{\overrightarrow{v_1} \cdot \overrightarrow{v_2}}{|\overrightarrow{v_1}| |\overrightarrow{v_2}|} = \frac{4\sqrt{3} + 3}{\sqrt{13} \cdot \sqrt{13}} = \frac{4\sqrt{3} + 3}{13}$

Step 3: Determine $\sin \theta$ using the direction of rotation and calculate $\tan \theta$.

The problem specifies that $\overrightarrow{v_2}$ is obtained by rotating $\overrightarrow{v_1}$ by an angle $\theta$ in the counter-clockwise direction. For 2D vectors $\overrightarrow{v_1} = x_1\widehat{i} + y_1\widehat{j}$ and $\overrightarrow{v_2} = x_2\widehat{i} + y_2\widehat{j}$, the sine of the counter-clockwise angle $\theta$ is given by $\sin \theta = \frac{x_1y_2 - x_2y_1}{|\overrightarrow{v_1}||\overrightarrow{v_2}|}$.

Using $\overrightarrow{v_1} = 2\sqrt{3}\widehat{i} + \widehat{j}$ and $\overrightarrow{v_2} = 2\widehat{i} + 3\widehat{j}$: $x_1 = 2\sqrt{3}, \quad y_1 = 1$ $x_2 = 2, \quad y_2 = 3$

Calculate the numerator for $\sin \theta$: $x_1y_2 - x_2y_1 = (2\sqrt{3})(3) - (1)(2) = 6\sqrt{3} - 2$

Now, calculate $\sin \theta$: $\sin \theta = \frac{6\sqrt{3} - 2}{13}$ Since $6\sqrt{3} \approx 6 \times 1.732 = 10.392$, the numerator $6\sqrt{3} - 2$ is positive, so $\sin \theta > 0$. This is consistent with a counter-clockwise rotation.

Calculate $\tan \theta$: $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{6\sqrt{3} - 2}{13}}{\frac{4\sqrt{3} + 3}{13}} = \frac{6\sqrt{3} - 2}{4\sqrt{3} + 3}$

Step 4: Compare the derived $\tan \theta$ with the given expression to find $\alpha$.

We are given the expression for $\tan \theta$: $\tan \theta = {{\left( {\alpha \sqrt 3 - 2} \right)} \over {\left( {4\sqrt 3 + 3} \right)}}$

Equate our derived expression for $\tan \theta$ with the given one: $\frac{6\sqrt{3} - 2}{4\sqrt{3} + 3} = \frac{\alpha\sqrt{3} - 2}{4\sqrt{3} + 3}$

Since the denominators are the same and non-zero, the numerators must be equal: $6\sqrt{3} - 2 = \alpha\sqrt{3} - 2$

Add 2 to both sides: $6\sqrt{3} = \alpha\sqrt{3}$

Divide both sides by $\sqrt{3}$: $\alpha = 6$

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Common Mistakes & Tips

• Forgetting the $p>0$ constraint: This can lead to using the incorrect value of $p$, resulting in an incorrect final answer. Always check problem constraints carefully.
• Sign errors in $\sin \theta$: The direction of rotation (counter-clockwise) is crucial for determining the sign of $\sin \theta$. Using the cross product formula for $\sin \theta$ is a reliable way to handle this.
• Algebraic simplification errors: Be meticulous when simplifying expressions, especially nested radicals. Double-checking the factorization or quadratic formula application is recommended.

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Summary

The problem required us to first find the value of $p$ by utilizing the property that vector rotation preserves magnitude. After determining $p=2$, we explicitly defined the vectors and used the dot product to find $\cos \theta$. The direction of rotation provided the necessary information to determine the sign of $\sin \theta$. Finally, by calculating $\tan \theta$ and equating it to the given expression, we solved for the unknown $\alpha$.

The final answer is \boxed{6}.