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(B) Let $V = [\vec{a} \vec{b} \vec{c}] = \vec{a} \cdot (\vec{b} 	imes \vec{c})$.Then $\vec{p} = \frac{\vec{b} 	imes \vec{c}}{V}, \vec{q} = \frac{\ve

Vedclass · Accepted Answer

$\frac{1}{[\vec{a} \vec{b} \vec{c}]}$

Vedclass · Answer

(B) Let $V = [\vec{a} \vec{b} \vec{c}] = \vec{a} \cdot (\vec{b} 	imes \vec{c})$.Then $\vec{p} = \frac{\vec{b} 	imes \vec{c}}{V}, \vec{q} = \frac{\vec{c} 	imes \vec{a}}{V}, \vec{r} = \frac{\vec{a} 	imes \vec{b}}{V}$.$[\vec{p} \vec{q} \vec{r}] = \vec{p} \cdot (\vec{q} 	imes \vec{r}) = \frac{\vec{b} 	imes \vec{c}}{V} \cdot \left( \frac{\vec{c} 	imes \vec{a}}{V} 	imes \frac{\vec{a} 	imes \vec{b}}{V} ight)$.Using the vector triple product identity $(\vec{u} 	imes \vec{v}) 	imes \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$,we have $(\vec{c} 	imes \vec{a}) 	imes (\vec{a} 	imes \vec{b}) = [(\vec{c} 	imes \vec{a}) \cdot \vec{b}]\vec{a} - [(\vec{c} 	imes \vec{a}) \cdot \vec{a}]\vec{b}$.Since $(\vec{c} 	imes \vec{a}) \cdot \vec{a} = 0$,this simplifies to $[\vec{c} \vec{a} \vec{b}]\vec{a} = [\vec{a} \vec{b} \vec{c}]\vec{a} = V\vec{a}$.Thus,$[\vec{p} \vec{q} \vec{r}] = \frac{1}{V^3} (\vec{b} 	imes \vec{c}) \cdot (V\vec{a}) = \frac{V}{V^3} (\vec{b} 	imes \vec{c}) \cdot \vec{a} = \frac{V^2}{V^3} = \frac{1}{V} = \frac{1}{[\vec{a} \vec{b} \vec{c}]}$.

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