Question

If the volume of the parallelepiped whose adjacent edges are \( \vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k} \), \( \vec{b} = \hat{i} + \alpha\hat{j} + 2\hat{k} \), \( \vec{c} = \hat{i} + 2\hat{j} + \alpha\hat{k} \) is 15, then \( \alpha \) is equal to

• A. 1
• B. 5/2
• C. 9/2
• D. 0

Hint:

The volume of a parallelepiped is the absolute value of the scalar triple product (the determinant). Remember that \( |D| = V \) leads to two equations: \( D = V \) and \( D = -V \). You must check both for real solutions.

Answer 1

The Correct Option is C

The volume of a parallelepiped formed by three vectors is given by the absolute value of their scalar triple product, which is calculated as the determinant of the matrix formed by their components. \[ V = |[\vec{a} \vec{b} \vec{c}]| = \left| \begin{vmatrix} 2 & 3 & 4 \\ 1 & \alpha & 2 \\ 1 & 2 & \alpha \end{vmatrix} \right| \] We are given that the volume is 15. \[ \left| \begin{vmatrix} 2 & 3 & 4 \\ 1 & \alpha & 2 \\ 1 & 2 & \alpha \end{vmatrix} \right| = 15 \] Let's evaluate the determinant: \[ 2(\alpha \cdot \alpha - 2 \cdot 2) - 3(1 \cdot \alpha - 2 \cdot 1) + 4(1 \cdot 2 - \alpha \cdot 1) \] \[ = 2(\alpha^2 - 4) - 3(\alpha - 2) + 4(2 - \alpha) \] \[ = 2\alpha^2 - 8 - 3\alpha + 6 + 8 - 4\alpha \] \[ = 2\alpha^2 - 7\alpha + 6 \] 

So, we have the equation \( |2\alpha^2 - 7\alpha + 6| = 15 \). This gives two possibilities: \[\begin{array}{rl} 1. & \text{\( 2\alpha^2 - 7\alpha + 6 = 15 \Rightarrow 2\alpha^2 - 7\alpha - 9 = 0 \)} \\ 2. & \text{\( 2\alpha^2 - 7\alpha + 6 = -15 \Rightarrow 2\alpha^2 - 7\alpha + 21 = 0 \)} \\ \end{array}\] 

For equation (2), the discriminant is \( D = b^2 - 4ac = (-7)^2 - 4(2)(21) = 49 - 168 < 0 \), so it has no real solutions. 

For equation (1), \( 2\alpha^2 - 7\alpha - 9 = 0 \), we can factor it: \[ (2\alpha - 9)(\alpha + 1) = 0 \] This gives two possible values for \( \alpha \): \( \alpha = 9/2 \) or \( \alpha = -1 \). From the given options, \( \alpha = 9/2 \) is a valid solution.