Key Concepts and Formulas

1. Area of a Parallelogram: The area of a parallelogram with adjacent sides given by vectors $\overrightarrow a$ and $\overrightarrow b$ is equal to the magnitude of their cross product: $\text{Area} = |\overrightarrow a \times \overrightarrow b|$
2. Cross Product of Two Vectors: For vectors $\overrightarrow a = a_1\widehat i + a_2\widehat j + a_3\widehat k$ and $\overrightarrow b = b_1\widehat i + b_2\widehat j + b_3\widehat k$, the cross product is: $\overrightarrow a \times \overrightarrow b = (a_2b_3 - a_3b_2)\widehat i - (a_1b_3 - a_3b_1)\widehat j + (a_1b_2 - a_2b_1)\widehat k$
3. Magnitude of a Vector: The magnitude of a vector $\overrightarrow v = v_1\widehat i + v_2\widehat j + v_3\widehat k$ is: $|\overrightarrow v| = \sqrt{v_1^2 + v_2^2 + v_3^2}$
4. Dot Product of Two Vectors: For vectors $\overrightarrow a = a_1\widehat i + a_2\widehat j + a_3\widehat k$ and $\overrightarrow b = b_1\widehat i + b_2\widehat j + b_3\widehat k$, the dot product is: $\overrightarrow a \cdot \overrightarrow b = a_1b_1 + a_2b_2 + a_3b_3$

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Step-by-Step Solution

1. Identify the Given Information We are given two vectors representing the adjacent sides of a parallelogram: $\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$ $\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$ The area of the parallelogram is $8\sqrt{3}$ square units. We need to find the value of $\overrightarrow a \cdot \overrightarrow b$.

2. Calculate the Cross Product $\overrightarrow a \times \overrightarrow b$ Why this step? The area of the parallelogram is directly related to the cross product of its adjacent sides. We need to compute this cross product to proceed. Using the determinant formula for the cross product: $\overrightarrow a \times \overrightarrow b = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 1 & \alpha & 3 \\ 3 & -\alpha & 1 \end{vmatrix}$ Expanding the determinant: $\overrightarrow a \times \overrightarrow b = \widehat i ((\alpha)(1) - (3)(-\alpha)) - \widehat j ((1)(1) - (3)(3)) + \widehat k ((1)(-\alpha) - (\alpha)(3))$ $\overrightarrow a \times \overrightarrow b = \widehat i (\alpha + 3\alpha) - \widehat j (1 - 9) + \widehat k (-\alpha - 3\alpha)$ $\overrightarrow a \times \overrightarrow b = (4\alpha)\widehat i - (-8)\widehat j + (-4\alpha)\widehat k$ $\overrightarrow a \times \overrightarrow b = 4\alpha\widehat i + 8\widehat j - 4\alpha\widehat k$

3. Calculate the Magnitude of the Cross Product Why this step? The magnitude of the cross product is equal to the area of the parallelogram, which is given in the problem. The magnitude of $\overrightarrow a \times \overrightarrow b = 4\alpha\widehat i + 8\widehat j - 4\alpha\widehat k$ is: $|\overrightarrow a \times \overrightarrow b| = \sqrt{(4\alpha)^2 + (8)^2 + (-4\alpha)^2}$ $|\overrightarrow a \times \overrightarrow b| = \sqrt{16\alpha^2 + 64 + 16\alpha^2}$ $|\overrightarrow a \times \overrightarrow b| = \sqrt{32\alpha^2 + 64}$

4. Use the Given Area to Solve for $\alpha^2$ Why this step? We are given the area of the parallelogram, and we have an expression for it in terms of $\alpha$. Equating these allows us to find the value of $\alpha^2$, which will be needed for the dot product. We are given that the Area $= 8\sqrt{3}$. $\sqrt{32\alpha^2 + 64} = 8\sqrt{3}$ Squaring both sides to remove the square root: $32\alpha^2 + 64 = (8\sqrt{3})^2$ $32\alpha^2 + 64 = 64 \times 3$ $32\alpha^2 + 64 = 192$ Subtract 64 from both sides: $32\alpha^2 = 192 - 64$ $32\alpha^2 = 128$ Divide by 32 to find $\alpha^2$: $\alpha^2 = \frac{128}{32}$ $\alpha^2 = 4$

5. Calculate the Dot Product $\overrightarrow a \cdot \overrightarrow b$ Why this step? This is the final quantity we need to compute, and we now have all the necessary information ($\alpha^2$). Recall the given vectors: $\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat k$ $\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat k$ Using the dot product formula $\overrightarrow a \cdot \overrightarrow b = a_1b_1 + a_2b_2 + a_3b_3$: $\overrightarrow a \cdot \overrightarrow b = (1)(3) + (\alpha)(-\alpha) + (3)(1)$ $\overrightarrow a \cdot \overrightarrow b = 3 - \alpha^2 + 3$ $\overrightarrow a \cdot \overrightarrow b = 6 - \alpha^2$ Substitute the value $\alpha^2 = 4$ that we found: $\overrightarrow a \cdot \overrightarrow b = 6 - 4$ $\overrightarrow a \cdot \overrightarrow b = 2$

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Common Mistakes & Tips

• Sign Errors in Cross Product: Be extremely careful with the signs when calculating the determinant for the cross product. A mistake in the $\widehat j$ component (which has a negative sign) is common.
• Squaring the Magnitude: Ensure that all components of the vector are squared correctly before summing them up for the magnitude calculation.
• Solving for $\alpha$ vs. $\alpha^2$: Notice that the dot product calculation only requires $\alpha^2$. This saves a step and avoids dealing with the ambiguity of $\alpha = \pm 2$.

Summary

The problem required us to find the dot product of two vectors given the area of the parallelogram formed by them. We first calculated the cross product of the vectors and then its magnitude. By equating this magnitude to the given area, we solved for $\alpha^2$. Finally, we computed the dot product of the vectors using the calculated value of $\alpha^2$.

The final answer is $\boxed{2}$.