Question

Let $\vec{a}=\hat{i}-\alpha\hat{j}+\beta\hat{k}$, $\vec{b}=3\hat{i}+\beta\hat{j}-\alpha\hat{k}$ and $\vec{c}=-\alpha\hat{i}-2\hat{j}+\hat{k}$, where $\alpha$ and $\beta$ are integers. If $\vec{a} \cdot \vec{b} = -1$ and $\vec{b} \cdot \vec{c} = 10$, then $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is equal to ________.

Hint:

The scalar triple product $(\vec{a} \times \vec{b}) \cdot \vec{c}$ represents the volume of the parallelepiped formed by the three vectors and can be efficiently calculated as the determinant of the matrix whose rows (or columns) are the components of the vectors.

Answer 1

Correct Answer: 9

We are given three vectors and two conditions. Let us use the conditions to find \( \alpha \) and \( \beta \).
\[ \vec{a} = (1, -\alpha, \beta), \quad \vec{b} = (3, \beta, -\alpha), \quad \vec{c} = (-\alpha, -2, 1) \] Condition 1: \( \vec{a} \cdot \vec{b} = -1 \).
\[ (1)(3) + (-\alpha)(\beta) + (\beta)(-\alpha) = -1 \] \[ 3 - 2\alpha\beta = -1 \Rightarrow \alpha\beta = 2 \] Condition 2: \( \vec{b} \cdot \vec{c} = 10 \).
\[ (3)(-\alpha) + (\beta)(-2) + (-\alpha)(1) = 10 \] \[ -4\alpha - 2\beta = 10 \Rightarrow 2\alpha + \beta = -5 \] Now solve: \[ \beta = -5 - 2\alpha \] \[ \alpha(-5 - 2\alpha) = 2 \] \[ 2\alpha^2 + 5\alpha + 2 = 0 \] \[ (2\alpha + 1)(\alpha + 2) = 0 \] Thus, \( \alpha = -\frac{1}{2} \) or \( \alpha = -2 \). Since \( \alpha \) is an integer, \( \alpha = -2 \).
\[ \beta = \frac{2}{\alpha} = -1 \] Now the vectors become: \[ \vec{a} = (1, 2, -1), \quad \vec{b} = (3, -1, 2), \quad \vec{c} = (2, -2, 1) \] Now compute the scalar triple product: \[ (\vec{a} \times \vec{b}) \cdot \vec{c} = \begin{vmatrix} 1 & 2 & -1 \\ 3 & -1 & 2 \\ 2 & -2 & 1 \end{vmatrix} \] \[ = 1((-1)(1) - (2)(-2)) - 2((3)(1) - (2)(2)) + (-1)((3)(-2) - (-1)(2)) \] \[ = 1(3) - 2(-1) - 1(-4) = 3 + 2 + 4 = 9 \]