If the volume of the parallelepiped whose adjacent edges are \(\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}\), \(\vec{b}=\hat{i}+\alpha \hat{j}+2\hat{k}\) and \(\vec{c}=\hat{i}+2\hat{j}+\alpha \hat{k}\) is 15, then \(\alpha\) is equal to
Step-by-step Solution:
The volume of the parallelepiped is: \[ [a \, b \, c] = 15 \] \[ \begin{aligned} & \left\lvert \begin{array}{ccc} 2 & 3 & 4 \\ 1 & \alpha & 2 \\ 1 & 2 & \alpha \end{array} \right\rvert \\ & \Rightarrow 2(\alpha^2 - 4) - 3(\alpha - 2) + 4(2 - \alpha) = 15 \\ & \Rightarrow 2\alpha^2 - 7\alpha - 9 = 0 \\ & \Rightarrow (\alpha + 1)(2\alpha - 9) = 0. \end{aligned} \] Hence, the values of \( \alpha \) are \( \alpha = -1 \) and \( \alpha = \frac{9}{2} \).