Parallelepiped Volume Calculation Using Vectors

The volume of a parallelepiped defined by three adjacent edge vectors $\rm \vec a$, $\rm \vec b$, and $\rm \vec c$ is determined by the absolute value of their scalar triple product, often written as $\rm |\vec a \cdot (\vec b \times \vec c)|$. This value can be efficiently calculated using the determinant of a matrix formed by the components of these vectors.

Scalar Triple Product via Determinant

The given adjacent edge vectors are:

• $\rm \vec a = 2\hat{i} + 3\hat{j} + 4\hat{k}$
• $\rm \vec b = \hat{i} + \alpha\hat{j} + 2\hat{k}$
• $\rm \vec c = \hat{i} + 2\hat{j} + \alpha\hat{k}$

The scalar triple product is computed as the determinant of the matrix formed by these vector components:

Volume $V = \left| \begin{vmatrix} 2 & 3 & 4 \\ 1 & \alpha & 2 \\ 1 & 2 & \alpha \end{vmatrix} \right|$

We are given that the volume $V$ is 15.

Determinant Expansion for Adjacent Edges

To find the value of $\alpha$, we first calculate the determinant:

Determinant $= 2 \begin{vmatrix} \alpha & 2 \\ 2 & \alpha \end{vmatrix} - 3 \begin{vmatrix} 1 & 2 \\ 1 & \alpha \end{vmatrix} + 4 \begin{vmatrix} 1 & \alpha \\ 1 & 2 \end{vmatrix}$

Expanding the $2 \times 2$ determinants:

Determinant $= 2 (\alpha \cdot \alpha - 2 \cdot 2) - 3 (1 \cdot \alpha - 2 \cdot 1) + 4 (1 \cdot 2 - \alpha \cdot 1)$

Simplifying the terms:

Determinant $= 2 (\alpha^2 - 4) - 3 (\alpha - 2) + 4 (2 - \alpha)$

Distributing the coefficients:

Determinant $= 2\alpha^2 - 8 - 3\alpha + 6 + 8 - 4\alpha$

Combining like terms yields:

Determinant $= 2\alpha^2 - 7\alpha + 6$

Solving Volume Equation for Alpha

The volume is the absolute value of this determinant. Therefore, we set the determinant equal to 15:

$|2\alpha^2 - 7\alpha + 6| = 15$

This equation breaks down into two separate cases:

Case 1: $2\alpha^2 - 7\alpha + 6 = 15$

Rearranging the terms to form a standard quadratic equation:

$2\alpha^2 - 7\alpha - 9 = 0$

Using the quadratic formula $\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=2$, $b=-7$, and $c=-9$:

$\alpha = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(-9)}}{2(2)}$

$\alpha = \frac{7 \pm \sqrt{49 + 72}}{4}$

$\alpha = \frac{7 \pm \sqrt{121}}{4}$

$\alpha = \frac{7 \pm 11}{4}$

This case yields two possible values for $\alpha$:

• $\alpha_1 = \frac{7 + 11}{4} = \frac{18}{4} = \dfrac92$
• $\alpha_2 = \frac{7 - 11}{4} = \frac{-4}{4} = -1$

Case 2: $2\alpha^2 - 7\alpha + 6 = -15$

Rearranging the terms:

$2\alpha^2 - 7\alpha + 21 = 0$

To determine if there are real solutions, we check the discriminant ($\Delta = b^2 - 4ac$) for this quadratic equation ($a=2$, $b=-7$, $c=21$):

$\Delta = (-7)^2 - 4(2)(21)$

$\Delta = 49 - 168$

$\Delta = -119$

Since the discriminant $\Delta$ is negative, there are no real solutions for $\alpha$ in this case.

Final Alpha Value Selection

The real values obtained for $\alpha$ are $\dfrac92$ and $-1$. Comparing these results with the provided multiple-choice options:

• Option 1: 1
• Option 2: $\dfrac52$
• Option 3: $\dfrac92$
• Option 4: 0

The value $\dfrac92$ corresponds to Option 3.