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

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(B) Let the scalar triple product be $[\vec{a} \vec{b} \vec{c}] = \vec{a} \cdot (\vec{b} 	imes \vec{c})$. Since $\vec{a}, \vec{b}, \vec{c}$ are non-c

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$3$

Vedclass · Answer

(B) Let the scalar triple product be $[\vec{a} \vec{b} \vec{c}] = \vec{a} \cdot (\vec{b} 	imes \vec{c})$. Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar, $[\vec{a} \vec{b} \vec{c}] 
eq 0$. Given $\vec{p} = \frac{\vec{b} 	imes \vec{c}}{[\vec{a} \vec{b} \vec{c}]}, \vec{q} = \frac{\vec{c} 	imes \vec{a}}{[\vec{a} \vec{b} \vec{c}]}, \vec{r} = \frac{\vec{a} 	imes \vec{b}}{[\vec{a} \vec{b} \vec{c}]}$. We need to evaluate $S = (\vec{a}+\vec{b}) \cdot \vec{p} + (\vec{b}+\vec{c}) \cdot \vec{q} + (\vec{c}+\vec{a}) \cdot \vec{r}$. Substituting the expressions: $S = \frac{(\vec{a}+\vec{b}) \cdot (\vec{b} 	imes \vec{c})}{[\vec{a} \vec{b} \vec{c}]} + \frac{(\vec{b}+\vec{c}) \cdot (\vec{c} 	imes \vec{a})}{[\vec{a} \vec{b} \vec{c}]} + \frac{(\vec{c}+\vec{a}) \cdot (\vec{a} 	imes \vec{b})}{[\vec{a} \vec{b} \vec{c}]}$. Using the property $\vec{a} \cdot (\vec{b} 	imes \vec{c}) = [\vec{a} \vec{b} \vec{c}]$ and $\vec{b} \cdot (\vec{b} 	imes \vec{c}) = 0$: $S = \frac{[\vec{a} \vec{b} \vec{c}] + 0}{[\vec{a} \vec{b} \vec{c}]} + \frac{[\vec{b} \vec{c} \vec{a}] + 0}{[\vec{a} \vec{b} \vec{c}]} + \frac{[\vec{c} \vec{a} \vec{b}] + 0}{[\vec{a} \vec{b} \vec{c}]}$. Since $[\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{a}] = [\vec{c} \vec{a} \vec{b}]$, we get: $S = 1 + 1 + 1 = 3$.

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