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Step-by-Step Solution
Step 1: Write down the given vectors
We have
\overrightarrow{a} = \hat{i} + \alpha \,\hat{j} + 3\,\hat{k}
and
\overrightarrow{b} = 3\,\hat{i} - \alpha \,\hat{j} + \hat{k}.
Step 2: Express the area of the parallelogram
The area of the parallelogram formed by vectors \overrightarrow{a} and \overrightarrow{b} is given by the magnitude of their cross product:
\text{Area} = \bigl|\overrightarrow{a} \times \overrightarrow{b}\bigr| = 8\sqrt{3}.
Step 3: Compute the cross product \overrightarrow{a} \times \overrightarrow{b}
Let us first compute \overrightarrow{a} \times \overrightarrow{b} in determinant form:
\overrightarrow{a} \times \overrightarrow{b} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & \alpha & 3 \\
3 & -\alpha & 1
\end{vmatrix}.
Evaluating this determinant (or by using the component-wise cross product formula):
\overrightarrow{a} \times \overrightarrow{b} = ( \alpha \cdot 1 - 3 \cdot (-\alpha) ) \,\hat{i} \;-\; ( 1 \cdot 1 - 3 \cdot 3 ) \,\hat{j} \;+\; ( 1 \cdot (-\alpha) - \alpha \cdot 3 )\,\hat{k}.
Simplifying each component:
• \hat{i} component: \alpha + 3\alpha = 4\alpha
• \hat{j} component: - \bigl( 1 - 9 \bigr) = -(-8) = 8
• \hat{k} component: -(\alpha + 3\alpha) = -4\alpha
So,
\overrightarrow{a} \times \overrightarrow{b}
= 4\alpha \,\hat{i} + 8\,\hat{j} - 4\alpha \,\hat{k}.
Step 4: Use the given area to find \alpha
Since \bigl|\overrightarrow{a} \times \overrightarrow{b}\bigr| = 8\sqrt{3}, we calculate the magnitude:
\bigl|\overrightarrow{a} \times \overrightarrow{b}\bigr|
= \sqrt{(4\alpha)^2 + 8^2 + (-4\alpha)^2}
= \sqrt{16\alpha^2 + 64 + 16\alpha^2}
= \sqrt{32\alpha^2 + 64}.
We know this quantity must equal 8\sqrt{3}. Hence,
8\sqrt{3} = \sqrt{32\alpha^2 + 64}.
Squaring both sides:
64 \cdot 3 = 32\alpha^2 + 64.
That is
192 = 32\alpha^2 + 64.
Subtract 64 from both sides:
128 = 32\alpha^2.
Dividing by 32:
4 = \alpha^2.
Hence,
\alpha = \pm 2.
Step 5: Compute the dot product \overrightarrow{a} \cdot \overrightarrow{b}
The dot product is:
\overrightarrow{a} \cdot \overrightarrow{b}
= (1)(3) + (\alpha)(-\alpha) + (3)(1).
Simplify the expression:
\overrightarrow{a} \cdot \overrightarrow{b}
= 3 - \alpha^2 + 3
= 6 - \alpha^2.
Since \alpha^2 = 4, we get
\overrightarrow{a} \cdot \overrightarrow{b} = 6 - 4 = 2.
Answer
\overrightarrow{a} \cdot \overrightarrow{b} = 2.
