Key Concepts and Formulas

• Area of a Parallelogram: The area of a parallelogram with adjacent sides represented by vectors $\overrightarrow u$ and $\overrightarrow v$ is given by the magnitude of their cross product: $\text{Area} = |\overrightarrow u \times \overrightarrow v|$.
• Cross Product of Two Vectors: For $\overrightarrow a = a_1 \widehat i + a_2 \widehat j + a_3 \widehat k$ and $\overrightarrow b = b_1 \widehat i + b_2 \widehat j + b_3 \widehat k$, their cross product is: $\overrightarrow a \times \overrightarrow b = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = (a_2 b_3 - a_3 b_2)\widehat i - (a_1 b_3 - a_3 b_1)\widehat j + (a_1 b_2 - a_2 b_1)\widehat k$
• Magnitude of a Vector: The magnitude of a vector $\overrightarrow v = x\widehat i + y\widehat j + z\widehat k$ is $|\overrightarrow v| = \sqrt{x^2 + y^2 + z^2}$.
• Dot Product of Two Vectors: For $\overrightarrow a = a_1 \widehat i + a_2 \widehat j + a_3 \widehat k$ and $\overrightarrow b = b_1 \widehat i + b_2 \widehat j + b_3 \widehat k$, their dot product is: $\overrightarrow a \cdot \overrightarrow b = a_1 b_1 + a_2 b_2 + a_3 b_3$
• Magnitude Squared of a Vector: $|\overrightarrow v|^2 = x^2 + y^2 + z^2$.

Step-by-Step Solution

Step 1: Calculate the Cross Product $\overrightarrow a \times \overrightarrow b$ We are given the vectors $\overrightarrow a = \alpha \widehat i + 2\widehat j - \widehat k$ and $\overrightarrow b = - 2\widehat i + \alpha \widehat j + \widehat k$. To find the area of the parallelogram, we first need to compute their cross product. $\overrightarrow a \times \overrightarrow b = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ \alpha & 2 & { - 1} \\ { - 2} & \alpha & 1 \end{vmatrix}$ Expanding the determinant: $\overrightarrow a \times \overrightarrow b = ((2)(1) - (-1)(\alpha))\widehat i - ((\alpha)(1) - (-1)(-2))\widehat j + ((\alpha)(\alpha) - (2)(-2))\widehat k$ $\overrightarrow a \times \overrightarrow b = (2 + \alpha)\widehat i - (\alpha - 2)\widehat j + (\alpha^2 + 4)\widehat k$ $\overrightarrow a \times \overrightarrow b = (2 + \alpha)\widehat i + (2 - \alpha)\widehat j + (\alpha^2 + 4)\widehat k$

Step 2: Calculate the Magnitude of the Cross Product and Use the Given Area The area of the parallelogram is given by $|\overrightarrow a \times \overrightarrow b|$. $|\overrightarrow a \times \overrightarrow b| = \sqrt{(2 + \alpha)^2 + (2 - \alpha)^2 + (\alpha^2 + 4)^2}$ Let's simplify the terms under the square root: $(2 + \alpha)^2 = 4 + 4\alpha + \alpha^2$ $(2 - \alpha)^2 = 4 - 4\alpha + \alpha^2$ Adding these two: $(2 + \alpha)^2 + (2 - \alpha)^2 = (4 + 4\alpha + \alpha^2) + (4 - 4\alpha + \alpha^2) = 8 + 2\alpha^2$ So, the square of the magnitude is: $|\overrightarrow a \times \overrightarrow b|^2 = (8 + 2\alpha^2) + (\alpha^2 + 4)^2$ We are given that the area is $\sqrt{15(\alpha^2 + 4)}$. Squaring this, we get: $\text{Area}^2 = 15(\alpha^2 + 4)$ Equating the square of the magnitude with the square of the given area: $(8 + 2\alpha^2) + (\alpha^2 + 4)^2 = 15(\alpha^2 + 4)$ To simplify this equation, let $y = \alpha^2 + 4$. Since $\alpha \in R$, $\alpha^2 \ge 0$, so $y \ge 4$. From $y = \alpha^2 + 4$, we have $\alpha^2 = y - 4$. Substituting into the equation: $8 + 2(y - 4) + y^2 = 15y$ $8 + 2y - 8 + y^2 = 15y$ $y^2 + 2y = 15y$ $y^2 - 13y = 0$ $y(y - 13) = 0$ This gives $y = 0$ or $y = 13$. Since $y = \alpha^2 + 4$ and $\alpha^2 \ge 0$, $y$ must be at least 4. Therefore, $y = 0$ is not a valid solution. So, we must have $y = 13$. Substituting back $y = \alpha^2 + 4$: $\alpha^2 + 4 = 13$ $\alpha^2 = 9$

Step 3: Calculate the Magnitudes Squared and the Dot Product Now that we have $\alpha^2 = 9$, we can compute the required terms for the final expression.

• Magnitude squared of $\overrightarrow a$: $|\overrightarrow a|^2 = (\alpha)^2 + (2)^2 + (-1)^2 = \alpha^2 + 4 + 1 = \alpha^2 + 5$ Substituting $\alpha^2 = 9$: $|\overrightarrow a|^2 = 9 + 5 = 14$

• Magnitude squared of $\overrightarrow b$: $|\overrightarrow b|^2 = (-2)^2 + (\alpha)^2 + (1)^2 = 4 + \alpha^2 + 1 = \alpha^2 + 5$ Substituting $\alpha^2 = 9$: $|\overrightarrow b|^2 = 9 + 5 = 14$

• Dot product $\overrightarrow a \cdot \overrightarrow b$: $\overrightarrow a \cdot \overrightarrow b = (\alpha)(-2) + (2)(\alpha) + (-1)(1)$ $\overrightarrow a \cdot \overrightarrow b = -2\alpha + 2\alpha - 1$ $\overrightarrow a \cdot \overrightarrow b = -1$

Step 4: Evaluate the Final Expression We need to find the value of $2|\overrightarrow a|^2 + (\overrightarrow a \cdot \overrightarrow b)|\overrightarrow b|^2$. Substitute the calculated values: $2|\overrightarrow a|^2 + (\overrightarrow a \cdot \overrightarrow b)|\overrightarrow b|^2 = 2(14) + (-1)(14)$ $= 28 - 14$ $= 14$

Common Mistakes & Tips

• Sign Errors in Cross Product: Be extremely careful with the signs when calculating the components of the cross product, especially the $\widehat j$ component.
• Algebraic Simplification: Using substitutions like $y = \alpha^2 + 4$ can greatly simplify complex polynomial equations, reducing the likelihood of errors.
• Valid Solutions for $\alpha^2$: Always check if the derived values for $\alpha^2$ (or related variables) are consistent with the domain of $\alpha$ (e.g., $\alpha^2 \ge 0$).

Summary

The problem required us to first find the value of $\alpha^2$ by utilizing the given area of the parallelogram, which is calculated using the magnitude of the cross product of the adjacent side vectors. After determining $\alpha^2 = 9$, we computed the magnitudes squared of the vectors $\overrightarrow a$ and $\overrightarrow b$, and their dot product. Finally, these values were substituted into the target expression $2|\overrightarrow a|^2 + (\overrightarrow a \cdot \overrightarrow b)|\overrightarrow b|^2$ to obtain the result.

The final answer is $\boxed{14}$.