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Question: Prove that any vector $\vec{d}$ can be expressed as, $$\vec{d}=\dfrac{\left\{ \vec{a}\cdot\vec{d}(\vec{b}\times \vec{c})\right\}+\left\{ \vec{b}\cdot \vec{d}(\vec{c}\times \vec{a})\right\}+\left\{ \vec{c}\cdot \vec{d}(\vec{a}\times \vec{b})\right\}}{[\vec{a}\vec{b}\vec{c}]}$$

My effort: Any three non-co-planar vectors $\vec{a},\vec{b}$ and $\vec{c}$ can be expressed as $\vec{d}=x\vec{a}+y\vec{b}+z\vec{c}$. Then
$\vec{d}=x\vec{a}+y\vec{b}+z\vec{c}\implies \vec{d}\cdot(\vec{b}\times \vec{c})=x[\vec{a}\vec{b}\vec{c}]\implies x=\dfrac{\vec{d}\cdot(\vec{b}\times \vec{c})}{[\vec{a}\vec{b}\vec{c}]}$. In this way we have
$$\vec{d}=\dfrac{\left\{ \vec{d}\cdot(\vec{b}\times \vec{c})\right\}\vec{a}+\left\{ \vec{d}\cdot(\vec{c}\times \vec{a})\right\}\vec{b}+\left\{ \vec{d}\cdot(\vec{a}\times \vec{b})\right\}\vec{c}}{[\vec{a}\vec{b}\vec{c}]}$$ which is different from the required result. What to do now to show the required result?

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    $\begingroup$ By the way, what you did is not exactly wrong. It was just not what the problem asked. Your line of reasoning here will still be useful in other situations. $\endgroup$ – Ivo Terek Dec 10 '19 at 3:43
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If $\vec{a}$, $\vec{b}$ and $\vec{c}$ are linearly independent, so are $\vec{a}\times\vec{b}$, $\vec{b}\times \vec{c}$ and $\vec{c}\times \vec{a}$. Then write $$\vec{d} = \alpha (\vec{b}\times \vec{c}) +\beta(\vec{c}\times\vec{a})+\gamma(\vec{a}\times \vec{b})$$for some $\alpha,\beta$ and $\gamma$. The goal is to solve for those coefficients. Dotting the above with $\vec{a}$, $\vec{b}$ and $\vec{c}$ we obtain $$\vec{d}\cdot \vec{a} = \alpha[\vec{a}\vec{b}\vec{c}], \qquad \vec{d}\cdot\vec{b} = \beta[\vec{a}\vec{b}\vec{c}] \quad \mbox{and}\quad \vec{d}\cdot \vec{c} = \gamma[\vec{a}\vec{b}\vec{c}].$$This means that $$\vec{d} = \frac{\vec{d}\cdot\vec{a}}{[\vec{a}\vec{b}\vec{c}]}(\vec{b}\times \vec{c}) + \frac{\vec{d}\cdot\vec{b}}{[\vec{a}\vec{b}\vec{c}]}(\vec{c}\times\vec{a}) + \frac{\vec{d}\cdot\vec{c}}{[\vec{a}\vec{b}\vec{c}]}(\vec{a}\times\vec{b}),$$as wanted.

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