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If \vec{a}=\alpha \hat{i}+\beta \hat{j}+3\hat{k}, \vec{b}=-\beta \hat{i}-\alpha \hat{j}-\hat{k} and \vec{c}= \hat{i}-2\hat{j}-\hat{k} such that \vec{a}.\vec{b}=1 and \vec{b}.\vec{c}=-3, then \frac{1}{3}\left ( \left ( \vec{a}\times \vec{b} \right ).\vec{c} \right ) is equal to __________.
Option: 1 0
Option: 2 1
Option: 3 2
Option: 4 3

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\begin{array}{l} \overrightarrow{ a }=\alpha \hat{ i }+\beta \hat{ j }+3 \hat{ k }, \\ \overrightarrow{ b }=-\beta \hat{ i }-\alpha \hat{ j }-\hat{ k } \text { and } \\ \overrightarrow{ c }=\hat{ i }-2 \hat{ j }-\hat{ k } \end{array}

\\\overrightarrow{ a } \cdot \overrightarrow{ b }=1 \qquad\\\Rightarrow\qquad-\alpha \beta-\alpha \beta-3=1\\\Rightarrow\qquad-2 \alpha \beta=4 \Rightarrow \alpha \beta=-2\qquad\ldots(1)

\begin{aligned} &\overrightarrow{ b } \cdot \overrightarrow{ c }=-3\\ & \Rightarrow -\beta+2 \alpha+1=-3 \\ & \beta-2 \alpha=4 \qquad\ldots(2)\end{aligned}

\begin{aligned} &\text { Solving }(1)\quad \&\quad(2)\\ &(\alpha, \beta)=(-1,2) \end{aligned}

\begin{aligned} \frac{1}{3}[\vec{a} \vec{b} \vec{c}] &=\frac{1}{3}\left|\begin{array}{ccc} \alpha & \beta & 3 \\ -\beta & -\alpha & -1 \\ 1 & -2 & -1 \end{array}\right| \\ &=\frac{1}{3}\left|\begin{array}{ccc} -1 & 2 & 3 \\ -2 & 1 & -1 \\ 1 & -2 & -1 \end{array}\right| \\ &=\frac{1}{3}\left|\begin{array}{ccc} 0 & 0 & 2 \\ -2 & 1 & -1 \\ 1 & -2 & -1 \end{array}\right|\\&=\frac{1}{3}[2(4-1)]=2 \end{aligned}

 

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Suraj Bhandari

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