If overrightarrow{a} = alpha hat{i} + beta ha | vedclass

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(C) Given $\vec{a} \cdot \vec{b} = 1 \Rightarrow (\alpha \hat{i} + \beta \hat{j} + 3 \hat{k}) \cdot (-\beta \hat{i} - \alpha \hat{j} - \hat{k}) = 1$.T

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$2$

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(C) Given $\vec{a} \cdot \vec{b} = 1 \Rightarrow (\alpha \hat{i} + \beta \hat{j} + 3 \hat{k}) \cdot (-\beta \hat{i} - \alpha \hat{j} - \hat{k}) = 1$.This simplifies to $-\alpha \beta - \alpha \beta - 3 = 1 \Rightarrow -2 \alpha \beta = 4 \Rightarrow \alpha \beta = -2$ $(1)$.Given $\vec{b} \cdot \vec{c} = -3 \Rightarrow (-\beta \hat{i} - \alpha \hat{j} - \hat{k}) \cdot (\hat{i} - 2 \hat{j} - \hat{k}) = -3$.This simplifies to $-\beta + 2 \alpha + 1 = -3 \Rightarrow 2 \alpha - \beta = -4$ $(2)$.From $(1)$,$\beta = -2/\alpha$. Substituting into $(2)$: $2 \alpha - (-2/\alpha) = -4 \Rightarrow 2 \alpha^2 + 2 = -4 \alpha \Rightarrow \alpha^2 + 2 \alpha + 1 = 0 \Rightarrow (\alpha + 1)^2 = 0 \Rightarrow \alpha = -1$.Then $\beta = -2/(-1) = 2$.Now,the scalar triple product is $\frac{1}{3}[\vec{a} \vec{b} \vec{c}] = \frac{1}{3} \begin{vmatrix} \alpha & \beta & 3 \ -\beta & -\alpha & -1 \ 1 & -2 & -1 \end{vmatrix} = \frac{1}{3} \begin{vmatrix} -1 & 2 & 3 \ -2 & 1 & -1 \ 1 & -2 & -1 \end{vmatrix}$.Performing row operations $R_1 	o R_1 + R_3$: $\frac{1}{3} \begin{vmatrix} 0 & 0 & 2 \ -2 & 1 & -1 \ 1 & -2 & -1 \end{vmatrix} = \frac{1}{3} [2(4 - 1)] = \frac{1}{3} 	imes 6 = 2$.

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