Let $\alpha = {\overrightarrow{\rm OA}}, \; \beta = \overrightarrow{\rm OB}, \; \gamma = \overrightarrow{\rm O\Gamma}$ be unitary vectors that form equal angle $\dfrac{\pi}{3}$, that is:
$$\bigl({\widehat{\vec{\alpha},\vec{\beta}}\,}\bigr)=\bigl({\widehat{\vec{\beta},\vec{\gamma}}\,}\bigr)=\bigl({\widehat{\vec{\gamma},\vec{\alpha}}\,}\bigr)=\dfrac{\pi}{3}$$
Evaluate the value of
$$A=\bigl({\overrightarrow{\alpha}\times\bigl({\overrightarrow{\alpha}\times\bigl({\overrightarrow{\alpha}\times\overrightarrow{\gamma}}\bigr)}\bigr)}\bigr)\cdot\bigl({\overrightarrow{\beta}\times\overrightarrow{\gamma}}\bigr)$$
Solution
It holds that:
$$\overrightarrow{\alpha}\times(\overrightarrow{\alpha}\times\overrightarrow{\gamma})=(\overrightarrow{\alpha}\cdot\overrightarrow{\gamma})\overrightarrow{\alpha}-(\overrightarrow{\alpha}\cdot\overrightarrow{\alpha})\overrightarrow{\gamma}=\dfrac{1}{2}\overrightarrow{\alpha}-\overrightarrow{\gamma}$$
Hence:
$$\overrightarrow{\alpha}\times(\overrightarrow{\alpha}\times(\overrightarrow{\alpha}\times\overrightarrow{\gamma}))=\overrightarrow{\alpha}\times\left(\dfrac{1}{2}\overrightarrow{\alpha}-\overrightarrow{\gamma}\right)=\dfrac{1}{2}\overrightarrow{\alpha}\times\overrightarrow{\alpha}-\overrightarrow{\alpha}\times\overrightarrow{\gamma}=\overrightarrow{\gamma}\times\overrightarrow{\alpha}$$
Therefore:
$$\begin{aligned}
A=(\overrightarrow{\gamma}\times\overrightarrow{\alpha})\cdot(\overrightarrow{\beta}\times\overrightarrow{\gamma}) &=((\overrightarrow{\gamma}\times\overrightarrow{\alpha})\times\overrightarrow{\beta})\cdot\overrightarrow{\gamma} \\
&= (\overrightarrow{\beta}\times(\overrightarrow{\alpha}\times\overrightarrow{\gamma}))\cdot\overrightarrow{\gamma}\\
&= \left[(\overrightarrow{\beta}\cdot\overrightarrow{\gamma})\overrightarrow{\alpha}-(\overrightarrow{\beta}\cdot\overrightarrow{\alpha})\overrightarrow{\gamma}\right]\cdot\overrightarrow{\gamma}\\
&= \left(\dfrac{1}{2}\overrightarrow{\alpha}-\dfrac{1}{2}\overrightarrow{\gamma}\right)\cdot\overrightarrow{\gamma} \\
&= \dfrac{1}{2}\overrightarrow{\alpha}\cdot\overrightarrow{\gamma}-\dfrac{1}{2}\overrightarrow{\gamma}\cdot\overrightarrow{\gamma} \\
&= \dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{4}
\end{aligned}$$
$$\bigl({\widehat{\vec{\alpha},\vec{\beta}}\,}\bigr)=\bigl({\widehat{\vec{\beta},\vec{\gamma}}\,}\bigr)=\bigl({\widehat{\vec{\gamma},\vec{\alpha}}\,}\bigr)=\dfrac{\pi}{3}$$
Evaluate the value of
$$A=\bigl({\overrightarrow{\alpha}\times\bigl({\overrightarrow{\alpha}\times\bigl({\overrightarrow{\alpha}\times\overrightarrow{\gamma}}\bigr)}\bigr)}\bigr)\cdot\bigl({\overrightarrow{\beta}\times\overrightarrow{\gamma}}\bigr)$$
Solution
It holds that:
$$\overrightarrow{\alpha}\times(\overrightarrow{\alpha}\times\overrightarrow{\gamma})=(\overrightarrow{\alpha}\cdot\overrightarrow{\gamma})\overrightarrow{\alpha}-(\overrightarrow{\alpha}\cdot\overrightarrow{\alpha})\overrightarrow{\gamma}=\dfrac{1}{2}\overrightarrow{\alpha}-\overrightarrow{\gamma}$$
Hence:
$$\overrightarrow{\alpha}\times(\overrightarrow{\alpha}\times(\overrightarrow{\alpha}\times\overrightarrow{\gamma}))=\overrightarrow{\alpha}\times\left(\dfrac{1}{2}\overrightarrow{\alpha}-\overrightarrow{\gamma}\right)=\dfrac{1}{2}\overrightarrow{\alpha}\times\overrightarrow{\alpha}-\overrightarrow{\alpha}\times\overrightarrow{\gamma}=\overrightarrow{\gamma}\times\overrightarrow{\alpha}$$
Therefore:
$$\begin{aligned}
A=(\overrightarrow{\gamma}\times\overrightarrow{\alpha})\cdot(\overrightarrow{\beta}\times\overrightarrow{\gamma}) &=((\overrightarrow{\gamma}\times\overrightarrow{\alpha})\times\overrightarrow{\beta})\cdot\overrightarrow{\gamma} \\
&= (\overrightarrow{\beta}\times(\overrightarrow{\alpha}\times\overrightarrow{\gamma}))\cdot\overrightarrow{\gamma}\\
&= \left[(\overrightarrow{\beta}\cdot\overrightarrow{\gamma})\overrightarrow{\alpha}-(\overrightarrow{\beta}\cdot\overrightarrow{\alpha})\overrightarrow{\gamma}\right]\cdot\overrightarrow{\gamma}\\
&= \left(\dfrac{1}{2}\overrightarrow{\alpha}-\dfrac{1}{2}\overrightarrow{\gamma}\right)\cdot\overrightarrow{\gamma} \\
&= \dfrac{1}{2}\overrightarrow{\alpha}\cdot\overrightarrow{\gamma}-\dfrac{1}{2}\overrightarrow{\gamma}\cdot\overrightarrow{\gamma} \\
&= \dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{4}
\end{aligned}$$
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