If vec{a}, vec{b}, vec{c} are three non-copla | vedclass

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(A) Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,they form a basis for the space of vectors. Any vector $\vec{r}$ can be expressed as a linear c

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$[\vec{a} \, \vec{b} \, \vec{c}] \vec{r}$

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(A) Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,they form a basis for the space of vectors. Any vector $\vec{r}$ can be expressed as a linear combination of $\vec{a}, \vec{b}, \vec{c}$:$\vec{r} = x_1 \vec{a} + x_2 \vec{b} + x_3 \vec{c}$Taking the scalar triple product with $\vec{b}$ and $\vec{c}$:$\vec{r} \cdot (\vec{b} 	imes \vec{c}) = x_1 \vec{a} \cdot (\vec{b} 	imes \vec{c}) + x_2 \vec{b} \cdot (\vec{b} 	imes \vec{c}) + x_3 \vec{c} \cdot (\vec{b} 	imes \vec{c})$Since $\vec{b} \cdot (\vec{b} 	imes \vec{c}) = 0$ and $\vec{c} \cdot (\vec{b} 	imes \vec{c}) = 0$,we get:$[\vec{r} \, \vec{b} \, \vec{c}] = x_1 [\vec{a} \, \vec{b} \, \vec{c}] \implies x_1 = \frac{[\vec{r} \, \vec{b} \, \vec{c}]}{[\vec{a} \, \vec{b} \, \vec{c}]} = \frac{[\vec{b} \, \vec{c} \, \vec{r}]}{[\vec{a} \, \vec{b} \, \vec{c}]}$Similarly,$x_2 = \frac{[\vec{c} \, \vec{a} \, \vec{r}]}{[\vec{a} \, \vec{b} \, \vec{c}]}$ and $x_3 = \frac{[\vec{a} \, \vec{b} \, \vec{r}]}{[\vec{a} \, \vec{b} \, \vec{c}]}$Substituting these into the expression for $\vec{r}$:$\vec{r} = \frac{[\vec{b} \, \vec{c} \, \vec{r}]}{[\vec{a} \, \vec{b} \, \vec{c}]} \vec{a} + \frac{[\vec{c} \, \vec{a} \, \vec{r}]}{[\vec{a} \, \vec{b} \, \vec{c}]} \vec{b} + \frac{[\vec{a} \, \vec{b} \, \vec{r}]}{[\vec{a} \, \vec{b} \, \vec{c}]} \vec{c}$Multiplying by $[\vec{a} \, \vec{b} \, \vec{c}]$:$[\vec{b} \, \vec{c} \, \vec{r}] \vec{a} + [\vec{c} \, \vec{a} \, \vec{r}] \vec{b} + [\vec{a} \, \vec{b} \, \vec{r}] \vec{c} = [\vec{a} \, \vec{b} \, \vec{c}] \vec{r}$

If $\vec{a}, \vec{b}, \vec{c}$ are three non-coplanar vectors and $\vec{r}$ is any vector,then $[\vec{b} \, \vec{c} \, \vec{r}] \vec{a} + [\vec{c} \, \vec{a} \, \vec{r}] \vec{b} + [\vec{a} \, \vec{b} \, \vec{r}] \vec{c} = \dots$

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