Let `vec(a) = alpha hat(i) + 2hat(j) - hat(k) and vec(b) = -2 hat(i) + alpha hat(j) + hat(k)`, where `alpha in R`. If the area of the parallelogram whose adjacent sides are represented by the vectors `vec(a) and vec(b)` is `sqrt(15 (alpha^(2)+ 4))`, then the value of `2|vec(a)|^(2) + (vec(a).vec(b))|vec(b)|^(2)` is equal to

Options

A. 10
B. 7
C. 9
D. 14

Answer

To solve the problem, we need to find the value of \( 2|\vec{a}|^2 + (\vec{a} \cdot \vec{b}) |\vec{b}|^2 \) given the vectors \( \vec{a} \) and \( \vec{b} \) and the area of the parallelogram formed by these vectors.

### Step 1: Define the vectors
Given:
\[ \vec{a} = \alpha \hat{i} + 2 \hat{j} - \hat{k} \]
\[ \vec{b} = -2 \hat{i} + \alpha \hat{j} + \hat{k} \]

### Step 2: Calculate the cross product \( \vec{a} \times \vec{b} \)
The area of the parallelogram is given by the magnitude of the cross product \( |\vec{a} \times \vec{b}| \).

Using the determinant method:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 2 & -1 \\ -2 & \alpha & 1 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 2 & -1 \\ \alpha & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} \alpha & -1 \\ -2 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} \alpha & 2 \\ -2 & \alpha \end{vmatrix} \]
\[ = \hat{i} (2 \cdot 1 - (-1) \cdot \alpha) - \hat{j} (\alpha \cdot 1 - (-1) \cdot -2) + \hat{k} (\alpha^2 + 4) \]
\[ = \hat{i} (2 + \alpha) - \hat{j} (\alpha - 2) + \hat{k} (\alpha^2 + 4) \]

Thus,
\[ \vec{a} \times \vec{b} = (2 + \alpha) \hat{i} - (\alpha - 2) \hat{j} + (\alpha^2 + 4) \hat{k} \]

### Step 3: Calculate the magnitude of the cross product
\[ |\vec{a} \times \vec{b}| = \sqrt{(2 + \alpha)^2 + (-(\alpha - 2))^2 + (\alpha^2 + 4)^2} \]
\[ = \sqrt{(2 + \alpha)^2 + (\alpha - 2)^2 + (\alpha^2 + 4)^2} \]

### Step 4: Set the area equal to the given expression
The area is given as:
\[ \sqrt{15(\alpha^2 + 4)} \]
Thus,
\[ \sqrt{(2 + \alpha)^2 + (\alpha - 2)^2 + (\alpha^2 + 4)^2} = \sqrt{15(\alpha^2 + 4)} \]

### Step 5: Square both sides
Squaring both sides gives:
\[ (2 + \alpha)^2 + (\alpha - 2)^2 + (\alpha^2 + 4)^2 = 15(\alpha^2 + 4) \]

### Step 6: Expand and simplify
Expanding the left-hand side:
\[ (2 + \alpha)^2 = 4 + 4\alpha + \alpha^2 \]
\[ (\alpha - 2)^2 = \alpha^2 - 4\alpha + 4 \]
\[ (\alpha^2 + 4)^2 = \alpha^4 + 8\alpha^2 + 16 \]
Combining:
\[ 4 + 4\alpha + \alpha^2 + \alpha^2 - 4\alpha + 4 + \alpha^4 + 8\alpha^2 + 16 = 15\alpha^2 + 60 \]
\[ \alpha^4 + 10\alpha^2 + 24 = 15\alpha^2 + 60 \]
Rearranging gives:
\[ \alpha^4 - 5\alpha^2 - 36 = 0 \]

### Step 7: Solve the quadratic in \( \alpha^2 \)
Let \( x = \alpha^2 \):
\[ x^2 - 5x - 36 = 0 \]
Using the quadratic formula:
\[ x = \frac{5 \pm \sqrt{25 + 144}}{2} = \frac{5 \pm 13}{2} \]
Thus,
\[ x = 9 \quad \text{or} \quad x = -4 \quad (\text{neglect since } x \geq 0) \]
So, \( \alpha^2 = 9 \) implies \( \alpha = 3 \) or \( \alpha = -3 \).

### Step 8: Calculate \( 2|\vec{a}|^2 + (\vec{a} \cdot \vec{b})|\vec{b}|^2 \)
Calculate \( |\vec{a}|^2 \):
\[ |\vec{a}|^2 = \alpha^2 + 4 + 1 = 9 + 4 + 1 = 14 \]
Thus,
\[ 2|\vec{a}|^2 = 2 \cdot 14 = 28 \]

Now calculate \( \vec{a} \cdot \vec{b} \):
\[ \vec{a} \cdot \vec{b} = \alpha \cdot (-2) + 2 \cdot \alpha + (-1) \cdot 1 = -2\alpha + 2\alpha - 1 = -1 \]

Calculate \( |\vec{b}|^2 \):
\[ |\vec{b}|^2 = (-2)^2 + \alpha^2 + 1 = 4 + 9 + 1 = 14 \]

Putting it all together:
\[ 2|\vec{a}|^2 + (\vec{a} \cdot \vec{b})|\vec{b}|^2 = 28 + (-1) \cdot 14 = 28 - 14 = 14 \]

### Final Answer
The value is:
\[ \boxed{14} \]