Let vec{a}=hat{i}-alpha hat{j}+beta hat{k},ve | vedclass

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(C) Given vectors are $\vec{a}=(1, -\alpha, \beta)$,$\vec{b}=(3, \beta, -\alpha)$,and $\vec{c}=(-\alpha, -2, 1)$,where $\alpha, \beta \in \mathbb{Z}$.

Vedclass · Accepted Answer

$9$

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(C) Given vectors are $\vec{a}=(1, -\alpha, \beta)$,$\vec{b}=(3, \beta, -\alpha)$,and $\vec{c}=(-\alpha, -2, 1)$,where $\alpha, \beta \in \mathbb{Z}$.From $\vec{a} \cdot \vec{b} = -1$:$(1)(3) + (-\alpha)(\beta) + (\beta)(-\alpha) = -1$$3 - 2\alpha\beta = -1 \Rightarrow 2\alpha\beta = 4 \Rightarrow \alpha\beta = 2$.From $\vec{b} \cdot \vec{c} = 10$:$(3)(-\alpha) + (\beta)(-2) + (-\alpha)(1) = 10$$-3\alpha - 2\beta - \alpha = 10 \Rightarrow -4\alpha - 2\beta = 10 \Rightarrow 2\alpha + \beta = -5$.Since $\alpha\beta = 2$ and $\beta = -5 - 2\alpha$,we substitute $\beta$:$\alpha(-5 - 2\alpha) = 2 \Rightarrow -5\alpha - 2\alpha^2 = 2 \Rightarrow 2\alpha^2 + 5\alpha + 2 = 0$.$(2\alpha + 1)(\alpha + 2) = 0$. Since $\alpha$ is an integer,$\alpha = -2$.Then $\beta = -5 - 2(-2) = -5 + 4 = -1$.Now,calculate the scalar triple product $[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1 & -(-2) & -1 \ 3 & -1 & -(-2) \ -(-2) & -2 & 1 \end{vmatrix} = \begin{vmatrix} 1 & 2 & -1 \ 3 & -1 & 2 \ 2 & -2 & 1 \end{vmatrix}$.$= 1(-1 + 4) - 2(3 - 4) - 1(-6 + 2)$$= 1(3) - 2(-1) - 1(-4) = 3 + 2 + 4 = 9$.

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