Let `veca= hati+alpha hatj+3hatk and vecb = 3hati - alpha hatj+hatk`. If the area of the parallelogram whose adjacent sides are represented by the vectors `veca and vecb` is `8sqrt3` square , then `veca.vecb` is equal to _______.

Answer

To solve the problem, we need to find the dot product \(\vec{a} \cdot \vec{b}\) given the vectors \(\vec{a} = \hat{i} + \alpha \hat{j} + 3\hat{k}\) and \(\vec{b} = 3\hat{i} - \alpha \hat{j} + \hat{k}\), and the area of the parallelogram formed by these vectors is \(8\sqrt{3}\).

### Step 1: Calculate the cross product \(\vec{a} \times \vec{b}\)

The area of the parallelogram formed by vectors \(\vec{a}\) and \(\vec{b}\) is given by the magnitude of their cross product:
\[ \text{Area} = |\vec{a} \times \vec{b}| \]
We can find \(\vec{a} \times \vec{b}\) using the determinant of a matrix formed by the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) and the components of \(\vec{a}\) and \(\vec{b}\).

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & \alpha & 3 \\ 3 & -\alpha & 1 \end{vmatrix} \]

### Step 2: Calculate the determinant

Calculating the determinant:
\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} \alpha & 3 \\ -\alpha & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ 3 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & \alpha \\ 3 & -\alpha \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \(\begin{vmatrix} \alpha & 3 \\ -\alpha & 1 \end{vmatrix} = \alpha(1) - 3(-\alpha) = \alpha + 3\alpha = 4\alpha\)
2. \(\begin{vmatrix} 1 & 3 \\ 3 & 1 \end{vmatrix} = 1(1) - 3(3) = 1 - 9 = -8\)
3. \(\begin{vmatrix} 1 & \alpha \\ 3 & -\alpha \end{vmatrix} = 1(-\alpha) - 3(\alpha) = -\alpha - 3\alpha = -4\alpha\)

Putting it all together:
\[ \vec{a} \times \vec{b} = \hat{i}(4\alpha) + \hat{j}(8) + \hat{k}(-4\alpha) = 4\alpha \hat{i} + 8 \hat{j} - 4\alpha \hat{k} \]

### Step 3: Find the magnitude of the cross product

Now, we find the magnitude:
\[ |\vec{a} \times \vec{b}| = \sqrt{(4\alpha)^2 + 8^2 + (-4\alpha)^2} = \sqrt{16\alpha^2 + 64 + 16\alpha^2} = \sqrt{32\alpha^2 + 64} \]

### Step 4: Set the magnitude equal to the area

We know the area is \(8\sqrt{3}\):
\[ \sqrt{32\alpha^2 + 64} = 8\sqrt{3} \]

Squaring both sides:
\[ 32\alpha^2 + 64 = 192 \]
\[ 32\alpha^2 = 128 \quad \Rightarrow \quad \alpha^2 = 4 \quad \Rightarrow \quad \alpha = 2 \text{ or } -2 \]

### Step 5: Calculate the dot product \(\vec{a} \cdot \vec{b}\)

Now we substitute \(\alpha = 2\) into \(\vec{a}\) and \(\vec{b}\):
\[ \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} \]
\[ \vec{b} = 3\hat{i} - 2\hat{j} + \hat{k} \]

Now calculate the dot product:
\[ \vec{a} \cdot \vec{b} = (1)(3) + (2)(-2) + (3)(1) = 3 - 4 + 3 = 2 \]

Thus, the value of \(\vec{a} \cdot \vec{b}\) is \(2\).

### Final Answer:
\(\vec{a} \cdot \vec{b} = 2\)

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