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If $r=\alpha \mathrm{b} \times \mathrm{c}+\beta \mathrm{c} \times \mathrm{a}+\gamma \mathrm{a} \times \mathrm{b}$ and [a b c] $=2$, then $\alpha+\beta+\gamma$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{2} \mathrm{r} \cdot(\mathrm{a}+\mathrm{b}+\mathrm{c})$
$\begin{array}{l}
\mathbf{r} \cdot \mathbf{a}=\alpha(\mathbf{a} \cdot \mathbf{b} \times \mathbf{c})+\beta(\mathbf{a} \cdot \mathbf{c} \times \mathbf{a})+\gamma(\mathbf{a} \cdot \mathbf{a} \times \mathbf{b}) \\
=\alpha[\mathbf{a b c}]+0+0
\end{array}$
Similarly, $\mathbf{r} . \mathbf{b}=\beta[\mathbf{a b c}]$ and $\mathbf{r} \cdot \mathbf{c}=\gamma[\mathbf{a b c}]$
$\begin{array}{l}
\therefore \frac{1}{2} \mathbf{r} \cdot(\mathbf{a}+\mathbf{b}+\mathbf{c})=\frac{1}{2}(\mathbf{r} \cdot \mathbf{a}+\mathbf{r} \cdot \mathbf{b}+\mathbf{r} \cdot \mathbf{c}) \\
=\frac{1}{2}(\alpha+\beta+\gamma)[\mathbf{a b c}] \\
=\frac{1}{2}(\alpha+\beta+\gamma) \times 2=\alpha+\beta+\gamma
\end{array}$
\mathbf{r} \cdot \mathbf{a}=\alpha(\mathbf{a} \cdot \mathbf{b} \times \mathbf{c})+\beta(\mathbf{a} \cdot \mathbf{c} \times \mathbf{a})+\gamma(\mathbf{a} \cdot \mathbf{a} \times \mathbf{b}) \\
=\alpha[\mathbf{a b c}]+0+0
\end{array}$
Similarly, $\mathbf{r} . \mathbf{b}=\beta[\mathbf{a b c}]$ and $\mathbf{r} \cdot \mathbf{c}=\gamma[\mathbf{a b c}]$
$\begin{array}{l}
\therefore \frac{1}{2} \mathbf{r} \cdot(\mathbf{a}+\mathbf{b}+\mathbf{c})=\frac{1}{2}(\mathbf{r} \cdot \mathbf{a}+\mathbf{r} \cdot \mathbf{b}+\mathbf{r} \cdot \mathbf{c}) \\
=\frac{1}{2}(\alpha+\beta+\gamma)[\mathbf{a b c}] \\
=\frac{1}{2}(\alpha+\beta+\gamma) \times 2=\alpha+\beta+\gamma
\end{array}$
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